{"id":806,"date":"2021-07-23T09:20:46","date_gmt":"2021-07-23T13:20:46","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/buffers\/"},"modified":"2022-06-23T09:21:29","modified_gmt":"2022-06-23T13:21:29","slug":"buffers","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/buffers\/","title":{"raw":"14.6 Buffers","rendered":"14.6 Buffers"},"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>Describe the composition and function of acid\u2013base buffers<\/li>\r\n \t<li>Calculate the pH of a buffer before and after the addition of added acid or base<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm117447344\">A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a <strong>buffer<\/strong>. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (<a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_compare\">(Figure)<\/a>). A solution of acetic acid and sodium acetate (CH<sub>3<\/sub>COOH + NaCH<sub>3<\/sub>COO) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH<sub>3<\/sub>(<em data-effect=\"italics\">aq<\/em>) + NH<sub>4<\/sub>Cl(<em data-effect=\"italics\">aq<\/em>)).<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_14_06_compare\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">(a) The buffered solution on the left and the unbuffered solution on the right have the same pH (pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH. (b) After the addition of 1 mL of a 0.01-<em data-effect=\"italics\">M<\/em> HCl solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)<\/div>\r\n<span id=\"fs-idm104574592\" data-type=\"media\" data-alt=\"Two images are shown. Image a on the left shows two beakers that each contain yellow solutions. The beaker on the left is labeled \u201cUnbuffered\u201d and the beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d Image b similarly shows 2 beakers. The beaker on the left contains a bright orange solution and is labeled \u201cUnbuffered.\u201d The beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_compare-1.jpg\" alt=\"Two images are shown. Image a on the left shows two beakers that each contain yellow solutions. The beaker on the left is labeled \u201cUnbuffered\u201d and the beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d Image b similarly shows 2 beakers. The beaker on the left contains a bright orange solution and is labeled \u201cUnbuffered.\u201d The beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idm112960288\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>How Buffers Work<\/strong><\/h3>\r\n<p id=\"fs-idm127493136\">To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, adding strong base to this solution will neutralize hydronium ion and shift the acetic acid ionization equilibrium to the right, partially restoring the decreased H<sub>3<\/sub>O<sup>+<\/sup> concentration:<\/p>\r\n\r\n<div id=\"fs-idm105688000\" style=\"padding-left: 40px\" data-type=\"equation\">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>)<\/div>\r\n<p id=\"fs-idp41915456\">Likewise, adding strong acid to this buffer solution will neutralize acetate ion, shifting the above ionization equilibrium left and returning [H<sub>3<\/sub>O<sup>+<\/sup>] to near its original value. <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_bufferchrt\">(Figure)<\/a> provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_14_06_bufferchrt\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">Buffering action in a mixture of acetic acid and acetate salt.<\/div>\r\n<span id=\"fs-idp38705984\" data-type=\"media\" data-alt=\"This figure begins with a chemical reaction at the top: C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O superscript positive sign ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below this equation are two arrows: one pointing left and other pointing right. The arrow pointing left has this phrase written above it, \u201cH subscript 3 O superscript positive sign added, equilibrium position shifts to the left.\u201d Below the arrow is the reaction: C H subscript 3 C O O H ( a q ) left-facing arrow C H subscript 3 C O O superscript negative sign ( a q ) plus H subscript 3 O superscript positive sign. The arrow pointing right has this phrase written above it, \u201cO H subscript negative sign added, equilibrium position shifts to the right.\u201d Below the arrow is the reaction: O H superscript negative sign plus C H subscript 3 C O O H ( a q ) right-facing arrow H subscript 2 O ( l ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below all the text is a figure that resembles a bar graph. In the middle are two bars of equal height. One is labeled, \u201cC H subscript 3 C O O H,\u201d and the other is labeled, \u201cC H subscript 3 C O O superscript negative sign.\u201d There is a dotted line at the same height of the bars which extends to the left and right. Above these two bars is the phrase, \u201cBuffer solution equimolar in acid and base.\u201d There is an arrow pointing to the right which is labeled, \u201cAdd O H superscript negative sign.\u201d The arrow points to two bars again, but this time the C H subscript 3 C O O H bar is shorter than that C H subscript 3 C O O superscript negative sign bar. Above these two bars is the phrase, \u201cBuffer solution after addition of strong base.\u201d From the middle bars again, there is an arrow that points left. The arrow is labeled, \u201cAdd H subscript 3 O superscript positive sign.\u201d This arrow points to two bars again, but this time the C H subscript 3 C O O H bar is taller than the C H subscript 3 C O O superscript negative sign bar. These two bars are labeled, \u201cBuffer solution after addition of strong acid.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_bufferchrt-1.jpg\" alt=\"This figure begins with a chemical reaction at the top: C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O superscript positive sign ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below this equation are two arrows: one pointing left and other pointing right. The arrow pointing left has this phrase written above it, \u201cH subscript 3 O superscript positive sign added, equilibrium position shifts to the left.\u201d Below the arrow is the reaction: C H subscript 3 C O O H ( a q ) left-facing arrow C H subscript 3 C O O superscript negative sign ( a q ) plus H subscript 3 O superscript positive sign. The arrow pointing right has this phrase written above it, \u201cO H subscript negative sign added, equilibrium position shifts to the right.\u201d Below the arrow is the reaction: O H superscript negative sign plus C H subscript 3 C O O H ( a q ) right-facing arrow H subscript 2 O ( l ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below all the text is a figure that resembles a bar graph. In the middle are two bars of equal height. One is labeled, \u201cC H subscript 3 C O O H,\u201d and the other is labeled, \u201cC H subscript 3 C O O superscript negative sign.\u201d There is a dotted line at the same height of the bars which extends to the left and right. Above these two bars is the phrase, \u201cBuffer solution equimolar in acid and base.\u201d There is an arrow pointing to the right which is labeled, \u201cAdd O H superscript negative sign.\u201d The arrow points to two bars again, but this time the C H subscript 3 C O O H bar is shorter than that C H subscript 3 C O O superscript negative sign bar. Above these two bars is the phrase, \u201cBuffer solution after addition of strong base.\u201d From the middle bars again, there is an arrow that points left. The arrow is labeled, \u201cAdd H subscript 3 O superscript positive sign.\u201d This arrow points to two bars again, but this time the C H subscript 3 C O O H bar is taller than the C H subscript 3 C O O superscript negative sign bar. These two bars are labeled, \u201cBuffer solution after addition of strong acid.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idm144695456\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm145736240\"><strong>pH Changes in Buffered and Unbuffered Solutions <\/strong><\/p>\r\nAcetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.\r\n<p id=\"fs-idp41841344\">(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 <em data-effect=\"italics\">M<\/em> acetic acid and 0.10 <em data-effect=\"italics\">M<\/em> sodium acetate.<\/p>\r\n<p id=\"fs-idm197032080\">(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100.0 mL of this buffer.<\/p>\r\n<p id=\"fs-idm215059040\">(c) For comparison, calculate the pH after 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH is added to 100.0 mL of a solution of an unbuffered solution with a pH of 4.74.<\/p>\r\n<p id=\"fs-idm105719280\"><strong>Solution:<\/strong><\/p>\r\n<p id=\"fs-idm224056512\">(a) Following the ICE approach to this equilibrium calculation yields the following:<\/p>\r\n<span id=\"fs-idm123715040\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 2 O ] equilibrium arrow H subscript 3 O superscript plus sign [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_ICETable16_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 2 O ] equilibrium arrow H subscript 3 O superscript plus sign [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idm192051584\">Substituting the equilibrium concentration terms into the <em data-effect=\"italics\">K<\/em><sub>a<\/sub> expression, assuming <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.10 M, i.e. <em>x<\/em> &lt; 0.0050 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> = 1.8 \u00d7 10<sup>-5<\/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(1.8 \u00d7 10<sup>-5 <\/sup>M) = 4.74<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp98075040\">(b) Calculate the pH after 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH is added to 100.0 mL of this buffer.<\/p>\r\n<p id=\"fs-idm196351552\">Adding strong acid will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.<\/p>\r\n\r\n<div id=\"fs-idm110133872\" style=\"padding-left: 40px\" data-type=\"equation\">0.0010 L \u00d7 0.10 mol NaOH\/L = 1.0 \u00d7 10<sup>-4<\/sup> mol NaOH<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm223581568\">The initial molar amount of acetic acid is<\/p>\r\n\r\n<div id=\"fs-idm53635376\" style=\"padding-left: 40px\" data-type=\"equation\">0.1000 L \u00d7 0.10 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H\/L = 1.0 \u00d7 10<sup>-2<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm162724496\">The amount of acetic acid remaining after some is neutralized by the added base is:<\/p>\r\n\r\n<div id=\"fs-idm91870896\" style=\"padding-left: 40px\" data-type=\"equation\">1.0 \u00d7 10<sup>-2<\/sup> mol - 1.0 \u00d7 10<sup>-4<\/sup> mol = 0.<span style=\"text-decoration: underline\">9<\/span>9 \u00d7 10<sup>-2<\/sup>\u00a0 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm178487280\">The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of<\/p>\r\n\r\n<div id=\"fs-idp82833792\" style=\"padding-left: 40px\" data-type=\"equation\">1.0 \u00d7 10<sup>-2<\/sup> mol + 1.0 \u00d7 10<sup>-4<\/sup> mol = 1.<span style=\"text-decoration: underline\">0<\/span>1\u00a0\u00d7 10<sup>-2<\/sup>\u00a0 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm213828416\">Compute molar concentrations for the two buffer components:<\/p>\r\n\r\n<div id=\"fs-idm165714384\" style=\"padding-left: 40px\" data-type=\"equation\">[CH<sub>3<\/sub>CO<sub>2<\/sub>H] = (0.<span style=\"text-decoration: underline\">9<\/span>9 \u00d7 10<sup>-2<\/sup>\u00a0 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H)\/(0.1010 L) = 0.0<span style=\"text-decoration: underline\">9<\/span>8M<\/div>\r\n<div id=\"fs-idp22032080\" style=\"padding-left: 40px\" data-type=\"equation\">[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup>] = (1.<span style=\"text-decoration: underline\">0<\/span>1 \u00d7 10<sup>-2<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>-<\/sup>)\/(0.1010 L) = 0.10 M<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm198587408\">Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.<span style=\"text-decoration: underline\">7<\/span>5 (only slightly different from that prior to adding the strong base).<\/p>\r\n<p id=\"fs-idm140802864\">(c) For comparison, calculate the pH after 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH is added to 100.0 mL of a solution of an unbuffered solution with a pH of 4.74.<\/p>\r\n<p id=\"fs-idm216667360\">The amount of hydrogen ion initially present in the solution is<\/p>\r\n\r\n<div id=\"fs-idm181266816\" style=\"padding-left: 40px\" data-type=\"equation\">[H<sub>3<\/sub>O<sup>+<\/sup>] = 10<sup>-4.74<\/sup> = 1.8 \u00d7 10<sup>-5<\/sup> M<\/div>\r\n<div id=\"fs-idm226340192\" style=\"padding-left: 40px\" data-type=\"equation\">n<sub>H3O+<\/sub> = (0.1000L)(1.8 \u00d7 10<sup>-5<\/sup> mol\/L) = 1.8 \u00d7 10<sup>-6 <\/sup>mol H<sub>3<\/sub>O<sup>+<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm213735984\">The amount of hydroxide ion added to the solution is<\/p>\r\n\r\n<div id=\"fs-idm167813376\" style=\"padding-left: 40px\" data-type=\"equation\">n<sub>OH-<\/sub> = (0.0010 L)(0.10 mol\/L) = 1.0 \u00d7 10<sup>-4 <\/sup>mol OH<sup>-<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm184173168\">The added hydroxide will neutralize hydronium ion via the reaction<\/p>\r\n\r\n<div id=\"fs-idm213939888\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>-<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>l<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm174460304\">The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).<\/p>\r\n<p id=\"fs-idm201883456\">The amount of hydroxide ion remaining is<\/p>\r\n\r\n<div id=\"fs-idm215027392\" style=\"padding-left: 40px\" data-type=\"equation\">1.0 \u00d7 10<sup>-4 <\/sup>mol - 1.8 \u00d7 10<sup>-6 <\/sup>mol = <span style=\"text-decoration: underline\">9<\/span>.8 \u00d7 10<sup>-5<\/sup>mol OH<sup>-<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm225014800\">corresponding to a hydroxide molarity of<\/p>\r\n\r\n<div id=\"fs-idm215810016\" style=\"padding-left: 40px\" data-type=\"equation\">[OH<sup>-<\/sup>] = (<span style=\"text-decoration: underline\">9<\/span>.8 \u00d7 10<sup> -5<\/sup> mol OH<sup>-<\/sup>)\/(0.1010 L) = <span style=\"text-decoration: underline\">9<\/span>.7 \u00d7 10<sup>-4<\/sup> M OH<sup>-<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm198420016\">The pH of the solution is then calculated to be<\/p>\r\n\r\n<div id=\"fs-idm226155632\" style=\"padding-left: 40px\" data-type=\"equation\">pH = 14.00 - pOH = 14.00 - (-log(<span style=\"text-decoration: underline\">9<\/span>.7 \u00d7 10<sup>-4<\/sup> M) = 14.00 + 3.0 = 11.0<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm215525104\">In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 11.0) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.<span style=\"text-decoration: underline\">7<\/span>5).<\/p>\r\n<p id=\"fs-idm159189968\"><strong>Check Your Learning: <\/strong><\/p>\r\nShow that adding 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl changes the pH of 100.0 mL of a 1.8 \u00d7 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl solution from 4.74 to 3.00.\r\n\r\n&nbsp;\r\n<div id=\"fs-idm145837936\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm119337184\">Initial pH of 1.8 \u00d7 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl; pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log[1.8 \u00d710<sup>\u22125<\/sup>M] = 4.74<span data-type=\"newline\">\r\n<\/span>n<sub>H3O+<\/sub> in 100.0 mL 1.8 \u00d7 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl = (1.8 \u00d7 10<sup>\u22125<\/sup> mol\/L)(0.1000 L) = 1.8 \u00d7 10<sup>\u22126 <\/sup>mol<span data-type=\"newline\">\r\n<\/span>n<sub>H3O+ <\/sub>added by addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl = (0.10 mol\/L)(0.0010 L) = 1.0 \u00d7 10<sup>\u22124<\/sup> mol; final pH after addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm103691008\" style=\"padding-left: 40px\" data-type=\"equation\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(total moles H<sub>3<\/sub>O<sup>+<\/sup>\/total volume) = \u2212log[(1.8 \u00d7 10<sup>\u22126 <\/sup>mol + 1.0 \u00d7 10<sup>\u22124 <\/sup>mol)\/0.1010 L] = 3.00<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp107911744\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Buffer Capacity<\/strong><\/h3>\r\n<p id=\"fs-idm68764768\">Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (<a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_exhaust\">(Figure)<\/a>). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_14_06_exhaust\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)<\/div>\r\n<span id=\"fs-idm195031712\" data-type=\"media\" data-alt=\"No Alt Text\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_exhaust-1.jpg\" alt=\"No Alt Text\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm103629216\">The <strong>buffer capacity <\/strong>is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 <em data-effect=\"italics\">M<\/em> in acetic acid and 1.0 <em data-effect=\"italics\">M<\/em> in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 <em data-effect=\"italics\">M<\/em> in acetic acid and 0.10 <em data-effect=\"italics\">M<\/em> in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm124351376\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Selection of Suitable Buffer Mixtures<\/strong><\/h3>\r\n<p id=\"fs-idp2820144\">There are two useful rules of thumb for selecting buffer mixtures:<\/p>\r\n\r\n<ol id=\"fs-idm128259280\" type=\"1\">\r\n \t<li>A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_buffer\">(Figure)<\/a> shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Chem_14_06_buffer\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Change in pH as an increasing amount of a 0.10-<em data-effect=\"italics\">M<\/em> NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 0.10 <em data-effect=\"italics\">M<\/em> and [CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>] = 0.10 M. Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.<\/div>\r\n<span id=\"fs-idm112112448\" data-type=\"media\" data-alt=\"A graph is shown with a horizontal axis labeled \u201cAdded m L of 0.10 M N a O H\u201d which has markings and vertical gridlines every 10 units from 0 to 110. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unit beginning at 0 extending to 11. A break is shown in the vertical axis between 0 and 4. A red curve is drawn on the graph which increases gradually from the point (0, 4.8) up to about (100, 7) after which the graph has a vertical section up to about (100, 11). The curve is labeled [ C H subscript 3 C O subscript 2 H ] is 11 percent of [ C H subscript 3 CO subscript 2 superscript negative].\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_buffer-1.jpg\" alt=\"A graph is shown with a horizontal axis labeled \u201cAdded m L of 0.10 M N a O H\u201d which has markings and vertical gridlines every 10 units from 0 to 110. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unit beginning at 0 extending to 11. A break is shown in the vertical axis between 0 and 4. A red curve is drawn on the graph which increases gradually from the point (0, 4.8) up to about (100, 7) after which the graph has a vertical section up to about (100, 11). The curve is labeled [ C H subscript 3 C O subscript 2 H ] is 11 percent of [ C H subscript 3 CO subscript 2 superscript negative].\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li>Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.<\/li>\r\n<\/ol>\r\n<p id=\"fs-idm53419232\">Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>, and the hydrogen carbonate ion, HCO<sub>3<\/sub><sup>\u2212<\/sup>. When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:<\/p>\r\n\r\n<div id=\"fs-idm106059312\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) \u27f6 H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/div>\r\n<p id=\"fs-idm109728256\">An added hydroxide ion is removed by the reaction:<\/p>\r\n\r\n<div id=\"fs-idm145990368\" style=\"padding-left: 40px\" data-type=\"equation\">OH<sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) \u27f6 HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/div>\r\n<p id=\"fs-idp65661488\">The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H<sub>3<\/sub>O<sup>+<\/sup> is converted to H<sub>2<\/sub>CO<sub>3<\/sub> and OH<sup>-<\/sup> is converted to HCO<sub>3<\/sub><sup>-<\/sup>). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm153676976\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>The Henderson-Hasselbalch Equation<\/strong><\/h3>\r\n<p id=\"fs-idm154866768\">The ionization-constant expression for a solution of a weak acid can be written as:<\/p>\r\n\r\n<div id=\"fs-idm43593456\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1904\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a-300x107.png\" alt=\"\" width=\"152\" height=\"54\" \/><\/div>\r\n<p id=\"fs-idm87196592\">Rearranging to solve for [H<sub>3<\/sub>O<sup>+<\/sup>] yields:<\/p>\r\n\r\n<div id=\"fs-idm97934592\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1905\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b-300x90.png\" alt=\"\" width=\"177\" height=\"53\" \/><\/div>\r\n<p id=\"fs-idp10558896\">Taking the negative logarithm of both sides of this equation gives<\/p>\r\n\r\n<div id=\"fs-idp26801296\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1906\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-300x58.png\" alt=\"\" width=\"285\" height=\"55\" \/><\/div>\r\n<p id=\"fs-idp7990848\">which can be written as<\/p>\r\n\r\n<div id=\"fs-idp2699408\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1907\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-300x89.png\" alt=\"\" width=\"175\" height=\"52\" \/><\/div>\r\n<p id=\"fs-idp97287200\">where p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> is the negative of the logarithm of the ionization constant of the weak acid (p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = \u2212log <em data-effect=\"italics\">K<\/em><sub>a<\/sub>). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the <span data-type=\"term\">Henderson-Hasselbalch equation<\/span>, to calculate the pH of buffer solutions. It is important to note that the \u201c<em data-effect=\"italics\">x<\/em> is small\u201d assumption must be valid to use this equation.<\/p>\r\n&nbsp;\r\n<div id=\"fs-idp1420208\" class=\"chemistry chemist-portrait\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Lawrence Joseph Henderson and Karl Albert Hasselbalch<\/strong><\/div>\r\n<p id=\"fs-idm68026080\">Lawrence Joseph <span class=\"no-emphasis\" data-type=\"term\">Henderson<\/span> (1878\u20131942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent two years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.<\/p>\r\n<p id=\"fs-idp6637456\">In 1916, Karl Albert <span class=\"no-emphasis\" data-type=\"term\">Hasselbalch<\/span> (1874\u20131962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, S\u00f8rensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson\u2019s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-idm2834272\" class=\"chemistry sciences-interconnect\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\"><strong>Medicine: The Buffer System in Blood<\/strong><\/div>\r\n<p id=\"fs-idm108618160\">The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:<\/p>\r\n\r\n<div id=\"fs-idm2229488\" style=\"padding-left: 40px\" data-type=\"equation\">CO<sub>2<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>)<\/div>\r\n<p id=\"fs-idm43954064\">The concentration of carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub> is approximately 0.0012 <em data-effect=\"italics\">M<\/em>, and the concentration of the hydrogen carbonate ion, HCO<sub>3<\/sub><sup>\u2212<\/sup>, is around 0.024 <em data-effect=\"italics\">M<\/em>. Using the Henderson-Hasselbalch equation and the p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> of carbonic acid at body temperature, we can calculate the pH of blood:<\/p>\r\n\r\n<div id=\"fs-idp15032160\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1908\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-300x41.png\" alt=\"\" width=\"388\" height=\"53\" \/><\/div>\r\n<p id=\"fs-idm108864208\">The fact that the H<sub>2<\/sub>CO<sub>3<\/sub> concentration is significantly lower than that of the HCO<sub>3<\/sub><sup>\u2212<\/sup> ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.<\/p>\r\n<p id=\"fs-idm136676864\">Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the HCO<sub>3<\/sub><sup>\u2212<\/sup> ion, producing H<sub>2<\/sub>CO<sub>3<\/sub>. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO<sub>2<\/sub> from the blood through the lungs driving the equilibrium reaction such that [H<sub>3<\/sub>O<sup>+<\/sup>] is lowered. If the blood is too alkaline, a lower breath rate increases CO<sub>2<\/sub> concentration in the blood, driving the equilibrium reaction the other way, increasing [H<sub>3<\/sub>O<sup>+<\/sup>] and restoring an appropriate pH.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp49477328\" class=\"chemistry link-to-learning\" data-type=\"note\">\r\n<p id=\"fs-idm108605600\">View <a href=\"http:\/\/openstaxcollege.org\/l\/16BufferSystem\">information<\/a> on the buffer system encountered in natural waters.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm104045184\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idm115041904\">Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm101899904\" class=\"key-equations\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\r\n<ul id=\"fs-idp50124944\" data-bullet-style=\"bullet\">\r\n \t<li>p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = \u2212log <em data-effect=\"italics\">K<\/em><sub>a<\/sub><\/li>\r\n \t<li>p<em data-effect=\"italics\">K<\/em><sub>b<\/sub> = \u2212log <em data-effect=\"italics\">K<\/em><sub>b<\/sub><\/li>\r\n \t<li><img class=\"alignnone wp-image-1907\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-300x89.png\" alt=\"\" width=\"175\" height=\"52\" \/><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-idp119412416\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm2126064\" data-type=\"exercise\">\r\n<div id=\"fs-idm143472640\" data-type=\"problem\"><\/div>\r\n<\/div>\r\n<div id=\"fs-idm111629776\" data-type=\"exercise\">\r\n<div id=\"fs-idm104905504\" data-type=\"problem\">\r\n<p id=\"fs-idm105727920\"><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\r\n<dl id=\"fs-idp88494848\">\r\n \t<dt>buffer capacity<\/dt>\r\n \t<dd id=\"fs-idp88495360\">amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp88495904\">\r\n \t<dt>buffer<\/dt>\r\n \t<dd id=\"fs-idp88496416\">mixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm71412016\">\r\n \t<dt>Henderson-Hasselbalch equation<\/dt>\r\n \t<dd id=\"fs-idm71411504\">logarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions<\/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>Describe the composition and function of acid\u2013base buffers<\/li>\n<li>Calculate the pH of a buffer before and after the addition of added acid or base<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm117447344\">A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a <strong>buffer<\/strong>. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (<a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_compare\">(Figure)<\/a>). A solution of acetic acid and sodium acetate (CH<sub>3<\/sub>COOH + NaCH<sub>3<\/sub>COO) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH<sub>3<\/sub>(<em data-effect=\"italics\">aq<\/em>) + NH<sub>4<\/sub>Cl(<em data-effect=\"italics\">aq<\/em>)).<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_14_06_compare\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">(a) The buffered solution on the left and the unbuffered solution on the right have the same pH (pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH. (b) After the addition of 1 mL of a 0.01-<em data-effect=\"italics\">M<\/em> HCl solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)<\/div>\n<p><span id=\"fs-idm104574592\" data-type=\"media\" data-alt=\"Two images are shown. Image a on the left shows two beakers that each contain yellow solutions. The beaker on the left is labeled \u201cUnbuffered\u201d and the beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d Image b similarly shows 2 beakers. The beaker on the left contains a bright orange solution and is labeled \u201cUnbuffered.\u201d The beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_compare-1.jpg\" alt=\"Two images are shown. Image a on the left shows two beakers that each contain yellow solutions. The beaker on the left is labeled \u201cUnbuffered\u201d and the beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d Image b similarly shows 2 beakers. The beaker on the left contains a bright orange solution and is labeled \u201cUnbuffered.\u201d The beaker on the right is labeled \u201cp H equals 8.0 buffer.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idm112960288\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>How Buffers Work<\/strong><\/h3>\n<p id=\"fs-idm127493136\">To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, adding strong base to this solution will neutralize hydronium ion and shift the acetic acid ionization equilibrium to the right, partially restoring the decreased H<sub>3<\/sub>O<sup>+<\/sup> concentration:<\/p>\n<div id=\"fs-idm105688000\" style=\"padding-left: 40px\" data-type=\"equation\">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>)<\/div>\n<p id=\"fs-idp41915456\">Likewise, adding strong acid to this buffer solution will neutralize acetate ion, shifting the above ionization equilibrium left and returning [H<sub>3<\/sub>O<sup>+<\/sup>] to near its original value. <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_bufferchrt\">(Figure)<\/a> provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer&#8217;s conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_14_06_bufferchrt\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">Buffering action in a mixture of acetic acid and acetate salt.<\/div>\n<p><span id=\"fs-idp38705984\" data-type=\"media\" data-alt=\"This figure begins with a chemical reaction at the top: C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O superscript positive sign ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below this equation are two arrows: one pointing left and other pointing right. The arrow pointing left has this phrase written above it, \u201cH subscript 3 O superscript positive sign added, equilibrium position shifts to the left.\u201d Below the arrow is the reaction: C H subscript 3 C O O H ( a q ) left-facing arrow C H subscript 3 C O O superscript negative sign ( a q ) plus H subscript 3 O superscript positive sign. The arrow pointing right has this phrase written above it, \u201cO H subscript negative sign added, equilibrium position shifts to the right.\u201d Below the arrow is the reaction: O H superscript negative sign plus C H subscript 3 C O O H ( a q ) right-facing arrow H subscript 2 O ( l ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below all the text is a figure that resembles a bar graph. In the middle are two bars of equal height. One is labeled, \u201cC H subscript 3 C O O H,\u201d and the other is labeled, \u201cC H subscript 3 C O O superscript negative sign.\u201d There is a dotted line at the same height of the bars which extends to the left and right. Above these two bars is the phrase, \u201cBuffer solution equimolar in acid and base.\u201d There is an arrow pointing to the right which is labeled, \u201cAdd O H superscript negative sign.\u201d The arrow points to two bars again, but this time the C H subscript 3 C O O H bar is shorter than that C H subscript 3 C O O superscript negative sign bar. Above these two bars is the phrase, \u201cBuffer solution after addition of strong base.\u201d From the middle bars again, there is an arrow that points left. The arrow is labeled, \u201cAdd H subscript 3 O superscript positive sign.\u201d This arrow points to two bars again, but this time the C H subscript 3 C O O H bar is taller than the C H subscript 3 C O O superscript negative sign bar. These two bars are labeled, \u201cBuffer solution after addition of strong acid.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_bufferchrt-1.jpg\" alt=\"This figure begins with a chemical reaction at the top: C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O superscript positive sign ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below this equation are two arrows: one pointing left and other pointing right. The arrow pointing left has this phrase written above it, \u201cH subscript 3 O superscript positive sign added, equilibrium position shifts to the left.\u201d Below the arrow is the reaction: C H subscript 3 C O O H ( a q ) left-facing arrow C H subscript 3 C O O superscript negative sign ( a q ) plus H subscript 3 O superscript positive sign. The arrow pointing right has this phrase written above it, \u201cO H subscript negative sign added, equilibrium position shifts to the right.\u201d Below the arrow is the reaction: O H superscript negative sign plus C H subscript 3 C O O H ( a q ) right-facing arrow H subscript 2 O ( l ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below all the text is a figure that resembles a bar graph. In the middle are two bars of equal height. One is labeled, \u201cC H subscript 3 C O O H,\u201d and the other is labeled, \u201cC H subscript 3 C O O superscript negative sign.\u201d There is a dotted line at the same height of the bars which extends to the left and right. Above these two bars is the phrase, \u201cBuffer solution equimolar in acid and base.\u201d There is an arrow pointing to the right which is labeled, \u201cAdd O H superscript negative sign.\u201d The arrow points to two bars again, but this time the C H subscript 3 C O O H bar is shorter than that C H subscript 3 C O O superscript negative sign bar. Above these two bars is the phrase, \u201cBuffer solution after addition of strong base.\u201d From the middle bars again, there is an arrow that points left. The arrow is labeled, \u201cAdd H subscript 3 O superscript positive sign.\u201d This arrow points to two bars again, but this time the C H subscript 3 C O O H bar is taller than the C H subscript 3 C O O superscript negative sign bar. These two bars are labeled, \u201cBuffer solution after addition of strong acid.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idm144695456\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm145736240\"><strong>pH Changes in Buffered and Unbuffered Solutions <\/strong><\/p>\n<p>Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.<\/p>\n<p id=\"fs-idp41841344\">(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 <em data-effect=\"italics\">M<\/em> acetic acid and 0.10 <em data-effect=\"italics\">M<\/em> sodium acetate.<\/p>\n<p id=\"fs-idm197032080\">(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100.0 mL of this buffer.<\/p>\n<p id=\"fs-idm215059040\">(c) For comparison, calculate the pH after 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH is added to 100.0 mL of a solution of an unbuffered solution with a pH of 4.74.<\/p>\n<p id=\"fs-idm105719280\"><strong>Solution:<\/strong><\/p>\n<p id=\"fs-idm224056512\">(a) Following the ICE approach to this equilibrium calculation yields the following:<\/p>\n<p><span id=\"fs-idm123715040\" data-type=\"media\" data-alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 2 O ] equilibrium arrow H subscript 3 O superscript plus sign [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_ICETable16_img-1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 2 O ] equilibrium arrow H subscript 3 O superscript plus sign [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idm192051584\">Substituting the equilibrium concentration terms into the <em data-effect=\"italics\">K<\/em><sub>a<\/sub> expression, assuming <em data-effect=\"italics\">x<\/em> &lt; 0.05 \u00d7 0.10 M, i.e. <em>x<\/em> &lt; 0.0050 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> = 1.8 \u00d7 10<sup>-5<\/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(1.8 \u00d7 10<sup>-5 <\/sup>M) = 4.74<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp98075040\">(b) Calculate the pH after 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH is added to 100.0 mL of this buffer.<\/p>\n<p id=\"fs-idm196351552\">Adding strong acid will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.<\/p>\n<div id=\"fs-idm110133872\" style=\"padding-left: 40px\" data-type=\"equation\">0.0010 L \u00d7 0.10 mol NaOH\/L = 1.0 \u00d7 10<sup>-4<\/sup> mol NaOH<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm223581568\">The initial molar amount of acetic acid is<\/p>\n<div id=\"fs-idm53635376\" style=\"padding-left: 40px\" data-type=\"equation\">0.1000 L \u00d7 0.10 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H\/L = 1.0 \u00d7 10<sup>-2<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm162724496\">The amount of acetic acid remaining after some is neutralized by the added base is:<\/p>\n<div id=\"fs-idm91870896\" style=\"padding-left: 40px\" data-type=\"equation\">1.0 \u00d7 10<sup>-2<\/sup> mol &#8211; 1.0 \u00d7 10<sup>-4<\/sup> mol = 0.<span style=\"text-decoration: underline\">9<\/span>9 \u00d7 10<sup>-2<\/sup>\u00a0 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm178487280\">The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of<\/p>\n<div id=\"fs-idp82833792\" style=\"padding-left: 40px\" data-type=\"equation\">1.0 \u00d7 10<sup>-2<\/sup> mol + 1.0 \u00d7 10<sup>-4<\/sup> mol = 1.<span style=\"text-decoration: underline\">0<\/span>1\u00a0\u00d7 10<sup>-2<\/sup>\u00a0 mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm213828416\">Compute molar concentrations for the two buffer components:<\/p>\n<div id=\"fs-idm165714384\" style=\"padding-left: 40px\" data-type=\"equation\">[CH<sub>3<\/sub>CO<sub>2<\/sub>H] = (0.<span style=\"text-decoration: underline\">9<\/span>9 \u00d7 10<sup>-2<\/sup>\u00a0 mol CH<sub>3<\/sub>CO<sub>2<\/sub>H)\/(0.1010 L) = 0.0<span style=\"text-decoration: underline\">9<\/span>8M<\/div>\n<div id=\"fs-idp22032080\" style=\"padding-left: 40px\" data-type=\"equation\">[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup>] = (1.<span style=\"text-decoration: underline\">0<\/span>1 \u00d7 10<sup>-2<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub><sup>&#8211;<\/sup>)\/(0.1010 L) = 0.10 M<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm198587408\">Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.<span style=\"text-decoration: underline\">7<\/span>5 (only slightly different from that prior to adding the strong base).<\/p>\n<p id=\"fs-idm140802864\">(c) For comparison, calculate the pH after 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH is added to 100.0 mL of a solution of an unbuffered solution with a pH of 4.74.<\/p>\n<p id=\"fs-idm216667360\">The amount of hydrogen ion initially present in the solution is<\/p>\n<div id=\"fs-idm181266816\" style=\"padding-left: 40px\" data-type=\"equation\">[H<sub>3<\/sub>O<sup>+<\/sup>] = 10<sup>-4.74<\/sup> = 1.8 \u00d7 10<sup>-5<\/sup> M<\/div>\n<div id=\"fs-idm226340192\" style=\"padding-left: 40px\" data-type=\"equation\">n<sub>H3O+<\/sub> = (0.1000L)(1.8 \u00d7 10<sup>-5<\/sup> mol\/L) = 1.8 \u00d7 10<sup>-6 <\/sup>mol H<sub>3<\/sub>O<sup>+<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm213735984\">The amount of hydroxide ion added to the solution is<\/p>\n<div id=\"fs-idm167813376\" style=\"padding-left: 40px\" data-type=\"equation\">n<sub>OH-<\/sub> = (0.0010 L)(0.10 mol\/L) = 1.0 \u00d7 10<sup>-4 <\/sup>mol OH<sup>&#8211;<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm184173168\">The added hydroxide will neutralize hydronium ion via the reaction<\/p>\n<div id=\"fs-idm213939888\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + OH<sup>&#8211;<\/sup>(<em>aq<\/em>) \u2192 2H<sub>2<\/sub>O(<em>l<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm174460304\">The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).<\/p>\n<p id=\"fs-idm201883456\">The amount of hydroxide ion remaining is<\/p>\n<div id=\"fs-idm215027392\" style=\"padding-left: 40px\" data-type=\"equation\">1.0 \u00d7 10<sup>-4 <\/sup>mol &#8211; 1.8 \u00d7 10<sup>-6 <\/sup>mol = <span style=\"text-decoration: underline\">9<\/span>.8 \u00d7 10<sup>-5<\/sup>mol OH<sup>&#8211;<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm225014800\">corresponding to a hydroxide molarity of<\/p>\n<div id=\"fs-idm215810016\" style=\"padding-left: 40px\" data-type=\"equation\">[OH<sup>&#8211;<\/sup>] = (<span style=\"text-decoration: underline\">9<\/span>.8 \u00d7 10<sup> -5<\/sup> mol OH<sup>&#8211;<\/sup>)\/(0.1010 L) = <span style=\"text-decoration: underline\">9<\/span>.7 \u00d7 10<sup>-4<\/sup> M OH<sup>&#8211;<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm198420016\">The pH of the solution is then calculated to be<\/p>\n<div id=\"fs-idm226155632\" style=\"padding-left: 40px\" data-type=\"equation\">pH = 14.00 &#8211; pOH = 14.00 &#8211; (-log(<span style=\"text-decoration: underline\">9<\/span>.7 \u00d7 10<sup>-4<\/sup> M) = 14.00 + 3.0 = 11.0<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm215525104\">In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 11.0) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.<span style=\"text-decoration: underline\">7<\/span>5).<\/p>\n<p id=\"fs-idm159189968\"><strong>Check Your Learning: <\/strong><\/p>\n<p>Show that adding 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl changes the pH of 100.0 mL of a 1.8 \u00d7 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl solution from 4.74 to 3.00.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm145837936\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm119337184\">Initial pH of 1.8 \u00d7 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl; pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log[1.8 \u00d710<sup>\u22125<\/sup>M] = 4.74<span data-type=\"newline\"><br \/>\n<\/span>n<sub>H3O+<\/sub> in 100.0 mL 1.8 \u00d7 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl = (1.8 \u00d7 10<sup>\u22125<\/sup> mol\/L)(0.1000 L) = 1.8 \u00d7 10<sup>\u22126 <\/sup>mol<span data-type=\"newline\"><br \/>\n<\/span>n<sub>H3O+ <\/sub>added by addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl = (0.10 mol\/L)(0.0010 L) = 1.0 \u00d7 10<sup>\u22124<\/sup> mol; final pH after addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm103691008\" style=\"padding-left: 40px\" data-type=\"equation\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(total moles H<sub>3<\/sub>O<sup>+<\/sup>\/total volume) = \u2212log[(1.8 \u00d7 10<sup>\u22126 <\/sup>mol + 1.0 \u00d7 10<sup>\u22124 <\/sup>mol)\/0.1010 L] = 3.00<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp107911744\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Buffer Capacity<\/strong><\/h3>\n<p id=\"fs-idm68764768\">Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (<a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_exhaust\">(Figure)<\/a>). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_14_06_exhaust\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)<\/div>\n<p><span id=\"fs-idm195031712\" data-type=\"media\" data-alt=\"No Alt Text\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_exhaust-1.jpg\" alt=\"No Alt Text\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm103629216\">The <strong>buffer capacity <\/strong>is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 <em data-effect=\"italics\">M<\/em> in acetic acid and 1.0 <em data-effect=\"italics\">M<\/em> in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 <em data-effect=\"italics\">M<\/em> in acetic acid and 0.10 <em data-effect=\"italics\">M<\/em> in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.<\/p>\n<\/div>\n<div id=\"fs-idm124351376\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Selection of Suitable Buffer Mixtures<\/strong><\/h3>\n<p id=\"fs-idp2820144\">There are two useful rules of thumb for selecting buffer mixtures:<\/p>\n<ol id=\"fs-idm128259280\" type=\"1\">\n<li>A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. <a class=\"autogenerated-content\" href=\"#CNX_Chem_14_06_buffer\">(Figure)<\/a> shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Chem_14_06_buffer\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Change in pH as an increasing amount of a 0.10-<em data-effect=\"italics\">M<\/em> NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 0.10 <em data-effect=\"italics\">M<\/em> and [CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>] = 0.10 M. Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.<\/div>\n<p><span id=\"fs-idm112112448\" data-type=\"media\" data-alt=\"A graph is shown with a horizontal axis labeled \u201cAdded m L of 0.10 M N a O H\u201d which has markings and vertical gridlines every 10 units from 0 to 110. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unit beginning at 0 extending to 11. A break is shown in the vertical axis between 0 and 4. A red curve is drawn on the graph which increases gradually from the point (0, 4.8) up to about (100, 7) after which the graph has a vertical section up to about (100, 11). The curve is labeled [ C H subscript 3 C O subscript 2 H ] is 11 percent of [ C H subscript 3 CO subscript 2 superscript negative].\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_14_06_buffer-1.jpg\" alt=\"A graph is shown with a horizontal axis labeled \u201cAdded m L of 0.10 M N a O H\u201d which has markings and vertical gridlines every 10 units from 0 to 110. The vertical axis is labeled \u201cp H\u201d and is marked every 1 unit beginning at 0 extending to 11. A break is shown in the vertical axis between 0 and 4. A red curve is drawn on the graph which increases gradually from the point (0, 4.8) up to about (100, 7) after which the graph has a vertical section up to about (100, 11). The curve is labeled [ C H subscript 3 C O subscript 2 H ] is 11 percent of [ C H subscript 3 CO subscript 2 superscript negative].\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/li>\n<li>Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.<\/li>\n<\/ol>\n<p id=\"fs-idm53419232\">Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>, and the hydrogen carbonate ion, HCO<sub>3<\/sub><sup>\u2212<\/sup>. When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:<\/p>\n<div id=\"fs-idm106059312\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>) + HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) \u27f6 H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/div>\n<p id=\"fs-idm109728256\">An added hydroxide ion is removed by the reaction:<\/p>\n<div id=\"fs-idm145990368\" style=\"padding-left: 40px\" data-type=\"equation\">OH<sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) \u27f6 HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/div>\n<p id=\"fs-idp65661488\">The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H<sub>3<\/sub>O<sup>+<\/sup> is converted to H<sub>2<\/sub>CO<sub>3<\/sub> and OH<sup>&#8211;<\/sup> is converted to HCO<sub>3<\/sub><sup>&#8211;<\/sup>). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.<\/p>\n<\/div>\n<div id=\"fs-idm153676976\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>The Henderson-Hasselbalch Equation<\/strong><\/h3>\n<p id=\"fs-idm154866768\">The ionization-constant expression for a solution of a weak acid can be written as:<\/p>\n<div id=\"fs-idm43593456\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1904\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a-300x107.png\" alt=\"\" width=\"152\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a-300x107.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a-65x23.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a-225x80.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a-350x125.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6a.png 518w\" sizes=\"auto, (max-width: 152px) 100vw, 152px\" \/><\/div>\n<p id=\"fs-idm87196592\">Rearranging to solve for [H<sub>3<\/sub>O<sup>+<\/sup>] yields:<\/p>\n<div id=\"fs-idm97934592\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1905\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b-300x90.png\" alt=\"\" width=\"177\" height=\"53\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b-300x90.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b-225x67.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b-350x104.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6b.png 573w\" sizes=\"auto, (max-width: 177px) 100vw, 177px\" \/><\/div>\n<p id=\"fs-idp10558896\">Taking the negative logarithm of both sides of this equation gives<\/p>\n<div id=\"fs-idp26801296\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1906\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-300x58.png\" alt=\"\" width=\"285\" height=\"55\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-300x58.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-768x148.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-225x43.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c-350x67.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6c.png 893w\" sizes=\"auto, (max-width: 285px) 100vw, 285px\" \/><\/div>\n<p id=\"fs-idp7990848\">which can be written as<\/p>\n<div id=\"fs-idp2699408\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1907\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-300x89.png\" alt=\"\" width=\"175\" height=\"52\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-300x89.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-225x67.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-350x104.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d.png 572w\" sizes=\"auto, (max-width: 175px) 100vw, 175px\" \/><\/div>\n<p id=\"fs-idp97287200\">where p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> is the negative of the logarithm of the ionization constant of the weak acid (p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = \u2212log <em data-effect=\"italics\">K<\/em><sub>a<\/sub>). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the <span data-type=\"term\">Henderson-Hasselbalch equation<\/span>, to calculate the pH of buffer solutions. It is important to note that the \u201c<em data-effect=\"italics\">x<\/em> is small\u201d assumption must be valid to use this equation.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp1420208\" class=\"chemistry chemist-portrait\" data-type=\"note\">\n<div data-type=\"title\"><strong>Lawrence Joseph Henderson and Karl Albert Hasselbalch<\/strong><\/div>\n<p id=\"fs-idm68026080\">Lawrence Joseph <span class=\"no-emphasis\" data-type=\"term\">Henderson<\/span> (1878\u20131942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent two years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.<\/p>\n<p id=\"fs-idp6637456\">In 1916, Karl Albert <span class=\"no-emphasis\" data-type=\"term\">Hasselbalch<\/span> (1874\u20131962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, S\u00f8rensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson\u2019s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-idm2834272\" class=\"chemistry sciences-interconnect\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\"><strong>Medicine: The Buffer System in Blood<\/strong><\/div>\n<p id=\"fs-idm108618160\">The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:<\/p>\n<div id=\"fs-idm2229488\" style=\"padding-left: 40px\" data-type=\"equation\">CO<sub>2<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>) + H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>)<\/div>\n<p id=\"fs-idm43954064\">The concentration of carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub> is approximately 0.0012 <em data-effect=\"italics\">M<\/em>, and the concentration of the hydrogen carbonate ion, HCO<sub>3<\/sub><sup>\u2212<\/sup>, is around 0.024 <em data-effect=\"italics\">M<\/em>. Using the Henderson-Hasselbalch equation and the p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> of carbonic acid at body temperature, we can calculate the pH of blood:<\/p>\n<div id=\"fs-idp15032160\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1908\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-300x41.png\" alt=\"\" width=\"388\" height=\"53\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-300x41.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-1024x139.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-768x104.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e-350x48.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6e.png 1287w\" sizes=\"auto, (max-width: 388px) 100vw, 388px\" \/><\/div>\n<p id=\"fs-idm108864208\">The fact that the H<sub>2<\/sub>CO<sub>3<\/sub> concentration is significantly lower than that of the HCO<sub>3<\/sub><sup>\u2212<\/sup> ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.<\/p>\n<p id=\"fs-idm136676864\">Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the HCO<sub>3<\/sub><sup>\u2212<\/sup> ion, producing H<sub>2<\/sub>CO<sub>3<\/sub>. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO<sub>2<\/sub> from the blood through the lungs driving the equilibrium reaction such that [H<sub>3<\/sub>O<sup>+<\/sup>] is lowered. If the blood is too alkaline, a lower breath rate increases CO<sub>2<\/sub> concentration in the blood, driving the equilibrium reaction the other way, increasing [H<sub>3<\/sub>O<sup>+<\/sup>] and restoring an appropriate pH.<\/p>\n<\/div>\n<div id=\"fs-idp49477328\" class=\"chemistry link-to-learning\" data-type=\"note\">\n<p id=\"fs-idm108605600\">View <a href=\"http:\/\/openstaxcollege.org\/l\/16BufferSystem\">information<\/a> on the buffer system encountered in natural waters.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm104045184\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idm115041904\">Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.<\/p>\n<\/div>\n<div id=\"fs-idm101899904\" class=\"key-equations\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\n<ul id=\"fs-idp50124944\" data-bullet-style=\"bullet\">\n<li>p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = \u2212log <em data-effect=\"italics\">K<\/em><sub>a<\/sub><\/li>\n<li>p<em data-effect=\"italics\">K<\/em><sub>b<\/sub> = \u2212log <em data-effect=\"italics\">K<\/em><sub>b<\/sub><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1907\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-300x89.png\" alt=\"\" width=\"175\" height=\"52\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-300x89.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-225x67.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d-350x104.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/14.6d.png 572w\" sizes=\"auto, (max-width: 175px) 100vw, 175px\" \/><\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-idp119412416\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm2126064\" data-type=\"exercise\">\n<div id=\"fs-idm143472640\" data-type=\"problem\"><\/div>\n<\/div>\n<div id=\"fs-idm111629776\" data-type=\"exercise\">\n<div id=\"fs-idm104905504\" data-type=\"problem\">\n<p id=\"fs-idm105727920\">\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\n<dl id=\"fs-idp88494848\">\n<dt>buffer capacity<\/dt>\n<dd id=\"fs-idp88495360\">amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)<\/dd>\n<\/dl>\n<dl id=\"fs-idp88495904\">\n<dt>buffer<\/dt>\n<dd id=\"fs-idp88496416\">mixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added<\/dd>\n<\/dl>\n<dl id=\"fs-idm71412016\">\n<dt>Henderson-Hasselbalch equation<\/dt>\n<dd id=\"fs-idm71411504\">logarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":7,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-806","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":766,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/806","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":8,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/806\/revisions"}],"predecessor-version":[{"id":2175,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/806\/revisions\/2175"}],"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\/806\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=806"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=806"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=806"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}