{"id":810,"date":"2021-07-23T09:20:47","date_gmt":"2021-07-23T13:20:47","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/acid-base-titrations\/"},"modified":"2022-06-23T09:21:42","modified_gmt":"2022-06-23T13:21:42","slug":"acid-base-titrations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/acid-base-titrations\/","title":{"raw":"14.7 Acid-Base Titrations","rendered":"14.7 Acid-Base Titrations"},"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>Interpret titration curves for strong and weak acid-base systems<\/li>\r\n \t<li>Compute sample pH at important stages of a titration<\/li>\r\n \t<li>Explain the function of acid-base indicators<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm43847120\">As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique.<\/p>\r\n\r\n<div id=\"fs-idm84795552\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Titration Curves<\/strong><\/h3>\r\n<p id=\"fs-idp135875056\">A <strong>titration curve<\/strong> is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration\u2019s progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH. The second example addresses a weak acid titration requiring equilibrium calculations.<\/p>\r\n\r\n<div id=\"fs-idm49336656\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp245005248\"><strong>Calculating pH for Titration Solutions: Strong Acid\/Strong Base <\/strong><\/p>\r\nA titration is carried out for 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCl (strong acid) with 0.100 <em data-effect=\"italics\">M<\/em> of a strong base NaOH (the titration curve is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration\">(Figure)<\/a>). Calculate the pH at these volumes of added base solution:\r\n<p id=\"fs-idp67154976\">(a) 0.00 mL<\/p>\r\n<p id=\"fs-idm80154320\">(b) 12.50 mL<\/p>\r\n<p id=\"fs-idm39390080\">(c) 25.00 mL<\/p>\r\n<p id=\"fs-idp81988928\">(d) 37.50 mL<\/p>\r\n<p id=\"fs-idp22584832\"><strong>Solution: <\/strong><\/p>\r\n(a) Titrant volume = 0.00 mL. The solution pH is due to the acid ionization of HCl. Because this is a strong acid, the ionization is complete and the hydronium ion molarity is 0.100 <em data-effect=\"italics\">M<\/em>. The pH of the solution is then\r\n<div id=\"fs-idm213177360\" style=\"padding-left: 40px\" data-type=\"equation\">pH = -log(0.100 M) = 1.000<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm214456576\">(b) Titrant volume = 12.50 mL.<\/p>\r\n<p style=\"padding-left: 40px\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>-<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>aq<\/em>)<\/p>\r\nSince the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the intial amount and then dividing by the solution volume:\r\n<p style=\"padding-left: 40px\">(0.02500 L)(0.100 mol\/L) = 0.00250 mol H<sub>3<\/sub>O<sup>+ <\/sup>initially<\/p>\r\n<p style=\"padding-left: 40px\">reacts with (0.01250 L)(0.100 mol\/L) = 0.00125 mol OH- in a 1:1 ratio<\/p>\r\n<p style=\"padding-left: 40px\">leaving 0.00250 mol - 0.00125 mol = 0.00125 mol H<sub>3<\/sub>O<sup>+<\/sup><\/p>\r\n&nbsp;\r\n<div id=\"fs-idp80308976\" style=\"padding-left: 40px\" data-type=\"equation\">[H<sub>3<\/sub>O<sup>+<\/sup>] = (0.00125 mol)\/(0.03750 L) = 0.0333 M<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">pH = -log(0.0333 M) = 1.477<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n(c) Titrant volume = 25.00 mL.\r\n<p style=\"padding-left: 40px\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>-<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>aq<\/em>)<\/p>\r\nThis titrant addition involves a stoichiometric amount of base (the <em data-effect=\"italics\">equivalence point<\/em>), and so only products of the neutralization reaction are in solution (water and NaCl). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autohydrolysis of water. The solution is neutral, having a pH = 7.00.\r\n<p id=\"fs-idm173154048\">(d) Titrant volume = 37.50 mL.<\/p>\r\n<p style=\"padding-left: 40px\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>-<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>aq<\/em>)<\/p>\r\nThis involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion:\r\n<div id=\"fs-idp66422816\" data-type=\"equation\">\r\n<p style=\"padding-left: 40px\">0.00250 mol H<sub>3<\/sub>O<sup>+ <\/sup>initially, as calculated in part (b)<\/p>\r\n<p style=\"padding-left: 40px\">(0.03750 L)(0.100 mol\/L) = 0.00375 mol OH<sup>-<\/sup> total added<\/p>\r\n<p style=\"padding-left: 40px\">reacts with H<sub>3<\/sub>O<sup>+<\/sup> in a 1:1 ratio<\/p>\r\n<p style=\"padding-left: 40px\">leaving 0.00375 mol - 0.00250 mol = 0.00125 mol OH<sup>-<\/sup><\/p>\r\n<p style=\"padding-left: 40px\">[OH-] = (0.00125 mol)\/(0.06250 L) = 0.0200 M<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idm41259968\" style=\"padding-left: 40px\">pH = 14.00 \u2212 pOH = 14.00 + log[OH<sup>\u2212<\/sup>] = 14.00 + log(0.0200 M) = 12.30<\/p>\r\n<p id=\"fs-idm15929056\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate the pH for the strong acid\/strong base titration between 50.0 mL of 0.100 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub>(<em data-effect=\"italics\">aq<\/em>) and 0.200 <em data-effect=\"italics\">M<\/em> NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.\r\n\r\n&nbsp;\r\n<div id=\"fs-idm44981872\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm89702512\">0.00 mL: 1.000; 15.0 mL: 1.51; 25.0 mL: 7.00; 40.0 mL: 12.52<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm99459616\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp114806608\"><strong>Titration of a Weak Acid with a Strong Base <\/strong><\/p>\r\n<span style=\"text-align: initial;font-size: 1em\">Consider the titration of 25.00 mL of 0.100 <\/span><em style=\"text-align: initial;font-size: 1em\" data-effect=\"italics\">M<\/em><span style=\"text-align: initial;font-size: 1em\"> CH<\/span><sub style=\"text-align: initial\">3<\/sub><span style=\"text-align: initial;font-size: 1em\">CO<\/span><sub style=\"text-align: initial\">2<\/sub><span style=\"text-align: initial;font-size: 1em\">H with 0.100 <\/span><em style=\"text-align: initial;font-size: 1em\" data-effect=\"italics\">M<\/em><span style=\"text-align: initial;font-size: 1em\"> NaOH. The reaction can be represented as:<\/span>\r\n<div id=\"fs-idp77362464\" style=\"padding-left: 40px\" data-type=\"equation\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>- <\/sup>+ H<sub>2<\/sub>O<\/div>\r\n<p id=\"fs-idm41606112\">Calculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:<\/p>\r\n<p id=\"fs-idm192768656\">(a) 0.00 mL<\/p>\r\n<p id=\"fs-idm178030592\">(b) 25.00 mL<\/p>\r\n<p id=\"fs-idm214305872\">(c) 12.50 mL<\/p>\r\n<p id=\"fs-idm223741312\">(d) 37.50 mL<\/p>\r\n<p id=\"fs-idp3678224\"><strong>Solution: <\/strong><\/p>\r\n(a)\u00a0 \u00a0 Titrant volume = 0.00 mL.\r\n<p style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>)<\/p>\r\nThe initial pH is computed for the acetic acid solution in the usual ICE approach:\r\n<p id=\"fs-idp30610784\" style=\"padding-left: 40px\"><img class=\"alignnone wp-image-1917\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a-300x68.png\" alt=\"\" width=\"207\" height=\"47\" \/> = 1.8 x 10<sup>-5<\/sup><\/p>\r\n<p id=\"fs-idm192051584\" style=\"padding-left: 40px\">assuming <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.100 M, i.e. <em>x<\/em> &lt; 0.00500 M, and solving the simplified equation for <em data-effect=\"italics\">x<\/em> yields<\/p>\r\n\r\n<div id=\"fs-idm7012176\" style=\"padding-left: 40px\" data-type=\"equation\"><em>x<\/em> = 1.3 \u00d7 10<sup>-3 <\/sup>M = [H<sub>3<\/sub>O<sup>+<\/sup>]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ASSUMPTION VALID<\/div>\r\n<div id=\"fs-idm109618832\" data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm165278336\" style=\"padding-left: 40px\" data-type=\"equation\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(1.3 \u00d7 10<sup>-3 <\/sup>M) = 2.87<\/div>\r\n<div id=\"fs-idm76922320\" data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm935760\">(b)\u00a0 \u00a0 Titrant volume = 25.00 mL.<\/p>\r\n<p style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>- <\/sup>+ H<sub>2<\/sub>O<\/p>\r\n<p style=\"padding-left: 40px\">(0.02500 L)(0.100 mol\/L) = 0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup>\u00a0<\/sup>initially<\/p>\r\n<p style=\"padding-left: 40px\">reacts with (0.02500 L)(0.100 mol\/L) = 0.00250 mol OH- in a 1:1 ratio<\/p>\r\n<p style=\"padding-left: 40px\">leaving 0.00250 mol - 0.00250 mol = zero mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/p>\r\n<p style=\"padding-left: 40px\">but producing 0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup><\/p>\r\n<p style=\"padding-left: 40px\">(This volume of titrant represents the equivalence point!)<\/p>\r\nUnlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of\r\n<div id=\"fs-idm33154592\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1922\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-300x62.png\" alt=\"\" width=\"271\" height=\"56\" \/><\/div>\r\n<p id=\"fs-idm2004176\">Base ionization of acetate is represented by the equation<\/p>\r\n\r\n<div id=\"fs-idp77139280\" style=\"padding-left: 40px\" data-type=\"equation\">CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc CH<sub>3<\/sub>CO<sub>2<\/sub>H(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idp75313664\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1921\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-300x34.png\" alt=\"\" width=\"371\" height=\"42\" \/><\/div>\r\n<p id=\"fs-idp58734736\">Assuming <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.0500 M, i.e. <em data-effect=\"italics\">x<\/em> &lt; 0.00250 M , the pH may be calculated via the usual ICE approach:<\/p>\r\n<p style=\"padding-left: 40px\"><img class=\"alignnone wp-image-1923\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d-300x96.png\" alt=\"\" width=\"134\" height=\"43\" \/><\/p>\r\n<p style=\"padding-left: 40px\"><span style=\"font-size: 1em\"><em>x = <\/em>[OH<sup>\u2212<\/sup>] = 5.3 \u00d7 10<sup>-6<\/sup> M\u00a0 \u00a0 \u00a0 ASSUMPTION VALID<\/span><\/p>\r\n\r\n<div id=\"fs-idp1478112\" style=\"padding-left: 40px\" data-type=\"equation\">pOH = \u2212log(5.3 \u00d7 10<sup>-6<\/sup> M) = 5.28<\/div>\r\n<div id=\"fs-idm54099632\" style=\"padding-left: 40px\" data-type=\"equation\">pH = 14.00 - 5.28 = 8.72<\/div>\r\n<p id=\"fs-idm40620912\">Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.<\/p>\r\n<p id=\"fs-idm15954576\">(c) Titrant volume = 12.50 mL.<\/p>\r\n<p id=\"fs-idm935760\" style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>- <\/sup>+ H<sub>2<\/sub>O<\/p>\r\n<p style=\"padding-left: 40px\">0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup>\u00a0<\/sup>initially, as calculated in part (b)<\/p>\r\n<p style=\"padding-left: 40px\">reacts with (0.01250 L)(0.100 mol\/L) = 0.00125 mol OH<sup>-<\/sup> in a 1:1 ratio<\/p>\r\n<p style=\"padding-left: 40px\">leaving 0.00250 mol - 0.00125 mol = 0.00125 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/p>\r\n<p style=\"padding-left: 40px\">and producing 0.00125 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup><\/p>\r\nThis volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. This is a buffer! A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:\r\n<div id=\"fs-idp121571008\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1926\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-300x26.png\" alt=\"\" width=\"508\" height=\"44\" \/><\/div>\r\n<div id=\"fs-idp163867552\" style=\"padding-left: 40px\" data-type=\"equation\">pH = \u2212log (1.8 \u00d7 10<sup>-5<\/sup>) = 4.74<\/div>\r\n<p id=\"fs-idm42058672\">(pH = p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> at the half-equivalence point in a titration of a weak acid)<\/p>\r\n<p id=\"fs-idp85997072\">(d) Titrant volume = 37.50 mL.<\/p>\r\n<p id=\"fs-idm935760\" style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>- <\/sup>+ H<sub>2<\/sub>O<\/p>\r\n<p style=\"padding-left: 40px\">0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup>\u00a0<\/sup>initially, as calculated in part (b)<\/p>\r\n<p style=\"padding-left: 40px\">(0.03750 L)(0.100 mol\/L) = 0.00375 mol OH<sup>-<\/sup><\/p>\r\nWe are past the equivalence point; OH<sup>-<\/sup> is in excess and CH<sub>3<\/sub>CO<sub>2<\/sub>H is completely consumed.\r\n<p style=\"padding-left: 40px\">0.00375 mol - 0.00250 mol = 0.00125 mol OH<sup>- <\/sup>remaining<\/p>\r\n<p style=\"padding-left: 40px\">and 0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup> produced<\/p>\r\nThis volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is controlled by the excess strong base:\r\n<div id=\"fs-idm52967552\" style=\"padding-left: 40px\" data-type=\"equation\">[OH<sup>\u2212<\/sup>] = (0.00125 mol)\/(0.06250 L) = 2.00 \u00d7 10<sup>-2 <\/sup>M<\/div>\r\n<div id=\"fs-idm21431696\" style=\"padding-left: 40px\" data-type=\"equation\">pOH = \u2212log (2.00 \u00d7 10<sup>-2<\/sup> M) = 1.70, and pH =14.00 - 1.70 = 12.30<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm41682336\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate the pH for the weak acid\/strong base titration between 50.0 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCOOH(<em data-effect=\"italics\">aq<\/em>) (formic acid) and 0.200 <em data-effect=\"italics\">M<\/em> NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp68668928\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm37877472\">0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.10<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm198420960\">Performing additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH\/volume data pairs for the strong and weak acid titrations is provided in <a class=\"autogenerated-content\" href=\"#fs-idm87178400\">(Figure)<\/a> and plotted as titration curves in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration\">(Figure)<\/a>. A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:<\/p>\r\n<p id=\"fs-idm201863856\">initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH<\/p>\r\n<p id=\"fs-idm444217232\">pre-equivalence point (0 mL &lt; <em data-effect=\"italics\">V<\/em> &lt; 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base<\/p>\r\n<p id=\"fs-idm183388016\">equivalence point (<em data-effect=\"italics\">V<\/em> = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid<\/p>\r\n<p id=\"fs-idm183509856\">post-equivalence point (<em data-effect=\"italics\">V<\/em> &gt; 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.<\/p>\r\n\r\n<table id=\"fs-idm87178400\" class=\"top-titled\" summary=\"This table has four columns and twenty rows. The first row is a header row, and it labels each column, \u201cVolume of 0.100 M N a O H Added ( m L ),\u201d \u201cMoles of N a O H Added,\u201d \u201cp H Values 0.100 M H C l footnote one,\u201d \u201cp H Values 0.100 M C H subscript 3 C O subscript 2 H footnote 2.\u201d Under the \u201cVolume of 0.100 M N a O H Added ( m L )\u201d column are the following values: 0.0, 5.0, 10.0, 15.0, 20.0, 22.0, 24.0, 24.5, 24.9, 25.0, 25.1, 25.5, 26.0, 28.0, 30.0, 35.0, 40.0, 45.0, and 50.0. Under the \u201cMoles of N a O H Added\u201d column are the following values: 0.0, 0.00050, 0.00100, 0.00150, 0.00200, 0.00220, 0.00240, 0.00245, 0.00249, 0.00250, 0.00251, 0.00255, 0.00260, 0.00280, 0.00300, 0.00350, 0.00400, 0.00450, and 0.00500. Under the \u201cp H Values 0.100 M H C l footnote one\u201d column are the following values: 1.00, 1.18, 1.37, 1.60, 1.95, 2.20, 2.69, 3.00, 3.70, 7.00, 10.30, 11.00, 11.29, 11.75, 11.96, 12.22, 12.36, 12.46, and 12.52. Foot note one reads, \u201cTitration of 25.00 m L of 0.100 M H C l ( 0.00250 mol of H C I ) with 0.100 M N a O H.\u201d Under the \u201cp H Values 0.100 M C H subscript 3 C O subscript 2 H footnote 2\u201d column are the following values: 2.87, 4.14, 4.57, 4.92, 5.35, 5.61, 6.13, 6.44, 7.17, 8.72, 10.30, 11.00, 11.29, 11.75, 11.96, 12.22, 12.36, 12.46, and 12.52. Footnote two reads, \u201cTitration of 25.00 m L of 0.100 M C H subscript 3 C O subscript 2 H ( 0.00250 mol of C H subscript 3C O subscript 2 H) with 0.100 M N a O H.\u201d\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"4\" data-align=\"center\">pH Values in the Titrations of a Strong Acid and of a Weak Acid<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<th data-align=\"left\">Volume of 0.100 <em data-effect=\"italics\">M<\/em> NaOH Added (mL)<\/th>\r\n<th data-align=\"left\">Moles of NaOH Added<\/th>\r\n<th data-align=\"left\">pH Values 0.100 <em data-effect=\"italics\">M<\/em> HCl<sup data-type=\"footnote-number\"><a href=\"#footnote1\" data-type=\"footnote-link\">1<\/a><\/sup><\/th>\r\n<th data-align=\"left\">pH Values 0.100 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup id=\"footnote-ref2\" data-type=\"footnote-number\"><a href=\"#footnote2\" data-type=\"footnote-link\">2<\/a><\/sup><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">0.0<\/td>\r\n<td data-align=\"left\">0.0<\/td>\r\n<td data-align=\"left\">1.00<\/td>\r\n<td data-align=\"left\">2.87<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">5.0<\/td>\r\n<td data-align=\"left\">0.00050<\/td>\r\n<td data-align=\"left\">1.18<\/td>\r\n<td data-align=\"left\">4.14<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">10.0<\/td>\r\n<td data-align=\"left\">0.00100<\/td>\r\n<td data-align=\"left\">1.37<\/td>\r\n<td data-align=\"left\">4.57<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">15.0<\/td>\r\n<td data-align=\"left\">0.00150<\/td>\r\n<td data-align=\"left\">1.60<\/td>\r\n<td data-align=\"left\">4.92<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">20.0<\/td>\r\n<td data-align=\"left\">0.00200<\/td>\r\n<td data-align=\"left\">1.95<\/td>\r\n<td data-align=\"left\">5.35<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">22.0<\/td>\r\n<td data-align=\"left\">0.00220<\/td>\r\n<td data-align=\"left\">2.20<\/td>\r\n<td data-align=\"left\">5.61<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">24.0<\/td>\r\n<td data-align=\"left\">0.00240<\/td>\r\n<td data-align=\"left\">2.69<\/td>\r\n<td data-align=\"left\">6.13<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">24.5<\/td>\r\n<td data-align=\"left\">0.00245<\/td>\r\n<td data-align=\"left\">3.00<\/td>\r\n<td data-align=\"left\">6.44<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">24.9<\/td>\r\n<td data-align=\"left\">0.00249<\/td>\r\n<td data-align=\"left\">3.70<\/td>\r\n<td data-align=\"left\">7.14<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">25.0<\/td>\r\n<td data-align=\"left\">0.00250<\/td>\r\n<td data-align=\"left\">7.00<\/td>\r\n<td data-align=\"left\">8.72<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">25.1<\/td>\r\n<td data-align=\"left\">0.00251<\/td>\r\n<td data-align=\"left\">10.30<\/td>\r\n<td data-align=\"left\">10.30<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">25.5<\/td>\r\n<td data-align=\"left\">0.00255<\/td>\r\n<td data-align=\"left\">11.00<\/td>\r\n<td data-align=\"left\">11.00<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">26.0<\/td>\r\n<td data-align=\"left\">0.00260<\/td>\r\n<td data-align=\"left\">11.29<\/td>\r\n<td data-align=\"left\">11.29<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">28.0<\/td>\r\n<td data-align=\"left\">0.00280<\/td>\r\n<td data-align=\"left\">11.75<\/td>\r\n<td data-align=\"left\">11.75<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">30.0<\/td>\r\n<td data-align=\"left\">0.00300<\/td>\r\n<td data-align=\"left\">11.96<\/td>\r\n<td data-align=\"left\">11.96<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">35.0<\/td>\r\n<td data-align=\"left\">0.00350<\/td>\r\n<td data-align=\"left\">12.22<\/td>\r\n<td data-align=\"left\">12.22<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">40.0<\/td>\r\n<td data-align=\"left\">0.00400<\/td>\r\n<td data-align=\"left\">12.36<\/td>\r\n<td data-align=\"left\">12.36<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">45.0<\/td>\r\n<td data-align=\"left\">0.00450<\/td>\r\n<td data-align=\"left\">12.46<\/td>\r\n<td data-align=\"left\">12.46<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">50.0<\/td>\r\n<td data-align=\"left\">0.00500<\/td>\r\n<td data-align=\"left\">12.52<\/td>\r\n<td data-align=\"left\">12.52<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"CNX_Chem_14_07_titration\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">(a) The titration curve for the titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCl (strong acid) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH (strong base) has an equivalence point of 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> acetic acid (weak acid) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH (strong base) has an equivalence point of 8.72 pH.<\/div>\r\n<span id=\"fs-idp134166880\" data-type=\"media\" data-alt=\"Two graphs are shown. The first graph on the left is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unis beginning at 0 extending to 14. A red curve is drawn on the graph which increases steadily from the point (0, 3) up to about (20, 5.5) after which the graph has a vertical section from (25, 7) up to (25, 11). The graph then levels off to a value of about 12.5 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d The second graph on the right is titled \u201cTitration of Strong Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 units beginning at 0 extending to 14. A red curve is drawn on the graph which increases gradually from the point (0, 1) up to about (22.5, 2.2) after which the graph has a vertical section from (25, 4) up to nearly (25, 11). The graph then levels off to a value of about 12.4 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 7.00.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_07_titration-1.jpg\" alt=\"Two graphs are shown. The first graph on the left is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unis beginning at 0 extending to 14. A red curve is drawn on the graph which increases steadily from the point (0, 3) up to about (20, 5.5) after which the graph has a vertical section from (25, 7) up to (25, 11). The graph then levels off to a value of about 12.5 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d The second graph on the right is titled \u201cTitration of Strong Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 units beginning at 0 extending to 14. A red curve is drawn on the graph which increases gradually from the point (0, 1) up to about (22.5, 2.2) after which the graph has a vertical section from (25, 4) up to nearly (25, 11). The graph then levels off to a value of about 12.4 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 7.00.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp22610576\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Acid-Base Indicators<\/strong><\/h3>\r\n<p id=\"fs-idm39085312\">Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 \u00d7 10<sup>\u22129<\/sup><em data-effect=\"italics\">M<\/em> (pH &lt; 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 \u00d7 10<sup>\u22129<\/sup><em data-effect=\"italics\">M<\/em> (pH &gt; 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called <strong>acid-base indicators<\/strong>. Acid-base indicators are either weak organic acids or weak organic bases.<\/p>\r\n<p id=\"fs-idm28417392\">The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:<\/p>\r\n\r\n<div id=\"fs-idp80468176\" style=\"padding-left: 40px\" data-type=\"equation\">HIn(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + In<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">red\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0yellow<\/div>\r\n<div id=\"fs-idm87967824\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1927\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-300x62.png\" alt=\"\" width=\"261\" height=\"54\" \/><\/div>\r\n<p id=\"fs-idm68547888\">The anion of methyl orange, In<sup>\u2212<\/sup>, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Ch\u00e2telier\u2019s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.<\/p>\r\n<p id=\"fs-idp64938256\">The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species In<sup>\u2212<\/sup> and HIn. If most of the indicator (typically about 60\u221290% or more) is present as In<sup>\u2212<\/sup>, the perceived color of the solution is yellow. If most is present as HIn, then the solution color appears red. The Henderson-Hasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):<\/p>\r\n\r\n<div id=\"fs-idp80076272\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1928\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g-300x77.png\" alt=\"\" width=\"211\" height=\"54\" \/><\/div>\r\n<p id=\"fs-idm188576256\">In solutions where pH &gt; p<em data-effect=\"italics\">K<\/em><sub>a<\/sub>, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When pH &gt; p<em data-effect=\"italics\">K<\/em><sub>a<\/sub>, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa, appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The <span data-type=\"term\">color change interval<\/span> (or <em data-effect=\"italics\">pH interval<\/em>) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> \u00b1 1.<\/p>\r\n<p id=\"fs-idm92451888\">There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_indicators\">(Figure)<\/a> presents several indicators, their colors, and their color-change intervals.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_14_07_indicators\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">This chart illustrates the color change intervals for several acid-base indicators.<\/div>\r\n<span id=\"fs-idp112102048\" data-type=\"media\" data-alt=\"This figure provides a graphical representation of indicators and color ranges. A horizontal axis is labeled \u201cp H.\u201d This axis begins at zero and increases by ones up to 13. The left side of the graphic provides a column with the names of indicators. To the right of each indicator name is either one or two colored bars that are shaded according to the indicator color at various p H ranges. From the top, the first row is labeled \u201cCrystal violet.\u201d The associated colored bar is yellow at its left end at a p H of 0 and changes to green and blue moving right to its endpoint at a p H of 1.8. The second row is labeled \u201cCresol red.\u201d The associated colored bar is red at its left end at a p H of 1 and changes to orange and yellow moving right to its endpoint at a p H of just over 2. A second bar to its right is yellow at a p H of around 7 and proceeds through orange to red at a p H of about 9. The third row is labeled \u201cThymol blue.\u201d The associated colored bar is red at its left end at a p H of nearly 1.2 and changes to orange and red moving right to its endpoint at a p H of 2.8. A second bar begins in yellow at a p H of 8 and proceeds through green and blue to its end at a p H of around 9.1. The fourth row is labeled \u201cErythrosin B.\u201d The associated colored bar is red from a p H of 2.2 to its endpoint at a p H of 3.6. The fifth row is labeled \u201c2 comma 4 dash Dinitrophenol.\u201d The associated colored bar is white at its left end at a p H of 2.6 and changes to yellow at its endpoint at a p H of 4. The sixth row is labeled \u201cBromophenol blue.\u201d The associated colored bar is yellow at its left end at a p H of 3 and changes to green and blue moving right to its endpoint at a p H of 4.5. The seventh row is labeled \u201cMethyl orange.\u201d The associated colored bar is red-orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The eighth row is labeled \u201cBromocresol green.\u201d The associated colored bar is yellow at its left end at a p H of 3.8 and changes to green and blue moving right to its endpoint at a p H of 5.4. The ninth row is labeled \u201cMethyl red.\u201d The associated colored bar is orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The tenth row is labeled \u201cEriochrome * Black T.\u201d The associated colored bar is red at its left end at a p H of 5 and changes to purple and blue moving right to its endpoint at a p H of 6.5. The eleventh row is labeled \u201cBromocresol purple.\u201d The associated colored bar is yellow at its left end at a p H of 5.2 and changes to purple moving right to its endpoint at a p H of 6.8. The twelfth row is labeled \u201cAlizarin.\u201d The first associated colored bar is yellow-orange at its left end at a p H of 5.7 and changes to red moving right to its endpoint at a p H of 7.2. A second bar begins in red at a p H of 11 and changes to purple, then dark blue at its right end at a p H of 12.4. The thirteenth row is labeled \u201cBromothymol blue.\u201d The associated colored bar is yellow at its left end at a p H of 6 and changes to green and blue moving right to its endpoint at a p H of 7.6. The fourteenth row is labeled \u201cPhenol red.\u201d The associated colored bar is yellow-orange at its left end at a p H of 6.8 and changes to orange and red moving right to its endpoint at a p H of 8.2. The fifteenth row is labeled \u201cm dash Nitrophenol.\u201d The associated colored bar is white at its left end at a p H of 6.8 and changes to yellow moving right to its endpoint at a p H of 8.6. The sixteenth row is labeled \u201co dash Cresolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8.3 and changes to red moving right to its endpoint at a p H of 9.8. The seventeenth row is labeled \u201cPhenolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8 and changes to pink moving right to its endpoint at a p H of 10. The eighteenth row is labeled \u201cThymolphthalein.\u201d The associated colored bar is light blue at its left end at a p H of 9.3 and changes to a deep, dark blue moving right to its endpoint at a p H of 10.5. The nineteenth row is labeled \u201cAlizarin yellow R.\u201d The associated colored bar is yellow-orange at its left end at a p H of 10 and changes to red moving right to its endpoint at a p H of 12.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_07_indicators-1.jpg\" alt=\"This figure provides a graphical representation of indicators and color ranges. A horizontal axis is labeled \u201cp H.\u201d This axis begins at zero and increases by ones up to 13. The left side of the graphic provides a column with the names of indicators. To the right of each indicator name is either one or two colored bars that are shaded according to the indicator color at various p H ranges. From the top, the first row is labeled \u201cCrystal violet.\u201d The associated colored bar is yellow at its left end at a p H of 0 and changes to green and blue moving right to its endpoint at a p H of 1.8. The second row is labeled \u201cCresol red.\u201d The associated colored bar is red at its left end at a p H of 1 and changes to orange and yellow moving right to its endpoint at a p H of just over 2. A second bar to its right is yellow at a p H of around 7 and proceeds through orange to red at a p H of about 9. The third row is labeled \u201cThymol blue.\u201d The associated colored bar is red at its left end at a p H of nearly 1.2 and changes to orange and red moving right to its endpoint at a p H of 2.8. A second bar begins in yellow at a p H of 8 and proceeds through green and blue to its end at a p H of around 9.1. The fourth row is labeled \u201cErythrosin B.\u201d The associated colored bar is red from a p H of 2.2 to its endpoint at a p H of 3.6. The fifth row is labeled \u201c2 comma 4 dash Dinitrophenol.\u201d The associated colored bar is white at its left end at a p H of 2.6 and changes to yellow at its endpoint at a p H of 4. The sixth row is labeled \u201cBromophenol blue.\u201d The associated colored bar is yellow at its left end at a p H of 3 and changes to green and blue moving right to its endpoint at a p H of 4.5. The seventh row is labeled \u201cMethyl orange.\u201d The associated colored bar is red-orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The eighth row is labeled \u201cBromocresol green.\u201d The associated colored bar is yellow at its left end at a p H of 3.8 and changes to green and blue moving right to its endpoint at a p H of 5.4. The ninth row is labeled \u201cMethyl red.\u201d The associated colored bar is orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The tenth row is labeled \u201cEriochrome * Black T.\u201d The associated colored bar is red at its left end at a p H of 5 and changes to purple and blue moving right to its endpoint at a p H of 6.5. The eleventh row is labeled \u201cBromocresol purple.\u201d The associated colored bar is yellow at its left end at a p H of 5.2 and changes to purple moving right to its endpoint at a p H of 6.8. The twelfth row is labeled \u201cAlizarin.\u201d The first associated colored bar is yellow-orange at its left end at a p H of 5.7 and changes to red moving right to its endpoint at a p H of 7.2. A second bar begins in red at a p H of 11 and changes to purple, then dark blue at its right end at a p H of 12.4. The thirteenth row is labeled \u201cBromothymol blue.\u201d The associated colored bar is yellow at its left end at a p H of 6 and changes to green and blue moving right to its endpoint at a p H of 7.6. The fourteenth row is labeled \u201cPhenol red.\u201d The associated colored bar is yellow-orange at its left end at a p H of 6.8 and changes to orange and red moving right to its endpoint at a p H of 8.2. The fifteenth row is labeled \u201cm dash Nitrophenol.\u201d The associated colored bar is white at its left end at a p H of 6.8 and changes to yellow moving right to its endpoint at a p H of 8.6. The sixteenth row is labeled \u201co dash Cresolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8.3 and changes to red moving right to its endpoint at a p H of 9.8. The seventeenth row is labeled \u201cPhenolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8 and changes to pink moving right to its endpoint at a p H of 10. The eighteenth row is labeled \u201cThymolphthalein.\u201d The associated colored bar is light blue at its left end at a p H of 9.3 and changes to a deep, dark blue moving right to its endpoint at a p H of 10.5. The nineteenth row is labeled \u201cAlizarin yellow R.\u201d The associated colored bar is yellow-orange at its left end at a p H of 10 and changes to red moving right to its endpoint at a p H of 12.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"CNX_Chem_14_07_titration2\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">Titration curves for strong and weak acids illustrating the proper choice of acid-base indicator. Any of the three indicators will exhibit a reasonably sharp color change at the equivalence point of the strong acid titration, but only phenolphthalein is suitable for use in the weak acid titration.<\/div>\r\n<span id=\"fs-idm5594336\" data-type=\"media\" data-alt=\"A graph is shown which is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L)\u201d and begins at 0 with markings every 5 units up to 50. The vertical axis is labeled \u201cp H\u201d and begins at 0 and increases by single units up to 14. A red curve is drawn on the graph. The curve begins at (0, 3) and passes through the points (5, 4.1), (10, 4.7), (15, 5), (20, 5.5), and (22.5, 6), after which it rapidly increases, forming a vertical section centered at the point (25, 8.7). The rapid increase of the curve then levels off and the curve passes through the points (30, 12), (35, 12.4), (40, 12.5), (45, 12.6), and (50, 12.6). A brown rectangle extends horizontally across the graph covering the p H of 3 to 4.2 range. To the right, this rectangle is labeled \u201cMethyl orange p H range.\u201d A blue rectangle extends horizontally across the graph covering the p H of 4.6 to 8 range. To the right, this rectangle is labeled \u201cLitmus p H range.\u201d A purple rectangle extends horizontally across the graph covering the p H of 8.4 to 10 range. To the right, this rectangle is labeled \u201cPhenolphthalein p H range.\u201d The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_07_titration2-1.jpg\" alt=\"A graph is shown which is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L)\u201d and begins at 0 with markings every 5 units up to 50. The vertical axis is labeled \u201cp H\u201d and begins at 0 and increases by single units up to 14. A red curve is drawn on the graph. The curve begins at (0, 3) and passes through the points (5, 4.1), (10, 4.7), (15, 5), (20, 5.5), and (22.5, 6), after which it rapidly increases, forming a vertical section centered at the point (25, 8.7). The rapid increase of the curve then levels off and the curve passes through the points (30, 12), (35, 12.4), (40, 12.5), (45, 12.6), and (50, 12.6). A brown rectangle extends horizontally across the graph covering the p H of 3 to 4.2 range. To the right, this rectangle is labeled \u201cMethyl orange p H range.\u201d A blue rectangle extends horizontally across the graph covering the p H of 4.6 to 8 range. To the right, this rectangle is labeled \u201cLitmus p H range.\u201d A purple rectangle extends horizontally across the graph covering the p H of 8.4 to 10 range. To the right, this rectangle is labeled \u201cPhenolphthalein p H range.\u201d The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm188916592\">The titration curves shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration2\">(Figure)<\/a> illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of ~24 mL of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.<\/p>\r\n<p id=\"fs-idm224614176\">The weak acid titration curve in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration2\">(Figure)<\/a> shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-to-yellow color change over a relatively large volume interval (0\u20136 mL), completing the color change well before the equivalence point (25 mL) has been reached. Use of litmus would show a color change that begins after adding 7\u20138 mL of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp81138704\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idp77329840\">The titration curve for an acid-base titration is typically a plot of pH versus volume of added titrant. These curves are useful in selecting appropriate acid-base indicators that will permit accurate determinations of titration end points.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp63641264\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm5595328\" data-type=\"exercise\">\r\n<div id=\"fs-idp5936080\" data-type=\"problem\"><\/div>\r\n<\/div>\r\n<div id=\"fs-idp105330480\" data-type=\"exercise\">\r\n<div id=\"fs-idp105330736\" data-type=\"problem\">\r\n<p id=\"fs-idp105330992\"><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"footnote-refs\">\r\n<h3 data-type=\"footnote-refs-title\"><strong>Footnotes<\/strong><\/h3>\r\n<ul data-list-type=\"bulleted\" data-bullet-style=\"none\">\r\n \t<li data-type=\"footnote-ref\"><a href=\"#footnote-ref1\" data-type=\"footnote-ref-link\">1<\/a><span data-type=\"footnote-ref-content\">Titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCl (0.00250 mol of HCI) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH.<\/span><\/li>\r\n \t<li id=\"footnote2\" data-type=\"footnote-ref\"><a href=\"#footnote-ref2\" data-type=\"footnote-ref-link\">2<\/a><span data-type=\"footnote-ref-content\">Titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H (0.00250 mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH.<\/span><\/li>\r\n<\/ul>\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-idp67176144\">\r\n \t<dt>acid-base indicator<\/dt>\r\n \t<dd id=\"fs-idp67176784\">weak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp67179744\">\r\n \t<dt>titration curve<\/dt>\r\n \t<dd id=\"fs-idp67180384\">plot of some sample property (such as pH) versus volume of added titrant<\/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>Interpret titration curves for strong and weak acid-base systems<\/li>\n<li>Compute sample pH at important stages of a titration<\/li>\n<li>Explain the function of acid-base indicators<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm43847120\">As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique.<\/p>\n<div id=\"fs-idm84795552\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Titration Curves<\/strong><\/h3>\n<p id=\"fs-idp135875056\">A <strong>titration curve<\/strong> is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration\u2019s progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH. The second example addresses a weak acid titration requiring equilibrium calculations.<\/p>\n<div id=\"fs-idm49336656\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp245005248\"><strong>Calculating pH for Titration Solutions: Strong Acid\/Strong Base <\/strong><\/p>\n<p>A titration is carried out for 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCl (strong acid) with 0.100 <em data-effect=\"italics\">M<\/em> of a strong base NaOH (the titration curve is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration\">(Figure)<\/a>). Calculate the pH at these volumes of added base solution:<\/p>\n<p id=\"fs-idp67154976\">(a) 0.00 mL<\/p>\n<p id=\"fs-idm80154320\">(b) 12.50 mL<\/p>\n<p id=\"fs-idm39390080\">(c) 25.00 mL<\/p>\n<p id=\"fs-idp81988928\">(d) 37.50 mL<\/p>\n<p id=\"fs-idp22584832\"><strong>Solution: <\/strong><\/p>\n<p>(a) Titrant volume = 0.00 mL. The solution pH is due to the acid ionization of HCl. Because this is a strong acid, the ionization is complete and the hydronium ion molarity is 0.100 <em data-effect=\"italics\">M<\/em>. The pH of the solution is then<\/p>\n<div id=\"fs-idm213177360\" style=\"padding-left: 40px\" data-type=\"equation\">pH = -log(0.100 M) = 1.000<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm214456576\">(b) Titrant volume = 12.50 mL.<\/p>\n<p style=\"padding-left: 40px\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>&#8211;<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>aq<\/em>)<\/p>\n<p>Since the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the intial amount and then dividing by the solution volume:<\/p>\n<p style=\"padding-left: 40px\">(0.02500 L)(0.100 mol\/L) = 0.00250 mol H<sub>3<\/sub>O<sup>+ <\/sup>initially<\/p>\n<p style=\"padding-left: 40px\">reacts with (0.01250 L)(0.100 mol\/L) = 0.00125 mol OH- in a 1:1 ratio<\/p>\n<p style=\"padding-left: 40px\">leaving 0.00250 mol &#8211; 0.00125 mol = 0.00125 mol H<sub>3<\/sub>O<sup>+<\/sup><\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp80308976\" style=\"padding-left: 40px\" data-type=\"equation\">[H<sub>3<\/sub>O<sup>+<\/sup>] = (0.00125 mol)\/(0.03750 L) = 0.0333 M<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">pH = -log(0.0333 M) = 1.477<\/div>\n<div data-type=\"equation\"><\/div>\n<p>(c) Titrant volume = 25.00 mL.<\/p>\n<p style=\"padding-left: 40px\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>&#8211;<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>aq<\/em>)<\/p>\n<p>This titrant addition involves a stoichiometric amount of base (the <em data-effect=\"italics\">equivalence point<\/em>), and so only products of the neutralization reaction are in solution (water and NaCl). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autohydrolysis of water. The solution is neutral, having a pH = 7.00.<\/p>\n<p id=\"fs-idm173154048\">(d) Titrant volume = 37.50 mL.<\/p>\n<p style=\"padding-left: 40px\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>&#8211;<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>aq<\/em>)<\/p>\n<p>This involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion:<\/p>\n<div id=\"fs-idp66422816\" data-type=\"equation\">\n<p style=\"padding-left: 40px\">0.00250 mol H<sub>3<\/sub>O<sup>+ <\/sup>initially, as calculated in part (b)<\/p>\n<p style=\"padding-left: 40px\">(0.03750 L)(0.100 mol\/L) = 0.00375 mol OH<sup>&#8211;<\/sup> total added<\/p>\n<p style=\"padding-left: 40px\">reacts with H<sub>3<\/sub>O<sup>+<\/sup> in a 1:1 ratio<\/p>\n<p style=\"padding-left: 40px\">leaving 0.00375 mol &#8211; 0.00250 mol = 0.00125 mol OH<sup>&#8211;<\/sup><\/p>\n<p style=\"padding-left: 40px\">[OH-] = (0.00125 mol)\/(0.06250 L) = 0.0200 M<\/p>\n<\/div>\n<p id=\"fs-idm41259968\" style=\"padding-left: 40px\">pH = 14.00 \u2212 pOH = 14.00 + log[OH<sup>\u2212<\/sup>] = 14.00 + log(0.0200 M) = 12.30<\/p>\n<p id=\"fs-idm15929056\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate the pH for the strong acid\/strong base titration between 50.0 mL of 0.100 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub>(<em data-effect=\"italics\">aq<\/em>) and 0.200 <em data-effect=\"italics\">M<\/em> NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm44981872\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm89702512\">0.00 mL: 1.000; 15.0 mL: 1.51; 25.0 mL: 7.00; 40.0 mL: 12.52<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm99459616\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp114806608\"><strong>Titration of a Weak Acid with a Strong Base <\/strong><\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Consider the titration of 25.00 mL of 0.100 <\/span><em style=\"text-align: initial;font-size: 1em\" data-effect=\"italics\">M<\/em><span style=\"text-align: initial;font-size: 1em\"> CH<\/span><sub style=\"text-align: initial\">3<\/sub><span style=\"text-align: initial;font-size: 1em\">CO<\/span><sub style=\"text-align: initial\">2<\/sub><span style=\"text-align: initial;font-size: 1em\">H with 0.100 <\/span><em style=\"text-align: initial;font-size: 1em\" data-effect=\"italics\">M<\/em><span style=\"text-align: initial;font-size: 1em\"> NaOH. The reaction can be represented as:<\/span><\/p>\n<div id=\"fs-idp77362464\" style=\"padding-left: 40px\" data-type=\"equation\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211; <\/sup>+ H<sub>2<\/sub>O<\/div>\n<p id=\"fs-idm41606112\">Calculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:<\/p>\n<p id=\"fs-idm192768656\">(a) 0.00 mL<\/p>\n<p id=\"fs-idm178030592\">(b) 25.00 mL<\/p>\n<p id=\"fs-idm214305872\">(c) 12.50 mL<\/p>\n<p id=\"fs-idm223741312\">(d) 37.50 mL<\/p>\n<p id=\"fs-idp3678224\"><strong>Solution: <\/strong><\/p>\n<p>(a)\u00a0 \u00a0 Titrant volume = 0.00 mL.<\/p>\n<p style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>)<\/p>\n<p>The initial pH is computed for the acetic acid solution in the usual ICE approach:<\/p>\n<p id=\"fs-idp30610784\" style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1917\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a-300x68.png\" alt=\"\" width=\"207\" height=\"47\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a-300x68.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a-225x51.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a-350x79.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7a.png 521w\" sizes=\"auto, (max-width: 207px) 100vw, 207px\" \/> = 1.8 x 10<sup>-5<\/sup><\/p>\n<p id=\"fs-idm192051584\" style=\"padding-left: 40px\">assuming <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.100 M, i.e. <em>x<\/em> &lt; 0.00500 M, and solving the simplified equation for <em data-effect=\"italics\">x<\/em> yields<\/p>\n<div id=\"fs-idm7012176\" style=\"padding-left: 40px\" data-type=\"equation\"><em>x<\/em> = 1.3 \u00d7 10<sup>-3 <\/sup>M = [H<sub>3<\/sub>O<sup>+<\/sup>]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ASSUMPTION VALID<\/div>\n<div id=\"fs-idm109618832\" data-type=\"equation\"><\/div>\n<div id=\"fs-idm165278336\" style=\"padding-left: 40px\" data-type=\"equation\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(1.3 \u00d7 10<sup>-3 <\/sup>M) = 2.87<\/div>\n<div id=\"fs-idm76922320\" data-type=\"equation\"><\/div>\n<p id=\"fs-idm935760\">(b)\u00a0 \u00a0 Titrant volume = 25.00 mL.<\/p>\n<p style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211; <\/sup>+ H<sub>2<\/sub>O<\/p>\n<p style=\"padding-left: 40px\">(0.02500 L)(0.100 mol\/L) = 0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup>\u00a0<\/sup>initially<\/p>\n<p style=\"padding-left: 40px\">reacts with (0.02500 L)(0.100 mol\/L) = 0.00250 mol OH- in a 1:1 ratio<\/p>\n<p style=\"padding-left: 40px\">leaving 0.00250 mol &#8211; 0.00250 mol = zero mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/p>\n<p style=\"padding-left: 40px\">but producing 0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup><\/p>\n<p style=\"padding-left: 40px\">(This volume of titrant represents the equivalence point!)<\/p>\n<p>Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of<\/p>\n<div id=\"fs-idm33154592\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1922\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-300x62.png\" alt=\"\" width=\"271\" height=\"56\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-300x62.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-768x159.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-225x47.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b-350x73.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7b.png 886w\" sizes=\"auto, (max-width: 271px) 100vw, 271px\" \/><\/div>\n<p id=\"fs-idm2004176\">Base ionization of acetate is represented by the equation<\/p>\n<div id=\"fs-idp77139280\" style=\"padding-left: 40px\" data-type=\"equation\">CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc CH<sub>3<\/sub>CO<sub>2<\/sub>H(<em>aq<\/em>) + OH<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idp75313664\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1921\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-300x34.png\" alt=\"\" width=\"371\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-300x34.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-1024x114.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-768x86.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-225x25.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1-350x39.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7c-1.png 1504w\" sizes=\"auto, (max-width: 371px) 100vw, 371px\" \/><\/div>\n<p id=\"fs-idp58734736\">Assuming <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.0500 M, i.e. <em data-effect=\"italics\">x<\/em> &lt; 0.00250 M , the pH may be calculated via the usual ICE approach:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1923\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d-300x96.png\" alt=\"\" width=\"134\" height=\"43\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d-300x96.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d-65x21.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d-225x72.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d-350x112.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7d.png 366w\" sizes=\"auto, (max-width: 134px) 100vw, 134px\" \/><\/p>\n<p style=\"padding-left: 40px\"><span style=\"font-size: 1em\"><em>x = <\/em>[OH<sup>\u2212<\/sup>] = 5.3 \u00d7 10<sup>-6<\/sup> M\u00a0 \u00a0 \u00a0 ASSUMPTION VALID<\/span><\/p>\n<div id=\"fs-idp1478112\" style=\"padding-left: 40px\" data-type=\"equation\">pOH = \u2212log(5.3 \u00d7 10<sup>-6<\/sup> M) = 5.28<\/div>\n<div id=\"fs-idm54099632\" style=\"padding-left: 40px\" data-type=\"equation\">pH = 14.00 &#8211; 5.28 = 8.72<\/div>\n<p id=\"fs-idm40620912\">Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.<\/p>\n<p id=\"fs-idm15954576\">(c) Titrant volume = 12.50 mL.<\/p>\n<p id=\"fs-idm935760\" style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211; <\/sup>+ H<sub>2<\/sub>O<\/p>\n<p style=\"padding-left: 40px\">0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup>\u00a0<\/sup>initially, as calculated in part (b)<\/p>\n<p style=\"padding-left: 40px\">reacts with (0.01250 L)(0.100 mol\/L) = 0.00125 mol OH<sup>&#8211;<\/sup> in a 1:1 ratio<\/p>\n<p style=\"padding-left: 40px\">leaving 0.00250 mol &#8211; 0.00125 mol = 0.00125 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/p>\n<p style=\"padding-left: 40px\">and producing 0.00125 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup><\/p>\n<p>This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. This is a buffer! A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:<\/p>\n<div id=\"fs-idp121571008\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1926\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-300x26.png\" alt=\"\" width=\"508\" height=\"44\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-300x26.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-1024x89.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-768x67.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-1536x134.png 1536w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-2048x179.png 2048w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7e-350x31.png 350w\" sizes=\"auto, (max-width: 508px) 100vw, 508px\" \/><\/div>\n<div id=\"fs-idp163867552\" style=\"padding-left: 40px\" data-type=\"equation\">pH = \u2212log (1.8 \u00d7 10<sup>-5<\/sup>) = 4.74<\/div>\n<p id=\"fs-idm42058672\">(pH = p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> at the half-equivalence point in a titration of a weak acid)<\/p>\n<p id=\"fs-idp85997072\">(d) Titrant volume = 37.50 mL.<\/p>\n<p id=\"fs-idm935760\" style=\"padding-left: 40px\">CH<sub>3<\/sub>CO<sub>2<\/sub>H + OH<sup>\u2212<\/sup> \u27f6 CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211; <\/sup>+ H<sub>2<\/sub>O<\/p>\n<p style=\"padding-left: 40px\">0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup>\u00a0<\/sup>initially, as calculated in part (b)<\/p>\n<p style=\"padding-left: 40px\">(0.03750 L)(0.100 mol\/L) = 0.00375 mol OH<sup>&#8211;<\/sup><\/p>\n<p>We are past the equivalence point; OH<sup>&#8211;<\/sup> is in excess and CH<sub>3<\/sub>CO<sub>2<\/sub>H is completely consumed.<\/p>\n<p style=\"padding-left: 40px\">0.00375 mol &#8211; 0.00250 mol = 0.00125 mol OH<sup>&#8211; <\/sup>remaining<\/p>\n<p style=\"padding-left: 40px\">and 0.00250 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup> produced<\/p>\n<p>This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is controlled by the excess strong base:<\/p>\n<div id=\"fs-idm52967552\" style=\"padding-left: 40px\" data-type=\"equation\">[OH<sup>\u2212<\/sup>] = (0.00125 mol)\/(0.06250 L) = 2.00 \u00d7 10<sup>-2 <\/sup>M<\/div>\n<div id=\"fs-idm21431696\" style=\"padding-left: 40px\" data-type=\"equation\">pOH = \u2212log (2.00 \u00d7 10<sup>-2<\/sup> M) = 1.70, and pH =14.00 &#8211; 1.70 = 12.30<\/div>\n<div data-type=\"equation\"><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm41682336\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate the pH for the weak acid\/strong base titration between 50.0 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCOOH(<em data-effect=\"italics\">aq<\/em>) (formic acid) and 0.200 <em data-effect=\"italics\">M<\/em> NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp68668928\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm37877472\">0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.10<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm198420960\">Performing additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH\/volume data pairs for the strong and weak acid titrations is provided in <a class=\"autogenerated-content\" href=\"#fs-idm87178400\">(Figure)<\/a> and plotted as titration curves in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration\">(Figure)<\/a>. A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:<\/p>\n<p id=\"fs-idm201863856\">initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH<\/p>\n<p id=\"fs-idm444217232\">pre-equivalence point (0 mL &lt; <em data-effect=\"italics\">V<\/em> &lt; 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base<\/p>\n<p id=\"fs-idm183388016\">equivalence point (<em data-effect=\"italics\">V<\/em> = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid<\/p>\n<p id=\"fs-idm183509856\">post-equivalence point (<em data-effect=\"italics\">V<\/em> &gt; 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.<\/p>\n<table id=\"fs-idm87178400\" class=\"top-titled\" summary=\"This table has four columns and twenty rows. The first row is a header row, and it labels each column, \u201cVolume of 0.100 M N a O H Added ( m L ),\u201d \u201cMoles of N a O H Added,\u201d \u201cp H Values 0.100 M H C l footnote one,\u201d \u201cp H Values 0.100 M C H subscript 3 C O subscript 2 H footnote 2.\u201d Under the \u201cVolume of 0.100 M N a O H Added ( m L )\u201d column are the following values: 0.0, 5.0, 10.0, 15.0, 20.0, 22.0, 24.0, 24.5, 24.9, 25.0, 25.1, 25.5, 26.0, 28.0, 30.0, 35.0, 40.0, 45.0, and 50.0. Under the \u201cMoles of N a O H Added\u201d column are the following values: 0.0, 0.00050, 0.00100, 0.00150, 0.00200, 0.00220, 0.00240, 0.00245, 0.00249, 0.00250, 0.00251, 0.00255, 0.00260, 0.00280, 0.00300, 0.00350, 0.00400, 0.00450, and 0.00500. Under the \u201cp H Values 0.100 M H C l footnote one\u201d column are the following values: 1.00, 1.18, 1.37, 1.60, 1.95, 2.20, 2.69, 3.00, 3.70, 7.00, 10.30, 11.00, 11.29, 11.75, 11.96, 12.22, 12.36, 12.46, and 12.52. Foot note one reads, \u201cTitration of 25.00 m L of 0.100 M H C l ( 0.00250 mol of H C I ) with 0.100 M N a O H.\u201d Under the \u201cp H Values 0.100 M C H subscript 3 C O subscript 2 H footnote 2\u201d column are the following values: 2.87, 4.14, 4.57, 4.92, 5.35, 5.61, 6.13, 6.44, 7.17, 8.72, 10.30, 11.00, 11.29, 11.75, 11.96, 12.22, 12.36, 12.46, and 12.52. Footnote two reads, \u201cTitration of 25.00 m L of 0.100 M C H subscript 3 C O subscript 2 H ( 0.00250 mol of C H subscript 3C O subscript 2 H) with 0.100 M N a O H.\u201d\">\n<thead>\n<tr>\n<th colspan=\"4\" data-align=\"center\">pH Values in the Titrations of a Strong Acid and of a Weak Acid<\/th>\n<\/tr>\n<tr valign=\"top\">\n<th data-align=\"left\">Volume of 0.100 <em data-effect=\"italics\">M<\/em> NaOH Added (mL)<\/th>\n<th data-align=\"left\">Moles of NaOH Added<\/th>\n<th data-align=\"left\">pH Values 0.100 <em data-effect=\"italics\">M<\/em> HCl<sup data-type=\"footnote-number\"><a href=\"#footnote1\" data-type=\"footnote-link\">1<\/a><\/sup><\/th>\n<th data-align=\"left\">pH Values 0.100 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H<sup id=\"footnote-ref2\" data-type=\"footnote-number\"><a href=\"#footnote2\" data-type=\"footnote-link\">2<\/a><\/sup><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-align=\"left\">0.0<\/td>\n<td data-align=\"left\">0.0<\/td>\n<td data-align=\"left\">1.00<\/td>\n<td data-align=\"left\">2.87<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">5.0<\/td>\n<td data-align=\"left\">0.00050<\/td>\n<td data-align=\"left\">1.18<\/td>\n<td data-align=\"left\">4.14<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">10.0<\/td>\n<td data-align=\"left\">0.00100<\/td>\n<td data-align=\"left\">1.37<\/td>\n<td data-align=\"left\">4.57<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">15.0<\/td>\n<td data-align=\"left\">0.00150<\/td>\n<td data-align=\"left\">1.60<\/td>\n<td data-align=\"left\">4.92<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">20.0<\/td>\n<td data-align=\"left\">0.00200<\/td>\n<td data-align=\"left\">1.95<\/td>\n<td data-align=\"left\">5.35<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">22.0<\/td>\n<td data-align=\"left\">0.00220<\/td>\n<td data-align=\"left\">2.20<\/td>\n<td data-align=\"left\">5.61<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">24.0<\/td>\n<td data-align=\"left\">0.00240<\/td>\n<td data-align=\"left\">2.69<\/td>\n<td data-align=\"left\">6.13<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">24.5<\/td>\n<td data-align=\"left\">0.00245<\/td>\n<td data-align=\"left\">3.00<\/td>\n<td data-align=\"left\">6.44<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">24.9<\/td>\n<td data-align=\"left\">0.00249<\/td>\n<td data-align=\"left\">3.70<\/td>\n<td data-align=\"left\">7.14<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">25.0<\/td>\n<td data-align=\"left\">0.00250<\/td>\n<td data-align=\"left\">7.00<\/td>\n<td data-align=\"left\">8.72<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">25.1<\/td>\n<td data-align=\"left\">0.00251<\/td>\n<td data-align=\"left\">10.30<\/td>\n<td data-align=\"left\">10.30<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">25.5<\/td>\n<td data-align=\"left\">0.00255<\/td>\n<td data-align=\"left\">11.00<\/td>\n<td data-align=\"left\">11.00<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">26.0<\/td>\n<td data-align=\"left\">0.00260<\/td>\n<td data-align=\"left\">11.29<\/td>\n<td data-align=\"left\">11.29<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">28.0<\/td>\n<td data-align=\"left\">0.00280<\/td>\n<td data-align=\"left\">11.75<\/td>\n<td data-align=\"left\">11.75<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">30.0<\/td>\n<td data-align=\"left\">0.00300<\/td>\n<td data-align=\"left\">11.96<\/td>\n<td data-align=\"left\">11.96<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">35.0<\/td>\n<td data-align=\"left\">0.00350<\/td>\n<td data-align=\"left\">12.22<\/td>\n<td data-align=\"left\">12.22<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">40.0<\/td>\n<td data-align=\"left\">0.00400<\/td>\n<td data-align=\"left\">12.36<\/td>\n<td data-align=\"left\">12.36<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">45.0<\/td>\n<td data-align=\"left\">0.00450<\/td>\n<td data-align=\"left\">12.46<\/td>\n<td data-align=\"left\">12.46<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">50.0<\/td>\n<td data-align=\"left\">0.00500<\/td>\n<td data-align=\"left\">12.52<\/td>\n<td data-align=\"left\">12.52<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"CNX_Chem_14_07_titration\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">(a) The titration curve for the titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCl (strong acid) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH (strong base) has an equivalence point of 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> acetic acid (weak acid) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH (strong base) has an equivalence point of 8.72 pH.<\/div>\n<p><span id=\"fs-idp134166880\" data-type=\"media\" data-alt=\"Two graphs are shown. The first graph on the left is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unis beginning at 0 extending to 14. A red curve is drawn on the graph which increases steadily from the point (0, 3) up to about (20, 5.5) after which the graph has a vertical section from (25, 7) up to (25, 11). The graph then levels off to a value of about 12.5 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d The second graph on the right is titled \u201cTitration of Strong Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 units beginning at 0 extending to 14. A red curve is drawn on the graph which increases gradually from the point (0, 1) up to about (22.5, 2.2) after which the graph has a vertical section from (25, 4) up to nearly (25, 11). The graph then levels off to a value of about 12.4 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 7.00.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_07_titration-1.jpg\" alt=\"Two graphs are shown. The first graph on the left is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unis beginning at 0 extending to 14. A red curve is drawn on the graph which increases steadily from the point (0, 3) up to about (20, 5.5) after which the graph has a vertical section from (25, 7) up to (25, 11). The graph then levels off to a value of about 12.5 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d The second graph on the right is titled \u201cTitration of Strong Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L).\u201d Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled \u201cp H\u201d and is marked every 1 units beginning at 0 extending to 14. A red curve is drawn on the graph which increases gradually from the point (0, 1) up to about (22.5, 2.2) after which the graph has a vertical section from (25, 4) up to nearly (25, 11). The graph then levels off to a value of about 12.4 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 7.00.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp22610576\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Acid-Base Indicators<\/strong><\/h3>\n<p id=\"fs-idm39085312\">Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 \u00d7 10<sup>\u22129<\/sup><em data-effect=\"italics\">M<\/em> (pH &lt; 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 \u00d7 10<sup>\u22129<\/sup><em data-effect=\"italics\">M<\/em> (pH &gt; 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called <strong>acid-base indicators<\/strong>. Acid-base indicators are either weak organic acids or weak organic bases.<\/p>\n<p id=\"fs-idm28417392\">The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:<\/p>\n<div id=\"fs-idp80468176\" style=\"padding-left: 40px\" data-type=\"equation\">HIn(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + In<sup>\u2212<\/sup>(<em>aq<\/em>)<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">red\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0yellow<\/div>\n<div id=\"fs-idm87967824\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1927\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-300x62.png\" alt=\"\" width=\"261\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-300x62.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-768x158.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-225x46.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f-350x72.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7f.png 882w\" sizes=\"auto, (max-width: 261px) 100vw, 261px\" \/><\/div>\n<p id=\"fs-idm68547888\">The anion of methyl orange, In<sup>\u2212<\/sup>, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Ch\u00e2telier\u2019s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.<\/p>\n<p id=\"fs-idp64938256\">The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species In<sup>\u2212<\/sup> and HIn. If most of the indicator (typically about 60\u221290% or more) is present as In<sup>\u2212<\/sup>, the perceived color of the solution is yellow. If most is present as HIn, then the solution color appears red. The Henderson-Hasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):<\/p>\n<div id=\"fs-idp80076272\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1928\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g-300x77.png\" alt=\"\" width=\"211\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g-300x77.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g-65x17.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g-225x58.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g-350x90.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.7g.png 665w\" sizes=\"auto, (max-width: 211px) 100vw, 211px\" \/><\/div>\n<p id=\"fs-idm188576256\">In solutions where pH &gt; p<em data-effect=\"italics\">K<\/em><sub>a<\/sub>, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When pH &gt; p<em data-effect=\"italics\">K<\/em><sub>a<\/sub>, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa, appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The <span data-type=\"term\">color change interval<\/span> (or <em data-effect=\"italics\">pH interval<\/em>) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> \u00b1 1.<\/p>\n<p id=\"fs-idm92451888\">There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_indicators\">(Figure)<\/a> presents several indicators, their colors, and their color-change intervals.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_14_07_indicators\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">This chart illustrates the color change intervals for several acid-base indicators.<\/div>\n<p><span id=\"fs-idp112102048\" data-type=\"media\" data-alt=\"This figure provides a graphical representation of indicators and color ranges. A horizontal axis is labeled \u201cp H.\u201d This axis begins at zero and increases by ones up to 13. The left side of the graphic provides a column with the names of indicators. To the right of each indicator name is either one or two colored bars that are shaded according to the indicator color at various p H ranges. From the top, the first row is labeled \u201cCrystal violet.\u201d The associated colored bar is yellow at its left end at a p H of 0 and changes to green and blue moving right to its endpoint at a p H of 1.8. The second row is labeled \u201cCresol red.\u201d The associated colored bar is red at its left end at a p H of 1 and changes to orange and yellow moving right to its endpoint at a p H of just over 2. A second bar to its right is yellow at a p H of around 7 and proceeds through orange to red at a p H of about 9. The third row is labeled \u201cThymol blue.\u201d The associated colored bar is red at its left end at a p H of nearly 1.2 and changes to orange and red moving right to its endpoint at a p H of 2.8. A second bar begins in yellow at a p H of 8 and proceeds through green and blue to its end at a p H of around 9.1. The fourth row is labeled \u201cErythrosin B.\u201d The associated colored bar is red from a p H of 2.2 to its endpoint at a p H of 3.6. The fifth row is labeled \u201c2 comma 4 dash Dinitrophenol.\u201d The associated colored bar is white at its left end at a p H of 2.6 and changes to yellow at its endpoint at a p H of 4. The sixth row is labeled \u201cBromophenol blue.\u201d The associated colored bar is yellow at its left end at a p H of 3 and changes to green and blue moving right to its endpoint at a p H of 4.5. The seventh row is labeled \u201cMethyl orange.\u201d The associated colored bar is red-orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The eighth row is labeled \u201cBromocresol green.\u201d The associated colored bar is yellow at its left end at a p H of 3.8 and changes to green and blue moving right to its endpoint at a p H of 5.4. The ninth row is labeled \u201cMethyl red.\u201d The associated colored bar is orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The tenth row is labeled \u201cEriochrome * Black T.\u201d The associated colored bar is red at its left end at a p H of 5 and changes to purple and blue moving right to its endpoint at a p H of 6.5. The eleventh row is labeled \u201cBromocresol purple.\u201d The associated colored bar is yellow at its left end at a p H of 5.2 and changes to purple moving right to its endpoint at a p H of 6.8. The twelfth row is labeled \u201cAlizarin.\u201d The first associated colored bar is yellow-orange at its left end at a p H of 5.7 and changes to red moving right to its endpoint at a p H of 7.2. A second bar begins in red at a p H of 11 and changes to purple, then dark blue at its right end at a p H of 12.4. The thirteenth row is labeled \u201cBromothymol blue.\u201d The associated colored bar is yellow at its left end at a p H of 6 and changes to green and blue moving right to its endpoint at a p H of 7.6. The fourteenth row is labeled \u201cPhenol red.\u201d The associated colored bar is yellow-orange at its left end at a p H of 6.8 and changes to orange and red moving right to its endpoint at a p H of 8.2. The fifteenth row is labeled \u201cm dash Nitrophenol.\u201d The associated colored bar is white at its left end at a p H of 6.8 and changes to yellow moving right to its endpoint at a p H of 8.6. The sixteenth row is labeled \u201co dash Cresolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8.3 and changes to red moving right to its endpoint at a p H of 9.8. The seventeenth row is labeled \u201cPhenolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8 and changes to pink moving right to its endpoint at a p H of 10. The eighteenth row is labeled \u201cThymolphthalein.\u201d The associated colored bar is light blue at its left end at a p H of 9.3 and changes to a deep, dark blue moving right to its endpoint at a p H of 10.5. The nineteenth row is labeled \u201cAlizarin yellow R.\u201d The associated colored bar is yellow-orange at its left end at a p H of 10 and changes to red moving right to its endpoint at a p H of 12.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_07_indicators-1.jpg\" alt=\"This figure provides a graphical representation of indicators and color ranges. A horizontal axis is labeled \u201cp H.\u201d This axis begins at zero and increases by ones up to 13. The left side of the graphic provides a column with the names of indicators. To the right of each indicator name is either one or two colored bars that are shaded according to the indicator color at various p H ranges. From the top, the first row is labeled \u201cCrystal violet.\u201d The associated colored bar is yellow at its left end at a p H of 0 and changes to green and blue moving right to its endpoint at a p H of 1.8. The second row is labeled \u201cCresol red.\u201d The associated colored bar is red at its left end at a p H of 1 and changes to orange and yellow moving right to its endpoint at a p H of just over 2. A second bar to its right is yellow at a p H of around 7 and proceeds through orange to red at a p H of about 9. The third row is labeled \u201cThymol blue.\u201d The associated colored bar is red at its left end at a p H of nearly 1.2 and changes to orange and red moving right to its endpoint at a p H of 2.8. A second bar begins in yellow at a p H of 8 and proceeds through green and blue to its end at a p H of around 9.1. The fourth row is labeled \u201cErythrosin B.\u201d The associated colored bar is red from a p H of 2.2 to its endpoint at a p H of 3.6. The fifth row is labeled \u201c2 comma 4 dash Dinitrophenol.\u201d The associated colored bar is white at its left end at a p H of 2.6 and changes to yellow at its endpoint at a p H of 4. The sixth row is labeled \u201cBromophenol blue.\u201d The associated colored bar is yellow at its left end at a p H of 3 and changes to green and blue moving right to its endpoint at a p H of 4.5. The seventh row is labeled \u201cMethyl orange.\u201d The associated colored bar is red-orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The eighth row is labeled \u201cBromocresol green.\u201d The associated colored bar is yellow at its left end at a p H of 3.8 and changes to green and blue moving right to its endpoint at a p H of 5.4. The ninth row is labeled \u201cMethyl red.\u201d The associated colored bar is orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The tenth row is labeled \u201cEriochrome * Black T.\u201d The associated colored bar is red at its left end at a p H of 5 and changes to purple and blue moving right to its endpoint at a p H of 6.5. The eleventh row is labeled \u201cBromocresol purple.\u201d The associated colored bar is yellow at its left end at a p H of 5.2 and changes to purple moving right to its endpoint at a p H of 6.8. The twelfth row is labeled \u201cAlizarin.\u201d The first associated colored bar is yellow-orange at its left end at a p H of 5.7 and changes to red moving right to its endpoint at a p H of 7.2. A second bar begins in red at a p H of 11 and changes to purple, then dark blue at its right end at a p H of 12.4. The thirteenth row is labeled \u201cBromothymol blue.\u201d The associated colored bar is yellow at its left end at a p H of 6 and changes to green and blue moving right to its endpoint at a p H of 7.6. The fourteenth row is labeled \u201cPhenol red.\u201d The associated colored bar is yellow-orange at its left end at a p H of 6.8 and changes to orange and red moving right to its endpoint at a p H of 8.2. The fifteenth row is labeled \u201cm dash Nitrophenol.\u201d The associated colored bar is white at its left end at a p H of 6.8 and changes to yellow moving right to its endpoint at a p H of 8.6. The sixteenth row is labeled \u201co dash Cresolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8.3 and changes to red moving right to its endpoint at a p H of 9.8. The seventeenth row is labeled \u201cPhenolphthalein.\u201d The associated colored bar is white at its left end at a p H of 8 and changes to pink moving right to its endpoint at a p H of 10. The eighteenth row is labeled \u201cThymolphthalein.\u201d The associated colored bar is light blue at its left end at a p H of 9.3 and changes to a deep, dark blue moving right to its endpoint at a p H of 10.5. The nineteenth row is labeled \u201cAlizarin yellow R.\u201d The associated colored bar is yellow-orange at its left end at a p H of 10 and changes to red moving right to its endpoint at a p H of 12.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"CNX_Chem_14_07_titration2\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">Titration curves for strong and weak acids illustrating the proper choice of acid-base indicator. Any of the three indicators will exhibit a reasonably sharp color change at the equivalence point of the strong acid titration, but only phenolphthalein is suitable for use in the weak acid titration.<\/div>\n<p><span id=\"fs-idm5594336\" data-type=\"media\" data-alt=\"A graph is shown which is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L)\u201d and begins at 0 with markings every 5 units up to 50. The vertical axis is labeled \u201cp H\u201d and begins at 0 and increases by single units up to 14. A red curve is drawn on the graph. The curve begins at (0, 3) and passes through the points (5, 4.1), (10, 4.7), (15, 5), (20, 5.5), and (22.5, 6), after which it rapidly increases, forming a vertical section centered at the point (25, 8.7). The rapid increase of the curve then levels off and the curve passes through the points (30, 12), (35, 12.4), (40, 12.5), (45, 12.6), and (50, 12.6). A brown rectangle extends horizontally across the graph covering the p H of 3 to 4.2 range. To the right, this rectangle is labeled \u201cMethyl orange p H range.\u201d A blue rectangle extends horizontally across the graph covering the p H of 4.6 to 8 range. To the right, this rectangle is labeled \u201cLitmus p H range.\u201d A purple rectangle extends horizontally across the graph covering the p H of 8.4 to 10 range. To the right, this rectangle is labeled \u201cPhenolphthalein p H range.\u201d The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_07_titration2-1.jpg\" alt=\"A graph is shown which is titled \u201cTitration of Weak Acid.\u201d The horizontal axis is labeled \u201cVolume of 0.100 M N a O H added (m L)\u201d and begins at 0 with markings every 5 units up to 50. The vertical axis is labeled \u201cp H\u201d and begins at 0 and increases by single units up to 14. A red curve is drawn on the graph. The curve begins at (0, 3) and passes through the points (5, 4.1), (10, 4.7), (15, 5), (20, 5.5), and (22.5, 6), after which it rapidly increases, forming a vertical section centered at the point (25, 8.7). The rapid increase of the curve then levels off and the curve passes through the points (30, 12), (35, 12.4), (40, 12.5), (45, 12.6), and (50, 12.6). A brown rectangle extends horizontally across the graph covering the p H of 3 to 4.2 range. To the right, this rectangle is labeled \u201cMethyl orange p H range.\u201d A blue rectangle extends horizontally across the graph covering the p H of 4.6 to 8 range. To the right, this rectangle is labeled \u201cLitmus p H range.\u201d A purple rectangle extends horizontally across the graph covering the p H of 8.4 to 10 range. To the right, this rectangle is labeled \u201cPhenolphthalein p H range.\u201d The midpoint of the vertical segment of the curve is labeled \u201cEquivalence point p H, 8.72.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm188916592\">The titration curves shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration2\">(Figure)<\/a> illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of ~24 mL of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration&#8217;s end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.<\/p>\n<p id=\"fs-idm224614176\">The weak acid titration curve in <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_07_titration2\">(Figure)<\/a> shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-to-yellow color change over a relatively large volume interval (0\u20136 mL), completing the color change well before the equivalence point (25 mL) has been reached. Use of litmus would show a color change that begins after adding 7\u20138 mL of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration&#8217;s equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.<\/p>\n<\/div>\n<div id=\"fs-idp81138704\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idp77329840\">The titration curve for an acid-base titration is typically a plot of pH versus volume of added titrant. These curves are useful in selecting appropriate acid-base indicators that will permit accurate determinations of titration end points.<\/p>\n<\/div>\n<div id=\"fs-idp63641264\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm5595328\" data-type=\"exercise\">\n<div id=\"fs-idp5936080\" data-type=\"problem\"><\/div>\n<\/div>\n<div id=\"fs-idp105330480\" data-type=\"exercise\">\n<div id=\"fs-idp105330736\" data-type=\"problem\">\n<p id=\"fs-idp105330992\">\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"footnote-refs\">\n<h3 data-type=\"footnote-refs-title\"><strong>Footnotes<\/strong><\/h3>\n<ul data-list-type=\"bulleted\" data-bullet-style=\"none\">\n<li data-type=\"footnote-ref\"><a href=\"#footnote-ref1\" data-type=\"footnote-ref-link\">1<\/a><span data-type=\"footnote-ref-content\">Titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> HCl (0.00250 mol of HCI) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH.<\/span><\/li>\n<li id=\"footnote2\" data-type=\"footnote-ref\"><a href=\"#footnote-ref2\" data-type=\"footnote-ref-link\">2<\/a><span data-type=\"footnote-ref-content\">Titration of 25.00 mL of 0.100 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H (0.00250 mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H) with 0.100 <em data-effect=\"italics\">M<\/em> NaOH.<\/span><\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\n<dl id=\"fs-idp67176144\">\n<dt>acid-base indicator<\/dt>\n<dd id=\"fs-idp67176784\">weak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH<\/dd>\n<\/dl>\n<dl id=\"fs-idp67179744\">\n<dt>titration curve<\/dt>\n<dd id=\"fs-idp67180384\">plot of some sample property (such as pH) versus volume of added titrant<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":8,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-810","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":766,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/810","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":13,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/810\/revisions"}],"predecessor-version":[{"id":2176,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/810\/revisions\/2176"}],"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\/810\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=810"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=810"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=810"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=810"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}