{"id":1019,"date":"2022-06-23T13:14:40","date_gmt":"2022-06-23T17:14:40","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/chapter\/14-7-acid-base-titrations-chemistry-2e\/"},"modified":"2022-07-07T08:14:00","modified_gmt":"2022-07-07T12:14:00","slug":"14-7-acid-base-titrations-chemistry-2e","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/chapter\/14-7-acid-base-titrations-chemistry-2e\/","title":{"raw":"14.7 Acid-Base Titrations","rendered":"14.7 Acid-Base Titrations"},"content":{"raw":"<div>\r\n\r\n&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 acid\/strong base systems<\/li>\r\n \t<li>Compute sample pH at important stages of a titration<\/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\">\r\n<h3><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 demonstrates the computation of pH for titration of a strong acid after additions of several specified volumes of strong base.<\/p>\r\n\r\n<div id=\"fs-idm49336656\" class=\"textbox textbox--examples\">\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>M<\/em> HCl (strong acid) with 0.100 <em>M<\/em> of a strong base NaOH. 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>M<\/em>. The pH of the solution is then\r\n<div id=\"fs-idm213177360\" style=\"padding-left: 40px\">pH = -log(0.100 M) = 1.000<\/div>\r\n<div><\/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>\u2013<\/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 \u2013 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\">[H<sub>3<\/sub>O<sup>+<\/sup>] = (0.00125 mol)\/(0.03750 L) = 0.0333 M<\/div>\r\n<div style=\"padding-left: 40px\">pH = -log(0.0333 M) = 1.477<\/div>\r\n<div><\/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>\u2013<\/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>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>\u2013<\/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\">\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>\u2013<\/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 \u2013 0.00250 mol = 0.00125 mol OH<sup>\u2013<\/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>M<\/em> HNO<sub>3<\/sub>(<em>aq<\/em>) and 0.200 <em>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\">\r\n<div><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<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 acid titrations is provided in <a class=\"autogenerated-content\" href=\"#fs-idm87178400\">(Figure)<\/a> and plotted as a titration curve. A consideration of the titration curve 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<\/p>\r\n<p id=\"fs-idm444217232\">pre-equivalence point (0 mL &lt; <em>V<\/em> &lt; 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant<\/p>\r\n<p id=\"fs-idm183388016\">equivalence point (<em>V<\/em> = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to neutral<\/p>\r\n<p id=\"fs-idm183509856\">post-equivalence point (<em>V<\/em> &gt; 25 mL): pH is determined by the amount of excess strong base titrant added<\/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\">pH Values in the Titrations of a Strong Acid<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<th>Volume of 0.100 <em>M<\/em> NaOH Added (mL)<\/th>\r\n<th>Moles of NaOH Added<\/th>\r\n<th>pH Values 0.100 <em>M<\/em> HCl<sup><a href=\"#footnote1\">1<\/a><\/sup><\/th>\r\n<th><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0.0<\/td>\r\n<td>0.0<\/td>\r\n<td>1.00<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5.0<\/td>\r\n<td>0.00050<\/td>\r\n<td>1.18<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>10.0<\/td>\r\n<td>0.00100<\/td>\r\n<td>1.37<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>15.0<\/td>\r\n<td>0.00150<\/td>\r\n<td>1.60<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>20.0<\/td>\r\n<td>0.00200<\/td>\r\n<td>1.95<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>22.0<\/td>\r\n<td>0.00220<\/td>\r\n<td>2.20<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>24.0<\/td>\r\n<td>0.00240<\/td>\r\n<td>2.69<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>24.5<\/td>\r\n<td>0.00245<\/td>\r\n<td>3.00<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>24.9<\/td>\r\n<td>0.00249<\/td>\r\n<td>3.70<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>25.0<\/td>\r\n<td>0.00250<\/td>\r\n<td>7.00<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>25.1<\/td>\r\n<td>0.00251<\/td>\r\n<td>10.30<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>25.5<\/td>\r\n<td>0.00255<\/td>\r\n<td>11.00<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>26.0<\/td>\r\n<td>0.00260<\/td>\r\n<td>11.29<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>28.0<\/td>\r\n<td>0.00280<\/td>\r\n<td>11.75<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>30.0<\/td>\r\n<td>0.00300<\/td>\r\n<td>11.96<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>35.0<\/td>\r\n<td>0.00350<\/td>\r\n<td>12.22<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>40.0<\/td>\r\n<td>0.00400<\/td>\r\n<td>12.36<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>45.0<\/td>\r\n<td>0.00450<\/td>\r\n<td>12.46<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>50.0<\/td>\r\n<td>0.00500<\/td>\r\n<td>12.52<\/td>\r\n<td><\/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\">The titration curve for the titration of 25.00 mL of 0.100 <em>M<\/em> HCl (strong acid) with 0.100 <em>M<\/em> NaOH (strong base) has an equivalence point of 7.00 pH.<\/div>\r\n<div><img class=\"alignnone size-medium wp-image-1570\" src=\"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve-260x300.png\" alt=\"\" width=\"260\" height=\"300\" \/><\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp81138704\" class=\"summary\">\r\n<h3><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.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp63641264\" class=\"exercises\">\r\n<div id=\"fs-idm5595328\">\r\n<div id=\"fs-idp5936080\"><\/div>\r\n<\/div>\r\n<div id=\"fs-idp105330480\">\r\n<div id=\"fs-idp105330736\">\r\n<p id=\"fs-idp105330992\"><strong style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Footnotes<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<ul>\r\n \t<li><a href=\"#footnote-ref1\">1<\/a>Titration of 25.00 mL of 0.100 <em>M<\/em> HCl (0.00250 mol of HCI) with 0.100 <em>M<\/em> NaOH.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\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>\r\n<\/div>\r\n<div><\/div>","rendered":"<div>\n<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 acid\/strong base systems<\/li>\n<li>Compute sample pH at important stages of a titration<\/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\">\n<h3><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 demonstrates the computation of pH for titration of a strong acid after additions of several specified volumes of strong base.<\/p>\n<div id=\"fs-idm49336656\" class=\"textbox textbox--examples\">\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>M<\/em> HCl (strong acid) with 0.100 <em>M<\/em> of a strong base NaOH. 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>M<\/em>. The pH of the solution is then<\/p>\n<div id=\"fs-idm213177360\" style=\"padding-left: 40px\">pH = -log(0.100 M) = 1.000<\/div>\n<div><\/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>\u2013<\/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 \u2013 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\">[H<sub>3<\/sub>O<sup>+<\/sup>] = (0.00125 mol)\/(0.03750 L) = 0.0333 M<\/div>\n<div style=\"padding-left: 40px\">pH = -log(0.0333 M) = 1.477<\/div>\n<div><\/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>\u2013<\/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>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>\u2013<\/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\">\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>\u2013<\/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 \u2013 0.00250 mol = 0.00125 mol OH<sup>\u2013<\/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>M<\/em> HNO<sub>3<\/sub>(<em>aq<\/em>) and 0.200 <em>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\">\n<div><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<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 acid titrations is provided in <a class=\"autogenerated-content\" href=\"#fs-idm87178400\">(Figure)<\/a> and plotted as a titration curve. A consideration of the titration curve 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<\/p>\n<p id=\"fs-idm444217232\">pre-equivalence point (0 mL &lt; <em>V<\/em> &lt; 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant<\/p>\n<p id=\"fs-idm183388016\">equivalence point (<em>V<\/em> = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to neutral<\/p>\n<p id=\"fs-idm183509856\">post-equivalence point (<em>V<\/em> &gt; 25 mL): pH is determined by the amount of excess strong base titrant added<\/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\">pH Values in the Titrations of a Strong Acid<\/th>\n<\/tr>\n<tr valign=\"top\">\n<th>Volume of 0.100 <em>M<\/em> NaOH Added (mL)<\/th>\n<th>Moles of NaOH Added<\/th>\n<th>pH Values 0.100 <em>M<\/em> HCl<sup><a href=\"#footnote1\">1<\/a><\/sup><\/th>\n<th><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0.0<\/td>\n<td>0.0<\/td>\n<td>1.00<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5.0<\/td>\n<td>0.00050<\/td>\n<td>1.18<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>10.0<\/td>\n<td>0.00100<\/td>\n<td>1.37<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>15.0<\/td>\n<td>0.00150<\/td>\n<td>1.60<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>20.0<\/td>\n<td>0.00200<\/td>\n<td>1.95<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>22.0<\/td>\n<td>0.00220<\/td>\n<td>2.20<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>24.0<\/td>\n<td>0.00240<\/td>\n<td>2.69<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>24.5<\/td>\n<td>0.00245<\/td>\n<td>3.00<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>24.9<\/td>\n<td>0.00249<\/td>\n<td>3.70<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>25.0<\/td>\n<td>0.00250<\/td>\n<td>7.00<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>25.1<\/td>\n<td>0.00251<\/td>\n<td>10.30<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>25.5<\/td>\n<td>0.00255<\/td>\n<td>11.00<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>26.0<\/td>\n<td>0.00260<\/td>\n<td>11.29<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>28.0<\/td>\n<td>0.00280<\/td>\n<td>11.75<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>30.0<\/td>\n<td>0.00300<\/td>\n<td>11.96<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>35.0<\/td>\n<td>0.00350<\/td>\n<td>12.22<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>40.0<\/td>\n<td>0.00400<\/td>\n<td>12.36<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>45.0<\/td>\n<td>0.00450<\/td>\n<td>12.46<\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>50.0<\/td>\n<td>0.00500<\/td>\n<td>12.52<\/td>\n<td><\/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\">The titration curve for the titration of 25.00 mL of 0.100 <em>M<\/em> HCl (strong acid) with 0.100 <em>M<\/em> NaOH (strong base) has an equivalence point of 7.00 pH.<\/div>\n<div><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1570\" src=\"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve-260x300.png\" alt=\"\" width=\"260\" height=\"300\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve-260x300.png 260w, https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve-65x75.png 65w, https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve-225x260.png 225w, https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve-350x404.png 350w, https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-content\/uploads\/sites\/1718\/2022\/06\/titration-curve.png 682w\" sizes=\"auto, (max-width: 260px) 100vw, 260px\" \/><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp81138704\" class=\"summary\">\n<h3><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.<\/p>\n<\/div>\n<div id=\"fs-idp63641264\" class=\"exercises\">\n<div id=\"fs-idm5595328\">\n<div id=\"fs-idp5936080\"><\/div>\n<\/div>\n<div id=\"fs-idp105330480\">\n<div id=\"fs-idp105330736\">\n<p id=\"fs-idp105330992\"><strong style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Footnotes<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<ul>\n<li><a href=\"#footnote-ref1\">1<\/a>Titration of 25.00 mL of 0.100 <em>M<\/em> HCl (0.00250 mol of HCI) with 0.100 <em>M<\/em> NaOH.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\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<\/div>\n<div><\/div>\n","protected":false},"author":1392,"menu_order":37,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[49],"contributor":[],"license":[],"class_list":["post-1019","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":1467,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/chapters\/1019","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":8,"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/chapters\/1019\/revisions"}],"predecessor-version":[{"id":1574,"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/chapters\/1019\/revisions\/1574"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/parts\/1467"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/chapters\/1019\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/wp\/v2\/media?parent=1019"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=1019"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/wp\/v2\/contributor?post=1019"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/wp-json\/wp\/v2\/license?post=1019"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}