{"id":773,"date":"2021-07-23T09:20:40","date_gmt":"2021-07-23T13:20:40","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/bronsted-lowry-acids-and-bases\/"},"modified":"2022-06-23T09:20:21","modified_gmt":"2022-06-23T13:20:21","slug":"bronsted-lowry-acids-and-bases","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/bronsted-lowry-acids-and-bases\/","title":{"raw":"14.1 Br\u00f8nsted-Lowry Acids and Bases","rendered":"14.1 Br\u00f8nsted-Lowry Acids and Bases"},"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>Identify acids, bases, and conjugate acid-base pairs according to the Br\u00f8nsted-Lowry definition<\/li>\r\n \t<li>Write equations for acid and base ionization reactions<\/li>\r\n \t<li>Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations<\/li>\r\n \t<li>Describe the acid-base behavior of amphiprotic substances<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm196456848\">The acid-base reaction class has been studied for quite some time. In 1680, Robert <span class=\"no-emphasis\" data-type=\"term\">Boyle<\/span> reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO<sub>2<\/sub>), and interact with alkalis to form neutral substances. In 1815, Humphry <span class=\"no-emphasis\" data-type=\"term\">Davy<\/span> contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante <span class=\"no-emphasis\" data-type=\"term\">Arrhenius<\/span> defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.<\/p>\r\n<p id=\"fs-idm26037424\">Johannes Br\u00f8nsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H<sup>+<\/sup>. (Note that these hydrogen ions are often referred to simply as <em data-effect=\"italics\">protons<\/em>, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, <sup>1<\/sup>H.) A compound that donates a proton to another compound is called a <strong>Br\u00f8nsted-Lowry acid<\/strong>, and a compound that accepts a proton is called a <strong>Br\u00f8nsted-Lowry base<\/strong>. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).<\/p>\r\n<p id=\"fs-idm56373376\">The concept of <em data-effect=\"italics\">conjugate pairs<\/em> is useful in describing Br\u00f8nsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates H<sup>+<\/sup>, the species that remains is called the <strong data-effect=\"bold\">conjugate base<\/strong> of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts H<sup>+<\/sup>, it is converted to its <strong data-effect=\"bold\">conjugate acid<\/strong>. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OH<sup>\u2212<\/sup>, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, NH<sub>4<\/sub><sup>+<\/sup>, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.<\/p>\r\n<span id=\"fs-idm51052736\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, an H atom is connected to an F atom with a single bond. The F atom has pairs of electron dots at the top, right, and bottom. This is followed by a plus sign, which is followed in blue by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in blue, with blue H atoms singly bonded above and to the right. A pair of blue electron dots is on the lower side of the O atom. To the left of the blue O atom, a red H atom is singly bonded. This is followed by a plus sign and an F atom in red with pairs of electron dots above, right, below, and to the left. This atom also has a superscript negative sign. The reaction is written in symbolic form below. H F is labeled in red below as \u201cAcid.\u201d This is followed by plus H subscript 2 O, which is labeled in blue below as \u201cBase.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in blue below as \u201cAcid.\u201d This is followed by plus and F superscript negative. The label below in red reads, \u201cBase.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_conjugate_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, an H atom is connected to an F atom with a single bond. The F atom has pairs of electron dots at the top, right, and bottom. This is followed by a plus sign, which is followed in blue by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in blue, with blue H atoms singly bonded above and to the right. A pair of blue electron dots is on the lower side of the O atom. To the left of the blue O atom, a red H atom is singly bonded. This is followed by a plus sign and an F atom in red with pairs of electron dots above, right, below, and to the left. This atom also has a superscript negative sign. The reaction is written in symbolic form below. H F is labeled in red below as \u201cAcid.\u201d This is followed by plus H subscript 2 O, which is labeled in blue below as \u201cBase.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in blue below as \u201cAcid.\u201d This is followed by plus and F superscript negative. The label below in red reads, \u201cBase.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idm59705584\">The reaction between a Br\u00f8nsted-Lowry acid and water is called <span data-type=\"term\">acid ionization<\/span>. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:<\/p>\r\n<span id=\"fs-idm180061536\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_HF_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idm185234272\"><span data-type=\"term\">Base ionization<\/span> of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, C<sub>5<\/sub>NH<sub>5<\/sub>, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:<\/p>\r\n<span id=\"fs-idm159126864\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. To the right, is the equation: k equals [ C subscript 5 N H subscript 6 superscript positive sign ] [ O H superscript negative sign] all divided by [ C subscript 5 N H subscript 5 ]. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_NH3_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. To the right, is the equation: k equals [ C subscript 5 N H subscript 6 superscript positive sign ] [ O H superscript negative sign] all divided by [ C subscript 5 N H subscript 5 ]. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"eip-304\">The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called <span data-type=\"term\">amphiprotic<\/span>, or more generally, <span data-type=\"term\">amphoteric<\/span>, a term that may be used for acids and bases per definitions other than the Br\u00f8nsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:<\/p>\r\n\r\n<div id=\"eip-402\" class=\"unnumbered\" style=\"padding-left: 40px\" data-type=\"equation\">HCO<sub>3<\/sub><sup>\u2013<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)\u00a0 \u00a0 \u21cc \u00a0 \u00a0 CO<sub>3<\/sub><sup>2\u2013<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>)<\/div>\r\n<div id=\"eip-167\" class=\"unnumbered\" style=\"padding-left: 40px\" data-type=\"equation\">HCO<sub>3<\/sub><sup>\u2013<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)\u00a0 \u00a0 \u21cc \u00a0 \u00a0 H<sub>2<\/sub>CO<sub>3<\/sub>(aq) + OH<sup>\u2013<\/sup>(aq)<\/div>\r\n<p id=\"fs-idp176970528\">The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.<\/p>\r\n<p id=\"fs-idm206801664\">In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:<\/p>\r\n<span id=\"fs-idm27406976\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as \u201cAcid subscript 1.\u201d This is followed by plus H subscript 2 O, which is labeled in green below as \u201cBase subscript 2.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as \u201cAcid subscript 2.\u201d This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, \u201c Base subscript 1.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_Water_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as \u201cAcid subscript 1.\u201d This is followed by plus H subscript 2 O, which is labeled in green below as \u201cBase subscript 2.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as \u201cAcid subscript 2.\u201d This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, \u201c Base subscript 1.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idm114594272\">The process in which like molecules react to yield ions is called <span data-type=\"term\">autoionization<\/span>. Liquid water undergoes autoionization to a very slight extent; at 25\u00b0C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the <span data-type=\"term\">ion-product constant for water, <em data-effect=\"italics\">K<\/em><sub>w<\/sub><\/span>:<\/p>\r\n\r\n<div id=\"fs-idm57289696\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>]<\/div>\r\n<p id=\"fs-idp31161136\">The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25\u00b0C, <em data-effect=\"italics\">K<\/em><sub>w<\/sub> has a value of 1.0 \u00d7 10<sup>\u221214<\/sup>. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 \u00b0C, the value for <em data-effect=\"italics\">K<\/em><sub>w<\/sub> is about 5.6 \u00d7 10<sup>\u221213<\/sup>, roughly 50 times larger than the value at 25\u00b0C.<\/p>\r\n\r\n<div id=\"fs-idm123712192\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm112349216\"><strong>Ion Concentrations in Pure Water<\/strong><\/p>\r\nWhat are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25\u00b0C?\r\n<p id=\"fs-idm173174512\"><strong>Solution:<\/strong><\/p>\r\nThe autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>] = <em data-effect=\"italics\">x<\/em>. At 25\u00b0C:\r\n<div id=\"fs-idm124796224\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = <em>x<sup>2<\/sup><\/em> = 1.0\u00d710<sup>-14<\/sup><\/div>\r\n<p id=\"fs-idm95588864\">So:<\/p>\r\n\r\n<div id=\"fs-idm116116704\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1837\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-300x28.png\" alt=\"\" width=\"375\" height=\"35\" \/><\/div>\r\n<p id=\"fs-idm157262416\">The hydronium ion concentration and the hydroxide ion concentration are the same, 1.0 \u00d7 10<sup>\u22127 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idm86436992\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe ion product of water at 80\u00b0C is 2.4 \u00d7 10<sup>\u221213<\/sup>. What are the concentrations of hydronium and hydroxide ions in pure water at 80\u00b0C?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp5038208\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm88673520\">[H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>] = 4.9 \u00d7 10<sup>\u22127 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm136168496\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm176597552\"><strong>The Inverse Relation between [H<sub>3<\/sub>O<sup>+<\/sup>] and [OH<sup>\u2212<\/sup>]<\/strong><\/p>\r\nA solution of an acid in water has a hydronium ion concentration of 2.0 \u00d7 10<sup>\u22126 <\/sup><em data-effect=\"italics\">M<\/em>. What is the concentration of hydroxide ion at 25\u00b0C?\r\n<p id=\"fs-idm86437536\"><strong>Solution:<\/strong><\/p>\r\nUse the value of the ion-product constant for water at 25\u00b0C\r\n<div id=\"fs-idm153840352\" style=\"padding-left: 40px\" data-type=\"equation\">2H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = 1.0 \u00d7 10<sup>-14<\/sup><\/div>\r\n<p id=\"fs-idm132298320\">to calculate the missing equilibrium concentration.<\/p>\r\n<p id=\"fs-idm216521216\">Rearrangement of the <em data-effect=\"italics\">K<\/em><sub>w<\/sub> expression shows that [OH<sup>\u2212<\/sup>] is inversely proportional to [H<sub>3<\/sub>O<sup>+<\/sup>]:<\/p>\r\n\r\n<div id=\"fs-idm133309424\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1838\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-300x42.png\" alt=\"\" width=\"300\" height=\"42\" \/><\/div>\r\n<p id=\"fs-idm156698544\">Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Ch\u00e2telier\u2019s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.<\/p>\r\n<p id=\"fs-idm215254096\">Substituting the ion concentrations into the <em data-effect=\"italics\">K<\/em><sub>w<\/sub> expression confirms this calculation, resulting in the expected value:<\/p>\r\n\r\n<div id=\"fs-idm53700928\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = (2.0 \u00d7 10<sup>-6<\/sup> M)(5.0 \u00d7 10<sup>-9<\/sup> M) = 1.0 \u00d7 10<sup>-14<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm88829680\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 <em data-effect=\"italics\">M<\/em> at 25\u00b0C?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp123132928\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm107000608\">[H<sub>3<\/sub>O<sup>+<\/sup>] = 1 \u00d7 10<sup>\u221211 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp171180224\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm102224080\"><strong>Representing the Acid-Base Behavior of an Amphoteric Substance<\/strong><\/p>\r\nWrite separate equations representing the reaction of HSO<sub>3<\/sub><sup>\u2212<\/sup>\r\n<p id=\"fs-idm108563136\">(a) as an acid with OH<sup>\u2212<\/sup><\/p>\r\n<p id=\"fs-idm65200656\">(b) as a base with H<sub>3<\/sub>O<sup>+<\/sup><\/p>\r\n<p id=\"fs-idp53602512\"><strong>Solution:<\/strong><\/p>\r\n(a) HSO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) \u21cc SO<sub>3<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)\r\n<p id=\"fs-idp141113152\">(b) HSO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) \u21cc H<sub>2<\/sub>SO<sub>3<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\r\n<p id=\"fs-idm119568288\"><strong>Check Your Learning:<\/strong><\/p>\r\nWrite separate equations representing the reaction of H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>\r\n<p id=\"fs-idm102000320\">(a) as a base with H<sub>3<\/sub>O<sup>+<\/sup><\/p>\r\n<p id=\"fs-idm68486720\">(b) as an acid with OH<sup>\u2212<\/sup><\/p>\r\n&nbsp;\r\n<div id=\"fs-idp54496480\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp8583968\">(a) H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) \u21cc H<sub>3<\/sub>PO<sub>4<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\r\n(b) H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) \u21cc HPO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp108739760\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idm102966944\">A compound that can donate a proton (a hydrogen ion) to another compound is called a Br\u00f8nsted-Lowry acid. The compound that accepts the proton is called a Br\u00f8nsted-Lowry base. The species remaining after a Br\u00f8nsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Br\u00f8nsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H<sub>3<\/sub>O<sup>+<\/sup>, and the hydroxide ion, OH<sup>\u2212<\/sup> when it undergoes autoionization:<\/p>\r\n\r\n<div id=\"fs-idp99871024\" style=\"padding-left: 40px\" data-type=\"equation\">2H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idm41496496\">The ion product of water, <em data-effect=\"italics\">K<\/em><sub>w<\/sub> is the equilibrium constant for the autoionization reaction:<\/p>\r\n\r\n<div id=\"fs-idp99842048\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = 1.0 \u00d7 10<sup>-14<\/sup> at 25\u00b0C<\/div>\r\n<\/div>\r\n<div id=\"fs-idp36036608\" class=\"key-equations\" data-depth=\"1\"><\/div>\r\n<div id=\"fs-idp54308864\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idp42982384\" data-type=\"exercise\">\r\n<div id=\"fs-idp42982640\" data-type=\"problem\">\r\n<p id=\"fs-idp42982896\"><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\r\n<dl id=\"fs-idm26650176\">\r\n \t<dd id=\"fs-idm26649536\"><span style=\"color: #373d3f;font-weight: bold;font-size: 1em\">amphiprotic<\/span><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp140545888\">\r\n \t<dd id=\"fs-idp140546528\">species that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp12571968\">\r\n \t<dt>amphoteric<\/dt>\r\n \t<dd id=\"fs-idp12572608\">species that can act as either an acid or a base<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp12573120\">\r\n \t<dt>autoionization<\/dt>\r\n \t<dd id=\"fs-idp46882848\">reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp43874496\">\r\n \t<dt>Br\u00f8nsted-Lowry acid<\/dt>\r\n \t<dd id=\"fs-idp43875136\">proton donor<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp89677040\">\r\n \t<dt>Br\u00f8nsted-Lowry base<\/dt>\r\n \t<dd id=\"fs-idp89677680\">proton acceptor<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp41345504\">\r\n \t<dt>conjugate acid<\/dt>\r\n \t<dd id=\"fs-idp41346144\">substance formed when a base gains a proton<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp41346656\">\r\n \t<dt>conjugate base<\/dt>\r\n \t<dd id=\"fs-idp9778112\">substance formed when an acid loses a proton<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp9778624\">\r\n \t<dt>ion-product constant for water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>)<\/dt>\r\n \t<dd id=\"fs-idp108529200\">equilibrium constant for the autoionization of water<\/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>Identify acids, bases, and conjugate acid-base pairs according to the Br\u00f8nsted-Lowry definition<\/li>\n<li>Write equations for acid and base ionization reactions<\/li>\n<li>Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations<\/li>\n<li>Describe the acid-base behavior of amphiprotic substances<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm196456848\">The acid-base reaction class has been studied for quite some time. In 1680, Robert <span class=\"no-emphasis\" data-type=\"term\">Boyle<\/span> reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO<sub>2<\/sub>), and interact with alkalis to form neutral substances. In 1815, Humphry <span class=\"no-emphasis\" data-type=\"term\">Davy<\/span> contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante <span class=\"no-emphasis\" data-type=\"term\">Arrhenius<\/span> defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.<\/p>\n<p id=\"fs-idm26037424\">Johannes Br\u00f8nsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H<sup>+<\/sup>. (Note that these hydrogen ions are often referred to simply as <em data-effect=\"italics\">protons<\/em>, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, <sup>1<\/sup>H.) A compound that donates a proton to another compound is called a <strong>Br\u00f8nsted-Lowry acid<\/strong>, and a compound that accepts a proton is called a <strong>Br\u00f8nsted-Lowry base<\/strong>. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).<\/p>\n<p id=\"fs-idm56373376\">The concept of <em data-effect=\"italics\">conjugate pairs<\/em> is useful in describing Br\u00f8nsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates H<sup>+<\/sup>, the species that remains is called the <strong data-effect=\"bold\">conjugate base<\/strong> of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts H<sup>+<\/sup>, it is converted to its <strong data-effect=\"bold\">conjugate acid<\/strong>. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OH<sup>\u2212<\/sup>, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, NH<sub>4<\/sub><sup>+<\/sup>, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.<\/p>\n<p><span id=\"fs-idm51052736\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, an H atom is connected to an F atom with a single bond. The F atom has pairs of electron dots at the top, right, and bottom. This is followed by a plus sign, which is followed in blue by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in blue, with blue H atoms singly bonded above and to the right. A pair of blue electron dots is on the lower side of the O atom. To the left of the blue O atom, a red H atom is singly bonded. This is followed by a plus sign and an F atom in red with pairs of electron dots above, right, below, and to the left. This atom also has a superscript negative sign. The reaction is written in symbolic form below. H F is labeled in red below as \u201cAcid.\u201d This is followed by plus H subscript 2 O, which is labeled in blue below as \u201cBase.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in blue below as \u201cAcid.\u201d This is followed by plus and F superscript negative. The label below in red reads, \u201cBase.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_conjugate_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, an H atom is connected to an F atom with a single bond. The F atom has pairs of electron dots at the top, right, and bottom. This is followed by a plus sign, which is followed in blue by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in blue, with blue H atoms singly bonded above and to the right. A pair of blue electron dots is on the lower side of the O atom. To the left of the blue O atom, a red H atom is singly bonded. This is followed by a plus sign and an F atom in red with pairs of electron dots above, right, below, and to the left. This atom also has a superscript negative sign. The reaction is written in symbolic form below. H F is labeled in red below as \u201cAcid.\u201d This is followed by plus H subscript 2 O, which is labeled in blue below as \u201cBase.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in blue below as \u201cAcid.\u201d This is followed by plus and F superscript negative. The label below in red reads, \u201cBase.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idm59705584\">The reaction between a Br\u00f8nsted-Lowry acid and water is called <span data-type=\"term\">acid ionization<\/span>. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:<\/p>\n<p><span id=\"fs-idm180061536\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_HF_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idm185234272\"><span data-type=\"term\">Base ionization<\/span> of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, C<sub>5<\/sub>NH<sub>5<\/sub>, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:<\/p>\n<p><span id=\"fs-idm159126864\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. To the right, is the equation: k equals [ C subscript 5 N H subscript 6 superscript positive sign ] [ O H superscript negative sign] all divided by [ C subscript 5 N H subscript 5 ]. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_NH3_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in red, is an O atom which has H atoms singly bonded above and to the right. The O atom has lone pairs of electron dots on its left and lower sides. This is followed by a plus sign. The plus sign is followed, in blue, by an N atom with one lone pair of electron dots. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. An equilibrium arrow follows this structure. To the right, in brackets is a structure where an N atom bonded to an H atom, which is red, appears. The N atom forms a double bond with a C atom, which forms a single bond with a C atom. The second C atom forms a double bond with another C atom, which forms a single bond with another C atom. The fourth C atom forms a double bond with a fifth C atom, which forms a single bond with the N atom. This creates a ring structure. Each C atom is also bonded to an H atom. Outside the brackets, to the right, is a superscript positive sign. This structure is followed by a plus sign. Another structure that appears in brackets also appears. An O atom with three lone pairs of electron dots is bonded to an H atom. There is a superscript negative sign outside the brackets. To the right, is the equation: k equals [ C subscript 5 N H subscript 6 superscript positive sign ] [ O H superscript negative sign] all divided by [ C subscript 5 N H subscript 5 ]. Under the initial equation, is this equation: H subscript 2 plus C subscript 5 N H subscript 5 equilibrium arrow C subscript 5 N H subscript 6 superscript positive sign plus O H superscript negative sign. H subscript 2 O is labeled, \u201cacid,\u201d in red. C subscript 5 N H subscript 5 is labeled, \u201cbase,\u201d in blue. C subscript 5 N H subscript 6 superscript positive sign is labeled, \u201cacid\u201d in blue. O H superscript negative sign is labeled, \u201cbase,\u201d in red.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"eip-304\">The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called <span data-type=\"term\">amphiprotic<\/span>, or more generally, <span data-type=\"term\">amphoteric<\/span>, a term that may be used for acids and bases per definitions other than the Br\u00f8nsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:<\/p>\n<div id=\"eip-402\" class=\"unnumbered\" style=\"padding-left: 40px\" data-type=\"equation\">HCO<sub>3<\/sub><sup>\u2013<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)\u00a0 \u00a0 \u21cc \u00a0 \u00a0 CO<sub>3<\/sub><sup>2\u2013<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>)<\/div>\n<div id=\"eip-167\" class=\"unnumbered\" style=\"padding-left: 40px\" data-type=\"equation\">HCO<sub>3<\/sub><sup>\u2013<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)\u00a0 \u00a0 \u21cc \u00a0 \u00a0 H<sub>2<\/sub>CO<sub>3<\/sub>(aq) + OH<sup>\u2013<\/sup>(aq)<\/div>\n<p id=\"fs-idp176970528\">The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.<\/p>\n<p id=\"fs-idm206801664\">In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:<\/p>\n<p><span id=\"fs-idm27406976\" class=\"scaled-down\" data-type=\"media\" data-alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as \u201cAcid subscript 1.\u201d This is followed by plus H subscript 2 O, which is labeled in green below as \u201cBase subscript 2.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as \u201cAcid subscript 2.\u201d This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, \u201c Base subscript 1.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_01_Water_img-1.jpg\" alt=\"This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as \u201cAcid subscript 1.\u201d This is followed by plus H subscript 2 O, which is labeled in green below as \u201cBase subscript 2.\u201d A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as \u201cAcid subscript 2.\u201d This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, \u201c Base subscript 1.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idm114594272\">The process in which like molecules react to yield ions is called <span data-type=\"term\">autoionization<\/span>. Liquid water undergoes autoionization to a very slight extent; at 25\u00b0C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the <span data-type=\"term\">ion-product constant for water, <em data-effect=\"italics\">K<\/em><sub>w<\/sub><\/span>:<\/p>\n<div id=\"fs-idm57289696\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>]<\/div>\n<p id=\"fs-idp31161136\">The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25\u00b0C, <em data-effect=\"italics\">K<\/em><sub>w<\/sub> has a value of 1.0 \u00d7 10<sup>\u221214<\/sup>. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 \u00b0C, the value for <em data-effect=\"italics\">K<\/em><sub>w<\/sub> is about 5.6 \u00d7 10<sup>\u221213<\/sup>, roughly 50 times larger than the value at 25\u00b0C.<\/p>\n<div id=\"fs-idm123712192\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm112349216\"><strong>Ion Concentrations in Pure Water<\/strong><\/p>\n<p>What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25\u00b0C?<\/p>\n<p id=\"fs-idm173174512\"><strong>Solution:<\/strong><\/p>\n<p>The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>] = <em data-effect=\"italics\">x<\/em>. At 25\u00b0C:<\/p>\n<div id=\"fs-idm124796224\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = <em>x<sup>2<\/sup><\/em> = 1.0\u00d710<sup>-14<\/sup><\/div>\n<p id=\"fs-idm95588864\">So:<\/p>\n<div id=\"fs-idm116116704\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1837\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-300x28.png\" alt=\"\" width=\"375\" height=\"35\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-300x28.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-768x71.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-225x21.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a-350x32.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1a.png 950w\" sizes=\"auto, (max-width: 375px) 100vw, 375px\" \/><\/div>\n<p id=\"fs-idm157262416\">The hydronium ion concentration and the hydroxide ion concentration are the same, 1.0 \u00d7 10<sup>\u22127 <\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idm86436992\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The ion product of water at 80\u00b0C is 2.4 \u00d7 10<sup>\u221213<\/sup>. What are the concentrations of hydronium and hydroxide ions in pure water at 80\u00b0C?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp5038208\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm88673520\">[H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>] = 4.9 \u00d7 10<sup>\u22127 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm136168496\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm176597552\"><strong>The Inverse Relation between [H<sub>3<\/sub>O<sup>+<\/sup>] and [OH<sup>\u2212<\/sup>]<\/strong><\/p>\n<p>A solution of an acid in water has a hydronium ion concentration of 2.0 \u00d7 10<sup>\u22126 <\/sup><em data-effect=\"italics\">M<\/em>. What is the concentration of hydroxide ion at 25\u00b0C?<\/p>\n<p id=\"fs-idm86437536\"><strong>Solution:<\/strong><\/p>\n<p>Use the value of the ion-product constant for water at 25\u00b0C<\/p>\n<div id=\"fs-idm153840352\" style=\"padding-left: 40px\" data-type=\"equation\">2H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = 1.0 \u00d7 10<sup>-14<\/sup><\/div>\n<p id=\"fs-idm132298320\">to calculate the missing equilibrium concentration.<\/p>\n<p id=\"fs-idm216521216\">Rearrangement of the <em data-effect=\"italics\">K<\/em><sub>w<\/sub> expression shows that [OH<sup>\u2212<\/sup>] is inversely proportional to [H<sub>3<\/sub>O<sup>+<\/sup>]:<\/p>\n<div id=\"fs-idm133309424\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1838\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-300x42.png\" alt=\"\" width=\"300\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-300x42.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-768x106.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b-350x48.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.1b.png 831w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm156698544\">Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Ch\u00e2telier\u2019s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.<\/p>\n<p id=\"fs-idm215254096\">Substituting the ion concentrations into the <em data-effect=\"italics\">K<\/em><sub>w<\/sub> expression confirms this calculation, resulting in the expected value:<\/p>\n<div id=\"fs-idm53700928\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = (2.0 \u00d7 10<sup>-6<\/sup> M)(5.0 \u00d7 10<sup>-9<\/sup> M) = 1.0 \u00d7 10<sup>-14<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm88829680\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 <em data-effect=\"italics\">M<\/em> at 25\u00b0C?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp123132928\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm107000608\">[H<sub>3<\/sub>O<sup>+<\/sup>] = 1 \u00d7 10<sup>\u221211 <\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp171180224\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm102224080\"><strong>Representing the Acid-Base Behavior of an Amphoteric Substance<\/strong><\/p>\n<p>Write separate equations representing the reaction of HSO<sub>3<\/sub><sup>\u2212<\/sup><\/p>\n<p id=\"fs-idm108563136\">(a) as an acid with OH<sup>\u2212<\/sup><\/p>\n<p id=\"fs-idm65200656\">(b) as a base with H<sub>3<\/sub>O<sup>+<\/sup><\/p>\n<p id=\"fs-idp53602512\"><strong>Solution:<\/strong><\/p>\n<p>(a) HSO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) \u21cc SO<sub>3<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p id=\"fs-idp141113152\">(b) HSO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) \u21cc H<sub>2<\/sub>SO<sub>3<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p id=\"fs-idm119568288\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Write separate equations representing the reaction of H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup><\/p>\n<p id=\"fs-idm102000320\">(a) as a base with H<sub>3<\/sub>O<sup>+<\/sup><\/p>\n<p id=\"fs-idm68486720\">(b) as an acid with OH<sup>\u2212<\/sup><\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp54496480\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp8583968\">(a) H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) \u21cc H<sub>3<\/sub>PO<sub>4<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p>(b) H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>) \u21cc HPO<sub>4<\/sub><sup>2\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp108739760\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idm102966944\">A compound that can donate a proton (a hydrogen ion) to another compound is called a Br\u00f8nsted-Lowry acid. The compound that accepts the proton is called a Br\u00f8nsted-Lowry base. The species remaining after a Br\u00f8nsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Br\u00f8nsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H<sub>3<\/sub>O<sup>+<\/sup>, and the hydroxide ion, OH<sup>\u2212<\/sup> when it undergoes autoionization:<\/p>\n<div id=\"fs-idp99871024\" style=\"padding-left: 40px\" data-type=\"equation\">2H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idm41496496\">The ion product of water, <em data-effect=\"italics\">K<\/em><sub>w<\/sub> is the equilibrium constant for the autoionization reaction:<\/p>\n<div id=\"fs-idp99842048\" style=\"padding-left: 40px\" data-type=\"equation\">K<sub>w <\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>] = 1.0 \u00d7 10<sup>-14<\/sup> at 25\u00b0C<\/div>\n<\/div>\n<div id=\"fs-idp36036608\" class=\"key-equations\" data-depth=\"1\"><\/div>\n<div id=\"fs-idp54308864\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idp42982384\" data-type=\"exercise\">\n<div id=\"fs-idp42982640\" data-type=\"problem\">\n<p id=\"fs-idp42982896\">\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\n<dl id=\"fs-idm26650176\">\n<dd id=\"fs-idm26649536\"><span style=\"color: #373d3f;font-weight: bold;font-size: 1em\">amphiprotic<\/span><\/dd>\n<\/dl>\n<dl id=\"fs-idp140545888\">\n<dd id=\"fs-idp140546528\">species that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction<\/dd>\n<\/dl>\n<dl id=\"fs-idp12571968\">\n<dt>amphoteric<\/dt>\n<dd id=\"fs-idp12572608\">species that can act as either an acid or a base<\/dd>\n<\/dl>\n<dl id=\"fs-idp12573120\">\n<dt>autoionization<\/dt>\n<dd id=\"fs-idp46882848\">reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions<\/dd>\n<\/dl>\n<dl id=\"fs-idp43874496\">\n<dt>Br\u00f8nsted-Lowry acid<\/dt>\n<dd id=\"fs-idp43875136\">proton donor<\/dd>\n<\/dl>\n<dl id=\"fs-idp89677040\">\n<dt>Br\u00f8nsted-Lowry base<\/dt>\n<dd id=\"fs-idp89677680\">proton acceptor<\/dd>\n<\/dl>\n<dl id=\"fs-idp41345504\">\n<dt>conjugate acid<\/dt>\n<dd id=\"fs-idp41346144\">substance formed when a base gains a proton<\/dd>\n<\/dl>\n<dl id=\"fs-idp41346656\">\n<dt>conjugate base<\/dt>\n<dd id=\"fs-idp9778112\">substance formed when an acid loses a proton<\/dd>\n<\/dl>\n<dl id=\"fs-idp9778624\">\n<dt>ion-product constant for water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>)<\/dt>\n<dd id=\"fs-idp108529200\">equilibrium constant for the autoionization of water<\/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-773","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":766,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/773","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":6,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/773\/revisions"}],"predecessor-version":[{"id":2170,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/773\/revisions\/2170"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/766"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/773\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=773"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=773"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=773"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=773"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}