{"id":823,"date":"2021-07-23T09:20:49","date_gmt":"2021-07-23T13:20:49","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/precipitation-and-dissolution\/"},"modified":"2022-06-23T09:22:14","modified_gmt":"2022-06-23T13:22:14","slug":"precipitation-and-dissolution","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/precipitation-and-dissolution\/","title":{"raw":"15.1 Precipitation and Dissolution","rendered":"15.1 Precipitation and Dissolution"},"content":{"raw":"&nbsp;\r\n<div class=\"textbox textbox--learning-objectives\">\r\n<h3><strong>Learning Objectives<\/strong><\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Write chemical equations and equilibrium expressions representing solubility equilibria<\/li>\r\n \t<li>Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp15457056\">Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.<\/p>\r\n\r\n<div id=\"fs-idp15827408\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>The Solubility Product<\/strong><\/h3>\r\n<p id=\"fs-idm402040992\">Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (<em data-effect=\"italics\">insoluble<\/em> or <em data-effect=\"italics\">sparingly soluble<\/em>) to infinity (<em data-effect=\"italics\">miscible<\/em>). A solute with finite solubility can yield a <em data-effect=\"italics\">saturated<\/em> solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.<\/p>\r\n\r\n<div id=\"fs-idp15836912\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1935\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-300x53.png\" alt=\"\" width=\"283\" height=\"50\" \/><\/div>\r\n<p id=\"fs-idp292944\">In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_AgCl\">(Figure)<\/a>). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_15_01_AgCl\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in equilibrium with undissolved silver chloride.<\/div>\r\n<span id=\"fs-idp15452416\" data-type=\"media\" data-alt=\"Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.\"><img class=\"scaled-down\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_AgCl-1.jpg\" alt=\"Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp15439888\">The equilibrium constant for solubility equilibria such as this one is called the <strong data-effect=\"bold\">solubility product constant, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub><\/strong>, in this case<\/p>\r\n\r\n<div id=\"fs-idp15217968\" data-type=\"equation\"><\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag<sup>+<\/sup>][Cl<sup>\u2212<\/sup>]<\/div>\r\n<p id=\"fs-idp15825056\">Recall that only gases and solutes are represented in equilibrium constant expressions, so the <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.<\/p>\r\n\r\n<div id=\"fs-idm38560\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm38304\"><strong>Writing Equations and Solubility Products <\/strong><\/p>\r\nWrite the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:\r\n<p id=\"fs-idp204000\">(a) AgI, silver iodide, a solid with antiseptic properties<\/p>\r\n<p id=\"fs-idp15889200\">(b) CaCO<sub>3<\/sub>, calcium carbonate, the active ingredient in many over-the-counter chewable antacids<\/p>\r\n<p id=\"fs-idp681040\">(c) Mg(OH)<sub>2<\/sub>, magnesium hydroxide, the active ingredient in Milk of Magnesia<\/p>\r\n<p id=\"fs-idp11820496\">(d) Mg(NH<sub>4<\/sub>)PO<sub>4<\/sub>, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium<\/p>\r\n<p id=\"fs-idp291040\">(e) Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, the mineral apatite, a source of phosphate for fertilizers<\/p>\r\n<p id=\"fs-idp313008\"><strong>Solution:<\/strong><\/p>\r\n(a) AgI(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + I<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag<sup>+<\/sup>][I<sup>\u2212<\/sup>]\r\n\r\n(b) CaCO<sub>3<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + CO<sub>3<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> =[Ca<sup>2+<\/sup>][CO<sub>3<\/sub><sup>2\u2212<\/sup>]\r\n\r\n(c) Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(aq)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Ksp =[Mg2+][OH\u2212]<sup>2<\/sup>\r\n\r\n(d) Mg(NH<sub>4<\/sub>)PO<sub>4<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + NH<sub>4<\/sub><sup>+<\/sup>(<em>aq<\/em>) + PO<sub>4<\/sub><sup>3\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Mg<sup>2+<\/sup>][NH<sub>4<\/sub><sup>+<\/sup>][PO<sub>4<\/sub><sup>3\u2212<\/sup>]\r\n\r\n(e) Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH(<em>s<\/em>) \u21cc 5Ca<sup>2+<\/sup>(<em>aq<\/em>) + 3PO<sub>4<\/sub><sup>3\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> =[Ca<sup>2+<\/sup>]<sup>5<\/sup>[PO<sub>4<\/sub><sup>3\u2212<\/sup>]<sup>3<\/sup>[OH<sup>\u2212<\/sup>]\r\n\r\n&nbsp;\r\n<p id=\"fs-idp413568\"><strong>Check Your Learning:<\/strong><\/p>\r\nWrite the dissolution equation and the solubility product for each of the following slightly soluble compounds:\r\n<p id=\"fs-idp414352\">(a) BaSO<sub>4<\/sub><\/p>\r\n<p id=\"fs-idp414992\">(b) Ag<sub>2<\/sub>SO<sub>4<\/sub><\/p>\r\n<p id=\"fs-idp11828480\">(c) Al(OH)<sub>3<\/sub><\/p>\r\n<p id=\"fs-idp11829120\">(d) Pb(OH)Cl<\/p>\r\n&nbsp;\r\n<div id=\"fs-idp11829504\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp11829888\">(a) BaSO<sub>4<\/sub>(<em>s<\/em>) \u21cc Ba<sup>2+<\/sup>(<em>aq<\/em>) + SO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ba<sup>2+<\/sup>][SO<sub>4<\/sub><sup>2\u2212<\/sup>]<\/p>\r\n(b) Ag<sub>2<\/sub>SO<sub>4<\/sub>(<em>s<\/em>) \u21cc 2Ag<sup>+<\/sup>(<em>aq<\/em>) + SO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag<sup>+<\/sup>]<sup>2<\/sup>[SO<sub>4<\/sub><sup>2\u2212<\/sup>]\r\n\r\n(c) Al(OH)<sub>3<\/sub>(<em>s<\/em>) \u21cc Al<sup>3+<\/sup>(<em>aq<\/em>) + 3OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Al<sup>3+<\/sup>][OH<sup>\u2212<\/sup>]<sup>3<\/sup>\r\n\r\n(d) Pb(OH)Cl(s) \u21cc Pb<sup>2+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Pb<sup>2+<\/sup>][OH<sup>\u2212<\/sup>][Cl<sup>\u2212<\/sup>]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp11911328\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong><em data-effect=\"italics\">K<\/em><sub>sp<\/sub> and Solubility<\/strong><\/h3>\r\n<p id=\"fs-idp11912720\">The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:<\/p>\r\n\r\n<div id=\"fs-idp11914144\" style=\"padding-left: 40px\" data-type=\"equation\">M<sub>p<\/sub>X<sub>q<\/sub>(<em>s<\/em>) \u21cc pM<sup>m+<\/sup>(<em>aq<\/em>) + qX<sup>n\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idp843744\">For cases such as these, one may derive <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound\u2019s molar solubility, measured as moles of dissolved solute per liter of saturated solution.<\/p>\r\n&nbsp;\r\n<div id=\"fs-idp844816\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp845072\"><strong>Calculation of <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> from Equilibrium Concentrations <\/strong><\/p>\r\nFluorite, CaF<sub>2<\/sub>, is a slightly soluble solid that dissolves according to the equation:\r\n<div id=\"fs-idp847472\" style=\"padding-left: 40px\" data-type=\"equation\">CaF<sub>2<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + 2F<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idp15201792\">The concentration of Ca<sup>2+<\/sup> in a saturated solution of CaF<sub>2<\/sub> is 2.15 \u00d7 10<sup>\u20134 <\/sup><em data-effect=\"italics\">M<\/em>. What is the solubility product of fluorite?<\/p>\r\n<p id=\"fs-idp15206752\"><strong>Solution:<\/strong><\/p>\r\nAccording to the stoichiometry of the dissolution equation:\r\n<p style=\"padding-left: 40px\">[Ca<sup>2+<\/sup>] = 2.15 \u00d7 10<sup>-4<\/sup> M<\/p>\r\n\r\n<div id=\"fs-idm421605264\" style=\"padding-left: 40px\" data-type=\"equation\">[F<sup>-<\/sup>] = 2(2.15 \u00d7 10<sup>-4<\/sup> M) = 4.30 \u00d7 10<sup>-4<\/sup> M<\/div>\r\n<p id=\"fs-idm332364112\">Substituting the ion concentrations into the <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> expression gives<\/p>\r\n\r\n<div id=\"fs-idm55552\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][F<sup>\u2212<\/sup>]<sup>2<\/sup> = (2.15 \u00d7 10<sup>-4<\/sup> M)(4.30 \u00d7 10<sup>-4<\/sup> M)<sup>2<\/sup> = 3.98 \u00d7 10<sup>-11<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp16640\"><strong>Check Your Learning:<\/strong><\/p>\r\nIn a saturated solution of Mg(OH)<sub>2<\/sub>, the concentration of Mg<sup>2+<\/sup> is 1.31 \u00d7 10<sup>\u20134<\/sup><em data-effect=\"italics\">M<\/em>. What is the solubility product for Mg(OH)<sub>2<\/sub>?\r\n<div id=\"fs-idp19952\" style=\"padding-left: 40px\" data-type=\"equation\">Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idp259840\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp260224\">8.99 \u00d7 10<sup>\u201312<\/sup><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp261504\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp261760\"><strong>Determination of Molar Solubility from <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> <\/strong><\/p>\r\nThe <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of copper(I) bromide, CuBr, is 6.3 \u00d7 10<sup>\u20139<\/sup>. Calculate the molar solubility of copper bromide.\r\n<p id=\"fs-idp264928\"><strong>Solution:<\/strong><\/p>\r\nThe dissolution equation and solubility product expression are\r\n<div id=\"fs-idp266848\" style=\"padding-left: 40px\" data-type=\"equation\">CuBr(<em>s<\/em>) \u21cc Cu<sup>+<\/sup>(<em>aq<\/em>) + Br<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div id=\"fs-idp590896\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Cu<sup>+<\/sup>][Br<sup>\u2212<\/sup>]<\/div>\r\n<p id=\"fs-idp594736\">Following the ICE approach to this calculation yields the table<\/p>\r\n<span id=\"fs-idp596176\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive x, x.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable1_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive x, x.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idp597536\">Substituting the equilibrium concentration terms into the solubility product expression and solving for <em data-effect=\"italics\">x<\/em> yields<\/p>\r\n\r\n<div id=\"fs-idp597920\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Cu+][Br<sup>\u2212<\/sup>]<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">6.3 \u00d7 10<sup>-9<\/sup> = <em>x<\/em><sup>2<\/sup><\/div>\r\n<div id=\"fs-idp15508368\" style=\"padding-left: 40px\" data-type=\"equation\"><em>x<\/em> = 7.9 \u00d7 10<sup>-5<\/sup> M<\/div>\r\n<p id=\"fs-idp15515696\">Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each mole of CuBr dissolved, the molar solubility of CuBr is 7.9 \u00d7 10<sup>\u20135 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp15517600\"><strong>Check Your Learning: <\/strong><\/p>\r\nThe <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of AgI is 1.5 \u00d7\u00a0 10<sup>\u201316<\/sup>. Calculate the molar solubility of silver iodide.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp645648\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp646032\">1.2 \u00d7 10<sup>\u20138 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp647936\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp648192\"><strong>Determination of Molar Solubility from <em data-effect=\"italics\">K<\/em><sub>sp<\/sub><\/strong><\/p>\r\nThe <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of calcium hydroxide, Ca(OH)<sub>2<\/sub>, is 1.3 \u00d7 10<sup>\u20136<\/sup>. Calculate the molar solubility of calcium hydroxide.\r\n<p id=\"fs-idp651872\"><strong>Solution:<\/strong><\/p>\r\nThe dissolution equation and solubility product expression are\r\n<div id=\"fs-idp653824\" style=\"padding-left: 40px\" data-type=\"equation\">Ca(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div id=\"fs-idp662160\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][OH<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\r\n<p id=\"fs-idp15421424\">The ICE table for this system is<\/p>\r\n<span id=\"fs-idp15423072\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive 2 x, 2 x.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable7_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive 2 x, 2 x.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idp15424432\">Substituting terms for the equilibrium concentrations into the solubility product expression and solving for <em data-effect=\"italics\">x<\/em> gives<\/p>\r\n\r\n<div id=\"fs-idp15424816\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][OH<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\r\n<div id=\"fs-idp15429408\" style=\"padding-left: 40px\" data-type=\"equation\">1.3 \u00d7 10<sup>-6<\/sup> = (<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup>= (<em>x<\/em>)(4<em>x<\/em><sup>2<\/sup>) = 4<em>x<\/em><sup>3<\/sup><\/div>\r\n<div id=\"fs-idp820144\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1941\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-300x68.png\" alt=\"\" width=\"238\" height=\"54\" \/><\/div>\r\n<p id=\"fs-idp827984\">As defined in the ICE table, <em data-effect=\"italics\">x<\/em> is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)<sub>2<\/sub> is 6.9 \u00d7 10<sup>\u20133<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp830272\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of PbI<sub>2<\/sub> is 1.4 \u00d7 10<sup>\u20138<\/sup>. Calculate the molar solubility of lead(II) iodide.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp833104\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp833488\">1.5 \u00d7 10<sup>\u20133<\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp836192\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp836448\"><strong>Determination of <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> from Gram Solubility<\/strong><\/p>\r\nMany of the pigments used by artists in oil-based paints (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_OilPaints\">(Figure)<\/a>) are sparingly soluble in water. For example, the solubility of the artist\u2019s pigment chrome yellow, PbCrO<sub>4<\/sub>, is 4.6 \u00d7 10<sup>\u20136<\/sup> g\/L. Determine the solubility product for PbCrO<sub>4<\/sub>.\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_15_01_OilPaints\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO<sub>4<\/sub>), examples include Prussian blue (Fe<sub>7<\/sub>(CN)<sub>18<\/sub>), the reddish-orange color vermilion (HgS), and green color veridian (Cr<sub>2<\/sub>O<sub>3<\/sub>). (credit: Sonny Abesamis)<\/div>\r\n<span id=\"fs-idp840928\" data-type=\"media\" data-alt=\"A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_OilPaints-1.jpg\" alt=\"A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp612896\"><strong>Solution:<\/strong><\/p>\r\n<p id=\"fs-idp619872\">Before calculating the solubility product, the provided solubility must be converted to molarity:<\/p>\r\n\r\n<div id=\"fs-idp120984032\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1942\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-300x86.png\" alt=\"\" width=\"317\" height=\"91\" \/><\/div>\r\n<p id=\"fs-idp15351552\">The dissolution equation for this compound is<\/p>\r\n\r\n<div id=\"fs-idp100842208\" style=\"padding-left: 40px\" data-type=\"equation\">PbCrO<sub>4<\/sub>(<em>s<\/em>) \u21cc Pb<sup>2+<\/sup>(<em>aq<\/em>) + CrO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idp53410000\">The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both [Pb<sup>2+<\/sup>] and [CrO<sub>4<\/sub><sup>2\u2212<\/sup>] are equal to the molar solubility of PbCrO<sub>4<\/sub>:<\/p>\r\n\r\n<div id=\"fs-idp34481696\" style=\"padding-left: 40px\" data-type=\"equation\">[Pb<sup>2+<\/sup>] = [CrO<sub>4<\/sub><sup>2\u2212<\/sup>] = 1.4 \u00d7 10<sup>-8<\/sup> M<\/div>\r\n<p id=\"fs-idp83072\" style=\"padding-left: 40px\"><em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][CrO<sub>4<\/sub><sup>2\u2212<\/sup>] = (1.4 \u00d7 10<sup>\u20138 <\/sup>M)(1.4 \u00d7 10<sup>\u20138 <\/sup>M) = 2.0 \u00d7 10<sup>\u201316<\/sup><\/p>\r\n<p id=\"fs-idp87280\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at 20\u00b0C. What is its solubility product?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp88320\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp88704\">2.08 \u00d7 10<sup>\u20134<\/sup><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp91888\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp92144\"><strong>Calculating the Solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> <\/strong><\/p>\r\nCalomel, Hg<sub>2<\/sub>Cl<sub>2<\/sub>, is a compound composed of the diatomic ion of mercury(I), Hg<sub>2<\/sub><sup>2+<\/sup>, and chloride ions, Cl<sup>\u2013<\/sup>. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>:\r\n<div id=\"fs-idp97168\" style=\"padding-left: 40px\" data-type=\"equation\">Hg<sub>2<\/sub>Cl<sub>2<\/sub>(<em>s<\/em>) \u21cc Hg<sub>2<\/sub><sup>2+<\/sup>(<em>aq<\/em>) + 2Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp <\/sub>= 1.1 \u00d7 10<sup>-18<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp110192\">Calculate the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub>.<\/p>\r\n<p id=\"fs-idp15521184\"><strong>Solution:<\/strong><\/p>\r\nThe dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to the concentration of Hg<sub>2<\/sub><sup>2+<\/sup> ions.\r\n<p id=\"fs-idm324614160\">Following the ICE approach results in<\/p>\r\n<span id=\"fs-idp15533712\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, \u201cH g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following: 0, positive 2 x, 2 x.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable2_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, \u201cH g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following: 0, positive 2 x, 2 x.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idm324611760\">Substituting the equilibrium concentration terms into the solubility product expression and solving for <em data-effect=\"italics\">x<\/em> gives<\/p>\r\n\r\n<div id=\"fs-idp53869216\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Hg<sub>2<\/sub><sup>2+<\/sup>][Cl<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\r\n<div id=\"fs-idm14694048\" style=\"padding-left: 40px\" data-type=\"equation\">1.1 \u00d7 10<sup>-18<\/sup> = (<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup><\/div>\r\n<div id=\"fs-idp127117360\" style=\"padding-left: 40px\" data-type=\"equation\">4<em>x<\/em><sup>3<\/sup> = 1.1 \u00d7 10<sup>-18<\/sup><\/div>\r\n<div id=\"fs-idp29132112\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1944\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-300x58.png\" alt=\"\" width=\"274\" height=\"53\" \/><\/div>\r\n<div id=\"fs-idp32484816\" style=\"padding-left: 40px\" data-type=\"equation\">[Hg<sub>2<\/sub><sup>2+<\/sup>] = 6.5 \u00d7 10<sup>-7<\/sup> M<\/div>\r\n<div id=\"fs-idm84487648\" style=\"padding-left: 40px\" data-type=\"equation\">[Cl<sup>\u2212<\/sup>] = 2x = 2(6.5 \u00d7 10<sup>-7<\/sup> M) = 1.3 \u00d7 10<sup>-6<\/sup> M<\/div>\r\n<p id=\"fs-idp15393408\">The dissolution stoichiometry shows the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to [Hg<sub>2<\/sub><sup>2+<\/sup>] or 6.5 \u00d7\u00a0 10<sup>\u20137<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp751088\"><strong>Check Your Learning:<\/strong><\/p>\r\nDetermine the molar solubility of MgF<sub>2<\/sub> from its solubility product: <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = 6.4 \u00d7 10<sup>\u20139<\/sup>.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp753920\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp754304\">1.2 \u00d7 10<sup>\u20133 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp765232\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Using Barium Sulfate for Medical Imaging<\/strong><\/div>\r\n<p id=\"fs-idp766160\">Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of barium sulfate is 2.3 \u00d7 10<sup>\u20138<\/sup>, very little of it dissolves as it coats the lining of the patient\u2019s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_BariumXray\">(Figure)<\/a>).<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_15_01_BariumXray\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by \u201cglitzy queen00\u201d\/Wikimedia Commons)<\/div>\r\n<span id=\"fs-idp770240\" data-type=\"media\" data-alt=\"This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_BariumXray-1.jpg\" alt=\"This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp772848\">Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn\u2019s disease, and ulcers in addition to other conditions.<\/p>\r\n<p id=\"fs-idp774080\">Visit this <a href=\"http:\/\/openstaxcollege.org\/l\/16barium\">website<\/a> for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp775600\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Predicting Precipitation<\/strong><\/h3>\r\n<p id=\"fs-idp776288\">The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:<\/p>\r\n\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idp776784\" style=\"padding-left: 40px\" data-type=\"equation\">CaCO<sub>3<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + CO<sub>3<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][CO<sub>3<\/sub><sup>2-<\/sup>] = 8.7 \u00d7 10<sup>-9<\/sup><\/div>\r\n<p id=\"fs-idp785408\">It is important to realize that this equilibrium is established in any aqueous solution containing Ca<sup>2+<\/sup> and CO<sub>3<\/sub><sup>2\u2013<\/sup> ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, <em data-effect=\"italics\">Q<sub>sp<\/sub><\/em>, that exceeds the solubility product, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (<em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> = <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>). The comparison of <em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> to <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:<\/p>\r\n<p id=\"fs-idm425158560\"><em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> &lt; <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)<\/p>\r\n<p id=\"fs-idm425156768\"><em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> &gt; <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)<\/p>\r\n<p id=\"fs-idm425154976\">This predictive strategy and related calculations are demonstrated in the next few example exercises.<\/p>\r\n\r\n<div id=\"fs-idp69099520\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp15157136\"><strong>Precipitation of Mg(OH)<sub>2<\/sub> <\/strong><\/p>\r\nThe first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of lime, Ca(OH)<sub>2<\/sub>, a readily available inexpensive source of OH<sup>\u2013<\/sup> ion:\r\n<div id=\"fs-idp15159216\" style=\"padding-left: 40px\" data-type=\"equation\">Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 8.9 \u00d7 10<sup>-12<\/sup><\/div>\r\n<p id=\"fs-idp15170960\">The concentration of Mg<sup>2+<\/sup>(<em data-effect=\"italics\">aq<\/em>) in sea water is 0.0537 <em data-effect=\"italics\">M<\/em>. Will Mg(OH)<sub>2<\/sub> precipitate when enough Ca(OH)<sub>2<\/sub> is added to give a [OH<sup>\u2013<\/sup>] of 0.0010 <em data-effect=\"italics\">M<\/em>?<\/p>\r\n<p id=\"fs-idp15174912\"><strong>Solution:<\/strong><\/p>\r\n<p id=\"fs-idp15183504\">Calculation of the reaction quotient under these conditions is shown here:<\/p>\r\n\r\n<div id=\"fs-idp15187872\" data-type=\"equation\">Q = [Mg<sup>2+<\/sup>]<sub>0<\/sub>[OH<sup>\u2212<\/sup>]<sub>0<\/sub><sup>2<\/sup> = (0.0537 M)(0.0010 M<sup>)2<\/sup> = 5.4 \u00d7 10<sup>-8<\/sup><\/div>\r\n<p id=\"fs-idp11854608\">Because <em data-effect=\"italics\">Q<\/em> is greater than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> (<em data-effect=\"italics\">Q<\/em> = 5.4 \u00d7 10<sup>\u20138<\/sup> is larger than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = 8.9 \u00d7 10<sup>\u201312<\/sup>), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that <em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> = <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>.<\/p>\r\n<p id=\"fs-idp11862608\"><strong>Check Your Learning:<\/strong><\/p>\r\nPredict whether CaHPO<sub>4<\/sub> will precipitate from a solution with [Ca<sup>2+<\/sup>] = 0.0001 <em data-effect=\"italics\">M<\/em> and [HPO<sub>4<\/sub><sup>2\u2212<\/sup>] = 0.001 <em data-effect=\"italics\">M<\/em>.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp11869072\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp11869456\">No precipitation of CaHPO<sub>4<\/sub>; <em data-effect=\"italics\">Q<\/em> = 1 \u00d7 10<sup>\u20137<\/sup>, which is less than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> (7 \u00d7 10<sup>\u20137<\/sup>)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp11872704\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp11872960\"><strong>Precipitation of AgCl <\/strong><\/p>\r\nDoes silver chloride precipitate when equal volumes of a 2.0 \u00d7 10<sup>\u20134<\/sup>-<em data-effect=\"italics\">M<\/em> solution of AgNO<sub>3<\/sub> and a 2.0 \u00d7 10<sup>\u20134<\/sup>-<em data-effect=\"italics\">M<\/em> solution of NaCl are mixed?\r\n<p id=\"fs-idp11879728\"><strong>Solution:<\/strong><\/p>\r\nThe equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:\r\n<div id=\"fs-idp11880480\" style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idp11887792\">The solubility product is 1.6 \u00d7 10<sup>\u201310<\/sup> (see Appendix J).<\/p>\r\n<p id=\"fs-idp11889984\">AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO<sub>3<\/sub> and NaCl is greater than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>. Because the volume doubles when equal volumes of AgNO<sub>3<\/sub> and NaCl solutions are mixed, each concentration is reduced to half its initial value<\/p>\r\n\r\n<div id=\"fs-idp11893136\" style=\"padding-left: 40px\" data-type=\"equation\">\u00bd(2.0 \u00d7 10<sup>-4<\/sup> M) = 1.0 \u00d7 10<sup>-4<\/sup> M<\/div>\r\n<p id=\"fs-idp11902464\">The reaction quotient, <em data-effect=\"italics\">Q<\/em>, is greater than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> for AgCl, so a supersaturated solution is formed:<\/p>\r\n\r\n<div id=\"fs-idp456352\" style=\"padding-left: 40px\" data-type=\"equation\">Q = [Ag+]<sub>0<\/sub>[Cl\u2212]<sub>0<\/sub> = (1.0 \u00d7 10<sup>-4\u00a0 <\/sup>M)(1.0 \u00d7 10<sup>-4<\/sup> M) = 1.0 \u00d7 10<sup>-8<\/sup><\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">Q &gt; K<sub>sp<\/sub><\/div>\r\n<p id=\"fs-idp470144\">AgCl will precipitate from the mixture until the dissolution equilibrium is established, with <em data-effect=\"italics\">Q<\/em> equal to <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>.<\/p>\r\n<p id=\"fs-idp472176\"><strong>Check Your Learning:<\/strong><\/p>\r\nWill KClO<sub>4<\/sub> precipitate when 20 mL of a 0.050-<em data-effect=\"italics\">M<\/em> solution of K<sup>+<\/sup> is added to 80 mL of a 0.50-<em data-effect=\"italics\">M<\/em> solution of ClO<sub>4<\/sub><sup>\u2212<\/sup>? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)\r\n\r\n&nbsp;\r\n<div id=\"fs-idp477040\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp477424\">No, <em data-effect=\"italics\">Q<\/em> = 4.0 \u00d7 10<sup>\u20133<\/sup>, which is less than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = 1.05 \u00d7 10<sup>\u20132<\/sup><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp488672\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp488928\"><strong>Precipitation of Calcium Oxalate <\/strong><\/p>\r\nBlood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>, for this purpose (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_Blood\">(Figure)<\/a>). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O (calcium oxalate monohydrate). The concentration of Ca<sup>2+<\/sup> in a sample of blood serum is 2.2 \u00d7 10<sup>\u20133 <\/sup><em data-effect=\"italics\">M<\/em>. What concentration of C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212 <\/sup>ion must be established before CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O begins to precipitate?\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_15_01_Blood\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Anticoagulants can be added to blood that will combine with the Ca<sup>2+<\/sup> ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)<\/div>\r\n<span id=\"fs-idp500240\" data-type=\"media\" data-alt=\"A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_Blood-1.jpg\" alt=\"A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp502656\"><strong>Solution:<\/strong><\/p>\r\nThe equilibrium expression is:\r\n<div id=\"fs-idp503344\" style=\"padding-left: 40px\" data-type=\"equation\">CaC<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp513120\">For this reaction:<\/p>\r\n\r\n<div id=\"eip-478\" class=\"unnumered\" style=\"padding-left: 40px\" data-type=\"equation\" data-label=\"\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>] = 1.96 \u00d7 10<sup>-8<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(see Appendix J)<\/div>\r\n<p id=\"fs-idp340688\">Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:<\/p>\r\n\r\n<div id=\"fs-idp347152\" style=\"padding-left: 40px\" data-type=\"equation\">Q = K<sub>sp<\/sub> =[Ca<sup>2+<\/sup>][C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>]= 1.96 \u00d7 10<sup>-8<\/sup><\/div>\r\n<div id=\"fs-idp356336\" style=\"padding-left: 40px\" data-type=\"equation\">(2.2 \u00d7 10<sup>-3<\/sup> M)[C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>] = 1.96 \u00d7 10<sup>-8<\/sup><\/div>\r\n<div id=\"fs-idp365472\" style=\"padding-left: 40px\" data-type=\"equation\">[C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>] = 8.9 \u00d7 10<sup>-6<\/sup> M<\/div>\r\n<p id=\"fs-idp377504\">A concentration of 8.9 \u00d7 10<sup>\u20136 <\/sup><em data-effect=\"italics\">M C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup><\/em> is necessary to initiate the precipitation of CaC<sub>2<\/sub>O<sub>4<\/sub> under these conditions.<\/p>\r\n<p id=\"fs-idp383648\"><strong>Check Your Learning:<\/strong><\/p>\r\nIf a solution contains 0.0020 mol CrO<sub>4<\/sub><sup>2\u2212<\/sup> per liter, what concentration of Ag<sup>+<\/sup> ion must be reached by adding solid AgNO<sub>3<\/sub> before Ag<sub>2<\/sub>CrO<sub>4<\/sub> begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp387952\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp388336\">6.7 \u00d7 10<sup>\u20135\u00a0<\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp392704\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp392960\"><strong>Concentrations Following Precipitation <\/strong><\/p>\r\nClothing washed in water that has [Mn<sup>2+<\/sup>] exceeding 0.1 mg\/L (1.8 \u00d7 10<sup>\u20136 <\/sup><em data-effect=\"italics\">M<\/em>) may be stained by the manganese upon oxidation, but the amount of Mn<sup>2+<\/sup> in the water can be decreased by adding a base to precipitate Mn(OH)<sub>2<\/sub>. What pH is required to keep [Mn<sup>2+<\/sup>] equal to 1.8 \u00d7 10<sup>\u20136 <\/sup><em data-effect=\"italics\">M<\/em>?\r\n<p id=\"fs-idp399216\"><strong>Solution:<\/strong><\/p>\r\nThe dissolution of Mn(OH)<sub>2<\/sub> is described by the equation:\r\n<div id=\"fs-idp400336\" style=\"padding-left: 40px\" data-type=\"equation\">Mn(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mn<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 2 \u00d7 10<sup>-13<\/sup><\/div>\r\n<p id=\"fs-idp1156752\">At equilibrium:<\/p>\r\n\r\n<div id=\"fs-idp1159600\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Mn<sup>2+<\/sup>][OH<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\r\n<div id=\"fs-idp1164576\" style=\"padding-left: 40px\" data-type=\"equation\">(1.8 \u00d7 10<sup>-6<\/sup> M)[OH<sup>\u2212<\/sup>]<sup>2<\/sup> = 2 \u00d7 10<sup>-13<\/sup><\/div>\r\n<p id=\"fs-idp1173408\" style=\"padding-left: 40px\">[ OH<sup>\u2212<\/sup>] = 3 \u00d7 10<sup>-4<\/sup> M<\/p>\r\n<p id=\"fs-idp1179472\">Calculate the pH from the pOH:<\/p>\r\n\r\n<div id=\"fs-idp1179904\" style=\"padding-left: 40px\" data-type=\"equation\">pOH = \u2212log [OH<sup>\u2212<\/sup>] = \u2212log(3 \u00d7 10<sup>-4<\/sup> M) = 3.5<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">pH = 14.00 - pOH = 14.00 - 3.5 = 10.5<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp1193504\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of Ca(OH)<sub>2<\/sub>. The concentration of Mg<sup>2+<\/sup>(<em data-effect=\"italics\">aq<\/em>) in sea water is 5.37 \u00d7 10<sup>\u20132 <\/sup><em data-effect=\"italics\">M<\/em>. Calculate the pH at which [Mg<sup>2+<\/sup>] is decreased to 1.0 \u00d7 10<sup>\u20135<\/sup><em data-effect=\"italics\">M.<\/em>\r\n\r\n&nbsp;\r\n<div id=\"fs-idp1200000\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp1200384\">10.97<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp1200896\">In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called <span data-type=\"term\">selective precipitation<\/span> may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound\u2019s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.<\/p>\r\n&nbsp;\r\n<div id=\"fs-idp1215552\" class=\"chemistry everyday-life\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>The Role of Precipitation in Wastewater Treatment<\/strong><\/div>\r\n<p id=\"fs-idp1216528\">Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_Wastewater\">(Figure)<\/a>). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions, PO<sub>4<\/sub><sup>3\u2212<\/sup>, are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_15_01_Wastewater\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: \u201ceutrophication&amp;hypoxia\u201d\/Wikimedia Commons)<\/div>\r\n<span id=\"fs-idp1220944\" class=\"scaled-down\" data-type=\"media\" data-alt=\"A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_Wastewater-1.jpg\" alt=\"A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp1223232\">One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)<sub>2<\/sub>. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, which then precipitates out of the solution:<\/p>\r\n\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idp1225152\" style=\"padding-left: 40px\" data-type=\"equation\">5Ca<sup>2+<\/sup>(<em>aq<\/em>) + 3PO<sub>4<\/sub><sup>3\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) \u2192 Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH(<em>s<\/em>)<\/div>\r\n<p id=\"fs-idp1234976\">Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO<sub>2<\/sub> in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.<\/p>\r\n<p id=\"fs-idp1236048\">View this <a href=\"http:\/\/openstaxcollege.org\/l\/16Wastewater\">site<\/a> for more information on how phosphorus is removed from wastewater.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp1242128\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp1242384\"><strong>Precipitation of Silver Halides <\/strong><\/p>\r\nA solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?\r\n<p id=\"fs-idp1243456\"><strong>Solution:<\/strong><\/p>\r\nThe two equilibria involved are:\r\n<div id=\"fs-idp1244144\" style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 1.6 \u00d7 10<sup>-10<\/sup><\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">AgBr(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Br<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 5.0 \u00d7 10<sup>-13<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp15236832\">If the solution contained about <em data-effect=\"italics\">equal<\/em> concentrations of Cl<sup>\u2013<\/sup> and Br<sup>\u2013<\/sup>, then the silver salt with the smaller <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag<sup>+<\/sup>] at which AgCl begins to precipitate and the [Ag<sup>+<\/sup>] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag<sup>+<\/sup>] precipitates first.<\/p>\r\n<p id=\"fs-idp15240992\">AgBr precipitates when <em data-effect=\"italics\">Q<\/em> equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> for AgBr:<\/p>\r\n\r\n<div id=\"fs-idp15244832\" data-type=\"equation\">Q = [Ag<sup>+<\/sup>]<sub>0<\/sub>[Br<sup>\u2212<\/sup>]<sub>0<\/sub> = [Ag<sup>+<\/sup>]<sub>0<\/sub>(0.00010 M) = 5.0 \u00d7 10<sup>-13<\/sup><\/div>\r\n<div id=\"fs-idp15252336\" data-type=\"equation\">[Ag<sup>+<\/sup>] = (5.0 \u00d7 10<sup>-13<\/sup>)\/(0.00010 M) = 5.0 \u00d7 10<sup>-9<\/sup> M<\/div>\r\n<p id=\"fs-idp15261056\">AgBr begins to precipitate when [Ag<sup>+<\/sup>] is 5.0 \u00d7 10<sup>\u20139 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp15263392\">For AgCl: AgCl precipitates when <em data-effect=\"italics\">Q<\/em> equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> for AgCl:<\/p>\r\n\r\n<div id=\"fs-idp15244832\" data-type=\"equation\">Q = [Ag<sup>+<\/sup>]<sub>0<\/sub>[Cl<sup>\u2212<\/sup>]<sub>0<\/sub> = [Ag<sup>+<\/sup>]<sub>0<\/sub>(0.10 M) = 1.6 \u00d7 10<sup>-10<\/sup><\/div>\r\n<div id=\"fs-idp15252336\" data-type=\"equation\">[Ag<sup>+<\/sup>] = (1.6 \u00d7 10<sup>-10<\/sup>)\/(0.10 M) = 1.6 \u00d7 10<sup>-9<\/sup>M<\/div>\r\n<p id=\"fs-idp15261056\">AgCl begins to precipitate when [Ag<sup>+<\/sup>] is 1.6 \u00d7 10<sup>\u20139 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n\r\n<div id=\"fs-idp15267232\" data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp15286672\">AgCl begins to precipitate at a lower [Ag<sup>+<\/sup>] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> greater than that of silver bromide.<\/p>\r\n<p id=\"fs-idp15287568\"><strong>Check Your Learning:<\/strong><\/p>\r\nIf silver nitrate solution is added to a solution which is 0.050 <em data-effect=\"italics\">M<\/em> in both Cl<sup>\u2013<\/sup> and Br<sup>\u2013<\/sup> ions, at what [Ag<sup>+<\/sup>] would precipitation begin, and what would be the formula of the precipitate?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp15290048\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp15290432\">[Ag<sup>+<\/sup>] = 1.0 \u00d7 10<sup>\u201311<\/sup><em data-effect=\"italics\">M<\/em>; AgBr precipitates first<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp15293024\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Common Ion Effect<\/strong><\/h3>\r\n<p id=\"fs-idp15293696\">Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a <em data-effect=\"italics\">common ion<\/em> (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the <strong data-effect=\"bold\">common ion effect<\/strong>, which may be explained using Le Ch\u00e2telier\u2019s principle. Consider the dissolution of silver iodide:<\/p>\r\n\r\n<div id=\"fs-idp15306976\" style=\"padding-left: 40px\" data-type=\"equation\">AgI(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + I<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idp15314160\">This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag<sup>+<\/sup> and I<sup>\u2013<\/sup>. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.<\/p>\r\n<p id=\"fs-idm386466432\">This effect may also be explained as represented in the solubility product expression:<\/p>\r\n\r\n<div id=\"fs-idm386465920\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ag<sup>+<\/sup>][I<sup>-<\/sup>]<\/div>\r\n<p id=\"fs-idm386458688\">The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture <em data-effect=\"italics\">regardless of the source of the ions<\/em>, and so an increase in one ion\u2019s concentration must be balanced by a proportional decrease in the other.<\/p>\r\n\r\n<div id=\"fs-idp11083040\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp11083296\"><strong>Common Ion Effect on Solubility <\/strong><\/p>\r\nWhat is the effect on the amount of solid Mg(OH)<sub>2<\/sub> and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a saturated solution of Mg(OH)<sub>2<\/sub>?\r\n<p id=\"fs-idp11085600\">(a) MgCl<sub>2<\/sub><\/p>\r\n<p id=\"fs-idp11086240\">(b) KOH<\/p>\r\n<p id=\"fs-idp11087008\">(c) NaNO<sub>3<\/sub><\/p>\r\n<p id=\"fs-idp11087648\">(d) Mg(OH)<sub>2<\/sub><\/p>\r\n<p id=\"fs-idp11088288\"><strong>Solution:<\/strong><\/p>\r\nThe solubility equilibrium is\r\n<div id=\"fs-idp141736144\" style=\"padding-left: 40px\" data-type=\"equation\">Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm536672\">(a) The reaction shifts to the left to relieve the stress produced by the additional Mg<sup>2+<\/sup> ion, in accordance with Le Ch\u00e2telier\u2019s principle. In quantitative terms, the added Mg<sup>2+<\/sup> causes the reaction quotient to be larger than the solubility product (<em data-effect=\"italics\">Q<\/em> &gt; <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>), and Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is less and [Mg<sup>2+<\/sup>] is greater than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/p>\r\n<p id=\"fs-idp15127072\">(b) The reaction shifts to the left to relieve the stress of the additional OH<sup>\u2013<\/sup> ion. Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is greater and [Mg<sup>2+<\/sup>] is less than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/p>\r\n<p id=\"fs-idp15130768\">(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.<\/p>\r\n<p id=\"fs-idm330455104\">(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.<\/p>\r\n\r\n<div id=\"fs-idm740800\" style=\"padding-left: 40px\" data-type=\"equation\">Q = [Mg<sup>2+<\/sup>]<sub>0<\/sub>[OH<sup>\u2212<\/sup>]<sub>0<\/sub><sup>2<\/sup><\/div>\r\n<p id=\"fs-idm735616\">Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of <em data-effect=\"italics\">Q<\/em>, and no shift is required to restore <em data-effect=\"italics\">Q<\/em> to the value of the equilibrium constant.<\/p>\r\n<p id=\"fs-idm701008\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the effect on the amount of solid NiCO<sub>3<\/sub> and the concentrations of Ni<sup>2+<\/sup> and CO<sub>3<\/sub><sup>2\u2212<\/sup> when each of the following are added to a saturated solution of NiCO<sub>3<\/sub>\r\n<p id=\"fs-idm697184\">(a) Ni(NO<sub>3<\/sub>)<sub>2<\/sub><\/p>\r\n<p id=\"fs-idm696160\">(b) KClO<sub>4<\/sub><\/p>\r\n<p id=\"fs-idm695520\">(c) NiCO<sub>3<\/sub><\/p>\r\n<p id=\"fs-idm694880\">(d) K<sub>2<\/sub>CO<sub>3<\/sub><\/p>\r\n&nbsp;\r\n<div id=\"fs-idm690608\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm690096\">(a) mass of NiCO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) increases, [Ni<sup>2+<\/sup>] increases, [CO<sub>3<\/sub><sup>2\u2212<\/sup>] decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO<sub>3<\/sub>; (d) mass of NiCO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) increases, [Ni<sup>2+<\/sup>] decreases, [CO<sub>3<\/sub><sup>2\u2212<\/sup>] increases.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp15318512\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp15318768\"><strong>Common Ion Effect <\/strong><\/p>\r\nCalculate the molar solubility of cadmium sulfide (CdS) in a 0.010-<em data-effect=\"italics\">M<\/em> solution of cadmium bromide (CdBr<sub>2<\/sub>). The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of CdS is 1.0 \u00d7 10<sup>\u201328<\/sup>.\r\n<p id=\"fs-idp15322272\"><strong>Solution:<\/strong><\/p>\r\nThis calculation can be performed using the ICE approach:\r\n<div id=\"fs-idp15323568\" style=\"padding-left: 40px\" data-type=\"equation\">CdS(<em>s<\/em>) \u21cc Cd<sup>2+<\/sup>(<em>aq<\/em>) + S<sup>2\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<span id=\"fs-idp15331040\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cC d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, positive x, 0.010 plus x. The third column has the following: 0, positive x, x.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable3_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cC d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, positive x, 0.010 plus x. The third column has the following: 0, positive x, x.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<div id=\"fs-idp15332448\" style=\"padding-left: 40px\" data-type=\"equation\">\u00a0K<sub>sp<\/sub> = [Cd<sup>2+<\/sup>][S<sup>2\u2212<\/sup>] = 1.0 \u00d7 10<sup>-28<\/sup><\/div>\r\n<div id=\"fs-idp13457936\" style=\"padding-left: 80px\" data-type=\"equation\">(0.010 M + <em>x<\/em>)(<em>x<\/em>) = 1.0 \u00d7 10<sup>-28<\/sup><\/div>\r\n<p id=\"fs-idm394907456\">Because <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> is very small, assume <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.010 M, i.e. <em data-effect=\"italics\">x<\/em> &lt; 0.00050 M and solve the simplified equation for <em data-effect=\"italics\">x<\/em>:<\/p>\r\n\r\n<div id=\"fs-idp13479456\" style=\"padding-left: 40px\" data-type=\"equation\">(0.010)(<em>x<\/em><span style=\"font-size: 1em\">) = 1.0 \u00d7 10<sup>-28<\/sup><\/span><\/div>\r\n<div id=\"fs-idp13485120\" style=\"padding-left: 40px\" data-type=\"equation\"><em>x<\/em> = 1.0 \u00d7 10<sup>-26<\/sup> M \u00a0 \u00a0 \u00a0 \u00a0 \u00a0ASSUMPTION VALID<\/div>\r\n<p id=\"fs-idp13488544\">The molar solubility of CdS in this solution is 1.0 \u00d7 10<sup>\u201326 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp13490448\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate the molar solubility of aluminum hydroxide, Al(OH)<sub>3<\/sub>, in a 0.015-<em data-effect=\"italics\">M<\/em> solution of aluminum nitrate, Al(NO<sub>3<\/sub>)<sub>3<\/sub>. The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of Al(OH)<sub>3<\/sub> is 2 \u00d7 10<sup>\u201332<\/sup>.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp13495104\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp13495488\">4 \u00d7 10<sup>\u201311 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp13497520\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idp13498448\">The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid M<sub><em data-effect=\"italics\">p<\/em><\/sub>X<sub><em data-effect=\"italics\">q<\/em><\/sub> and its ions M<sup>m+<\/sup> and X<sup>n\u2013<\/sup>:<\/p>\r\n\r\n<div id=\"fs-idp13501568\" style=\"padding-left: 40px\" data-type=\"equation\">M<sub>p<\/sub>X<sub>q<\/sub>(<em>s<\/em>) \u21cc pM<sup>m+<\/sup>(<em>aq<\/em>) + qX<sup>n\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idp13510416\">the solubility product expression is:<\/p>\r\n\r\n<div id=\"fs-idp13510864\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [M<sup>m+<\/sup>]<sup>p<\/sup>[X<sup>n\u2212<\/sup>]<sup>q<\/sup><\/div>\r\n<p id=\"fs-idp13516480\">The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.<\/p>\r\n<p id=\"fs-idp13518032\">A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp13519952\" class=\"key-equations\" data-depth=\"1\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-idp13535408\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idp3007504\" data-type=\"exercise\"><\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\r\n<dl id=\"fs-idp3050576\">\r\n \t<dt>common ion effect<\/dt>\r\n \t<dd id=\"fs-idp3050992\">effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp3051616\">\r\n \t<dt>molar solubility<\/dt>\r\n \t<dd id=\"fs-idp3052032\">solubility of a compound expressed in units of moles per liter (mol\/L)<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp3053344\">\r\n \t<dt>solubility product constant (<em data-effect=\"italics\">K<\/em><sub>sp<\/sub>)<\/dt>\r\n \t<dd id=\"fs-idp3054640\">equilibrium constant for the dissolution of an ionic compound<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p>&nbsp;<\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<h3><strong>Learning Objectives<\/strong><\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Write chemical equations and equilibrium expressions representing solubility equilibria<\/li>\n<li>Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp15457056\">Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.<\/p>\n<div id=\"fs-idp15827408\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>The Solubility Product<\/strong><\/h3>\n<p id=\"fs-idm402040992\">Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (<em data-effect=\"italics\">insoluble<\/em> or <em data-effect=\"italics\">sparingly soluble<\/em>) to infinity (<em data-effect=\"italics\">miscible<\/em>). A solute with finite solubility can yield a <em data-effect=\"italics\">saturated<\/em> solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.<\/p>\n<div id=\"fs-idp15836912\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1935\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-300x53.png\" alt=\"\" width=\"283\" height=\"50\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-300x53.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-768x136.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-65x11.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-225x40.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a-350x62.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1a.png 950w\" sizes=\"auto, (max-width: 283px) 100vw, 283px\" \/><\/div>\n<p id=\"fs-idp292944\">In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_AgCl\">(Figure)<\/a>). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_15_01_AgCl\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in equilibrium with undissolved silver chloride.<\/div>\n<p><span id=\"fs-idp15452416\" data-type=\"media\" data-alt=\"Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.\"><img decoding=\"async\" class=\"scaled-down\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_AgCl-1.jpg\" alt=\"Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp15439888\">The equilibrium constant for solubility equilibria such as this one is called the <strong data-effect=\"bold\">solubility product constant, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub><\/strong>, in this case<\/p>\n<div id=\"fs-idp15217968\" data-type=\"equation\"><\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag<sup>+<\/sup>][Cl<sup>\u2212<\/sup>]<\/div>\n<p id=\"fs-idp15825056\">Recall that only gases and solutes are represented in equilibrium constant expressions, so the <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.<\/p>\n<div id=\"fs-idm38560\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm38304\"><strong>Writing Equations and Solubility Products <\/strong><\/p>\n<p>Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:<\/p>\n<p id=\"fs-idp204000\">(a) AgI, silver iodide, a solid with antiseptic properties<\/p>\n<p id=\"fs-idp15889200\">(b) CaCO<sub>3<\/sub>, calcium carbonate, the active ingredient in many over-the-counter chewable antacids<\/p>\n<p id=\"fs-idp681040\">(c) Mg(OH)<sub>2<\/sub>, magnesium hydroxide, the active ingredient in Milk of Magnesia<\/p>\n<p id=\"fs-idp11820496\">(d) Mg(NH<sub>4<\/sub>)PO<sub>4<\/sub>, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium<\/p>\n<p id=\"fs-idp291040\">(e) Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, the mineral apatite, a source of phosphate for fertilizers<\/p>\n<p id=\"fs-idp313008\"><strong>Solution:<\/strong><\/p>\n<p>(a) AgI(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + I<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag<sup>+<\/sup>][I<sup>\u2212<\/sup>]<\/p>\n<p>(b) CaCO<sub>3<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + CO<sub>3<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> =[Ca<sup>2+<\/sup>][CO<sub>3<\/sub><sup>2\u2212<\/sup>]<\/p>\n<p>(c) Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(aq)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Ksp =[Mg2+][OH\u2212]<sup>2<\/sup><\/p>\n<p>(d) Mg(NH<sub>4<\/sub>)PO<sub>4<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + NH<sub>4<\/sub><sup>+<\/sup>(<em>aq<\/em>) + PO<sub>4<\/sub><sup>3\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Mg<sup>2+<\/sup>][NH<sub>4<\/sub><sup>+<\/sup>][PO<sub>4<\/sub><sup>3\u2212<\/sup>]<\/p>\n<p>(e) Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH(<em>s<\/em>) \u21cc 5Ca<sup>2+<\/sup>(<em>aq<\/em>) + 3PO<sub>4<\/sub><sup>3\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> =[Ca<sup>2+<\/sup>]<sup>5<\/sup>[PO<sub>4<\/sub><sup>3\u2212<\/sup>]<sup>3<\/sup>[OH<sup>\u2212<\/sup>]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp413568\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Write the dissolution equation and the solubility product for each of the following slightly soluble compounds:<\/p>\n<p id=\"fs-idp414352\">(a) BaSO<sub>4<\/sub><\/p>\n<p id=\"fs-idp414992\">(b) Ag<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p id=\"fs-idp11828480\">(c) Al(OH)<sub>3<\/sub><\/p>\n<p id=\"fs-idp11829120\">(d) Pb(OH)Cl<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp11829504\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp11829888\">(a) BaSO<sub>4<\/sub>(<em>s<\/em>) \u21cc Ba<sup>2+<\/sup>(<em>aq<\/em>) + SO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ba<sup>2+<\/sup>][SO<sub>4<\/sub><sup>2\u2212<\/sup>]<\/p>\n<p>(b) Ag<sub>2<\/sub>SO<sub>4<\/sub>(<em>s<\/em>) \u21cc 2Ag<sup>+<\/sup>(<em>aq<\/em>) + SO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag<sup>+<\/sup>]<sup>2<\/sup>[SO<sub>4<\/sub><sup>2\u2212<\/sup>]<\/p>\n<p>(c) Al(OH)<sub>3<\/sub>(<em>s<\/em>) \u21cc Al<sup>3+<\/sup>(<em>aq<\/em>) + 3OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Al<sup>3+<\/sup>][OH<sup>\u2212<\/sup>]<sup>3<\/sup><\/p>\n<p>(d) Pb(OH)Cl(s) \u21cc Pb<sup>2+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Pb<sup>2+<\/sup>][OH<sup>\u2212<\/sup>][Cl<sup>\u2212<\/sup>]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp11911328\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong><em data-effect=\"italics\">K<\/em><sub>sp<\/sub> and Solubility<\/strong><\/h3>\n<p id=\"fs-idp11912720\">The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:<\/p>\n<div id=\"fs-idp11914144\" style=\"padding-left: 40px\" data-type=\"equation\">M<sub>p<\/sub>X<sub>q<\/sub>(<em>s<\/em>) \u21cc pM<sup>m+<\/sup>(<em>aq<\/em>) + qX<sup>n\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idp843744\">For cases such as these, one may derive <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound\u2019s molar solubility, measured as moles of dissolved solute per liter of saturated solution.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp844816\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp845072\"><strong>Calculation of <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> from Equilibrium Concentrations <\/strong><\/p>\n<p>Fluorite, CaF<sub>2<\/sub>, is a slightly soluble solid that dissolves according to the equation:<\/p>\n<div id=\"fs-idp847472\" style=\"padding-left: 40px\" data-type=\"equation\">CaF<sub>2<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + 2F<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idp15201792\">The concentration of Ca<sup>2+<\/sup> in a saturated solution of CaF<sub>2<\/sub> is 2.15 \u00d7 10<sup>\u20134 <\/sup><em data-effect=\"italics\">M<\/em>. What is the solubility product of fluorite?<\/p>\n<p id=\"fs-idp15206752\"><strong>Solution:<\/strong><\/p>\n<p>According to the stoichiometry of the dissolution equation:<\/p>\n<p style=\"padding-left: 40px\">[Ca<sup>2+<\/sup>] = 2.15 \u00d7 10<sup>-4<\/sup> M<\/p>\n<div id=\"fs-idm421605264\" style=\"padding-left: 40px\" data-type=\"equation\">[F<sup>&#8211;<\/sup>] = 2(2.15 \u00d7 10<sup>-4<\/sup> M) = 4.30 \u00d7 10<sup>-4<\/sup> M<\/div>\n<p id=\"fs-idm332364112\">Substituting the ion concentrations into the <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> expression gives<\/p>\n<div id=\"fs-idm55552\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][F<sup>\u2212<\/sup>]<sup>2<\/sup> = (2.15 \u00d7 10<sup>-4<\/sup> M)(4.30 \u00d7 10<sup>-4<\/sup> M)<sup>2<\/sup> = 3.98 \u00d7 10<sup>-11<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp16640\"><strong>Check Your Learning:<\/strong><\/p>\n<p>In a saturated solution of Mg(OH)<sub>2<\/sub>, the concentration of Mg<sup>2+<\/sup> is 1.31 \u00d7 10<sup>\u20134<\/sup><em data-effect=\"italics\">M<\/em>. What is the solubility product for Mg(OH)<sub>2<\/sub>?<\/p>\n<div id=\"fs-idp19952\" style=\"padding-left: 40px\" data-type=\"equation\">Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idp259840\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp260224\">8.99 \u00d7 10<sup>\u201312<\/sup><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp261504\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp261760\"><strong>Determination of Molar Solubility from <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> <\/strong><\/p>\n<p>The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of copper(I) bromide, CuBr, is 6.3 \u00d7 10<sup>\u20139<\/sup>. Calculate the molar solubility of copper bromide.<\/p>\n<p id=\"fs-idp264928\"><strong>Solution:<\/strong><\/p>\n<p>The dissolution equation and solubility product expression are<\/p>\n<div id=\"fs-idp266848\" style=\"padding-left: 40px\" data-type=\"equation\">CuBr(<em>s<\/em>) \u21cc Cu<sup>+<\/sup>(<em>aq<\/em>) + Br<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div id=\"fs-idp590896\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Cu<sup>+<\/sup>][Br<sup>\u2212<\/sup>]<\/div>\n<p id=\"fs-idp594736\">Following the ICE approach to this calculation yields the table<\/p>\n<p><span id=\"fs-idp596176\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive x, x.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable1_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive x, x.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idp597536\">Substituting the equilibrium concentration terms into the solubility product expression and solving for <em data-effect=\"italics\">x<\/em> yields<\/p>\n<div id=\"fs-idp597920\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Cu+][Br<sup>\u2212<\/sup>]<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">6.3 \u00d7 10<sup>-9<\/sup> = <em>x<\/em><sup>2<\/sup><\/div>\n<div id=\"fs-idp15508368\" style=\"padding-left: 40px\" data-type=\"equation\"><em>x<\/em> = 7.9 \u00d7 10<sup>-5<\/sup> M<\/div>\n<p id=\"fs-idp15515696\">Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each mole of CuBr dissolved, the molar solubility of CuBr is 7.9 \u00d7 10<sup>\u20135 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp15517600\"><strong>Check Your Learning: <\/strong><\/p>\n<p>The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of AgI is 1.5 \u00d7\u00a0 10<sup>\u201316<\/sup>. Calculate the molar solubility of silver iodide.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp645648\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp646032\">1.2 \u00d7 10<sup>\u20138 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp647936\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp648192\"><strong>Determination of Molar Solubility from <em data-effect=\"italics\">K<\/em><sub>sp<\/sub><\/strong><\/p>\n<p>The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of calcium hydroxide, Ca(OH)<sub>2<\/sub>, is 1.3 \u00d7 10<sup>\u20136<\/sup>. Calculate the molar solubility of calcium hydroxide.<\/p>\n<p id=\"fs-idp651872\"><strong>Solution:<\/strong><\/p>\n<p>The dissolution equation and solubility product expression are<\/p>\n<div id=\"fs-idp653824\" style=\"padding-left: 40px\" data-type=\"equation\">Ca(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div id=\"fs-idp662160\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][OH<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\n<p id=\"fs-idp15421424\">The ICE table for this system is<\/p>\n<p><span id=\"fs-idp15423072\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive 2 x, 2 x.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable7_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive 2 x, 2 x.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idp15424432\">Substituting terms for the equilibrium concentrations into the solubility product expression and solving for <em data-effect=\"italics\">x<\/em> gives<\/p>\n<div id=\"fs-idp15424816\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][OH<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\n<div id=\"fs-idp15429408\" style=\"padding-left: 40px\" data-type=\"equation\">1.3 \u00d7 10<sup>-6<\/sup> = (<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup>= (<em>x<\/em>)(4<em>x<\/em><sup>2<\/sup>) = 4<em>x<\/em><sup>3<\/sup><\/div>\n<div id=\"fs-idp820144\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1941\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-300x68.png\" alt=\"\" width=\"238\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-300x68.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-768x175.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-225x51.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b-350x80.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1b.png 923w\" sizes=\"auto, (max-width: 238px) 100vw, 238px\" \/><\/div>\n<p id=\"fs-idp827984\">As defined in the ICE table, <em data-effect=\"italics\">x<\/em> is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)<sub>2<\/sub> is 6.9 \u00d7 10<sup>\u20133<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp830272\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of PbI<sub>2<\/sub> is 1.4 \u00d7 10<sup>\u20138<\/sup>. Calculate the molar solubility of lead(II) iodide.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp833104\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp833488\">1.5 \u00d7 10<sup>\u20133<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp836192\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp836448\"><strong>Determination of <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> from Gram Solubility<\/strong><\/p>\n<p>Many of the pigments used by artists in oil-based paints (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_OilPaints\">(Figure)<\/a>) are sparingly soluble in water. For example, the solubility of the artist\u2019s pigment chrome yellow, PbCrO<sub>4<\/sub>, is 4.6 \u00d7 10<sup>\u20136<\/sup> g\/L. Determine the solubility product for PbCrO<sub>4<\/sub>.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_15_01_OilPaints\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO<sub>4<\/sub>), examples include Prussian blue (Fe<sub>7<\/sub>(CN)<sub>18<\/sub>), the reddish-orange color vermilion (HgS), and green color veridian (Cr<sub>2<\/sub>O<sub>3<\/sub>). (credit: Sonny Abesamis)<\/div>\n<p><span id=\"fs-idp840928\" data-type=\"media\" data-alt=\"A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_OilPaints-1.jpg\" alt=\"A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp612896\"><strong>Solution:<\/strong><\/p>\n<p id=\"fs-idp619872\">Before calculating the solubility product, the provided solubility must be converted to molarity:<\/p>\n<div id=\"fs-idp120984032\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1942\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-300x86.png\" alt=\"\" width=\"317\" height=\"91\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-300x86.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-1024x292.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-768x219.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-225x64.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c-350x100.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1c.png 1138w\" sizes=\"auto, (max-width: 317px) 100vw, 317px\" \/><\/div>\n<p id=\"fs-idp15351552\">The dissolution equation for this compound is<\/p>\n<div id=\"fs-idp100842208\" style=\"padding-left: 40px\" data-type=\"equation\">PbCrO<sub>4<\/sub>(<em>s<\/em>) \u21cc Pb<sup>2+<\/sup>(<em>aq<\/em>) + CrO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idp53410000\">The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both [Pb<sup>2+<\/sup>] and [CrO<sub>4<\/sub><sup>2\u2212<\/sup>] are equal to the molar solubility of PbCrO<sub>4<\/sub>:<\/p>\n<div id=\"fs-idp34481696\" style=\"padding-left: 40px\" data-type=\"equation\">[Pb<sup>2+<\/sup>] = [CrO<sub>4<\/sub><sup>2\u2212<\/sup>] = 1.4 \u00d7 10<sup>-8<\/sup> M<\/div>\n<p id=\"fs-idp83072\" style=\"padding-left: 40px\"><em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][CrO<sub>4<\/sub><sup>2\u2212<\/sup>] = (1.4 \u00d7 10<sup>\u20138 <\/sup>M)(1.4 \u00d7 10<sup>\u20138 <\/sup>M) = 2.0 \u00d7 10<sup>\u201316<\/sup><\/p>\n<p id=\"fs-idp87280\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at 20\u00b0C. What is its solubility product?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp88320\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp88704\">2.08 \u00d7 10<sup>\u20134<\/sup><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp91888\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp92144\"><strong>Calculating the Solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> <\/strong><\/p>\n<p>Calomel, Hg<sub>2<\/sub>Cl<sub>2<\/sub>, is a compound composed of the diatomic ion of mercury(I), Hg<sub>2<\/sub><sup>2+<\/sup>, and chloride ions, Cl<sup>\u2013<\/sup>. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>:<\/p>\n<div id=\"fs-idp97168\" style=\"padding-left: 40px\" data-type=\"equation\">Hg<sub>2<\/sub>Cl<sub>2<\/sub>(<em>s<\/em>) \u21cc Hg<sub>2<\/sub><sup>2+<\/sup>(<em>aq<\/em>) + 2Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp <\/sub>= 1.1 \u00d7 10<sup>-18<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp110192\">Calculate the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub>.<\/p>\n<p id=\"fs-idp15521184\"><strong>Solution:<\/strong><\/p>\n<p>The dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to the concentration of Hg<sub>2<\/sub><sup>2+<\/sup> ions.<\/p>\n<p id=\"fs-idm324614160\">Following the ICE approach results in<\/p>\n<p><span id=\"fs-idp15533712\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, \u201cH g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following: 0, positive 2 x, 2 x.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable2_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, \u201cH g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following: 0, positive 2 x, 2 x.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idm324611760\">Substituting the equilibrium concentration terms into the solubility product expression and solving for <em data-effect=\"italics\">x<\/em> gives<\/p>\n<div id=\"fs-idp53869216\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Hg<sub>2<\/sub><sup>2+<\/sup>][Cl<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\n<div id=\"fs-idm14694048\" style=\"padding-left: 40px\" data-type=\"equation\">1.1 \u00d7 10<sup>-18<\/sup> = (<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup><\/div>\n<div id=\"fs-idp127117360\" style=\"padding-left: 40px\" data-type=\"equation\">4<em>x<\/em><sup>3<\/sup> = 1.1 \u00d7 10<sup>-18<\/sup><\/div>\n<div id=\"fs-idp29132112\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1944\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-300x58.png\" alt=\"\" width=\"274\" height=\"53\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-300x58.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-768x149.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-225x44.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d-350x68.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/15.1d.png 1005w\" sizes=\"auto, (max-width: 274px) 100vw, 274px\" \/><\/div>\n<div id=\"fs-idp32484816\" style=\"padding-left: 40px\" data-type=\"equation\">[Hg<sub>2<\/sub><sup>2+<\/sup>] = 6.5 \u00d7 10<sup>-7<\/sup> M<\/div>\n<div id=\"fs-idm84487648\" style=\"padding-left: 40px\" data-type=\"equation\">[Cl<sup>\u2212<\/sup>] = 2x = 2(6.5 \u00d7 10<sup>-7<\/sup> M) = 1.3 \u00d7 10<sup>-6<\/sup> M<\/div>\n<p id=\"fs-idp15393408\">The dissolution stoichiometry shows the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to [Hg<sub>2<\/sub><sup>2+<\/sup>] or 6.5 \u00d7\u00a0 10<sup>\u20137<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp751088\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Determine the molar solubility of MgF<sub>2<\/sub> from its solubility product: <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = 6.4 \u00d7 10<sup>\u20139<\/sup>.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp753920\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp754304\">1.2 \u00d7 10<sup>\u20133 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp765232\" class=\"chemistry sciences-interconnect\" data-type=\"note\">\n<div data-type=\"title\"><strong>Using Barium Sulfate for Medical Imaging<\/strong><\/div>\n<p id=\"fs-idp766160\">Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of barium sulfate is 2.3 \u00d7 10<sup>\u20138<\/sup>, very little of it dissolves as it coats the lining of the patient\u2019s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_BariumXray\">(Figure)<\/a>).<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_15_01_BariumXray\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by \u201cglitzy queen00\u201d\/Wikimedia Commons)<\/div>\n<p><span id=\"fs-idp770240\" data-type=\"media\" data-alt=\"This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_BariumXray-1.jpg\" alt=\"This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp772848\">Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn\u2019s disease, and ulcers in addition to other conditions.<\/p>\n<p id=\"fs-idp774080\">Visit this <a href=\"http:\/\/openstaxcollege.org\/l\/16barium\">website<\/a> for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp775600\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Predicting Precipitation<\/strong><\/h3>\n<p id=\"fs-idp776288\">The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:<\/p>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idp776784\" style=\"padding-left: 40px\" data-type=\"equation\">CaCO<sub>3<\/sub>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + CO<sub>3<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][CO<sub>3<\/sub><sup>2-<\/sup>] = 8.7 \u00d7 10<sup>-9<\/sup><\/div>\n<p id=\"fs-idp785408\">It is important to realize that this equilibrium is established in any aqueous solution containing Ca<sup>2+<\/sup> and CO<sub>3<\/sub><sup>2\u2013<\/sup> ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, <em data-effect=\"italics\">Q<sub>sp<\/sub><\/em>, that exceeds the solubility product, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (<em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> = <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>). The comparison of <em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> to <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:<\/p>\n<p id=\"fs-idm425158560\"><em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> &lt; <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)<\/p>\n<p id=\"fs-idm425156768\"><em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> &gt; <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)<\/p>\n<p id=\"fs-idm425154976\">This predictive strategy and related calculations are demonstrated in the next few example exercises.<\/p>\n<div id=\"fs-idp69099520\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp15157136\"><strong>Precipitation of Mg(OH)<sub>2<\/sub> <\/strong><\/p>\n<p>The first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of lime, Ca(OH)<sub>2<\/sub>, a readily available inexpensive source of OH<sup>\u2013<\/sup> ion:<\/p>\n<div id=\"fs-idp15159216\" style=\"padding-left: 40px\" data-type=\"equation\">Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 8.9 \u00d7 10<sup>-12<\/sup><\/div>\n<p id=\"fs-idp15170960\">The concentration of Mg<sup>2+<\/sup>(<em data-effect=\"italics\">aq<\/em>) in sea water is 0.0537 <em data-effect=\"italics\">M<\/em>. Will Mg(OH)<sub>2<\/sub> precipitate when enough Ca(OH)<sub>2<\/sub> is added to give a [OH<sup>\u2013<\/sup>] of 0.0010 <em data-effect=\"italics\">M<\/em>?<\/p>\n<p id=\"fs-idp15174912\"><strong>Solution:<\/strong><\/p>\n<p id=\"fs-idp15183504\">Calculation of the reaction quotient under these conditions is shown here:<\/p>\n<div id=\"fs-idp15187872\" data-type=\"equation\">Q = [Mg<sup>2+<\/sup>]<sub>0<\/sub>[OH<sup>\u2212<\/sup>]<sub>0<\/sub><sup>2<\/sup> = (0.0537 M)(0.0010 M<sup>)2<\/sup> = 5.4 \u00d7 10<sup>-8<\/sup><\/div>\n<p id=\"fs-idp11854608\">Because <em data-effect=\"italics\">Q<\/em> is greater than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> (<em data-effect=\"italics\">Q<\/em> = 5.4 \u00d7 10<sup>\u20138<\/sup> is larger than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = 8.9 \u00d7 10<sup>\u201312<\/sup>), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that <em data-effect=\"italics\">Q<sub>sp<\/sub><\/em> = <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>.<\/p>\n<p id=\"fs-idp11862608\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Predict whether CaHPO<sub>4<\/sub> will precipitate from a solution with [Ca<sup>2+<\/sup>] = 0.0001 <em data-effect=\"italics\">M<\/em> and [HPO<sub>4<\/sub><sup>2\u2212<\/sup>] = 0.001 <em data-effect=\"italics\">M<\/em>.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp11869072\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp11869456\">No precipitation of CaHPO<sub>4<\/sub>; <em data-effect=\"italics\">Q<\/em> = 1 \u00d7 10<sup>\u20137<\/sup>, which is less than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> (7 \u00d7 10<sup>\u20137<\/sup>)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp11872704\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp11872960\"><strong>Precipitation of AgCl <\/strong><\/p>\n<p>Does silver chloride precipitate when equal volumes of a 2.0 \u00d7 10<sup>\u20134<\/sup>&#8211;<em data-effect=\"italics\">M<\/em> solution of AgNO<sub>3<\/sub> and a 2.0 \u00d7 10<sup>\u20134<\/sup>&#8211;<em data-effect=\"italics\">M<\/em> solution of NaCl are mixed?<\/p>\n<p id=\"fs-idp11879728\"><strong>Solution:<\/strong><\/p>\n<p>The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:<\/p>\n<div id=\"fs-idp11880480\" style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idp11887792\">The solubility product is 1.6 \u00d7 10<sup>\u201310<\/sup> (see Appendix J).<\/p>\n<p id=\"fs-idp11889984\">AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO<sub>3<\/sub> and NaCl is greater than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>. Because the volume doubles when equal volumes of AgNO<sub>3<\/sub> and NaCl solutions are mixed, each concentration is reduced to half its initial value<\/p>\n<div id=\"fs-idp11893136\" style=\"padding-left: 40px\" data-type=\"equation\">\u00bd(2.0 \u00d7 10<sup>-4<\/sup> M) = 1.0 \u00d7 10<sup>-4<\/sup> M<\/div>\n<p id=\"fs-idp11902464\">The reaction quotient, <em data-effect=\"italics\">Q<\/em>, is greater than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> for AgCl, so a supersaturated solution is formed:<\/p>\n<div id=\"fs-idp456352\" style=\"padding-left: 40px\" data-type=\"equation\">Q = [Ag+]<sub>0<\/sub>[Cl\u2212]<sub>0<\/sub> = (1.0 \u00d7 10<sup>-4\u00a0 <\/sup>M)(1.0 \u00d7 10<sup>-4<\/sup> M) = 1.0 \u00d7 10<sup>-8<\/sup><\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">Q &gt; K<sub>sp<\/sub><\/div>\n<p id=\"fs-idp470144\">AgCl will precipitate from the mixture until the dissolution equilibrium is established, with <em data-effect=\"italics\">Q<\/em> equal to <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>.<\/p>\n<p id=\"fs-idp472176\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Will KClO<sub>4<\/sub> precipitate when 20 mL of a 0.050-<em data-effect=\"italics\">M<\/em> solution of K<sup>+<\/sup> is added to 80 mL of a 0.50-<em data-effect=\"italics\">M<\/em> solution of ClO<sub>4<\/sub><sup>\u2212<\/sup>? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp477040\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp477424\">No, <em data-effect=\"italics\">Q<\/em> = 4.0 \u00d7 10<sup>\u20133<\/sup>, which is less than <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> = 1.05 \u00d7 10<sup>\u20132<\/sup><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp488672\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp488928\"><strong>Precipitation of Calcium Oxalate <\/strong><\/p>\n<p>Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>, for this purpose (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_Blood\">(Figure)<\/a>). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O (calcium oxalate monohydrate). The concentration of Ca<sup>2+<\/sup> in a sample of blood serum is 2.2 \u00d7 10<sup>\u20133 <\/sup><em data-effect=\"italics\">M<\/em>. What concentration of C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212 <\/sup>ion must be established before CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O begins to precipitate?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_15_01_Blood\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Anticoagulants can be added to blood that will combine with the Ca<sup>2+<\/sup> ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)<\/div>\n<p><span id=\"fs-idp500240\" data-type=\"media\" data-alt=\"A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_Blood-1.jpg\" alt=\"A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp502656\"><strong>Solution:<\/strong><\/p>\n<p>The equilibrium expression is:<\/p>\n<div id=\"fs-idp503344\" style=\"padding-left: 40px\" data-type=\"equation\">CaC<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>(<em>s<\/em>) \u21cc Ca<sup>2+<\/sup>(<em>aq<\/em>) + C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp513120\">For this reaction:<\/p>\n<div id=\"eip-478\" class=\"unnumered\" style=\"padding-left: 40px\" data-type=\"equation\" data-label=\"\">K<sub>sp<\/sub> = [Ca<sup>2+<\/sup>][C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>] = 1.96 \u00d7 10<sup>-8<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(see Appendix J)<\/div>\n<p id=\"fs-idp340688\">Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:<\/p>\n<div id=\"fs-idp347152\" style=\"padding-left: 40px\" data-type=\"equation\">Q = K<sub>sp<\/sub> =[Ca<sup>2+<\/sup>][C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>]= 1.96 \u00d7 10<sup>-8<\/sup><\/div>\n<div id=\"fs-idp356336\" style=\"padding-left: 40px\" data-type=\"equation\">(2.2 \u00d7 10<sup>-3<\/sup> M)[C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>] = 1.96 \u00d7 10<sup>-8<\/sup><\/div>\n<div id=\"fs-idp365472\" style=\"padding-left: 40px\" data-type=\"equation\">[C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup>] = 8.9 \u00d7 10<sup>-6<\/sup> M<\/div>\n<p id=\"fs-idp377504\">A concentration of 8.9 \u00d7 10<sup>\u20136 <\/sup><em data-effect=\"italics\">M C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup><\/em> is necessary to initiate the precipitation of CaC<sub>2<\/sub>O<sub>4<\/sub> under these conditions.<\/p>\n<p id=\"fs-idp383648\"><strong>Check Your Learning:<\/strong><\/p>\n<p>If a solution contains 0.0020 mol CrO<sub>4<\/sub><sup>2\u2212<\/sup> per liter, what concentration of Ag<sup>+<\/sup> ion must be reached by adding solid AgNO<sub>3<\/sub> before Ag<sub>2<\/sub>CrO<sub>4<\/sub> begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp387952\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp388336\">6.7 \u00d7 10<sup>\u20135\u00a0<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp392704\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp392960\"><strong>Concentrations Following Precipitation <\/strong><\/p>\n<p>Clothing washed in water that has [Mn<sup>2+<\/sup>] exceeding 0.1 mg\/L (1.8 \u00d7 10<sup>\u20136 <\/sup><em data-effect=\"italics\">M<\/em>) may be stained by the manganese upon oxidation, but the amount of Mn<sup>2+<\/sup> in the water can be decreased by adding a base to precipitate Mn(OH)<sub>2<\/sub>. What pH is required to keep [Mn<sup>2+<\/sup>] equal to 1.8 \u00d7 10<sup>\u20136 <\/sup><em data-effect=\"italics\">M<\/em>?<\/p>\n<p id=\"fs-idp399216\"><strong>Solution:<\/strong><\/p>\n<p>The dissolution of Mn(OH)<sub>2<\/sub> is described by the equation:<\/p>\n<div id=\"fs-idp400336\" style=\"padding-left: 40px\" data-type=\"equation\">Mn(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mn<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 2 \u00d7 10<sup>-13<\/sup><\/div>\n<p id=\"fs-idp1156752\">At equilibrium:<\/p>\n<div id=\"fs-idp1159600\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Mn<sup>2+<\/sup>][OH<sup>\u2212<\/sup>]<sup>2<\/sup><\/div>\n<div id=\"fs-idp1164576\" style=\"padding-left: 40px\" data-type=\"equation\">(1.8 \u00d7 10<sup>-6<\/sup> M)[OH<sup>\u2212<\/sup>]<sup>2<\/sup> = 2 \u00d7 10<sup>-13<\/sup><\/div>\n<p id=\"fs-idp1173408\" style=\"padding-left: 40px\">[ OH<sup>\u2212<\/sup>] = 3 \u00d7 10<sup>-4<\/sup> M<\/p>\n<p id=\"fs-idp1179472\">Calculate the pH from the pOH:<\/p>\n<div id=\"fs-idp1179904\" style=\"padding-left: 40px\" data-type=\"equation\">pOH = \u2212log [OH<sup>\u2212<\/sup>] = \u2212log(3 \u00d7 10<sup>-4<\/sup> M) = 3.5<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">pH = 14.00 &#8211; pOH = 14.00 &#8211; 3.5 = 10.5<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp1193504\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of Ca(OH)<sub>2<\/sub>. The concentration of Mg<sup>2+<\/sup>(<em data-effect=\"italics\">aq<\/em>) in sea water is 5.37 \u00d7 10<sup>\u20132 <\/sup><em data-effect=\"italics\">M<\/em>. Calculate the pH at which [Mg<sup>2+<\/sup>] is decreased to 1.0 \u00d7 10<sup>\u20135<\/sup><em data-effect=\"italics\">M.<\/em><\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp1200000\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp1200384\">10.97<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idp1200896\">In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called <span data-type=\"term\">selective precipitation<\/span> may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound\u2019s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp1215552\" class=\"chemistry everyday-life\" data-type=\"note\">\n<div data-type=\"title\"><strong>The Role of Precipitation in Wastewater Treatment<\/strong><\/div>\n<p id=\"fs-idp1216528\">Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (<a class=\"autogenerated-content\" href=\"#CNX_Chem_15_01_Wastewater\">(Figure)<\/a>). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions, PO<sub>4<\/sub><sup>3\u2212<\/sup>, are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_15_01_Wastewater\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: \u201ceutrophication&amp;hypoxia\u201d\/Wikimedia Commons)<\/div>\n<p><span id=\"fs-idp1220944\" class=\"scaled-down\" data-type=\"media\" data-alt=\"A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_Wastewater-1.jpg\" alt=\"A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp1223232\">One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)<sub>2<\/sub>. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, which then precipitates out of the solution:<\/p>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idp1225152\" style=\"padding-left: 40px\" data-type=\"equation\">5Ca<sup>2+<\/sup>(<em>aq<\/em>) + 3PO<sub>4<\/sub><sup>3\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) \u2192 Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH(<em>s<\/em>)<\/div>\n<p id=\"fs-idp1234976\">Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO<sub>2<\/sub> in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.<\/p>\n<p id=\"fs-idp1236048\">View this <a href=\"http:\/\/openstaxcollege.org\/l\/16Wastewater\">site<\/a> for more information on how phosphorus is removed from wastewater.<\/p>\n<\/div>\n<div id=\"fs-idp1242128\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp1242384\"><strong>Precipitation of Silver Halides <\/strong><\/p>\n<p>A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?<\/p>\n<p id=\"fs-idp1243456\"><strong>Solution:<\/strong><\/p>\n<p>The two equilibria involved are:<\/p>\n<div id=\"fs-idp1244144\" style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 1.6 \u00d7 10<sup>-10<\/sup><\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">AgBr(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Br<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = 5.0 \u00d7 10<sup>-13<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp15236832\">If the solution contained about <em data-effect=\"italics\">equal<\/em> concentrations of Cl<sup>\u2013<\/sup> and Br<sup>\u2013<\/sup>, then the silver salt with the smaller <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag<sup>+<\/sup>] at which AgCl begins to precipitate and the [Ag<sup>+<\/sup>] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag<sup>+<\/sup>] precipitates first.<\/p>\n<p id=\"fs-idp15240992\">AgBr precipitates when <em data-effect=\"italics\">Q<\/em> equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> for AgBr:<\/p>\n<div id=\"fs-idp15244832\" data-type=\"equation\">Q = [Ag<sup>+<\/sup>]<sub>0<\/sub>[Br<sup>\u2212<\/sup>]<sub>0<\/sub> = [Ag<sup>+<\/sup>]<sub>0<\/sub>(0.00010 M) = 5.0 \u00d7 10<sup>-13<\/sup><\/div>\n<div id=\"fs-idp15252336\" data-type=\"equation\">[Ag<sup>+<\/sup>] = (5.0 \u00d7 10<sup>-13<\/sup>)\/(0.00010 M) = 5.0 \u00d7 10<sup>-9<\/sup> M<\/div>\n<p id=\"fs-idp15261056\">AgBr begins to precipitate when [Ag<sup>+<\/sup>] is 5.0 \u00d7 10<sup>\u20139 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp15263392\">For AgCl: AgCl precipitates when <em data-effect=\"italics\">Q<\/em> equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> for AgCl:<\/p>\n<div id=\"fs-idp15244832\" data-type=\"equation\">Q = [Ag<sup>+<\/sup>]<sub>0<\/sub>[Cl<sup>\u2212<\/sup>]<sub>0<\/sub> = [Ag<sup>+<\/sup>]<sub>0<\/sub>(0.10 M) = 1.6 \u00d7 10<sup>-10<\/sup><\/div>\n<div id=\"fs-idp15252336\" data-type=\"equation\">[Ag<sup>+<\/sup>] = (1.6 \u00d7 10<sup>-10<\/sup>)\/(0.10 M) = 1.6 \u00d7 10<sup>-9<\/sup>M<\/div>\n<p id=\"fs-idp15261056\">AgCl begins to precipitate when [Ag<sup>+<\/sup>] is 1.6 \u00d7 10<sup>\u20139 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<div id=\"fs-idp15267232\" data-type=\"equation\"><\/div>\n<p id=\"fs-idp15286672\">AgCl begins to precipitate at a lower [Ag<sup>+<\/sup>] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> greater than that of silver bromide.<\/p>\n<p id=\"fs-idp15287568\"><strong>Check Your Learning:<\/strong><\/p>\n<p>If silver nitrate solution is added to a solution which is 0.050 <em data-effect=\"italics\">M<\/em> in both Cl<sup>\u2013<\/sup> and Br<sup>\u2013<\/sup> ions, at what [Ag<sup>+<\/sup>] would precipitation begin, and what would be the formula of the precipitate?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp15290048\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp15290432\">[Ag<sup>+<\/sup>] = 1.0 \u00d7 10<sup>\u201311<\/sup><em data-effect=\"italics\">M<\/em>; AgBr precipitates first<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp15293024\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Common Ion Effect<\/strong><\/h3>\n<p id=\"fs-idp15293696\">Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a <em data-effect=\"italics\">common ion<\/em> (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the <strong data-effect=\"bold\">common ion effect<\/strong>, which may be explained using Le Ch\u00e2telier\u2019s principle. Consider the dissolution of silver iodide:<\/p>\n<div id=\"fs-idp15306976\" style=\"padding-left: 40px\" data-type=\"equation\">AgI(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + I<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idp15314160\">This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag<sup>+<\/sup> and I<sup>\u2013<\/sup>. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.<\/p>\n<p id=\"fs-idm386466432\">This effect may also be explained as represented in the solubility product expression:<\/p>\n<div id=\"fs-idm386465920\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [Ag<sup>+<\/sup>][I<sup>&#8211;<\/sup>]<\/div>\n<p id=\"fs-idm386458688\">The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture <em data-effect=\"italics\">regardless of the source of the ions<\/em>, and so an increase in one ion\u2019s concentration must be balanced by a proportional decrease in the other.<\/p>\n<div id=\"fs-idp11083040\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp11083296\"><strong>Common Ion Effect on Solubility <\/strong><\/p>\n<p>What is the effect on the amount of solid Mg(OH)<sub>2<\/sub> and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a saturated solution of Mg(OH)<sub>2<\/sub>?<\/p>\n<p id=\"fs-idp11085600\">(a) MgCl<sub>2<\/sub><\/p>\n<p id=\"fs-idp11086240\">(b) KOH<\/p>\n<p id=\"fs-idp11087008\">(c) NaNO<sub>3<\/sub><\/p>\n<p id=\"fs-idp11087648\">(d) Mg(OH)<sub>2<\/sub><\/p>\n<p id=\"fs-idp11088288\"><strong>Solution:<\/strong><\/p>\n<p>The solubility equilibrium is<\/p>\n<div id=\"fs-idp141736144\" style=\"padding-left: 40px\" data-type=\"equation\">Mg(OH)<sub>2<\/sub>(<em>s<\/em>) \u21cc Mg<sup>2+<\/sup>(<em>aq<\/em>) + 2OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm536672\">(a) The reaction shifts to the left to relieve the stress produced by the additional Mg<sup>2+<\/sup> ion, in accordance with Le Ch\u00e2telier\u2019s principle. In quantitative terms, the added Mg<sup>2+<\/sup> causes the reaction quotient to be larger than the solubility product (<em data-effect=\"italics\">Q<\/em> &gt; <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>), and Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is less and [Mg<sup>2+<\/sup>] is greater than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/p>\n<p id=\"fs-idp15127072\">(b) The reaction shifts to the left to relieve the stress of the additional OH<sup>\u2013<\/sup> ion. Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is greater and [Mg<sup>2+<\/sup>] is less than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/p>\n<p id=\"fs-idp15130768\">(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.<\/p>\n<p id=\"fs-idm330455104\">(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.<\/p>\n<div id=\"fs-idm740800\" style=\"padding-left: 40px\" data-type=\"equation\">Q = [Mg<sup>2+<\/sup>]<sub>0<\/sub>[OH<sup>\u2212<\/sup>]<sub>0<\/sub><sup>2<\/sup><\/div>\n<p id=\"fs-idm735616\">Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of <em data-effect=\"italics\">Q<\/em>, and no shift is required to restore <em data-effect=\"italics\">Q<\/em> to the value of the equilibrium constant.<\/p>\n<p id=\"fs-idm701008\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the effect on the amount of solid NiCO<sub>3<\/sub> and the concentrations of Ni<sup>2+<\/sup> and CO<sub>3<\/sub><sup>2\u2212<\/sup> when each of the following are added to a saturated solution of NiCO<sub>3<\/sub><\/p>\n<p id=\"fs-idm697184\">(a) Ni(NO<sub>3<\/sub>)<sub>2<\/sub><\/p>\n<p id=\"fs-idm696160\">(b) KClO<sub>4<\/sub><\/p>\n<p id=\"fs-idm695520\">(c) NiCO<sub>3<\/sub><\/p>\n<p id=\"fs-idm694880\">(d) K<sub>2<\/sub>CO<sub>3<\/sub><\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm690608\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm690096\">(a) mass of NiCO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) increases, [Ni<sup>2+<\/sup>] increases, [CO<sub>3<\/sub><sup>2\u2212<\/sup>] decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO<sub>3<\/sub>; (d) mass of NiCO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) increases, [Ni<sup>2+<\/sup>] decreases, [CO<sub>3<\/sub><sup>2\u2212<\/sup>] increases.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp15318512\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp15318768\"><strong>Common Ion Effect <\/strong><\/p>\n<p>Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-<em data-effect=\"italics\">M<\/em> solution of cadmium bromide (CdBr<sub>2<\/sub>). The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of CdS is 1.0 \u00d7 10<sup>\u201328<\/sup>.<\/p>\n<p id=\"fs-idp15322272\"><strong>Solution:<\/strong><\/p>\n<p>This calculation can be performed using the ICE approach:<\/p>\n<div id=\"fs-idp15323568\" style=\"padding-left: 40px\" data-type=\"equation\">CdS(<em>s<\/em>) \u21cc Cd<sup>2+<\/sup>(<em>aq<\/em>) + S<sup>2\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p><span id=\"fs-idp15331040\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cC d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, positive x, 0.010 plus x. The third column has the following: 0, positive x, x.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_15_01_ICETable3_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cC d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, positive x, 0.010 plus x. The third column has the following: 0, positive x, x.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<div id=\"fs-idp15332448\" style=\"padding-left: 40px\" data-type=\"equation\">\u00a0K<sub>sp<\/sub> = [Cd<sup>2+<\/sup>][S<sup>2\u2212<\/sup>] = 1.0 \u00d7 10<sup>-28<\/sup><\/div>\n<div id=\"fs-idp13457936\" style=\"padding-left: 80px\" data-type=\"equation\">(0.010 M + <em>x<\/em>)(<em>x<\/em>) = 1.0 \u00d7 10<sup>-28<\/sup><\/div>\n<p id=\"fs-idm394907456\">Because <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> is very small, assume <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.010 M, i.e. <em data-effect=\"italics\">x<\/em> &lt; 0.00050 M and solve the simplified equation for <em data-effect=\"italics\">x<\/em>:<\/p>\n<div id=\"fs-idp13479456\" style=\"padding-left: 40px\" data-type=\"equation\">(0.010)(<em>x<\/em><span style=\"font-size: 1em\">) = 1.0 \u00d7 10<sup>-28<\/sup><\/span><\/div>\n<div id=\"fs-idp13485120\" style=\"padding-left: 40px\" data-type=\"equation\"><em>x<\/em> = 1.0 \u00d7 10<sup>-26<\/sup> M \u00a0 \u00a0 \u00a0 \u00a0 \u00a0ASSUMPTION VALID<\/div>\n<p id=\"fs-idp13488544\">The molar solubility of CdS in this solution is 1.0 \u00d7 10<sup>\u201326 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp13490448\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate the molar solubility of aluminum hydroxide, Al(OH)<sub>3<\/sub>, in a 0.015-<em data-effect=\"italics\">M<\/em> solution of aluminum nitrate, Al(NO<sub>3<\/sub>)<sub>3<\/sub>. The <em data-effect=\"italics\">K<\/em><sub>sp<\/sub> of Al(OH)<sub>3<\/sub> is 2 \u00d7 10<sup>\u201332<\/sup>.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp13495104\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp13495488\">4 \u00d7 10<sup>\u201311 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp13497520\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idp13498448\">The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid M<sub><em data-effect=\"italics\">p<\/em><\/sub>X<sub><em data-effect=\"italics\">q<\/em><\/sub> and its ions M<sup>m+<\/sup> and X<sup>n\u2013<\/sup>:<\/p>\n<div id=\"fs-idp13501568\" style=\"padding-left: 40px\" data-type=\"equation\">M<sub>p<\/sub>X<sub>q<\/sub>(<em>s<\/em>) \u21cc pM<sup>m+<\/sup>(<em>aq<\/em>) + qX<sup>n\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idp13510416\">the solubility product expression is:<\/p>\n<div id=\"fs-idp13510864\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>sp<\/sub> = [M<sup>m+<\/sup>]<sup>p<\/sup>[X<sup>n\u2212<\/sup>]<sup>q<\/sup><\/div>\n<p id=\"fs-idp13516480\">The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.<\/p>\n<p id=\"fs-idp13518032\">A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.<\/p>\n<\/div>\n<div id=\"fs-idp13519952\" class=\"key-equations\" data-depth=\"1\">\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-idp13535408\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idp3007504\" data-type=\"exercise\"><\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\n<dl id=\"fs-idp3050576\">\n<dt>common ion effect<\/dt>\n<dd id=\"fs-idp3050992\">effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base<\/dd>\n<\/dl>\n<dl id=\"fs-idp3051616\">\n<dt>molar solubility<\/dt>\n<dd id=\"fs-idp3052032\">solubility of a compound expressed in units of moles per liter (mol\/L)<\/dd>\n<\/dl>\n<dl id=\"fs-idp3053344\">\n<dt>solubility product constant (<em data-effect=\"italics\">K<\/em><sub>sp<\/sub>)<\/dt>\n<dd id=\"fs-idp3054640\">equilibrium constant for the dissolution of an ionic compound<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-823","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":811,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/823","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/823\/revisions"}],"predecessor-version":[{"id":2177,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/823\/revisions\/2177"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/811"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/823\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=823"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=823"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=823"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=823"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}