{"id":3521,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=3521"},"modified":"2016-08-09T18:28:19","modified_gmt":"2016-08-09T18:28:19","slug":"buffers-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/buffers-2\/","title":{"raw":"Buffers","rendered":"Buffers"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this module, 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 mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a <span data-type=\"term\">buffer<\/span>. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure 1). A solution of acetic acid and sodium acetate (CH<sub>3<\/sub>COOH + CH<sub>3<\/sub>COONa) 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\r\n\r\n[caption id=\"attachment_5446\" align=\"aligncenter\" width=\"975\"]<img class=\"size-full wp-image-5446\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043335\/CNX_Chem_14_06_compare.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\" width=\"975\" height=\"347\" \/> Figure 1. (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-M 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)[\/caption]\r\n\r\n<section id=\"fs-idm112960288\" data-depth=\"1\">\r\n<h2 data-type=\"title\">How Buffers Work<\/h2>\r\n<p id=\"fs-idm127493136\">A mixture of acetic acid and sodium acetate is acidic because the <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of acetic acid is greater than the <em data-effect=\"italics\">K<\/em><sub>b<\/sub> of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. Then more of the acetic acid reacts with water, restoring the hydronium ion concentration almost to its original value:<\/p>\r\n\r\n<div id=\"fs-idm105688000\" data-type=\"equation\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right)[\/latex]<\/div>\r\n<p id=\"fs-idp41915456\">The pH changes very little. If we add an acid such as hydrochloric acid, most of the hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid molecules:<\/p>\r\n\r\n<div id=\"fs-idm115859376\" data-type=\"equation\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\r\n<p id=\"fs-idp6711056\">Thus, there is very little increase in the concentration of the hydronium ion, and the pH remains practically unchanged (Figure 2).<\/p>\r\n\r\n\r\n[caption id=\"attachment_5448\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-5448\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043433\/CNX_Chem_14_06_bufferchrt-1024x566-1024x566.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\" width=\"1024\" height=\"566\" \/> Figure 2. This diagram shows the buffer action of these reactions.[\/caption]\r\n<p id=\"fs-idm155475664\">A mixture of ammonia and ammonium chloride is basic because the <em data-effect=\"italics\">K<\/em><sub>b<\/sub> for ammonia is greater than the <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value:<\/p>\r\n\r\n<div id=\"fs-idm142306048\" data-type=\"equation\">[latex]{\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\r\n<p id=\"fs-idp98958624\">If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value:<\/p>\r\n\r\n<div id=\"fs-idm144711616\" data-type=\"equation\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)\\longrightarrow {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\r\n<p id=\"fs-idm72230384\">The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid.<\/p>\r\n\r\n<div id=\"fs-idm144695456\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idm145736240\"><span data-type=\"title\">pH Changes in Buffered and Unbuffered Solutions<\/span><\/h4>\r\nAcetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change 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\r\n<h4 id=\"fs-idm105719280\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nTo determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):\r\n\r\n<span id=\"fs-idp49040960\" data-type=\"media\" data-alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\"><img class=\"aligncenter size-full wp-image-5449\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043544\/CNX_Chem_14_06_steps1_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" width=\"1000\" height=\"177\" \/>\r\n<\/span>\r\n<ol id=\"fs-idm106943744\" class=\"stepwise\" data-number-style=\"arabic\">\r\n \t<li><em data-effect=\"italics\">Determine the direction of change.<\/em> The equilibrium in a mixture of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex], [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex], and CH<sub>3<\/sub>CO<sub>2<\/sub>H is:\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idp57064960\" data-type=\"equation\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right)[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\nThe equilibrium constant for CH<sub>3<\/sub>CO<sub>2<\/sub>H is not given, so we look it up in <a href=\".\/chapter\/ionization-constants-of-weak-acids-missing-formulas\/\" target=\"_blank\">Ionization Constants of Weak Acids<\/a>: <em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>. With [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] = 0.10 <em data-effect=\"italics\">M<\/em> and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = ~0 <em data-effect=\"italics\">M<\/em>, the reaction shifts to the right to form [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex].<\/li>\r\n \t<li><em data-effect=\"italics\">Determine<\/em> x <em data-effect=\"italics\">and equilibrium concentrations<\/em>. A table of changes and concentrations follows:\r\n\r\n<img class=\"aligncenter size-full wp-image-5450\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043645\/CNX_Chem_14_06_ICETable16_img.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 ( 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, x, x. The fourth column has the following: 0.10, x, 0.10 plus sign x.\" width=\"975\" height=\"264\" \/><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 ( 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, x, x. The fourth column has the following: 0.10, x, 0.10 plus sign x.\"> <\/span><\/li>\r\n \t<li><em data-effect=\"italics\">Solve for x and the equilibrium concentrations.<\/em> We find:\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm7012176\" data-type=\"equation\">[latex]x=1.8\\times {10}^{-5}M[\/latex]<\/div>\r\nand\r\n<div id=\"fs-idm109618832\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=0+x=1.8\\times {10}^{-5}M[\/latex]<\/div>\r\nThus:\r\n<div id=\"fs-idm165278336\" data-type=\"equation\">[latex]\\text{pH}=\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\text{-log}\\left(1.8\\times {10}^{-5}\\right)[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm111454032\" data-type=\"equation\">[latex]=4.74[\/latex]<\/div><\/li>\r\n \t<li><em data-effect=\"italics\">Check the work<\/em>. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, <em data-effect=\"italics\">Q<\/em> = <em data-effect=\"italics\">K<\/em><sub>a<\/sub>.<\/li>\r\n<\/ol>\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 mL of this buffer, giving a solution with a volume of 101 mL.<\/p>\r\n<p id=\"fs-idm85734032\">First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:\r\n<img class=\"aligncenter size-full wp-image-5451\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043740\/CNX_Chem_14_06_steps2_img.jpg\" alt=\"Eight tan rectangles are shown in four columns of two rectangles each that are connected with right pointing arrows. The first rectangle in the upper left is labeled \u201cVolume of N a O H solution.\u201d An arrow points right to a second rectangle labeled \u201cMoles of N a O H added.\u201d A second arrow points right to a third rectangle labeled \u201cAdditional moles of N a C H subscript 3 C O subscript 2.\u201d Just beneath the first rectangle in the upper left is a rectangle labeled \u201cVolume of buffer solution.\u201d An arrow points right to another rectangle labeled \u201cInitial moles of C H subscript 3 C O subscript 2 H.\u201d This rectangle points to the same third rectangle, which is labeled \u201c Additional moles of N a C H subscript 3 C O subscript 2.\u201d An arrow points right to a rectangle labeled \u201c Unreacted moles of C H subscript 3 C O subscript 2 H.\u201d An arrow points from this rectangle to a rectangle below labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d An arrow extends below the \u201cAdditional moles of N a C H subscript 3 C O subscript 2\u201d rectangle to a rectangle labeled \u201c[ C H subscript 3 C O subscript 2 ].\u201d This rectangle points right to the rectangle labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d\" width=\"1000\" height=\"475\" \/><span id=\"fs-idm125292992\" data-type=\"media\" data-alt=\"Eight tan rectangles are shown in four columns of two rectangles each that are connected with right pointing arrows. The first rectangle in the upper left is labeled \u201cVolume of N a O H solution.\u201d An arrow points right to a second rectangle labeled \u201cMoles of N a O H added.\u201d A second arrow points right to a third rectangle labeled \u201cAdditional moles of N a C H subscript 3 C O subscript 2.\u201d Just beneath the first rectangle in the upper left is a rectangle labeled \u201cVolume of buffer solution.\u201d An arrow points right to another rectangle labeled \u201cInitial moles of C H subscript 3 C O subscript 2 H.\u201d This rectangle points to the same third rectangle, which is labeled \u201c Additional moles of N a C H subscript 3 C O subscript 2.\u201d An arrow points right to a rectangle labeled \u201c Unreacted moles of C H subscript 3 C O subscript 2 H.\u201d An arrow points from this rectangle to a rectangle below labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d An arrow extends below the \u201cAdditional moles of N a C H subscript 3 C O subscript 2\u201d rectangle to a rectangle labeled \u201c[ C H subscript 3 C O subscript 2 ].\u201d This rectangle points right to the rectangle labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d\">\r\n<\/span><\/p>\r\n\r\n<ol id=\"fs-idm142618096\" class=\"stepwise\" data-number-style=\"arabic\">\r\n \t<li><em data-effect=\"italics\">Determine the moles of NaOH.<\/em> One milliliter (0.0010 L) of 0.10 <em data-effect=\"italics\">M<\/em> NaOH contains:\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm110133872\" data-type=\"equation\">[latex]0.0010\\cancel{\\text{L}}\\times \\left(\\frac{0.10\\text{mol NaOH}}{1\\cancel{\\text{L}}}\\right)=1.0\\times {10}^{-4}\\text{mol NaOH}[\/latex]<\/div><\/li>\r\n \t<li><em data-effect=\"italics\">Determine the moles of CH<sub>2<\/sub>CO<sub>2<\/sub>H.<\/em> Before reaction, 0.100 L of the buffer solution contains:\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm53635376\" data-type=\"equation\">[latex]0.100\\cancel{\\text{L}}\\times \\left(\\frac{0.100\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}}{1\\cancel{\\text{L}}}\\right)=1.00\\times {10}^{-2}\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}[\/latex]<\/div><\/li>\r\n \t<li><em data-effect=\"italics\">Solve for the amount of NaCH<sub>3<\/sub>CO<sub>2<\/sub> produced.<\/em> The 1.0 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> mol of NaOH neutralizes 1.0 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H, leaving:\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm91870896\" data-type=\"equation\">[latex]\\left(1.0\\times {10}^{-2}\\right)-\\left(0.01\\times {10}^{-2}\\right)=0.99\\times {10}^{-2}\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}[\/latex]<\/div>\r\nand producing 1.0 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> mol of NaCH<sub>3<\/sub>CO<sub>2<\/sub>. This makes a total of:\r\n<div id=\"fs-idp82833792\" data-type=\"equation\">[latex]\\left(1.0\\times {10}^{-2}\\right)+\\left(0.01\\times {10}^{-2}\\right)=1.01\\times {10}^{-2}\\text{mol}{\\text{NaCH}}_{3}{\\text{CO}}_{2}[\/latex]<\/div><\/li>\r\n \t<li><em data-effect=\"italics\">Find the molarity of the products.<\/em> After reaction, CH<sub>3<\/sub>CO<sub>2<\/sub>H and NaCH<sub>3<\/sub>CO<sub>2<\/sub> are contained in 101 mL of the intermediate solution, so:\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm165714384\" data-type=\"equation\">[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\frac{9.9\\times {10}^{-3}\\text{mol}}{0.101\\text{L}}=0.098M[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idp22032080\" data-type=\"equation\">[latex]\\left[{\\text{NaCH}}_{3}{\\text{CO}}_{2}\\right]=\\frac{1.01\\times {10}^{-2}\\text{mol}}{0.101\\text{L}}=0.100M[\/latex]<\/div>\r\nNow we calculate the pH after the intermediate solution, which is 0.098 <em data-effect=\"italics\">M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.100 <em data-effect=\"italics\">M<\/em> in NaCH<sub>3<\/sub>CO<sub>2<\/sub>, comes to equilibrium. The calculation is very similar to that in part (a) of this example:\r\n<img class=\"aligncenter size-full wp-image-5453\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043847\/CNX_Chem_14_06_steps3_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" width=\"1000\" height=\"177\" \/><span id=\"fs-idm159880832\" data-type=\"media\" data-alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\">\r\n<\/span>\r\n\r\nThis series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution (F).<\/li>\r\n<\/ol>\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 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>-<em data-effect=\"italics\">M<\/em> solution of HCl). The volume of the final solution is 101 mL.<\/p>\r\n\r\n<h4 id=\"fs-idm51120032\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThis 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>-<em data-effect=\"italics\">M<\/em> solution of HCl has the same hydronium ion concentration as the 0.10-<em data-effect=\"italics\">M<\/em> solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:\r\n<div id=\"fs-idp16142224\" data-type=\"equation\">[latex]0.100\\text{L}\\times \\left(\\frac{1.8\\times {10}^{-5}\\text{mol HCl}}{1\\text{L}}\\right)=1.8\\times {10}^{-6}\\text{mol HCl}[\/latex]<\/div>\r\n<p id=\"fs-idp35168928\">As shown in part (b), 1 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH contains 1.0 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:<\/p>\r\n\r\n<div id=\"fs-idp21695968\" data-type=\"equation\">[latex]\\left(1.0\\times {10}^{-4}\\right)-\\left(1.8\\times {10}^{-6}\\right)=9.8\\times {10}^{-5}M[\/latex]<\/div>\r\n<p id=\"fs-idm52812304\">The concentration of NaOH is:<\/p>\r\n\r\n<div id=\"fs-idm72079040\" data-type=\"equation\">[latex]\\frac{9.8\\times {10}^{-5}M\\text{NaOH}}{0.101\\text{L}}=9.7\\times {10}^{-4}M[\/latex]<\/div>\r\n<p id=\"fs-idm165600608\">The pOH of this solution is:<\/p>\r\n\r\n<div id=\"fs-idm97337872\" data-type=\"equation\">[latex]\\text{pOH}=\\text{-log}\\left[{\\text{OH}}^{\\text{-}}\\right]=\\text{-log}\\left(9.7\\times {10}^{-4}\\right)=3.01[\/latex]<\/div>\r\n<p id=\"fs-idm109107696\">The pH is:<\/p>\r\n\r\n<div id=\"fs-idm158590768\" data-type=\"equation\">[latex]\\text{pH}=14.00-\\text{pOH}=10.99[\/latex]<\/div>\r\n<p id=\"fs-idp53095888\">The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).<\/p>\r\n\r\n<h4 id=\"fs-idm159189968\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\r\nShow that adding 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl changes the pH of 100 mL of a 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl solution from 4.74 to 3.00.\r\n<div id=\"fs-idm145837936\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>Initial pH of 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl; pH = \u2212log [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = \u2212log[1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>] = 4.74<\/div>\r\n<p style=\"text-align: right;\">Moles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] in 100 mL 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl; 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup> moles\/L [latex]\\times [\/latex] 0.100 L = 1.8 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><\/p>\r\n<p style=\"text-align: right;\">Moles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] added by addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl: 0.10 moles\/L [latex]\\times [\/latex] 0.0010 L = 1.0 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> moles; final pH after addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl:<\/p>\r\n\r\n<div id=\"fs-idm103691008\" style=\"text-align: right;\" data-type=\"equation\">[latex]\\text{pH}=\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\text{-log}\\left(\\frac{\\text{total moles}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}}{\\text{total volume}}\\right)=\\text{-log}\\left(\\frac{1.0\\times {10}^{-4}\\text{mol}+1.8\\times {10}^{-6}\\text{mol}}{101\\text{mL}\\left(\\frac{1\\text{L}}{1000\\text{mL}}\\right)}\\right)=3.00[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm109632000\">If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.<\/p>\r\n\r\n<\/section><section id=\"fs-idp107911744\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Buffer Capacity<\/h2>\r\n<p id=\"fs-idm68764768\">Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 3). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5454\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-5454\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043956\/CNX_Chem_14_06_exhaust-1024x285.jpg\" alt=\"Figure 3. 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)\" width=\"1024\" height=\"285\" \/> Figure 3. 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)[\/caption]\r\n<p id=\"fs-idm103629216\">The <span data-type=\"term\">buffer capacity<\/span> 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<\/section><section id=\"fs-idm124351376\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Selection of Suitable Buffer Mixtures<\/h2>\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\" data-number-style=\"arabic\">\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. Figure 4\u00a0shows 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.\r\n[caption id=\"attachment_5456\" align=\"aligncenter\" width=\"1000\"]<img class=\"size-full wp-image-5456\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044040\/CNX_Chem_14_06_buffer.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].\" width=\"1000\" height=\"715\" \/> The graph, an illustration of buffering action, shows change of pH as an increasing amount of a 0.10-<em>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>M<\/em> and [CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]=0.10M.[\/caption]\r\n<div data-type=\"newline\"><\/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 bicarbonate ion, [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex]. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:<\/p>\r\n\r\n<div id=\"fs-idm106059312\" data-type=\"equation\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\r\n<p id=\"fs-idm109728256\">When an excess of the hydroxide ion is present, it is removed by the reaction:<\/p>\r\n\r\n<div id=\"fs-idm145990368\" data-type=\"equation\">[latex]{\\text{OH}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\longrightarrow {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\r\n<p id=\"fs-idp65661488\">The pH of human blood thus remains very near 7.35, that is, slightly basic. Variations are usually less than 0.1 of a pH unit. A change of 0.4 of a pH unit is likely to be fatal.<\/p>\r\n\r\n<\/section><section id=\"fs-idm153676976\" data-depth=\"1\">\r\n<h2 data-type=\"title\">The Henderson-Hasselbalch Equation<\/h2>\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\" data-type=\"equation\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{A}}^{\\text{-}}\\right]}{\\text{[HA]}}[\/latex]<\/div>\r\n<p id=\"fs-idm87196592\">Rearranging to solve for [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], we get:<\/p>\r\n\r\n<div id=\"fs-idm97934592\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]={K}_{\\text{a}}\\times \\frac{\\text{[HA]}}{\\left[{\\text{A}}^{\\text{-}}\\right]}[\/latex]<\/div>\r\n<p id=\"fs-idp10558896\">Taking the negative logarithm of both sides of this equation, we arrive at:<\/p>\r\n\r\n<div id=\"fs-idp26801296\" data-type=\"equation\">[latex]\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\text{-log}{K}_{\\text{a}}\\text{- log}\\frac{\\left[\\text{HA}\\right]}{\\left[{\\text{A}}^{\\text{-}}\\right]}[\/latex],<\/div>\r\n<p id=\"fs-idp7990848\">which can be written as<\/p>\r\n\r\n<div id=\"fs-idp2699408\" data-type=\"equation\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[{\\text{A}}^{\\text{-}}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/div>\r\n<p id=\"fs-idp97287200\">where p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> is the negative of the common 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 acid and its salt in a buffered solution. Scientists often use this expression, called the <strong><span data-type=\"term\">Henderson-Hasselbalch equation<\/span><\/strong>, 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\r\n<div id=\"fs-idp1420208\" class=\"chemistry chemist-portrait textbox shaded\" data-type=\"note\">\r\n<h3 data-type=\"title\">Lawrence Joseph Henderson and Karl Albert Hasselbalch<\/h3>\r\n<p id=\"fs-idm68026080\">Lawrence Joseph <strong><span class=\"no-emphasis\" data-type=\"term\">Henderson<\/span><\/strong> (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 2 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 <strong><span class=\"no-emphasis\" data-type=\"term\">Hasselbalch<\/span><\/strong> (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\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Medicine: The Buffer System in Blood<\/h3>\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[latex]{\\text{CO}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)[\/latex]\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, [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex], 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[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[\\text{base}\\right]}{\\left[\\text{acid}\\right]}=6.1+\\text{log}\\frac{0.024}{0.0012}=7.4[\/latex]\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 [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] 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 [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] 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 [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] 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<sup>+<\/sup>] and restoring an appropriate pH.<\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm104045184\" class=\"summary\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idm115041904\">A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).<\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm101899904\" class=\"key-equations\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/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>[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[{\\text{A}}^{\\text{-}}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><section id=\"fs-idp119412416\" class=\"exercises\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<div id=\"fs-idm2126064\" data-type=\"exercise\">\r\n<div id=\"fs-idm143472640\" data-type=\"problem\">\r\n<ol>\r\n \t<li id=\"fs-idp15824512\">Explain why a buffer can be prepared from a mixture of NH<sub>4<\/sub>Cl and NaOH but not from NH<sub>3<\/sub> and NaOH.<\/li>\r\n \t<li>Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H<sub>3<\/sub>PO<sub>4<\/sub> and a salt of its conjugate base NaH<sub>2<\/sub>PO<sub>4<\/sub>.<\/li>\r\n \t<li>Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH<sub>3<\/sub> and a salt of its conjugate acid NH<sub>4<\/sub>Cl.<\/li>\r\n \t<li>What is [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] in a solution of 0.25 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.030 <em data-effect=\"italics\">M<\/em> NaCH<sub>3<\/sub>CO<sub>2<\/sub>?\r\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/li>\r\n \t<li>What is [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] in a solution of 0.075 <em data-effect=\"italics\">M<\/em> HNO<sub>2<\/sub> and 0.030 <em data-effect=\"italics\">M<\/em> NaNO<sub>2<\/sub>?\r\n[latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=4.5\\times {10}^{-5}[\/latex]<\/li>\r\n \t<li>What is [OH<sup>\u2212<\/sup>] in a solution of 0.125 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>NH<sub>2<\/sub> and 0.130 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>NH<sub>3<\/sub>Cl?\r\n[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=4.4\\times {10}^{-4}[\/latex]<\/li>\r\n \t<li>What is [OH<sup>\u2212<\/sup>] in a solution of 1.25 <em data-effect=\"italics\">M<\/em> NH<sub>3<\/sub> and 0.78 <em data-effect=\"italics\">M<\/em> NH<sub>4<\/sub>NO<sub>3<\/sub>?\r\n[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]<\/li>\r\n \t<li>What concentration of NH<sub>4<\/sub>NO<sub>3<\/sub> is required to make [OH<sup>\u2212<\/sup>] = 1.0 [latex]\\times [\/latex] 10<sup>\u22125<\/sup> in a 0.200-<em data-effect=\"italics\">M<\/em> solution of NH<sub>3<\/sub>?<\/li>\r\n \t<li>What concentration of NaF is required to make [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 2.3 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> in a 0.300-<em data-effect=\"italics\">M<\/em> solution of HF?<\/li>\r\n \t<li>What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:\r\n<ol>\r\n \t<li>HCl<\/li>\r\n \t<li>KCH<sub>3<\/sub>CO<sub>2<\/sub><\/li>\r\n \t<li>NaCl<\/li>\r\n \t<li>KOH<\/li>\r\n \t<li>CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:\r\n<ol>\r\n \t<li>KI<\/li>\r\n \t<li>NH<sub>3<\/sub><\/li>\r\n \t<li>HI<\/li>\r\n \t<li>NaOH<\/li>\r\n \t<li>NH<sub>4<\/sub>Cl<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What will be the pH of a buffer solution prepared from 0.20 mol NH<sub>3<\/sub>, 0.40 mol NH<sub>4<\/sub>NO<sub>3<\/sub>, and just enough water to give 1.00 L of solution?<\/li>\r\n \t<li>Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH<sub>2<\/sub>PO<sub>4<\/sub>, and enough water to make 0.500 L of solution.<\/li>\r\n \t<li>How much solid NaCH<sub>3<\/sub>CO<sub>2<\/sub>\u20223H<sub>2<\/sub>O must be added to 0.300 L of a 0.50-<em data-effect=\"italics\">M<\/em> acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)<\/li>\r\n \t<li>What mass of NH<sub>4<\/sub>Cl must be added to 0.750 L of a 0.100-<em data-effect=\"italics\">M<\/em> solution of NH<sub>3<\/sub> to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)<\/li>\r\n \t<li>A buffer solution is prepared from equal volumes of 0.200 <em data-effect=\"italics\">M<\/em> acetic acid and 0.600 <em data-effect=\"italics\">M<\/em> sodium acetate. Use 1.80 [latex]\\times [\/latex] 10<sup>\u22125<\/sup> as <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for acetic acid.\r\n<ol>\r\n \t<li>What is the pH of the solution?<\/li>\r\n \t<li>Is the solution acidic or basic?<\/li>\r\n \t<li>What is the pH of a solution that results when 3.00 mL of 0.034 <em data-effect=\"italics\">M<\/em> HCl is added to 0.200 L of the original buffer?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A 5.36\u2013g sample of NH<sub>4<\/sub>Cl was added to 25.0 mL of 1.00 <em data-effect=\"italics\">M<\/em> NaOH and the resulting solution\u00a0diluted to 0.100 L.\r\n<ol>\r\n \t<li>What is the pH of this buffer solution?<\/li>\r\n \t<li>Is the solution acidic or basic?<\/li>\r\n \t<li>What is the pH of a solution that results when 3.00 mL of 0.034 <em data-effect=\"italics\">M<\/em> HCl is added to the solution?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which acid in Table 1 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.<\/li>\r\n \t<li>Which acid in Table 1 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0\u00a0is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.<\/li>\r\n \t<li>Which base in Table 2 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0\u00a0is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.<\/li>\r\n \t<li>Which base in <a class=\"autogenerated-content\" href=\"\/contents\/64b7990f-3343-41b2-b0cb-ddf487e91677@2#fs-idm84795184\">[link]<\/a> is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.<\/li>\r\n \t<li>Saccharin, C<sub>7<\/sub>H<sub>4<\/sub>NSO<sub>3<\/sub>H, is a weak acid (<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 2.1 [latex]\\times [\/latex] 10<sup>\u22122<\/sup>). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 [latex]\\times [\/latex] 10<sup>\u22123<\/sup> g of sodium saccharide, Na(C<sub>7<\/sub>H<sub>4<\/sub>NSO<sub>3<\/sub>), what are the final concentrations of saccharine and sodium saccharide in the solution?<\/li>\r\n \t<li>What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C<sub>5<\/sub>H<sub>9<\/sub>NO<sub>4<\/sub>, a diprotic acid; <em data-effect=\"italics\">K<\/em><sub>1<\/sub> = 8.5 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>, <em data-effect=\"italics\">K<\/em><sub>2<\/sub> = 3.39 [latex]\\times [\/latex] 10<sup>\u221210<\/sup>) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n2.\u00a0Excess [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] is removed primarily by the reaction:\r\n\r\n[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]\r\n\r\nExcess base is removed by the reaction:\r\n\r\n[latex]{\\text{OH}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]\r\n\r\n4. The equilibrium expression is:\r\n\r\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm144576784\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.25, negative x, 0.25 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.030, x, 0.030 plus sign x.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter size-full wp-image-5457\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044202\/CNX_Chem_14_06_ICETable1_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.25, negative x, 0.25 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.030, x, 0.030 plus sign x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.25 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.25 and (0.030 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.030, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(x\\right)\\left(0.030-x\\right)}{\\left(0.25-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.030\\right)}{0.25}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 1.50 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.25 and 0.030, our assumptions are correct. Therefore, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 1.5 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>.\r\n\r\nThis problem can also be solved using the Henderson-Hasselbalch equation: [latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[{\\text{A}}^{\\text{-}}\\right]}{\\left[\\text{HA}\\right]}[\/latex]; p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = \u2212log(<em data-effect=\"italics\">K<\/em><sub>a<\/sub>) = \u2212log(1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>) = 4.74; [HA] \u2248 [HA]<sub>0<\/sub> = [CH<sub>3<\/sub>CO<sub>2<\/sub>H]<sub>0<\/sub> = 0.25 <em data-effect=\"italics\">M<\/em>; [A<sup>\u2212<\/sup>] \u2248 [NaCH<sub>3<\/sub>CO<sub>2<\/sub>] = 0.030 <em data-effect=\"italics\">M<\/em>. Using these data: [latex]\\text{pH}=4.74\\text{-log}\\left(\\frac{0.030M}{0.25M}\\right)=3.82[\/latex]; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 10<sup>\u2212pH<\/sup><em data-effect=\"italics\">M<\/em> = 10<sup>\u22123.82<\/sup><em data-effect=\"italics\">M<\/em> = 1.5 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>\r\n\r\n6.\u00a0The equilibrium expression is:\r\n\r\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=4.4\\times {10}^{-4}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm283559344\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 N H subscript 2 ] [ C H subscript 3 N H subscript 3 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.125, negative sign x, 0.125 minus sign x. The second column has the following: 0.130, x, 0.130 plus sign x. The third column has the following: 0, x, x.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter size-full wp-image-5459\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044248\/CNX_Chem_14_06_ICETable3_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 N H subscript 2 ] [ C H subscript 3 N H subscript 3 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.125, negative sign x, 0.125 minus sign x. The second column has the following: 0.130, x, 0.130 plus sign x. The third column has the following: 0, x, x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.125 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.125 and (0.130 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.130, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=\\frac{\\left(0.130-x\\right)\\left(x\\right)}{\\left(0.125-x\\right)}\\approx \\frac{\\left(0.130\\right)\\left(x\\right)}{0.125}=4.4\\times {10}^{-4}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 4.23 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.125 and 0.130, our assumptions are correct. Therefore, [OH<sup>\u2212<\/sup>] = 4.2 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>.\r\n\r\n8. The reaction and equilibrium constant are:\r\n\r\n[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-4}[\/latex]\r\n\r\nThe equilibrium expression is:\r\n\r\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nLet <em data-effect=\"italics\">x<\/em> = the concentration of NH<sub>4<\/sub>NO<sub>3<\/sub> required. The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm69975744\" 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 ( M ). The second column has the header of \u201c[ N H subscript 3 ] [ N H subscript 4 superscript positive sign ] [ O H negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.200, negative sign x, 0.200 minus sign x. The second column has the following: 0.78, x plus sign x, x plus sign x. The third column has the following: 0, x, x equals 1.0 times 10 to the negative fifth power.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter wp-image-5460 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044427\/CNX_Chem_14_06_ICETable5_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 N H subscript 2 ] [ C H subscript 3 N H subscript 3 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.125, negative sign x, 0.125 minus sign x. The second column has the following: 0.130, x, 0.130 plus sign x. The third column has the following: 0, x, x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em data-effect=\"italics\">x<\/em> + <em data-effect=\"italics\">x<\/em>) \u2248 <em data-effect=\"italics\">x<\/em>, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(x-x\\right)\\left(1.0\\times {10}^{-5}\\right)}{\\left(0.200 - 1.0\\times {10}^{-5}\\right)}\\approx \\frac{\\left(x\\right)\\left(1.0\\times {10}^{-5}\\right)}{0.200}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 0.360 <em data-effect=\"italics\">M<\/em>. Because <em data-effect=\"italics\">x<\/em> is less than 5% of this value, our assumption is correct. Therefore, [latex]\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right][\/latex] = [NH<sub>4<\/sub>NO<sub>3<\/sub>] = 0.36 <em data-effect=\"italics\">M<\/em>.\r\n\r\n10.\u00a0The reaction and equilibrium constant are:\r\n\r\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\n(a) The added HCl will increase the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] slightly, which will react with [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex] and produce CH<sub>3<\/sub>CO<sub>2<\/sub>H in the process. Thus, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] decreases and [CH<sub>3<\/sub>CO<sub>2<\/sub>H] increases.\r\n\r\n(b) The added KCH<sub>3<\/sub>CO<sub>2<\/sub> will increase the concentration of [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] which will react with [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and produce CH<sub>3<\/sub>CO<sub>2<\/sub> H in the process. Thus, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] decreases slightly and [CH<sub>3<\/sub>CO<sub>2<\/sub>H] increases.\r\n\r\n(c) The added NaCl will have no effect on the concentration of the ions.\r\n\r\n(d) The added KOH will produce OH<sup>\u2212<\/sup> ions, which will react with the [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex], thus reducing [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex]. Some additional CH<sub>3<\/sub>CO<sub>2<\/sub>H will dissociate, producing [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] ions in the process. Thus, [CH<sub>3<\/sub>CO<sub>2<\/sub>H] decreases slightly and [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] increases.\r\n\r\n(e) The added CH<sub>3<\/sub>CO<sub>2<\/sub>H will increase its concentration, causing more of it to dissociate and producing more [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] and [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] in the process. Thus, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] increases slightly and [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] increases.\r\n\r\n12.\u00a0The reaction and equilibrium constant are:\r\n\r\n[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe equilibrium expression is:\r\n\r\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe initial concentrations of NH<sub>3<\/sub> and [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] are 0.20 <em data-effect=\"italics\">M<\/em> and 0.40 <em data-effect=\"italics\">M<\/em>, respectively. The equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm117581744\" 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 (M). The second column has the header of \u201c[ N H subscript 3 ] [ N H subscript 4 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, negative sign x, 0.20 minus sign x. The second column has the following: 0.40, x, 0.40 plus sign x. The third column has the following 0, x, x.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter size-full wp-image-5461\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044629\/CNX_Chem_14_06_ICETable7_img.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 (M). The second column has the header of \u201c[ N H subscript 3 ] [ N H subscript 4 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, negative sign x, 0.20 minus sign x. The second column has the following: 0.40, x, 0.40 plus sign x. The third column has the following 0, x, x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.20 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.20 and (0.40 + <em data-effect=\"italics\">x<\/em>) \u2248 0.40, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.40+x\\right)\\left(x\\right)}{\\left(0.20-x\\right)}\\approx \\frac{\\left(0.40\\right)\\left(x\\right)}{0.20}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 9.00 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.20 and 0.40, our assumptions are correct. Therefore, [OH<sup>\u2212<\/sup>] = 9.00 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Thus:\r\n\r\npOH = \u2212log(9.00 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>) = 5.046\r\n\r\npH = 14.000 \u2212 pOH = 14.000 \u2212 5.046 = 8.954 = 8.95\r\n\r\n14.\u00a0The reaction and equilibrium constant are:\r\n\r\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe equilibrium expression is:\r\n\r\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nLet <em data-effect=\"italics\">x<\/em> be the concentration of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex]. The hydronium ion concentration at equilibrium is:\r\n\r\n[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 10<sup>\u2212pH<\/sup> = 10<sup>\u22125.00<\/sup> = 1.00 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em>\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm35109248\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.50, negative sign x, 0.50 minus sign x. The second column has the following: 0, x, x equals 1.0 times 10 to the negative fifth power. The third column has the following: x, x, x plus sign x.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter size-full wp-image-5462\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044758\/CNX_Chem_14_06_ICETable9_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.50, negative sign x, 0.50 minus sign x. The second column has the following: 0, x, x equals 1.0 times 10 to the negative fifth power. The third column has the following: x, x, x plus sign x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em data-effect=\"italics\">x<\/em> + <em data-effect=\"italics\">x<\/em>) \u2248 <em data-effect=\"italics\">x<\/em>, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(1.0\\times {10}^{-5}\\right)\\left(x+x\\right)}{\\left(0.50 - 1.0\\times {10}^{-5}\\right)}\\approx \\frac{\\left(1.0\\times {10}^{-5}\\right)\\left(x\\right)}{0.50}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 0.900 <em data-effect=\"italics\">M<\/em>. Because <em data-effect=\"italics\">x<\/em> is less than 5% of this value, our assumption is correct. Therefore, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] = 0.900 <em data-effect=\"italics\">M<\/em>. Using the molar mass of NaC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>\u20223H<sub>2<\/sub>O (136.080 \/mol) and the volume gives the mass required:\r\n\r\n[latex]\\frac{0.900\\text{mol}}{1\\text{L}}\\times 0.300\\text{L}\\times \\frac{136.080\\text{g}}{1\\text{mol}}=36.7=37\\text{g}\\left(0.27\\text{mol}\\right)[\/latex]\r\n\r\n16.\u00a0(a) The reaction and equilibrium constant are:\r\n\r\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe equilibrium expression is:\r\n\r\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe molar mass of NH<sub>4<\/sub>Cl is 53.4912 g\/mol. The moles of NH<sub>4<\/sub>Cl are: [latex]\\frac{5.36\\text{g}}{53.4912\\text{g}{\\text{mol}}^{-1}}=0.1002\\text{mol}[\/latex]\r\n\r\nAssume 0.500 L of each solution is present The total volume is thus 1.000 L. The initial concentrations of the ions is obtained using <em data-effect=\"italics\">M<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">M<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub>, or:\r\n\r\n[latex]\\begin{array}{l}\\\\ \\\\ \\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]={M}_{1}\\times \\frac{{V}_{1}}{{V}_{2}}=\\left(0.200\\right)\\times \\frac{0.500\\text{L}}{1.000\\text{L}}=0.100M\\\\ \\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]={M}_{1}\\times \\frac{{V}_{1}}{{V}_{2}}=\\left(0.600\\right)\\times \\frac{0.500\\text{L}}{1.000\\text{L}}=0.300M\\end{array}[\/latex]\r\n\r\nThe initial and equilibrium concentrations of this system can be written as follows:<span id=\"fs-idp20201888\" 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 ( M ). The second column has the header of \u201c[ C H subscript C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.100, negative sign x, 0.100 negative sign x. The second column is the following: 0, x, x. The third column has the following: 0.300, x, 0.300 plus sign x.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter size-full wp-image-5463\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044850\/CNX_Chem_14_06_ICETable11_img.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 ( M ). The second column has the header of \u201c[ C H subscript C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.100, negative sign x, 0.100 negative sign x. The second column is the following: 0, x, x. The third column has the following: 0.300, x, 0.300 plus sign x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.100 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.100 and (0.300 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.300, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(x\\right)\\left(0.300+x\\right)}{\\left(0.100-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.300\\right)}{0.100}=1.80\\times {10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 6.000 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.100 and 0.300, our assumptions are correct. Therefore [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 6.000 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>:\r\n\r\npH = \u2212log(6.000 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>) = 5.2218 = 5.222;\r\n\r\n(b) The solution is acidic.\r\n\r\n(c) Assume that the added H<sup>+<\/sup> reacts completely with an equal amount of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex], forming an equal amount of CH<sub>3<\/sub>CO<sub>2<\/sub>H in the process. The moles of H<sup>+<\/sup> added equal 0.034 <em data-effect=\"italics\">M<\/em> [latex]\\times [\/latex] 0.00300 L = 1.02 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> mol. For the acetic acid, the initial moles present equal 0.2000 <em data-effect=\"italics\">M<\/em> [latex]\\times [\/latex] 0.500 L = 0.1000 mol, and for acetate ion, 0.600 <em data-effect=\"italics\">M<\/em> [latex]\\times [\/latex] 0.500 L = 0.3000 mol. Thus:\r\n\r\nmol CH<sub>3<\/sub>CO<sub>2<\/sub>H = 0.1000 + 1.02 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> = 0.1001 mol\r\n\r\n[latex]\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}=0.3000 - 1.02\\times {10}^{-4}=0.2999\\text{mol}[\/latex]\r\n\r\nFinal volume = 1.000 L + 3.00 [latex]\\times [\/latex] 10<sup>\u22123<\/sup> L = 1.0030 L\r\n\r\nThe initial concentrations are therefore:\r\n\r\n[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\frac{0.1001\\text{mol}}{1.0030\\text{L}}=0.09980M[\/latex]\r\n\r\n[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]=\\frac{0.2999\\text{mol}}{1.0030\\text{L}}=0.2990M[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm165224336\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.09980, negative x, 0.09980 minus sign x. The second column has the following: 0, x, x. The third column has the following: .2990, x, 0.2992 plus sign x.\">\r\n<\/span>\r\n\r\n<img class=\"aligncenter size-full wp-image-5465\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044926\/CNX_Chem_14_06_ICETable12_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.09980, negative x, 0.09980 minus sign x. The second column has the following: 0, x, x. The third column has the following: .2990, x, 0.2992 plus sign x.\" width=\"975\" height=\"264\" \/>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.09980 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.09980 and (0.2990 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.2990, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(x\\right)\\left(0.2990+x\\right)}{\\left(0.09980-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.2990\\right)}{0.09980}=1.80\\times {10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 6.008 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.09980 and 0.2990, our assumptions are correct. Therefore, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 6.008 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>.\r\n\r\npH = \u2212log(6.008 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>) = 5.2213 = 5.221\r\n\r\n18.\u00a0To prepare the best buffer for a weak acid HA and its salt, the ratio [latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{{K}_{\\text{a}}}[\/latex] should be as close to 1 as possible for effective buffer action. The [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] concentration in a buffer of pH 3.1 is [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 10<sup>\u22123.1<\/sup> = 7.94 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>\r\n\r\nWe can now solve for <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of the best acid as follows:\r\n\r\n[latex]\\begin{array}{l}\\\\ \\\\ \\\\ \\\\ \\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{{K}_{\\text{a}}}=1\\\\ {K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{1}=7.94\\times {10}^{-4}\\end{array}[\/latex]\r\n\r\nIn Table 1 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>, the acid with the closest <em data-effect=\"italics\">K<\/em><sub>a<\/sub> to 7.94 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> is HF, with a <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of 7.2 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>.\r\n\r\n20.\u00a0For buffers with pHs &gt; 7, you should use a weak base and its salt. The most effective buffer will have a ratio [latex]\\frac{\\left[{\\text{OH}}^{\\text{-}}\\right]}{{K}_{\\text{b}}}[\/latex] that is as close to 1 as possible. The pOH of the buffer is 14.00 \u2212 10.65 = 3.35. Therefore, [OH<sup>\u2212<\/sup>] is [OH<sup>\u2212<\/sup>] = 10<sup>\u2212pOH<\/sup> = 10<sup>\u22123.35<\/sup> = 4.467 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>.\r\n\r\nWe can now solve for <em data-effect=\"italics\">K<\/em><sub>b<\/sub> of the best base as follows:\r\n\r\n[latex]\\frac{\\left[{\\text{OH}}^{\\text{-}}\\right]}{{K}_{\\text{b}}}=1[\/latex]\r\n\r\n<em data-effect=\"italics\">K<\/em><sub>b<\/sub> = [OH<sup>\u2212<\/sup>] = 4.47 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>\r\n\r\nIn Table 2\u00a0of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>, the base with the closest <em data-effect=\"italics\">K<\/em><sub>b<\/sub> to 4.47 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> is CH<sub>3<\/sub>NH<sub>2<\/sub>, with a <em data-effect=\"italics\">K<\/em><sub>b<\/sub> = 4.4 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>.\r\n\r\n22. The molar mass of sodium saccharide is 205.169 g\/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:\r\n\r\n[latex]2.00\\times {10}^{-3}\\text{g}\\times \\frac{1\\text{mol}}{205.169\\text{g}}=9.75\\times {10}^{-6}\\text{mol}[\/latex]\r\n\r\nThis ionizes initially to form saccharin ions, A<sup>\u2212<\/sup>, with:\r\n\r\n[latex]\\left[{\\text{A}}^{\\text{-}}\\right]=\\frac{9.75\\times {10}^{-6}\\text{mol}}{0.250\\text{L}}=3.9\\times {10}^{-5}M[\/latex]\r\n\r\nbut A<sup>\u2212<\/sup> reacts with water:\r\n\r\n[latex]\\begin{array}{l}{\\text{A}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HA}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\\\ {K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{1.0\\times {10}^{-14}}{2.1\\times {10}^{-12}}=4.8\\times {10}^{-3}\\\\ =4.8\\times {10}^{-3}=\\frac{\\left[\\text{HA}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{A}}^{\\text{-}}\\right]}\\end{array}[\/latex]\r\n\r\nThe pH of the solution is 5.48, so:\r\n\r\npOH = 14.00 \u2212 5.48 = 8.52\r\n\r\nand\r\n\r\n[OH<sup>\u2212<\/sup>] = 10<sup>\u22128.52<\/sup> = 3.02 [latex]\\times [\/latex] 10<sup>\u22129<\/sup><em data-effect=\"italics\">M<\/em>\r\n\r\nBecause of the small size of <em data-effect=\"italics\">K<\/em><sub>b<\/sub>, almost all the A<sup>\u2212<\/sup> will be in the form of HA. Therefore:\r\n\r\n[latex]4.8\\times {10}^{-3}=\\frac{x\\left(3.02\\times {10}^{-9}\\right)}{3.9\\times {10}^{-5}-x}[\/latex]\r\n\r\nwhere\r\n\r\n<em data-effect=\"italics\">x<\/em> \u2248 3.9 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> = [HA] = [C<sub>7<\/sub>H<sub>4<\/sub>NSO<sub>3<\/sub>H]\r\n\r\nConsequently, [A<sup>\u2212<\/sup>] is extremely small. Therefore, solve for [A<sup>\u2212<\/sup>] from the equilibrium expression:\r\n\r\n[latex]\\left[{\\text{A}}^{\\text{-}}\\right]=\\frac{\\left[\\text{HA}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{{K}_{\\text{b}}}=\\frac{\\left(3.9\\times {10}^{-5}\\right)\\left(3.02\\times {10}^{-9}\\right)}{4.8\\times {10}^{-3}}=2.5\\times {10}^{-11}M=\\left[\\text{Na}\\left({\\text{C}}_{7}{\\text{H}}_{4}{\\text{NSO}}_{3}\\right)\\right][\/latex]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Glossary<\/h3>\r\n<div id=\"fs-idp88494848\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">buffer capacity\r\n<\/span><\/strong>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)\r\n\r\n<\/div>\r\n<div id=\"fs-idp88495904\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">buffer\r\n<\/span><\/strong>mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added\r\n\r\n<\/div>\r\n<div id=\"fs-idm71412016\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">Henderson-Hasselbalch equation\r\n<\/span><\/strong>equation used to calculate the pH of buffer solutions\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this module, 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 mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a <span data-type=\"term\">buffer<\/span>. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure 1). A solution of acetic acid and sodium acetate (CH<sub>3<\/sub>COOH + CH<sub>3<\/sub>COONa) 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<div id=\"attachment_5446\" style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5446\" class=\"size-full wp-image-5446\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043335\/CNX_Chem_14_06_compare.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\" width=\"975\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-5446\" class=\"wp-caption-text\">Figure 1. (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-M 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)<\/p>\n<\/div>\n<section id=\"fs-idm112960288\" data-depth=\"1\">\n<h2 data-type=\"title\">How Buffers Work<\/h2>\n<p id=\"fs-idm127493136\">A mixture of acetic acid and sodium acetate is acidic because the <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of acetic acid is greater than the <em data-effect=\"italics\">K<\/em><sub>b<\/sub> of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. Then more of the acetic acid reacts with water, restoring the hydronium ion concentration almost to its original value:<\/p>\n<div id=\"fs-idm105688000\" data-type=\"equation\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right)[\/latex]<\/div>\n<p id=\"fs-idp41915456\">The pH changes very little. If we add an acid such as hydrochloric acid, most of the hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid molecules:<\/p>\n<div id=\"fs-idm115859376\" data-type=\"equation\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\n<p id=\"fs-idp6711056\">Thus, there is very little increase in the concentration of the hydronium ion, and the pH remains practically unchanged (Figure 2).<\/p>\n<div id=\"attachment_5448\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5448\" class=\"size-large wp-image-5448\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043433\/CNX_Chem_14_06_bufferchrt-1024x566-1024x566.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\" width=\"1024\" height=\"566\" \/><\/p>\n<p id=\"caption-attachment-5448\" class=\"wp-caption-text\">Figure 2. This diagram shows the buffer action of these reactions.<\/p>\n<\/div>\n<p id=\"fs-idm155475664\">A mixture of ammonia and ammonium chloride is basic because the <em data-effect=\"italics\">K<\/em><sub>b<\/sub> for ammonia is greater than the <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value:<\/p>\n<div id=\"fs-idm142306048\" data-type=\"equation\">[latex]{\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\n<p id=\"fs-idp98958624\">If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value:<\/p>\n<div id=\"fs-idm144711616\" data-type=\"equation\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)\\longrightarrow {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\n<p id=\"fs-idm72230384\">The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid.<\/p>\n<div id=\"fs-idm144695456\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idm145736240\"><span data-type=\"title\">pH Changes in Buffered and Unbuffered Solutions<\/span><\/h4>\n<p>Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change 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<h4 id=\"fs-idm105719280\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):<\/p>\n<p><span id=\"fs-idp49040960\" data-type=\"media\" data-alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5449\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043544\/CNX_Chem_14_06_steps1_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" width=\"1000\" height=\"177\" \/><br \/>\n<\/span><\/p>\n<ol id=\"fs-idm106943744\" class=\"stepwise\" data-number-style=\"arabic\">\n<li><em data-effect=\"italics\">Determine the direction of change.<\/em> The equilibrium in a mixture of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex], [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex], and CH<sub>3<\/sub>CO<sub>2<\/sub>H is:\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idp57064960\" data-type=\"equation\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right)[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<p>The equilibrium constant for CH<sub>3<\/sub>CO<sub>2<\/sub>H is not given, so we look it up in <a href=\".\/chapter\/ionization-constants-of-weak-acids-missing-formulas\/\" target=\"_blank\">Ionization Constants of Weak Acids<\/a>: <em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>. With [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] = 0.10 <em data-effect=\"italics\">M<\/em> and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = ~0 <em data-effect=\"italics\">M<\/em>, the reaction shifts to the right to form [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex].<\/li>\n<li><em data-effect=\"italics\">Determine<\/em> x <em data-effect=\"italics\">and equilibrium concentrations<\/em>. A table of changes and concentrations follows:\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5450\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043645\/CNX_Chem_14_06_ICETable16_img.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 ( 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, x, x. The fourth column has the following: 0.10, x, 0.10 plus sign x.\" width=\"975\" height=\"264\" \/><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 ( 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, x, x. The fourth column has the following: 0.10, x, 0.10 plus sign x.\"> <\/span><\/li>\n<li><em data-effect=\"italics\">Solve for x and the equilibrium concentrations.<\/em> We find:\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm7012176\" data-type=\"equation\">[latex]x=1.8\\times {10}^{-5}M[\/latex]<\/div>\n<p>and<\/p>\n<div id=\"fs-idm109618832\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=0+x=1.8\\times {10}^{-5}M[\/latex]<\/div>\n<p>Thus:<\/p>\n<div id=\"fs-idm165278336\" data-type=\"equation\">[latex]\\text{pH}=\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\text{-log}\\left(1.8\\times {10}^{-5}\\right)[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm111454032\" data-type=\"equation\">[latex]=4.74[\/latex]<\/div>\n<\/li>\n<li><em data-effect=\"italics\">Check the work<\/em>. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, <em data-effect=\"italics\">Q<\/em> = <em data-effect=\"italics\">K<\/em><sub>a<\/sub>.<\/li>\n<\/ol>\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 mL of this buffer, giving a solution with a volume of 101 mL.<\/p>\n<p id=\"fs-idm85734032\">First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5451\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043740\/CNX_Chem_14_06_steps2_img.jpg\" alt=\"Eight tan rectangles are shown in four columns of two rectangles each that are connected with right pointing arrows. The first rectangle in the upper left is labeled \u201cVolume of N a O H solution.\u201d An arrow points right to a second rectangle labeled \u201cMoles of N a O H added.\u201d A second arrow points right to a third rectangle labeled \u201cAdditional moles of N a C H subscript 3 C O subscript 2.\u201d Just beneath the first rectangle in the upper left is a rectangle labeled \u201cVolume of buffer solution.\u201d An arrow points right to another rectangle labeled \u201cInitial moles of C H subscript 3 C O subscript 2 H.\u201d This rectangle points to the same third rectangle, which is labeled \u201c Additional moles of N a C H subscript 3 C O subscript 2.\u201d An arrow points right to a rectangle labeled \u201c Unreacted moles of C H subscript 3 C O subscript 2 H.\u201d An arrow points from this rectangle to a rectangle below labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d An arrow extends below the \u201cAdditional moles of N a C H subscript 3 C O subscript 2\u201d rectangle to a rectangle labeled \u201c[ C H subscript 3 C O subscript 2 ].\u201d This rectangle points right to the rectangle labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d\" width=\"1000\" height=\"475\" \/><span id=\"fs-idm125292992\" data-type=\"media\" data-alt=\"Eight tan rectangles are shown in four columns of two rectangles each that are connected with right pointing arrows. The first rectangle in the upper left is labeled \u201cVolume of N a O H solution.\u201d An arrow points right to a second rectangle labeled \u201cMoles of N a O H added.\u201d A second arrow points right to a third rectangle labeled \u201cAdditional moles of N a C H subscript 3 C O subscript 2.\u201d Just beneath the first rectangle in the upper left is a rectangle labeled \u201cVolume of buffer solution.\u201d An arrow points right to another rectangle labeled \u201cInitial moles of C H subscript 3 C O subscript 2 H.\u201d This rectangle points to the same third rectangle, which is labeled \u201c Additional moles of N a C H subscript 3 C O subscript 2.\u201d An arrow points right to a rectangle labeled \u201c Unreacted moles of C H subscript 3 C O subscript 2 H.\u201d An arrow points from this rectangle to a rectangle below labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d An arrow extends below the \u201cAdditional moles of N a C H subscript 3 C O subscript 2\u201d rectangle to a rectangle labeled \u201c[ C H subscript 3 C O subscript 2 ].\u201d This rectangle points right to the rectangle labeled \u201c[ C H subscript 3 C O subscript 2 H ].\u201d\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-idm142618096\" class=\"stepwise\" data-number-style=\"arabic\">\n<li><em data-effect=\"italics\">Determine the moles of NaOH.<\/em> One milliliter (0.0010 L) of 0.10 <em data-effect=\"italics\">M<\/em> NaOH contains:\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm110133872\" data-type=\"equation\">[latex]0.0010\\cancel{\\text{L}}\\times \\left(\\frac{0.10\\text{mol NaOH}}{1\\cancel{\\text{L}}}\\right)=1.0\\times {10}^{-4}\\text{mol NaOH}[\/latex]<\/div>\n<\/li>\n<li><em data-effect=\"italics\">Determine the moles of CH<sub>2<\/sub>CO<sub>2<\/sub>H.<\/em> Before reaction, 0.100 L of the buffer solution contains:\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm53635376\" data-type=\"equation\">[latex]0.100\\cancel{\\text{L}}\\times \\left(\\frac{0.100\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}}{1\\cancel{\\text{L}}}\\right)=1.00\\times {10}^{-2}\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}[\/latex]<\/div>\n<\/li>\n<li><em data-effect=\"italics\">Solve for the amount of NaCH<sub>3<\/sub>CO<sub>2<\/sub> produced.<\/em> The 1.0 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> mol of NaOH neutralizes 1.0 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H, leaving:\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm91870896\" data-type=\"equation\">[latex]\\left(1.0\\times {10}^{-2}\\right)-\\left(0.01\\times {10}^{-2}\\right)=0.99\\times {10}^{-2}\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}[\/latex]<\/div>\n<p>and producing 1.0 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> mol of NaCH<sub>3<\/sub>CO<sub>2<\/sub>. This makes a total of:<\/p>\n<div id=\"fs-idp82833792\" data-type=\"equation\">[latex]\\left(1.0\\times {10}^{-2}\\right)+\\left(0.01\\times {10}^{-2}\\right)=1.01\\times {10}^{-2}\\text{mol}{\\text{NaCH}}_{3}{\\text{CO}}_{2}[\/latex]<\/div>\n<\/li>\n<li><em data-effect=\"italics\">Find the molarity of the products.<\/em> After reaction, CH<sub>3<\/sub>CO<sub>2<\/sub>H and NaCH<sub>3<\/sub>CO<sub>2<\/sub> are contained in 101 mL of the intermediate solution, so:\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm165714384\" data-type=\"equation\">[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\frac{9.9\\times {10}^{-3}\\text{mol}}{0.101\\text{L}}=0.098M[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idp22032080\" data-type=\"equation\">[latex]\\left[{\\text{NaCH}}_{3}{\\text{CO}}_{2}\\right]=\\frac{1.01\\times {10}^{-2}\\text{mol}}{0.101\\text{L}}=0.100M[\/latex]<\/div>\n<p>Now we calculate the pH after the intermediate solution, which is 0.098 <em data-effect=\"italics\">M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.100 <em data-effect=\"italics\">M<\/em> in NaCH<sub>3<\/sub>CO<sub>2<\/sub>, comes to equilibrium. The calculation is very similar to that in part (a) of this example:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5453\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043847\/CNX_Chem_14_06_steps3_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" width=\"1000\" height=\"177\" \/><span id=\"fs-idm159880832\" data-type=\"media\" data-alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\"><br \/>\n<\/span><\/p>\n<p>This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution (F).<\/li>\n<\/ol>\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 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>&#8211;<em data-effect=\"italics\">M<\/em> solution of HCl). The volume of the final solution is 101 mL.<\/p>\n<h4 id=\"fs-idm51120032\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>This 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>&#8211;<em data-effect=\"italics\">M<\/em> solution of HCl has the same hydronium ion concentration as the 0.10-<em data-effect=\"italics\">M<\/em> solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:<\/p>\n<div id=\"fs-idp16142224\" data-type=\"equation\">[latex]0.100\\text{L}\\times \\left(\\frac{1.8\\times {10}^{-5}\\text{mol HCl}}{1\\text{L}}\\right)=1.8\\times {10}^{-6}\\text{mol HCl}[\/latex]<\/div>\n<p id=\"fs-idp35168928\">As shown in part (b), 1 mL of 0.10 <em data-effect=\"italics\">M<\/em> NaOH contains 1.0 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:<\/p>\n<div id=\"fs-idp21695968\" data-type=\"equation\">[latex]\\left(1.0\\times {10}^{-4}\\right)-\\left(1.8\\times {10}^{-6}\\right)=9.8\\times {10}^{-5}M[\/latex]<\/div>\n<p id=\"fs-idm52812304\">The concentration of NaOH is:<\/p>\n<div id=\"fs-idm72079040\" data-type=\"equation\">[latex]\\frac{9.8\\times {10}^{-5}M\\text{NaOH}}{0.101\\text{L}}=9.7\\times {10}^{-4}M[\/latex]<\/div>\n<p id=\"fs-idm165600608\">The pOH of this solution is:<\/p>\n<div id=\"fs-idm97337872\" data-type=\"equation\">[latex]\\text{pOH}=\\text{-log}\\left[{\\text{OH}}^{\\text{-}}\\right]=\\text{-log}\\left(9.7\\times {10}^{-4}\\right)=3.01[\/latex]<\/div>\n<p id=\"fs-idm109107696\">The pH is:<\/p>\n<div id=\"fs-idm158590768\" data-type=\"equation\">[latex]\\text{pH}=14.00-\\text{pOH}=10.99[\/latex]<\/div>\n<p id=\"fs-idp53095888\">The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).<\/p>\n<h4 id=\"fs-idm159189968\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\n<p>Show that adding 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl changes the pH of 100 mL of a 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl solution from 4.74 to 3.00.<\/p>\n<div id=\"fs-idm145837936\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>Initial pH of 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl; pH = \u2212log [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = \u2212log[1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>] = 4.74<\/div>\n<p style=\"text-align: right;\">Moles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] in 100 mL 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> HCl; 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup> moles\/L [latex]\\times[\/latex] 0.100 L = 1.8 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><\/p>\n<p style=\"text-align: right;\">Moles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] added by addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl: 0.10 moles\/L [latex]\\times[\/latex] 0.0010 L = 1.0 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> moles; final pH after addition of 1.0 mL of 0.10 <em data-effect=\"italics\">M<\/em> HCl:<\/p>\n<div id=\"fs-idm103691008\" style=\"text-align: right;\" data-type=\"equation\">[latex]\\text{pH}=\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\text{-log}\\left(\\frac{\\text{total moles}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}}{\\text{total volume}}\\right)=\\text{-log}\\left(\\frac{1.0\\times {10}^{-4}\\text{mol}+1.8\\times {10}^{-6}\\text{mol}}{101\\text{mL}\\left(\\frac{1\\text{L}}{1000\\text{mL}}\\right)}\\right)=3.00[\/latex]<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idm109632000\">If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.<\/p>\n<\/section>\n<section id=\"fs-idp107911744\" data-depth=\"1\">\n<h2 data-type=\"title\">Buffer Capacity<\/h2>\n<p id=\"fs-idm68764768\">Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 3). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.<\/p>\n<div id=\"attachment_5454\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5454\" class=\"size-large wp-image-5454\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09043956\/CNX_Chem_14_06_exhaust-1024x285.jpg\" alt=\"Figure 3. 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)\" width=\"1024\" height=\"285\" \/><\/p>\n<p id=\"caption-attachment-5454\" class=\"wp-caption-text\">Figure 3. 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)<\/p>\n<\/div>\n<p id=\"fs-idm103629216\">The <span data-type=\"term\">buffer capacity<\/span> 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<\/section>\n<section id=\"fs-idm124351376\" data-depth=\"1\">\n<h2 data-type=\"title\">Selection of Suitable Buffer Mixtures<\/h2>\n<p id=\"fs-idp2820144\">There are two useful rules of thumb for selecting buffer mixtures:<\/p>\n<ol id=\"fs-idm128259280\" data-number-style=\"arabic\">\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. Figure 4\u00a0shows 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.\n<div id=\"attachment_5456\" style=\"width: 1010px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5456\" class=\"size-full wp-image-5456\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044040\/CNX_Chem_14_06_buffer.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].\" width=\"1000\" height=\"715\" \/><\/p>\n<p id=\"caption-attachment-5456\" class=\"wp-caption-text\">The graph, an illustration of buffering action, shows change of pH as an increasing amount of a 0.10-<em>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>M<\/em> and [CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]=0.10M.<\/p>\n<\/div>\n<div data-type=\"newline\"><\/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 bicarbonate ion, [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex]. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:<\/p>\n<div id=\"fs-idm106059312\" data-type=\"equation\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\n<p id=\"fs-idm109728256\">When an excess of the hydroxide ion is present, it is removed by the reaction:<\/p>\n<div id=\"fs-idm145990368\" data-type=\"equation\">[latex]{\\text{OH}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\longrightarrow {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\n<p id=\"fs-idp65661488\">The pH of human blood thus remains very near 7.35, that is, slightly basic. Variations are usually less than 0.1 of a pH unit. A change of 0.4 of a pH unit is likely to be fatal.<\/p>\n<\/section>\n<section id=\"fs-idm153676976\" data-depth=\"1\">\n<h2 data-type=\"title\">The Henderson-Hasselbalch Equation<\/h2>\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\" data-type=\"equation\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{A}}^{\\text{-}}\\right]}{\\text{[HA]}}[\/latex]<\/div>\n<p id=\"fs-idm87196592\">Rearranging to solve for [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], we get:<\/p>\n<div id=\"fs-idm97934592\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]={K}_{\\text{a}}\\times \\frac{\\text{[HA]}}{\\left[{\\text{A}}^{\\text{-}}\\right]}[\/latex]<\/div>\n<p id=\"fs-idp10558896\">Taking the negative logarithm of both sides of this equation, we arrive at:<\/p>\n<div id=\"fs-idp26801296\" data-type=\"equation\">[latex]\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\text{-log}{K}_{\\text{a}}\\text{- log}\\frac{\\left[\\text{HA}\\right]}{\\left[{\\text{A}}^{\\text{-}}\\right]}[\/latex],<\/div>\n<p id=\"fs-idp7990848\">which can be written as<\/p>\n<div id=\"fs-idp2699408\" data-type=\"equation\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[{\\text{A}}^{\\text{-}}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/div>\n<p id=\"fs-idp97287200\">where p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> is the negative of the common 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 acid and its salt in a buffered solution. Scientists often use this expression, called the <strong><span data-type=\"term\">Henderson-Hasselbalch equation<\/span><\/strong>, 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<div id=\"fs-idp1420208\" class=\"chemistry chemist-portrait textbox shaded\" data-type=\"note\">\n<h3 data-type=\"title\">Lawrence Joseph Henderson and Karl Albert Hasselbalch<\/h3>\n<p id=\"fs-idm68026080\">Lawrence Joseph <strong><span class=\"no-emphasis\" data-type=\"term\">Henderson<\/span><\/strong> (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 2 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 <strong><span class=\"no-emphasis\" data-type=\"term\">Hasselbalch<\/span><\/strong> (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<\/div>\n<div class=\"textbox shaded\">\n<h3>Medicine: The Buffer System in Blood<\/h3>\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<p>[latex]{\\text{CO}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\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, [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex], 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<p>[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[\\text{base}\\right]}{\\left[\\text{acid}\\right]}=6.1+\\text{log}\\frac{0.024}{0.0012}=7.4[\/latex]<\/p>\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 [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] 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 [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] 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 [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] 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<sup>+<\/sup>] and restoring an appropriate pH.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm104045184\" class=\"summary\" data-depth=\"1\">\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idm115041904\">A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm101899904\" class=\"key-equations\" data-depth=\"1\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/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>[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[{\\text{A}}^{\\text{-}}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section id=\"fs-idp119412416\" class=\"exercises\" data-depth=\"1\">\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<div id=\"fs-idm2126064\" data-type=\"exercise\">\n<div id=\"fs-idm143472640\" data-type=\"problem\">\n<ol>\n<li id=\"fs-idp15824512\">Explain why a buffer can be prepared from a mixture of NH<sub>4<\/sub>Cl and NaOH but not from NH<sub>3<\/sub> and NaOH.<\/li>\n<li>Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H<sub>3<\/sub>PO<sub>4<\/sub> and a salt of its conjugate base NaH<sub>2<\/sub>PO<sub>4<\/sub>.<\/li>\n<li>Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH<sub>3<\/sub> and a salt of its conjugate acid NH<sub>4<\/sub>Cl.<\/li>\n<li>What is [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] in a solution of 0.25 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.030 <em data-effect=\"italics\">M<\/em> NaCH<sub>3<\/sub>CO<sub>2<\/sub>?<br \/>\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/li>\n<li>What is [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] in a solution of 0.075 <em data-effect=\"italics\">M<\/em> HNO<sub>2<\/sub> and 0.030 <em data-effect=\"italics\">M<\/em> NaNO<sub>2<\/sub>?<br \/>\n[latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=4.5\\times {10}^{-5}[\/latex]<\/li>\n<li>What is [OH<sup>\u2212<\/sup>] in a solution of 0.125 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>NH<sub>2<\/sub> and 0.130 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>NH<sub>3<\/sub>Cl?<br \/>\n[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=4.4\\times {10}^{-4}[\/latex]<\/li>\n<li>What is [OH<sup>\u2212<\/sup>] in a solution of 1.25 <em data-effect=\"italics\">M<\/em> NH<sub>3<\/sub> and 0.78 <em data-effect=\"italics\">M<\/em> NH<sub>4<\/sub>NO<sub>3<\/sub>?<br \/>\n[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]<\/li>\n<li>What concentration of NH<sub>4<\/sub>NO<sub>3<\/sub> is required to make [OH<sup>\u2212<\/sup>] = 1.0 [latex]\\times[\/latex] 10<sup>\u22125<\/sup> in a 0.200-<em data-effect=\"italics\">M<\/em> solution of NH<sub>3<\/sub>?<\/li>\n<li>What concentration of NaF is required to make [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 2.3 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> in a 0.300-<em data-effect=\"italics\">M<\/em> solution of HF?<\/li>\n<li>What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:\n<ol>\n<li>HCl<\/li>\n<li>KCH<sub>3<\/sub>CO<sub>2<\/sub><\/li>\n<li>NaCl<\/li>\n<li>KOH<\/li>\n<li>CH<sub>3<\/sub>CO<sub>2<\/sub>H<\/li>\n<\/ol>\n<\/li>\n<li>What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:\n<ol>\n<li>KI<\/li>\n<li>NH<sub>3<\/sub><\/li>\n<li>HI<\/li>\n<li>NaOH<\/li>\n<li>NH<sub>4<\/sub>Cl<\/li>\n<\/ol>\n<\/li>\n<li>What will be the pH of a buffer solution prepared from 0.20 mol NH<sub>3<\/sub>, 0.40 mol NH<sub>4<\/sub>NO<sub>3<\/sub>, and just enough water to give 1.00 L of solution?<\/li>\n<li>Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH<sub>2<\/sub>PO<sub>4<\/sub>, and enough water to make 0.500 L of solution.<\/li>\n<li>How much solid NaCH<sub>3<\/sub>CO<sub>2<\/sub>\u20223H<sub>2<\/sub>O must be added to 0.300 L of a 0.50-<em data-effect=\"italics\">M<\/em> acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)<\/li>\n<li>What mass of NH<sub>4<\/sub>Cl must be added to 0.750 L of a 0.100-<em data-effect=\"italics\">M<\/em> solution of NH<sub>3<\/sub> to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)<\/li>\n<li>A buffer solution is prepared from equal volumes of 0.200 <em data-effect=\"italics\">M<\/em> acetic acid and 0.600 <em data-effect=\"italics\">M<\/em> sodium acetate. Use 1.80 [latex]\\times[\/latex] 10<sup>\u22125<\/sup> as <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for acetic acid.\n<ol>\n<li>What is the pH of the solution?<\/li>\n<li>Is the solution acidic or basic?<\/li>\n<li>What is the pH of a solution that results when 3.00 mL of 0.034 <em data-effect=\"italics\">M<\/em> HCl is added to 0.200 L of the original buffer?<\/li>\n<\/ol>\n<\/li>\n<li>A 5.36\u2013g sample of NH<sub>4<\/sub>Cl was added to 25.0 mL of 1.00 <em data-effect=\"italics\">M<\/em> NaOH and the resulting solution\u00a0diluted to 0.100 L.\n<ol>\n<li>What is the pH of this buffer solution?<\/li>\n<li>Is the solution acidic or basic?<\/li>\n<li>What is the pH of a solution that results when 3.00 mL of 0.034 <em data-effect=\"italics\">M<\/em> HCl is added to the solution?<\/li>\n<\/ol>\n<\/li>\n<li>Which acid in Table 1 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.<\/li>\n<li>Which acid in Table 1 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0\u00a0is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.<\/li>\n<li>Which base in Table 2 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0\u00a0is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.<\/li>\n<li>Which base in <a class=\"autogenerated-content\" href=\"\/contents\/64b7990f-3343-41b2-b0cb-ddf487e91677@2#fs-idm84795184\">[link]<\/a> is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.<\/li>\n<li>Saccharin, C<sub>7<\/sub>H<sub>4<\/sub>NSO<sub>3<\/sub>H, is a weak acid (<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 2.1 [latex]\\times[\/latex] 10<sup>\u22122<\/sup>). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 [latex]\\times[\/latex] 10<sup>\u22123<\/sup> g of sodium saccharide, Na(C<sub>7<\/sub>H<sub>4<\/sub>NSO<sub>3<\/sub>), what are the final concentrations of saccharine and sodium saccharide in the solution?<\/li>\n<li>What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C<sub>5<\/sub>H<sub>9<\/sub>NO<sub>4<\/sub>, a diprotic acid; <em data-effect=\"italics\">K<\/em><sub>1<\/sub> = 8.5 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>, <em data-effect=\"italics\">K<\/em><sub>2<\/sub> = 3.39 [latex]\\times[\/latex] 10<sup>\u221210<\/sup>) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>2.\u00a0Excess [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] is removed primarily by the reaction:<\/p>\n<p>[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right)\\longrightarrow {\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>Excess base is removed by the reaction:<\/p>\n<p>[latex]{\\text{OH}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>4. The equilibrium expression is:<\/p>\n<p>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm144576784\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.25, negative x, 0.25 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.030, x, 0.030 plus sign x.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5457\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044202\/CNX_Chem_14_06_ICETable1_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.25, negative x, 0.25 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.030, x, 0.030 plus sign x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.25 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.25 and (0.030 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.030, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(x\\right)\\left(0.030-x\\right)}{\\left(0.25-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.030\\right)}{0.25}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 1.50 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.25 and 0.030, our assumptions are correct. Therefore, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 1.5 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p>This problem can also be solved using the Henderson-Hasselbalch equation: [latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\frac{\\left[{\\text{A}}^{\\text{-}}\\right]}{\\left[\\text{HA}\\right]}[\/latex]; p<em data-effect=\"italics\">K<\/em><sub>a<\/sub> = \u2212log(<em data-effect=\"italics\">K<\/em><sub>a<\/sub>) = \u2212log(1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>) = 4.74; [HA] \u2248 [HA]<sub>0<\/sub> = [CH<sub>3<\/sub>CO<sub>2<\/sub>H]<sub>0<\/sub> = 0.25 <em data-effect=\"italics\">M<\/em>; [A<sup>\u2212<\/sup>] \u2248 [NaCH<sub>3<\/sub>CO<sub>2<\/sub>] = 0.030 <em data-effect=\"italics\">M<\/em>. Using these data: [latex]\\text{pH}=4.74\\text{-log}\\left(\\frac{0.030M}{0.25M}\\right)=3.82[\/latex]; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 10<sup>\u2212pH<\/sup><em data-effect=\"italics\">M<\/em> = 10<sup>\u22123.82<\/sup><em data-effect=\"italics\">M<\/em> = 1.5 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<p>6.\u00a0The equilibrium expression is:<\/p>\n<p>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=4.4\\times {10}^{-4}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm283559344\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 N H subscript 2 ] [ C H subscript 3 N H subscript 3 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.125, negative sign x, 0.125 minus sign x. The second column has the following: 0.130, x, 0.130 plus sign x. The third column has the following: 0, x, x.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5459\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044248\/CNX_Chem_14_06_ICETable3_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 N H subscript 2 ] [ C H subscript 3 N H subscript 3 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.125, negative sign x, 0.125 minus sign x. The second column has the following: 0.130, x, 0.130 plus sign x. The third column has the following: 0, x, x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.125 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.125 and (0.130 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.130, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=\\frac{\\left(0.130-x\\right)\\left(x\\right)}{\\left(0.125-x\\right)}\\approx \\frac{\\left(0.130\\right)\\left(x\\right)}{0.125}=4.4\\times {10}^{-4}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 4.23 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.125 and 0.130, our assumptions are correct. Therefore, [OH<sup>\u2212<\/sup>] = 4.2 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p>8. The reaction and equilibrium constant are:<\/p>\n<p>[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-4}[\/latex]<\/p>\n<p>The equilibrium expression is:<\/p>\n<p>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Let <em data-effect=\"italics\">x<\/em> = the concentration of NH<sub>4<\/sub>NO<sub>3<\/sub> required. The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm69975744\" 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 ( M ). The second column has the header of \u201c[ N H subscript 3 ] [ N H subscript 4 superscript positive sign ] [ O H negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.200, negative sign x, 0.200 minus sign x. The second column has the following: 0.78, x plus sign x, x plus sign x. The third column has the following: 0, x, x equals 1.0 times 10 to the negative fifth power.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5460 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044427\/CNX_Chem_14_06_ICETable5_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 N H subscript 2 ] [ C H subscript 3 N H subscript 3 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.125, negative sign x, 0.125 minus sign x. The second column has the following: 0.130, x, 0.130 plus sign x. The third column has the following: 0, x, x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em data-effect=\"italics\">x<\/em> + <em data-effect=\"italics\">x<\/em>) \u2248 <em data-effect=\"italics\">x<\/em>, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(x-x\\right)\\left(1.0\\times {10}^{-5}\\right)}{\\left(0.200 - 1.0\\times {10}^{-5}\\right)}\\approx \\frac{\\left(x\\right)\\left(1.0\\times {10}^{-5}\\right)}{0.200}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 0.360 <em data-effect=\"italics\">M<\/em>. Because <em data-effect=\"italics\">x<\/em> is less than 5% of this value, our assumption is correct. Therefore, [latex]\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right][\/latex] = [NH<sub>4<\/sub>NO<sub>3<\/sub>] = 0.36 <em data-effect=\"italics\">M<\/em>.<\/p>\n<p>10.\u00a0The reaction and equilibrium constant are:<\/p>\n<p>[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>(a) The added HCl will increase the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] slightly, which will react with [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex] and produce CH<sub>3<\/sub>CO<sub>2<\/sub>H in the process. Thus, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] decreases and [CH<sub>3<\/sub>CO<sub>2<\/sub>H] increases.<\/p>\n<p>(b) The added KCH<sub>3<\/sub>CO<sub>2<\/sub> will increase the concentration of [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] which will react with [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and produce CH<sub>3<\/sub>CO<sub>2<\/sub> H in the process. Thus, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] decreases slightly and [CH<sub>3<\/sub>CO<sub>2<\/sub>H] increases.<\/p>\n<p>(c) The added NaCl will have no effect on the concentration of the ions.<\/p>\n<p>(d) The added KOH will produce OH<sup>\u2212<\/sup> ions, which will react with the [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex], thus reducing [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex]. Some additional CH<sub>3<\/sub>CO<sub>2<\/sub>H will dissociate, producing [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] ions in the process. Thus, [CH<sub>3<\/sub>CO<sub>2<\/sub>H] decreases slightly and [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] increases.<\/p>\n<p>(e) The added CH<sub>3<\/sub>CO<sub>2<\/sub>H will increase its concentration, causing more of it to dissociate and producing more [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] and [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] in the process. Thus, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] increases slightly and [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] increases.<\/p>\n<p>12.\u00a0The reaction and equilibrium constant are:<\/p>\n<p>[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The equilibrium expression is:<\/p>\n<p>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The initial concentrations of NH<sub>3<\/sub> and [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] are 0.20 <em data-effect=\"italics\">M<\/em> and 0.40 <em data-effect=\"italics\">M<\/em>, respectively. The equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm117581744\" 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 (M). The second column has the header of \u201c[ N H subscript 3 ] [ N H subscript 4 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, negative sign x, 0.20 minus sign x. The second column has the following: 0.40, x, 0.40 plus sign x. The third column has the following 0, x, x.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5461\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044629\/CNX_Chem_14_06_ICETable7_img.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 (M). The second column has the header of \u201c[ N H subscript 3 ] [ N H subscript 4 superscript positive sign ] [ O H superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, negative sign x, 0.20 minus sign x. The second column has the following: 0.40, x, 0.40 plus sign x. The third column has the following 0, x, x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.20 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.20 and (0.40 + <em data-effect=\"italics\">x<\/em>) \u2248 0.40, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.40+x\\right)\\left(x\\right)}{\\left(0.20-x\\right)}\\approx \\frac{\\left(0.40\\right)\\left(x\\right)}{0.20}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 9.00 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.20 and 0.40, our assumptions are correct. Therefore, [OH<sup>\u2212<\/sup>] = 9.00 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Thus:<\/p>\n<p>pOH = \u2212log(9.00 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>) = 5.046<\/p>\n<p>pH = 14.000 \u2212 pOH = 14.000 \u2212 5.046 = 8.954 = 8.95<\/p>\n<p>14.\u00a0The reaction and equilibrium constant are:<\/p>\n<p>[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The equilibrium expression is:<\/p>\n<p>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Let <em data-effect=\"italics\">x<\/em> be the concentration of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex]. The hydronium ion concentration at equilibrium is:<\/p>\n<p>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 10<sup>\u2212pH<\/sup> = 10<sup>\u22125.00<\/sup> = 1.00 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm35109248\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.50, negative sign x, 0.50 minus sign x. The second column has the following: 0, x, x equals 1.0 times 10 to the negative fifth power. The third column has the following: x, x, x plus sign x.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5462\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044758\/CNX_Chem_14_06_ICETable9_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.50, negative sign x, 0.50 minus sign x. The second column has the following: 0, x, x equals 1.0 times 10 to the negative fifth power. The third column has the following: x, x, x plus sign x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em data-effect=\"italics\">x<\/em> + <em data-effect=\"italics\">x<\/em>) \u2248 <em data-effect=\"italics\">x<\/em>, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(1.0\\times {10}^{-5}\\right)\\left(x+x\\right)}{\\left(0.50 - 1.0\\times {10}^{-5}\\right)}\\approx \\frac{\\left(1.0\\times {10}^{-5}\\right)\\left(x\\right)}{0.50}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 0.900 <em data-effect=\"italics\">M<\/em>. Because <em data-effect=\"italics\">x<\/em> is less than 5% of this value, our assumption is correct. Therefore, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right][\/latex] = 0.900 <em data-effect=\"italics\">M<\/em>. Using the molar mass of NaC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>\u20223H<sub>2<\/sub>O (136.080 \/mol) and the volume gives the mass required:<\/p>\n<p>[latex]\\frac{0.900\\text{mol}}{1\\text{L}}\\times 0.300\\text{L}\\times \\frac{136.080\\text{g}}{1\\text{mol}}=36.7=37\\text{g}\\left(0.27\\text{mol}\\right)[\/latex]<\/p>\n<p>16.\u00a0(a) The reaction and equilibrium constant are:<\/p>\n<p>[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The equilibrium expression is:<\/p>\n<p>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The molar mass of NH<sub>4<\/sub>Cl is 53.4912 g\/mol. The moles of NH<sub>4<\/sub>Cl are: [latex]\\frac{5.36\\text{g}}{53.4912\\text{g}{\\text{mol}}^{-1}}=0.1002\\text{mol}[\/latex]<\/p>\n<p>Assume 0.500 L of each solution is present The total volume is thus 1.000 L. The initial concentrations of the ions is obtained using <em data-effect=\"italics\">M<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">M<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub>, or:<\/p>\n<p>[latex]\\begin{array}{l}\\\\ \\\\ \\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]={M}_{1}\\times \\frac{{V}_{1}}{{V}_{2}}=\\left(0.200\\right)\\times \\frac{0.500\\text{L}}{1.000\\text{L}}=0.100M\\\\ \\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]={M}_{1}\\times \\frac{{V}_{1}}{{V}_{2}}=\\left(0.600\\right)\\times \\frac{0.500\\text{L}}{1.000\\text{L}}=0.300M\\end{array}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations of this system can be written as follows:<span id=\"fs-idp20201888\" 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 ( M ). The second column has the header of \u201c[ C H subscript C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.100, negative sign x, 0.100 negative sign x. The second column is the following: 0, x, x. The third column has the following: 0.300, x, 0.300 plus sign x.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5463\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044850\/CNX_Chem_14_06_ICETable11_img.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 ( M ). The second column has the header of \u201c[ C H subscript C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.100, negative sign x, 0.100 negative sign x. The second column is the following: 0, x, x. The third column has the following: 0.300, x, 0.300 plus sign x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.100 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.100 and (0.300 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.300, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(x\\right)\\left(0.300+x\\right)}{\\left(0.100-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.300\\right)}{0.100}=1.80\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 6.000 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.100 and 0.300, our assumptions are correct. Therefore [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 6.000 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>:<\/p>\n<p>pH = \u2212log(6.000 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>) = 5.2218 = 5.222;<\/p>\n<p>(b) The solution is acidic.<\/p>\n<p>(c) Assume that the added H<sup>+<\/sup> reacts completely with an equal amount of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex], forming an equal amount of CH<sub>3<\/sub>CO<sub>2<\/sub>H in the process. The moles of H<sup>+<\/sup> added equal 0.034 <em data-effect=\"italics\">M<\/em> [latex]\\times[\/latex] 0.00300 L = 1.02 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> mol. For the acetic acid, the initial moles present equal 0.2000 <em data-effect=\"italics\">M<\/em> [latex]\\times[\/latex] 0.500 L = 0.1000 mol, and for acetate ion, 0.600 <em data-effect=\"italics\">M<\/em> [latex]\\times[\/latex] 0.500 L = 0.3000 mol. Thus:<\/p>\n<p>mol CH<sub>3<\/sub>CO<sub>2<\/sub>H = 0.1000 + 1.02 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> = 0.1001 mol<\/p>\n<p>[latex]\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}=0.3000 - 1.02\\times {10}^{-4}=0.2999\\text{mol}[\/latex]<\/p>\n<p>Final volume = 1.000 L + 3.00 [latex]\\times[\/latex] 10<sup>\u22123<\/sup> L = 1.0030 L<\/p>\n<p>The initial concentrations are therefore:<\/p>\n<p>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\frac{0.1001\\text{mol}}{1.0030\\text{L}}=0.09980M[\/latex]<\/p>\n<p>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]=\\frac{0.2999\\text{mol}}{1.0030\\text{L}}=0.2990M[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm165224336\" 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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.09980, negative x, 0.09980 minus sign x. The second column has the following: 0, x, x. The third column has the following: .2990, x, 0.2992 plus sign x.\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5465\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09044926\/CNX_Chem_14_06_ICETable12_img.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 ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 3 O superscript positive sign ] [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.09980, negative x, 0.09980 minus sign x. The second column has the following: 0, x, x. The third column has the following: .2990, x, 0.2992 plus sign x.\" width=\"975\" height=\"264\" \/><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.09980 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.09980 and (0.2990 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.2990, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\frac{\\left(x\\right)\\left(0.2990+x\\right)}{\\left(0.09980-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.2990\\right)}{0.09980}=1.80\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 6.008 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>. Because this value is less than 5% of both 0.09980 and 0.2990, our assumptions are correct. Therefore, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 6.008 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p>pH = \u2212log(6.008 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>) = 5.2213 = 5.221<\/p>\n<p>18.\u00a0To prepare the best buffer for a weak acid HA and its salt, the ratio [latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{{K}_{\\text{a}}}[\/latex] should be as close to 1 as possible for effective buffer action. The [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] concentration in a buffer of pH 3.1 is [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 10<sup>\u22123.1<\/sup> = 7.94 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<p>We can now solve for <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of the best acid as follows:<\/p>\n<p>[latex]\\begin{array}{l}\\\\ \\\\ \\\\ \\\\ \\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{{K}_{\\text{a}}}=1\\\\ {K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{1}=7.94\\times {10}^{-4}\\end{array}[\/latex]<\/p>\n<p>In Table 1 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>, the acid with the closest <em data-effect=\"italics\">K<\/em><sub>a<\/sub> to 7.94 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> is HF, with a <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of 7.2 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>.<\/p>\n<p>20.\u00a0For buffers with pHs &gt; 7, you should use a weak base and its salt. The most effective buffer will have a ratio [latex]\\frac{\\left[{\\text{OH}}^{\\text{-}}\\right]}{{K}_{\\text{b}}}[\/latex] that is as close to 1 as possible. The pOH of the buffer is 14.00 \u2212 10.65 = 3.35. Therefore, [OH<sup>\u2212<\/sup>] is [OH<sup>\u2212<\/sup>] = 10<sup>\u2212pOH<\/sup> = 10<sup>\u22123.35<\/sup> = 4.467 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em data-effect=\"italics\">M<\/em>.<\/p>\n<p>We can now solve for <em data-effect=\"italics\">K<\/em><sub>b<\/sub> of the best base as follows:<\/p>\n<p>[latex]\\frac{\\left[{\\text{OH}}^{\\text{-}}\\right]}{{K}_{\\text{b}}}=1[\/latex]<\/p>\n<p><em data-effect=\"italics\">K<\/em><sub>b<\/sub> = [OH<sup>\u2212<\/sup>] = 4.47 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><\/p>\n<p>In Table 2\u00a0of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>, the base with the closest <em data-effect=\"italics\">K<\/em><sub>b<\/sub> to 4.47 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> is CH<sub>3<\/sub>NH<sub>2<\/sub>, with a <em data-effect=\"italics\">K<\/em><sub>b<\/sub> = 4.4 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>.<\/p>\n<p>22. The molar mass of sodium saccharide is 205.169 g\/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:<\/p>\n<p>[latex]2.00\\times {10}^{-3}\\text{g}\\times \\frac{1\\text{mol}}{205.169\\text{g}}=9.75\\times {10}^{-6}\\text{mol}[\/latex]<\/p>\n<p>This ionizes initially to form saccharin ions, A<sup>\u2212<\/sup>, with:<\/p>\n<p>[latex]\\left[{\\text{A}}^{\\text{-}}\\right]=\\frac{9.75\\times {10}^{-6}\\text{mol}}{0.250\\text{L}}=3.9\\times {10}^{-5}M[\/latex]<\/p>\n<p>but A<sup>\u2212<\/sup> reacts with water:<\/p>\n<p>[latex]\\begin{array}{l}{\\text{A}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HA}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\\\ {K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{1.0\\times {10}^{-14}}{2.1\\times {10}^{-12}}=4.8\\times {10}^{-3}\\\\ =4.8\\times {10}^{-3}=\\frac{\\left[\\text{HA}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{A}}^{\\text{-}}\\right]}\\end{array}[\/latex]<\/p>\n<p>The pH of the solution is 5.48, so:<\/p>\n<p>pOH = 14.00 \u2212 5.48 = 8.52<\/p>\n<p>and<\/p>\n<p>[OH<sup>\u2212<\/sup>] = 10<sup>\u22128.52<\/sup> = 3.02 [latex]\\times[\/latex] 10<sup>\u22129<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<p>Because of the small size of <em data-effect=\"italics\">K<\/em><sub>b<\/sub>, almost all the A<sup>\u2212<\/sup> will be in the form of HA. Therefore:<\/p>\n<p>[latex]4.8\\times {10}^{-3}=\\frac{x\\left(3.02\\times {10}^{-9}\\right)}{3.9\\times {10}^{-5}-x}[\/latex]<\/p>\n<p>where<\/p>\n<p><em data-effect=\"italics\">x<\/em> \u2248 3.9 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em> = [HA] = [C<sub>7<\/sub>H<sub>4<\/sub>NSO<sub>3<\/sub>H]<\/p>\n<p>Consequently, [A<sup>\u2212<\/sup>] is extremely small. Therefore, solve for [A<sup>\u2212<\/sup>] from the equilibrium expression:<\/p>\n<p>[latex]\\left[{\\text{A}}^{\\text{-}}\\right]=\\frac{\\left[\\text{HA}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{{K}_{\\text{b}}}=\\frac{\\left(3.9\\times {10}^{-5}\\right)\\left(3.02\\times {10}^{-9}\\right)}{4.8\\times {10}^{-3}}=2.5\\times {10}^{-11}M=\\left[\\text{Na}\\left({\\text{C}}_{7}{\\text{H}}_{4}{\\text{NSO}}_{3}\\right)\\right][\/latex]<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<div id=\"fs-idp88494848\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">buffer capacity<br \/>\n<\/span><\/strong>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)<\/p>\n<\/div>\n<div id=\"fs-idp88495904\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">buffer<br \/>\n<\/span><\/strong>mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added<\/p>\n<\/div>\n<div id=\"fs-idm71412016\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">Henderson-Hasselbalch equation<br \/>\n<\/span><\/strong>equation used to calculate the pH of buffer solutions<\/p>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3521\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3521","chapter","type-chapter","status-publish","hentry"],"part":2988,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3521","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3521\/revisions"}],"predecessor-version":[{"id":5517,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3521\/revisions\/5517"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/parts\/2988"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3521\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/media?parent=3521"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapter-type?post=3521"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/contributor?post=3521"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/license?post=3521"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}