{"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":"2020-12-31T20:05:12","modified_gmt":"2020-12-31T20:05:12","slug":"buffers-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/buffers-2\/","title":{"raw":"Buffers","rendered":"Buffers"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\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\nA solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a\u00a0<strong>buffer<\/strong>.\u00a0Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure\u00a01). 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>aq<\/em>) + NH<sub>4<\/sub>Cl(<em>aq<\/em>)).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"745\"]<img src=\"https:\/\/openstax.org\/resources\/e16a8da01b6fa05343118e53d6db9e980b36916b\" 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=\"745\" height=\"265\" \/> Figure 1. (a) The unbuffered solution on the left and the buffered 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<h2>How Buffers Work<\/h2>\r\nTo illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, adding strong base to this solution will neutralize hydronium ion and shift the acetic acid ionization equilibrium to the right, partially restoring the decreased H<sub>3<\/sub>O<sup>+<\/sup>\u00a0concentration:\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\leftrightharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nLikewise, adding strong acid to this buffer solution will neutralize acetate ion, shifting the above ionization equilibrium right and returning [H<sub>3<\/sub>O<sup>+<\/sup>] to near its original value. Figure 2 provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.\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\u00a02. Buffering action in a mixture of acetic acid and acetate salt.[\/caption]\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0pH Changes in Buffered and Unbuffered Solutions<\/h3>\r\nAcetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.\r\n<p id=\"fs-idp41841344\">(a) Calculate the pH of an acetate buffer that is a mixture with 0.10\u00a0<em data-effect=\"italics\">M<\/em>\u00a0acetic acid and 0.10\u00a0<em data-effect=\"italics\">M<\/em>\u00a0sodium acetate.<\/p>\r\n<p id=\"fs-idm197032080\">(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.<\/p>\r\n<p id=\"fs-idm215059040\">(c) For comparison, calculate the pH after 1.0 mL of 0.10\u00a0<em data-effect=\"italics\">M<\/em>\u00a0NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.<\/p>\r\n\r\n<h4>Part A<\/h4>\r\n[reveal-answer q=\"547454\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"547454\"]\r\n\r\nTo determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):\r\n\r\n<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\r\n<em>Step 1. Determine the direction of change.<\/em>\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}{}^{-}\\left(aq\\right)[\/latex]\r\n\r\nWe look it up in <a href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>: <em>K<\/em><sub>a<\/sub> = 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>.\r\n\r\nWith [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right][\/latex] = 0.10 <em>M<\/em> and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = ~0 <em>M<\/em>, the reaction shifts to the right to form [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex].\r\n\r\n<em>Step 2. Determine<\/em> x <em>and equilibrium concentrations<\/em>. A table of changes and concentrations follows:<img src=\"https:\/\/openstax.org\/resources\/c6cd395413486614bd489c119fd3913a299faa7e\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c[ C H subscript 3 C O subscript 2 H ] [ H subscript 2 O ] equilibrium arrow H subscript 3 O superscript plus sign [ C H subscript 3 C O subscript 2 superscript negative sign ].\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.\" \/>\r\n\r\n<em>Step 3. Solve for x and the equilibrium concentrations.<\/em>\r\n<p style=\"text-align: center;\">[latex]x=1.8\\times {10}^{-5}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=0+x=1.8\\times {10}^{-5}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=-\\text{log}\\left(1.8\\times {10}^{-5}\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]=4.74[\/latex]<\/p>\r\n<em>Step 4. Check the work<\/em>. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, <em>Q<\/em> = <em>K<\/em><sub>a<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4>Part B<\/h4>\r\n[reveal-answer q=\"935154\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"935154\"]\r\n\r\nAdding strong acid will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.<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\" \/>\r\n\r\n<em>Step 1. Determine the moles of NaOH.<\/em> One milliliter (0.0010 L) of 0.10 <em>M<\/em> NaOH contains\r\n<p style=\"text-align: center;\">[latex]0.0010\\cancel{\\text{L}}\\times \\left(\\dfrac{0.10\\text{mol NaOH}}{1\\cancel{\\text{L}}}\\right)=1.0\\times {10}^{-4}\\text{mol NaOH}[\/latex]<\/p>\r\n<em>Step 2. 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<p style=\"text-align: center;\">[latex]0.100\\cancel{\\text{L}}\\times \\left(\\dfrac{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]<\/p>\r\n<em>Step 3. 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<p style=\"text-align: center;\">[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]<\/p>\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<p style=\"text-align: center;\">[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]<\/p>\r\n<em>Step 4. 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<p style=\"text-align: center;\">[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\dfrac{9.9\\times {10}^{-3}\\text{mol}}{0.101\\text{L}}=0.098M[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{NaCH}}_{3}{\\text{CO}}_{2}\\right]=\\dfrac{1.01\\times {10}^{-2}\\text{mol}}{0.101\\text{L}}=0.100M[\/latex]<\/p>\r\nNow we calculate the pH after the intermediate solution, which is 0.098 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.100 <em>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\" \/>This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution (F).\r\n\r\n[\/hidden-answer]\r\n<h4>Part C<\/h4>\r\n[reveal-answer q=\"892953\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"892953\"]\r\n\r\nThis 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>-<em>M<\/em> solution of HCl has the same hydronium ion concentration as the 0.10-<em>M<\/em> solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:\r\n<p style=\"text-align: center;\">[latex]0.100\\text{L}\\times \\left(\\dfrac{1.8\\times {10}^{-5}\\text{mol HCl}}{1\\text{L}}\\right)=1.8\\times {10}^{-6}\\text{mol HCl}[\/latex]<\/p>\r\nAs shown in part (b), 1 mL of 0.10 <em>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:\r\n<p style=\"text-align: center;\">[latex]\\left(1.0\\times {10}^{-4}\\right)-\\left(1.8\\times {10}^{-6}\\right)=9.8\\times {10}^{-5}M[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>The concentration of NaOH is:\u00a0[latex]\\dfrac{9.8\\times {10}^{-5}M\\text{NaOH}}{0.101\\text{L}}=9.7\\times {10}^{-4}M[\/latex]<\/li>\r\n \t<li>The pOH of this solution is:\u00a0[latex]\\text{pOH}=-\\text{log}\\left[{\\text{OH}}^{-}\\right]=-\\text{log}\\left(9.7\\times {10}^{-4}\\right)=3.01[\/latex]<\/li>\r\n \t<li>The pH is:\u00a0[latex]\\text{pH}=14.00-\\text{pOH}=10.99[\/latex]<\/li>\r\n<\/ul>\r\nThe 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).\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm159189968\">Check Your Learning<\/h4>\r\nShow that adding 1.0 mL of 0.10 <em>M<\/em> HCl changes the pH of 100 mL of a 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em>M<\/em> HCl solution from 4.74 to 3.00.\r\n[reveal-answer q=\"337851\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"337851\"]\r\n\r\nInitial pH of 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em>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\r\n\r\nMoles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] in 100 mL 1.8 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em>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>\r\n\r\nMoles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] added by addition of 1.0 mL of 0.10 <em>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>M<\/em> HCl:\r\n\r\n[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=-\\text{log}\\left(\\dfrac{\\text{total moles}{\\text{ H}}_{3}{\\text{O}}^{\\text{+}}}{\\text{total volume}}\\right)=-\\text{log}\\left(\\dfrac{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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIf 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.\r\n<h2>Buffer Capacity<\/h2>\r\nBuffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure\u00a03). 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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1300\"]<img src=\"https:\/\/openstax.org\/resources\/3240e8dc0e74de8edb102fc89e62d8a6be06cd7d\" alt=\"Figure\u00a03. 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=\"1300\" height=\"362\" \/> 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\r\nThe <strong>buffer capacity<\/strong> is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 <em>M<\/em> in acetic acid and 1.0 <em>M<\/em> in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 <em>M<\/em> in acetic acid and 0.10 <em>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.\r\n<h2>Selection of Suitable Buffer Mixtures<\/h2>\r\nThere are two useful rules of thumb for selecting buffer mixtures:\r\n<ol>\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\u00a04\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=\"700\"]<img class=\" 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=\"700\" height=\"500\" \/> <strong>Figure 4.<\/strong> 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.\u00a0Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.[\/caption]<\/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\nBlood 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}{}^{-}[\/latex]. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\nWhen an excess of the hydroxide ion is present, it is removed by the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\longrightarrow {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\nThe added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H<sub>3<\/sub>O<sup>+<\/sup>\u00a0is converted to H<sub>2<\/sub>CO<sub>3<\/sub>\u00a0and OH<sup>-<\/sup>\u00a0is converted to HCO<sub>3<\/sub><sup>-<\/sup>). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.\r\n<h2>The Henderson-Hasselbalch Equation<\/h2>\r\nThe ionization-constant expression for a solution of a weak acid can be written as:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{A}}^{-}\\right]}{\\text{[HA]}}[\/latex]<\/p>\r\nRearranging to solve for [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], we get:\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]={K}_{\\text{a}}\\times \\dfrac{\\text{[HA]}}{\\left[{\\text{A}}^{-}\\right]}[\/latex]<\/p>\r\nTaking the negative logarithm of both sides of this equation, we arrive at:\r\n<p style=\"text-align: center;\">[latex]-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=-\\text{log}{K}_{\\text{a}}\\text{- log}\\dfrac{\\left[\\text{HA}\\right]}{\\left[{\\text{A}}^{-}\\right]}[\/latex],<\/p>\r\nwhich can be written as\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\dfrac{\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/p>\r\nwhere p<em>K<\/em><sub>a<\/sub> is the negative of the common logarithm of the ionization constant of the weak acid (p<em>K<\/em><sub>a<\/sub> = \u2212log <em>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>Henderson-Hasselbalch equation<\/strong>, to calculate the pH of buffer solutions. It is important to note that the \u201c<em>x<\/em> is small\u201d assumption must be valid to use this equation.\r\n<div class=\"textbox shaded\">\r\n<h3>Lawrence Joseph Henderson and Karl Albert Hasselbalch<\/h3>\r\nLawrence Joseph <strong>Henderson<\/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.\r\n\r\nIn 1916, Karl Albert <strong>Hasselbalch<\/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.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3><span style=\"font-size: 16px;\">Medicine: The Buffer System in Blood<\/span><\/h3>\r\nThe normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:\r\n<p style=\"text-align: center;\">[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}{}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\r\nThe concentration of carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub> is approximately 0.0012 <em>M<\/em>, and the concentration of the hydrogen carbonate ion, [latex]{\\text{HCO}}_{3}{}^{-}[\/latex], is around 0.024 <em>M<\/em>. Using the Henderson-Hasselbalch equation and the p<em>K<\/em><sub>a<\/sub> of carbonic acid at body temperature, we can calculate the pH of blood:\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\dfrac{\\left[\\text{base}\\right]}{\\left[\\text{acid}\\right]}=6.1+\\text{log}\\dfrac{0.024}{0.0012}=7.4[\/latex]<\/p>\r\nThe fact that the H<sub>2<\/sub>CO<sub>3<\/sub> concentration is significantly lower than that of the [latex]{\\text{HCO}}_{3}{}^{-}[\/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.\r\n\r\nLactic 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}{}^{-}[\/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.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nSolutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]\\text{p}K_{\\text{a}}=\u2212\\text{log}K_{\\text{a}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{p}K_{\\text{b}}=\u2212\\text{log}K_{\\text{b}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\dfrac{\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>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>M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.030 <em>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}{}^{-}\\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>M<\/em> HNO<sub>2<\/sub> and 0.030 <em>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}{}^{-}\\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>M<\/em> CH<sub>3<\/sub>NH<sub>2<\/sub> and 0.130 <em>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}}^{-}\\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>M<\/em> NH<sub>3<\/sub> and 0.78 <em>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}}^{-}\\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>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>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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\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>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>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<\/ol>\r\n[reveal-answer q=\"768271\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"768271\"]\r\n\r\n2.\u00a0Excess [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] is removed primarily by the reaction\u00a0[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}\\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\u00a0[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]\r\n\r\n4. The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.25<\/td>\r\n<td>0<\/td>\r\n<td>0.030<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.25\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>0.030 +\u00a0<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.25 \u2212 <em>x<\/em>) \u2248 0.25 and (0.030 \u2212 <em>x<\/em>) \u2248 0.030, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>\r\nSolving for <em>x<\/em> gives 1.50 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em>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>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}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]; p<em>K<\/em><sub>a<\/sub> = \u2212log(<em>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>M<\/em>; [A<sup>\u2212<\/sup>] \u2248 [NaCH<sub>3<\/sub>CO<sub>2<\/sub>] = 0.030 <em>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>M<\/em> = 10<sup>\u22123.82<\/sup><em>M<\/em> = 1.5 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em>M<\/em>\r\n\r\n6.\u00a0The equilibrium expression is:\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[CH<sub>3<\/sub>NH<sub>2<\/sub>]<\/th>\r\n<th>[CH<sub>3<\/sub>NH<sub>3<\/sub><sup>+<\/sup>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.125<\/td>\r\n<td>0.130<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<i>x<\/i><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.125\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td>0.130 +\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.125 \u2212 <em>x<\/em>) \u2248 0.125 and (0.130 \u2212 <em>x<\/em>) \u2248 0.130, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>\r\nSolving for <em>x<\/em> gives 4.23 [latex]\\times [\/latex] 10<sup>\u22124<\/sup><em>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>M<\/em>.\r\n\r\n8. The reaction and equilibrium constant are\u00a0[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}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{b}}=1.8\\times {10}^{-4}[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nLet <em>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:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[NH<sub>3<\/sub>]<\/th>\r\n<th>[NH<sub>4<\/sub><sup>+<\/sup>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.200<\/td>\r\n<td>0.78<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>\u00a0x + x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.200\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x + x<\/em><\/td>\r\n<td><em>x<\/em> = 1.0 \u00d7\u00a010<sup>\u22125<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em>x<\/em> + <em>x<\/em>) \u2248 <em>x<\/em>, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>\r\nSolving for <em>x<\/em> gives 0.360 <em>M<\/em>. Because <em>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>M<\/em>.\r\n\r\n10.\u00a0The reaction and equilibrium constant are\u00a0[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}{}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>(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}{}^{-}[\/latex] and produce CH<sub>3<\/sub>CO<sub>2<\/sub>H in the process. Thus, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right][\/latex] decreases and [CH<sub>3<\/sub>CO<sub>2<\/sub>H] increases.<\/li>\r\n \t<li>(b) The added KCH<sub>3<\/sub>CO<sub>2<\/sub> will increase the concentration of [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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.<\/li>\r\n \t<li>(c) The added NaCl will have no effect on the concentration of the ions.<\/li>\r\n \t<li>(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}{}^{-}\\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}{}^{-}\\right][\/latex] increases.<\/li>\r\n \t<li>(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}{}^{-}\\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}{}^{-}\\right][\/latex] increases.<\/li>\r\n<\/ol>\r\n12.\u00a0The reaction and equilibrium constant are:\u00a0[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}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe equilibrium expression is:\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>M<\/em> and 0.40 <em>M<\/em>, respectively. The equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[NH<sub>3<\/sub>]<\/th>\r\n<th>NH<sub>4<\/sub><sup>+<\/sup>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.20<\/td>\r\n<td>0.40<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td><em>\u2212x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.20\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td>0.40 +\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.20 \u2212 <em>x<\/em>) \u2248 0.20 and (0.40 + <em>x<\/em>) \u2248 0.40, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>\r\nSolving for <em>x<\/em> gives 9.00 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em>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>M<\/em>. Thus:\r\n<ul>\r\n \t<li>pOH = \u2212log(9.00 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>) = 5.046<\/li>\r\n \t<li>pH = 14.000 \u2212 pOH = 14.000 \u2212 5.046 = 8.954 = 8.95<\/li>\r\n<\/ul>\r\n14.\u00a0The reaction and equilibrium constant are:\u00a0[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}{}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>x<\/em> be the concentration of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}[\/latex]. The hydronium ion concentration at equilibrium is\u00a0[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>M<\/em>\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.50<\/td>\r\n<td>0<\/td>\r\n<td><i>x<\/i><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.50\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em> = 10 \u00d7 10<sup>\u22125<\/sup><\/td>\r\n<td><em>x + x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em>x<\/em> + <em>x<\/em>) \u2248 <em>x<\/em>, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>\r\nSolving for <em>x<\/em> gives 0.900 <em>M<\/em>. Because <em>x<\/em> is less than 5% of this value, our assumption is correct. Therefore, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right][\/latex] = 0.900 <em>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<p style=\"text-align: center;\">[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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>A buffer solution is prepared from equal volumes of 0.200 <em>M<\/em> acetic acid and 0.600 <em>M<\/em> sodium acetate. Use 1.80 \u00d7 10<sup>\u22125<\/sup> as <em>K<\/em><sub>a<\/sub> for acetic acid.\r\n<ol style=\"list-style-type: lower-alpha;\">\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>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>M<\/em> NaOH and the resulting solution\u00a0diluted to 0.100 L.\r\n<ol style=\"list-style-type: lower-alpha;\">\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>M<\/em> HCl is added to the solution?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"817908\"]Show Solution to Question 1[\/reveal-answer]\r\n[hidden-answer a=\"817908\"]\r\n<h4>Part A<\/h4>\r\nThe reaction and equilibrium constant are\u00a0[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}{}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>M<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub> = <em>M<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub>, or:\r\n<p style=\"text-align: center;\">[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}{}^{-}\\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>\r\nThe initial and equilibrium concentrations of this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.100<\/td>\r\n<td>0<\/td>\r\n<td>0.300<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<i>x<\/i><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.100\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>0.300 +\u00a0<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.100 \u2212 <em>x<\/em>) \u2248 0.100 and (0.300 \u2212 <em>x<\/em>) \u2248 0.300, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>\r\nSolving for <em>x<\/em> gives 6.000 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em>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>M<\/em>:\r\n<p style=\"text-align: center;\">pH = \u2212log(6.000 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>) = 5.2218 = 5.222;<\/p>\r\n\r\n<h4>Part B<\/h4>\r\nThe solution is acidic.\r\n<h4>Part C<\/h4>\r\nAssume that the added H<sup>+<\/sup> reacts completely with an equal amount of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}[\/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>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>M<\/em> [latex]\\times [\/latex] 0.500 L = 0.1000 mol, and for acetate ion, 0.600 <em>M<\/em> [latex]\\times [\/latex] 0.500 L = 0.3000 mol. Thus:\r\n<p style=\"text-align: center;\">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>\r\n<p style=\"text-align: center;\">[latex]\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}=0.3000 - 1.02\\times {10}^{-4}=0.2999\\text{mol}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Final volume = 1.000 L + 3.00 [latex]\\times [\/latex] 10<sup>\u22123<\/sup> L = 1.0030 L<\/p>\r\nThe initial concentrations are therefore:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\frac{0.1001\\text{mol}}{1.0030\\text{L}}=0.09980M[\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right]=\\frac{0.2999\\text{mol}}{1.0030\\text{L}}=0.2990M[\/latex]<\/li>\r\n<\/ul>\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.09980<\/td>\r\n<td>0<\/td>\r\n<td>0.2990<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<i>x<\/i><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.09980\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>0.2990\u00a0+\u00a0<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.09980 \u2212 <em>x<\/em>) \u2248 0.09980 and (0.2990 \u2212 <em>x<\/em>) \u2248 0.2990, gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>\r\nSolving for <em>x<\/em> gives 6.008 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em>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>M<\/em>.\r\n<p style=\"text-align: center;\">pH = \u2212log(6.008 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>) = 5.2213 = 5.221<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Which acid in Table\u00a01 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">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\u00a01 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>\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\u00a02 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>\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 Table\u00a02 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/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>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>K<\/em><sub>1<\/sub> = 8.5 [latex]\\times [\/latex] 10<sup>\u22125<\/sup>, <em>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[reveal-answer q=\"469090\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"469090\"]\r\n\r\n1.\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>M<\/em>\r\n\r\nWe can now solve for <em>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\u00a01 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>, the acid with the closest <em>K<\/em><sub>a<\/sub> to 7.94 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> is HF, with a <em>K<\/em><sub>a<\/sub> of 7.2 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>.\r\n\r\n3.\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}}^{-}\\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>M<\/em>.\r\n\r\nWe can now solve for <em>K<\/em><sub>b<\/sub> of the best base as follows:\r\n\r\n[latex]\\frac{\\left[{\\text{OH}}^{-}\\right]}{{K}_{\\text{b}}}=1[\/latex]\r\n\r\n<em>K<\/em><sub>b<\/sub> = [OH<sup>\u2212<\/sup>] = 4.47 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>\r\n\r\nIn Table\u00a02\u00a0of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>, the base with the closest <em>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>K<\/em><sub>b<\/sub> = 4.4 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>.\r\n\r\n5. 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:\u00a0[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<p style=\"text-align: center;\">[latex]\\left[{\\text{A}}^{-}\\right]=\\frac{9.75\\times {10}^{-6}\\text{mol}}{0.250\\text{L}}=3.9\\times {10}^{-5}M[\/latex]<\/p>\r\nbut A<sup>\u2212<\/sup> reacts with water:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{A}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HA}\\left(aq\\right)+{\\text{OH}}^{-}\\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}}^{-}\\right]}{\\left[{\\text{A}}^{-}\\right]}\\end{array}[\/latex]<\/p>\r\nThe pH of the solution is 5.48, so\u00a0pOH = 14.00 \u2212 5.48 = 8.52,\u00a0and\u00a0[OH<sup>\u2212<\/sup>] = 10<sup>\u22128.52<\/sup> = 3.02 [latex]\\times [\/latex] 10<sup>\u22129<\/sup><em>M<\/em>\r\n\r\nBecause of the small size of <em>K<\/em><sub>b<\/sub>, almost all the A<sup>\u2212<\/sup> will be in the form of HA. Therefore,\u00a0[latex]4.8\\times {10}^{-3}=\\frac{x\\left(3.02\\times {10}^{-9}\\right)}{3.9\\times {10}^{-5}-x}[\/latex],\u00a0where\u00a0<em>x<\/em> \u2248 3.9 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em>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<p style=\"text-align: center;\">[latex]\\left[{\\text{A}}^{-}\\right]=\\frac{\\left[\\text{HA}\\right]\\left[{\\text{OH}}^{-}\\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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>buffer capacity: <\/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<strong>buffer: <\/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<strong>Henderson-Hasselbalch equation: <\/strong>equation used to calculate the pH of buffer solutions","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\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>A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a\u00a0<strong>buffer<\/strong>.\u00a0Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure\u00a01). 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>aq<\/em>) + NH<sub>4<\/sub>Cl(<em>aq<\/em>)).<\/p>\n<div style=\"width: 755px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/e16a8da01b6fa05343118e53d6db9e980b36916b\" 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=\"745\" height=\"265\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. (a) The unbuffered solution on the left and the buffered 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<h2>How Buffers Work<\/h2>\n<p>To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, adding strong base to this solution will neutralize hydronium ion and shift the acetic acid ionization equilibrium to the right, partially restoring the decreased H<sub>3<\/sub>O<sup>+<\/sup>\u00a0concentration:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\leftrightharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>Likewise, adding strong acid to this buffer solution will neutralize acetate ion, shifting the above ionization equilibrium right and returning [H<sub>3<\/sub>O<sup>+<\/sup>] to near its original value. Figure 2 provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer&#8217;s conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.<\/p>\n<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\u00a02. Buffering action in a mixture of acetic acid and acetate salt.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0pH Changes in Buffered and Unbuffered Solutions<\/h3>\n<p>Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.<\/p>\n<p id=\"fs-idp41841344\">(a) Calculate the pH of an acetate buffer that is a mixture with 0.10\u00a0<em data-effect=\"italics\">M<\/em>\u00a0acetic acid and 0.10\u00a0<em data-effect=\"italics\">M<\/em>\u00a0sodium acetate.<\/p>\n<p id=\"fs-idm197032080\">(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.<\/p>\n<p id=\"fs-idm215059040\">(c) For comparison, calculate the pH after 1.0 mL of 0.10\u00a0<em data-effect=\"italics\">M<\/em>\u00a0NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.<\/p>\n<h4>Part A<\/h4>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q547454\">Show Solution<\/span><\/p>\n<div id=\"q547454\" class=\"hidden-answer\" style=\"display: none\">\n<p>To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):<\/p>\n<p><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\" \/><\/p>\n<p><em>Step 1. Determine the direction of change.<\/em><\/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}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>We look it up in <a href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>: <em>K<\/em><sub>a<\/sub> = 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>.<\/p>\n<p>With [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right][\/latex] = 0.10 <em>M<\/em> and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = ~0 <em>M<\/em>, the reaction shifts to the right to form [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex].<\/p>\n<p><em>Step 2. Determine<\/em> x <em>and equilibrium concentrations<\/em>. A table of changes and concentrations follows:<img decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/c6cd395413486614bd489c119fd3913a299faa7e\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c&#091; C H subscript 3 C O subscript 2 H &#093; &#091; H subscript 2 O &#093; equilibrium arrow H subscript 3 O superscript plus sign &#091; C H subscript 3 C O subscript 2 superscript negative sign &#093;.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.\" \/><\/p>\n<p><em>Step 3. Solve for x and the equilibrium concentrations.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]x=1.8\\times {10}^{-5}M[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=0+x=1.8\\times {10}^{-5}M[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=-\\text{log}\\left(1.8\\times {10}^{-5}\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]=4.74[\/latex]<\/p>\n<p><em>Step 4. Check the work<\/em>. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, <em>Q<\/em> = <em>K<\/em><sub>a<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4>Part B<\/h4>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q935154\">Show Solution<\/span><\/p>\n<div id=\"q935154\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding strong acid will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.<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&#091; C H subscript 3 C O subscript 2 H &#093;.\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&#091; C H subscript 3 C O subscript 2 &#093;.\u201d This rectangle points right to the rectangle labeled \u201c&#091; C H subscript 3 C O subscript 2 H &#093;.\u201d\" width=\"1000\" height=\"475\" \/><\/p>\n<p><em>Step 1. Determine the moles of NaOH.<\/em> One milliliter (0.0010 L) of 0.10 <em>M<\/em> NaOH contains<\/p>\n<p style=\"text-align: center;\">[latex]0.0010\\cancel{\\text{L}}\\times \\left(\\dfrac{0.10\\text{mol NaOH}}{1\\cancel{\\text{L}}}\\right)=1.0\\times {10}^{-4}\\text{mol NaOH}[\/latex]<\/p>\n<p><em>Step 2. Determine the moles of CH<sub>2<\/sub>CO<sub>2<\/sub>H.<\/em> Before reaction, 0.100 L of the buffer solution contains<\/p>\n<p style=\"text-align: center;\">[latex]0.100\\cancel{\\text{L}}\\times \\left(\\dfrac{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]<\/p>\n<p><em>Step 3. 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<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\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<p style=\"text-align: center;\">[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]<\/p>\n<p><em>Step 4. 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:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\dfrac{9.9\\times {10}^{-3}\\text{mol}}{0.101\\text{L}}=0.098M[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{NaCH}}_{3}{\\text{CO}}_{2}\\right]=\\dfrac{1.01\\times {10}^{-2}\\text{mol}}{0.101\\text{L}}=0.100M[\/latex]<\/p>\n<p>Now we calculate the pH after the intermediate solution, which is 0.098 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.100 <em>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\" \/>This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution (F).<\/p>\n<\/div>\n<\/div>\n<h4>Part C<\/h4>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q892953\">Show Solution<\/span><\/p>\n<div id=\"q892953\" class=\"hidden-answer\" style=\"display: none\">\n<p>This 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>&#8211;<em>M<\/em> solution of HCl has the same hydronium ion concentration as the 0.10-<em>M<\/em> solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:<\/p>\n<p style=\"text-align: center;\">[latex]0.100\\text{L}\\times \\left(\\dfrac{1.8\\times {10}^{-5}\\text{mol HCl}}{1\\text{L}}\\right)=1.8\\times {10}^{-6}\\text{mol HCl}[\/latex]<\/p>\n<p>As shown in part (b), 1 mL of 0.10 <em>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<p style=\"text-align: center;\">[latex]\\left(1.0\\times {10}^{-4}\\right)-\\left(1.8\\times {10}^{-6}\\right)=9.8\\times {10}^{-5}M[\/latex]<\/p>\n<ul>\n<li>The concentration of NaOH is:\u00a0[latex]\\dfrac{9.8\\times {10}^{-5}M\\text{NaOH}}{0.101\\text{L}}=9.7\\times {10}^{-4}M[\/latex]<\/li>\n<li>The pOH of this solution is:\u00a0[latex]\\text{pOH}=-\\text{log}\\left[{\\text{OH}}^{-}\\right]=-\\text{log}\\left(9.7\\times {10}^{-4}\\right)=3.01[\/latex]<\/li>\n<li>The pH is:\u00a0[latex]\\text{pH}=14.00-\\text{pOH}=10.99[\/latex]<\/li>\n<\/ul>\n<p>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<\/div>\n<\/div>\n<h4 id=\"fs-idm159189968\">Check Your Learning<\/h4>\n<p>Show that adding 1.0 mL of 0.10 <em>M<\/em> HCl changes the pH of 100 mL of a 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em>M<\/em> HCl solution from 4.74 to 3.00.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q337851\">Show Solution<\/span><\/p>\n<div id=\"q337851\" class=\"hidden-answer\" style=\"display: none\">\n<p>Initial pH of 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em>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<\/p>\n<p>Moles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] in 100 mL 1.8 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em>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>Moles of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] added by addition of 1.0 mL of 0.10 <em>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>M<\/em> HCl:<\/p>\n<p>[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=-\\text{log}\\left(\\dfrac{\\text{total moles}{\\text{ H}}_{3}{\\text{O}}^{\\text{+}}}{\\text{total volume}}\\right)=-\\text{log}\\left(\\dfrac{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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>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<h2>Buffer Capacity<\/h2>\n<p>Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure\u00a03). 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 style=\"width: 1310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/3240e8dc0e74de8edb102fc89e62d8a6be06cd7d\" alt=\"Figure\u00a03. 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=\"1300\" height=\"362\" \/><\/p>\n<p 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>The <strong>buffer capacity<\/strong> is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 <em>M<\/em> in acetic acid and 1.0 <em>M<\/em> in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 <em>M<\/em> in acetic acid and 0.10 <em>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<h2>Selection of Suitable Buffer Mixtures<\/h2>\n<p>There are two useful rules of thumb for selecting buffer mixtures:<\/p>\n<ol>\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\u00a04\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: 710px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5456\" class=\"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=\"700\" height=\"500\" \/><\/p>\n<p id=\"caption-attachment-5456\" class=\"wp-caption-text\"><strong>Figure 4.<\/strong> 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.\u00a0Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.<\/p>\n<\/div>\n<\/li>\n<li>Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.<\/li>\n<\/ol>\n<p>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}{}^{-}[\/latex]. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>When an excess of the hydroxide ion is present, it is removed by the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\longrightarrow {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H<sub>3<\/sub>O<sup>+<\/sup>\u00a0is converted to H<sub>2<\/sub>CO<sub>3<\/sub>\u00a0and OH<sup>&#8211;<\/sup>\u00a0is converted to HCO<sub>3<\/sub><sup>&#8211;<\/sup>). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.<\/p>\n<h2>The Henderson-Hasselbalch Equation<\/h2>\n<p>The ionization-constant expression for a solution of a weak acid can be written as:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{A}}^{-}\\right]}{\\text{[HA]}}[\/latex]<\/p>\n<p>Rearranging to solve for [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], we get:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]={K}_{\\text{a}}\\times \\dfrac{\\text{[HA]}}{\\left[{\\text{A}}^{-}\\right]}[\/latex]<\/p>\n<p>Taking the negative logarithm of both sides of this equation, we arrive at:<\/p>\n<p style=\"text-align: center;\">[latex]-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=-\\text{log}{K}_{\\text{a}}\\text{- log}\\dfrac{\\left[\\text{HA}\\right]}{\\left[{\\text{A}}^{-}\\right]}[\/latex],<\/p>\n<p>which can be written as<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\dfrac{\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/p>\n<p>where p<em>K<\/em><sub>a<\/sub> is the negative of the common logarithm of the ionization constant of the weak acid (p<em>K<\/em><sub>a<\/sub> = \u2212log <em>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>Henderson-Hasselbalch equation<\/strong>, to calculate the pH of buffer solutions. It is important to note that the \u201c<em>x<\/em> is small\u201d assumption must be valid to use this equation.<\/p>\n<div class=\"textbox shaded\">\n<h3>Lawrence Joseph Henderson and Karl Albert Hasselbalch<\/h3>\n<p>Lawrence Joseph <strong>Henderson<\/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>In 1916, Karl Albert <strong>Hasselbalch<\/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><span style=\"font-size: 16px;\">Medicine: The Buffer System in Blood<\/span><\/h3>\n<p>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 style=\"text-align: center;\">[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}{}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\n<p>The concentration of carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub> is approximately 0.0012 <em>M<\/em>, and the concentration of the hydrogen carbonate ion, [latex]{\\text{HCO}}_{3}{}^{-}[\/latex], is around 0.024 <em>M<\/em>. Using the Henderson-Hasselbalch equation and the p<em>K<\/em><sub>a<\/sub> of carbonic acid at body temperature, we can calculate the pH of blood:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\dfrac{\\left[\\text{base}\\right]}{\\left[\\text{acid}\\right]}=6.1+\\text{log}\\dfrac{0.024}{0.0012}=7.4[\/latex]<\/p>\n<p>The fact that the H<sub>2<\/sub>CO<sub>3<\/sub> concentration is significantly lower than that of the [latex]{\\text{HCO}}_{3}{}^{-}[\/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>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}{}^{-}[\/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<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]\\text{p}K_{\\text{a}}=\u2212\\text{log}K_{\\text{a}}[\/latex]<\/li>\n<li>[latex]\\text{p}K_{\\text{b}}=\u2212\\text{log}K_{\\text{b}}[\/latex]<\/li>\n<li>[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\text{log}\\dfrac{\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>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>M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H and 0.030 <em>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}{}^{-}\\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>M<\/em> HNO<sub>2<\/sub> and 0.030 <em>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}{}^{-}\\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>M<\/em> CH<sub>3<\/sub>NH<sub>2<\/sub> and 0.130 <em>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}}^{-}\\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>M<\/em> NH<sub>3<\/sub> and 0.78 <em>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}}^{-}\\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>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>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 style=\"list-style-type: lower-alpha;\">\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 style=\"list-style-type: lower-alpha;\">\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>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>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<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q768271\">Show Selected Solutions<\/span><\/p>\n<div id=\"q768271\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.\u00a0Excess [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] is removed primarily by the reaction\u00a0[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}\\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\u00a0[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>4. The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.25<\/td>\n<td>0<\/td>\n<td>0.030<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.25\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>0.030 +\u00a0<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.25 \u2212 <em>x<\/em>) \u2248 0.25 and (0.030 \u2212 <em>x<\/em>) \u2248 0.030, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>x<\/em> gives 1.50 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em>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>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}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]; p<em>K<\/em><sub>a<\/sub> = \u2212log(<em>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>M<\/em>; [A<sup>\u2212<\/sup>] \u2248 [NaCH<sub>3<\/sub>CO<sub>2<\/sub>] = 0.030 <em>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>M<\/em> = 10<sup>\u22123.82<\/sup><em>M<\/em> = 1.5 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em>M<\/em><\/p>\n<p>6.\u00a0The equilibrium expression is:\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[CH<sub>3<\/sub>NH<sub>2<\/sub>]<\/th>\n<th>[CH<sub>3<\/sub>NH<sub>3<\/sub><sup>+<\/sup>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.125<\/td>\n<td>0.130<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<i>x<\/i><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.125\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td>0.130 +\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.125 \u2212 <em>x<\/em>) \u2248 0.125 and (0.130 \u2212 <em>x<\/em>) \u2248 0.130, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>x<\/em> gives 4.23 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><em>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>M<\/em>.<\/p>\n<p>8. The reaction and equilibrium constant are\u00a0[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}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{b}}=1.8\\times {10}^{-4}[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Let <em>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:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[NH<sub>3<\/sub>]<\/th>\n<th>[NH<sub>4<\/sub><sup>+<\/sup>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.200<\/td>\n<td>0.78<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>\u00a0x + x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.200\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x + x<\/em><\/td>\n<td><em>x<\/em> = 1.0 \u00d7\u00a010<sup>\u22125<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em>x<\/em> + <em>x<\/em>) \u2248 <em>x<\/em>, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>x<\/em> gives 0.360 <em>M<\/em>. Because <em>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>M<\/em>.<\/p>\n<p>10.\u00a0The reaction and equilibrium constant are\u00a0[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}{}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>(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}{}^{-}[\/latex] and produce CH<sub>3<\/sub>CO<sub>2<\/sub>H in the process. Thus, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right][\/latex] decreases and [CH<sub>3<\/sub>CO<sub>2<\/sub>H] increases.<\/li>\n<li>(b) The added KCH<sub>3<\/sub>CO<sub>2<\/sub> will increase the concentration of [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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.<\/li>\n<li>(c) The added NaCl will have no effect on the concentration of the ions.<\/li>\n<li>(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}{}^{-}\\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}{}^{-}\\right][\/latex] increases.<\/li>\n<li>(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}{}^{-}\\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}{}^{-}\\right][\/latex] increases.<\/li>\n<\/ol>\n<p>12.\u00a0The reaction and equilibrium constant are:\u00a0[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}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The equilibrium expression is:\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>M<\/em> and 0.40 <em>M<\/em>, respectively. The equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[NH<sub>3<\/sub>]<\/th>\n<th>NH<sub>4<\/sub><sup>+<\/sup>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.20<\/td>\n<td>0.40<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td><em>\u2212x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.20\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td>0.40 +\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.20 \u2212 <em>x<\/em>) \u2248 0.20 and (0.40 + <em>x<\/em>) \u2248 0.40, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\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>x<\/em> gives 9.00 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em>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>M<\/em>. Thus:<\/p>\n<ul>\n<li>pOH = \u2212log(9.00 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>) = 5.046<\/li>\n<li>pH = 14.000 \u2212 pOH = 14.000 \u2212 5.046 = 8.954 = 8.95<\/li>\n<\/ul>\n<p>14.\u00a0The reaction and equilibrium constant are:\u00a0[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}{}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>x<\/em> be the concentration of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}[\/latex]. The hydronium ion concentration at equilibrium is\u00a0[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>M<\/em><\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.50<\/td>\n<td>0<\/td>\n<td><i>x<\/i><\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.50\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em> = 10 \u00d7 10<sup>\u22125<\/sup><\/td>\n<td><em>x + x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (<em>x<\/em> + <em>x<\/em>) \u2248 <em>x<\/em>, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>x<\/em> gives 0.900 <em>M<\/em>. Because <em>x<\/em> is less than 5% of this value, our assumption is correct. Therefore, [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right][\/latex] = 0.900 <em>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 style=\"text-align: center;\">[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<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>A buffer solution is prepared from equal volumes of 0.200 <em>M<\/em> acetic acid and 0.600 <em>M<\/em> sodium acetate. Use 1.80 \u00d7 10<sup>\u22125<\/sup> as <em>K<\/em><sub>a<\/sub> for acetic acid.\n<ol style=\"list-style-type: lower-alpha;\">\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>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>M<\/em> NaOH and the resulting solution\u00a0diluted to 0.100 L.\n<ol style=\"list-style-type: lower-alpha;\">\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>M<\/em> HCl is added to the solution?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q817908\">Show Solution to Question 1<\/span><\/p>\n<div id=\"q817908\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Part A<\/h4>\n<p>The reaction and equilibrium constant are\u00a0[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}{}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>M<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub> = <em>M<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub>, or:<\/p>\n<p style=\"text-align: center;\">[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}{}^{-}\\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:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.100<\/td>\n<td>0<\/td>\n<td>0.300<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<i>x<\/i><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.100\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>0.300 +\u00a0<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.100 \u2212 <em>x<\/em>) \u2248 0.100 and (0.300 \u2212 <em>x<\/em>) \u2248 0.300, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>x<\/em> gives 6.000 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em>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>M<\/em>:<\/p>\n<p style=\"text-align: center;\">pH = \u2212log(6.000 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>) = 5.2218 = 5.222;<\/p>\n<h4>Part B<\/h4>\n<p>The solution is acidic.<\/p>\n<h4>Part C<\/h4>\n<p>Assume that the added H<sup>+<\/sup> reacts completely with an equal amount of [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}[\/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>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>M<\/em> [latex]\\times[\/latex] 0.500 L = 0.1000 mol, and for acetate ion, 0.600 <em>M<\/em> [latex]\\times[\/latex] 0.500 L = 0.3000 mol. Thus:<\/p>\n<p style=\"text-align: center;\">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 style=\"text-align: center;\">[latex]\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}=0.3000 - 1.02\\times {10}^{-4}=0.2999\\text{mol}[\/latex]<\/p>\n<p style=\"text-align: center;\">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<ul>\n<li>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]=\\frac{0.1001\\text{mol}}{1.0030\\text{L}}=0.09980M[\/latex]<\/li>\n<li>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right]=\\frac{0.2999\\text{mol}}{1.0030\\text{L}}=0.2990M[\/latex]<\/li>\n<\/ul>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.09980<\/td>\n<td>0<\/td>\n<td>0.2990<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<i>x<\/i><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.09980\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>0.2990\u00a0+\u00a0<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.09980 \u2212 <em>x<\/em>) \u2248 0.09980 and (0.2990 \u2212 <em>x<\/em>) \u2248 0.2990, gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\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>x<\/em> gives 6.008 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em>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>M<\/em>.<\/p>\n<p style=\"text-align: center;\">pH = \u2212log(6.008 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>) = 5.2213 = 5.221<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Which acid in Table\u00a01 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">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\u00a01 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>\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\u00a02 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>\u00a0is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.<\/li>\n<li>Which base in Table\u00a02 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/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>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>K<\/em><sub>1<\/sub> = 8.5 [latex]\\times[\/latex] 10<sup>\u22125<\/sup>, <em>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q469090\">Show Selected Solutions<\/span><\/p>\n<div id=\"q469090\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\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>M<\/em><\/p>\n<p>We can now solve for <em>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\u00a01 of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>, the acid with the closest <em>K<\/em><sub>a<\/sub> to 7.94 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> is HF, with a <em>K<\/em><sub>a<\/sub> of 7.2 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>.<\/p>\n<p>3.\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}}^{-}\\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>M<\/em>.<\/p>\n<p>We can now solve for <em>K<\/em><sub>b<\/sub> of the best base as follows:<\/p>\n<p>[latex]\\frac{\\left[{\\text{OH}}^{-}\\right]}{{K}_{\\text{b}}}=1[\/latex]<\/p>\n<p><em>K<\/em><sub>b<\/sub> = [OH<sup>\u2212<\/sup>] = 4.47 [latex]\\times[\/latex] 10<sup>\u22124<\/sup><\/p>\n<p>In Table\u00a02\u00a0of\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\" rel=\"noopener\">Relative Strengths of Acids and Bases<\/a>, the base with the closest <em>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>K<\/em><sub>b<\/sub> = 4.4 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>.<\/p>\n<p>5. 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:\u00a0[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 style=\"text-align: center;\">[latex]\\left[{\\text{A}}^{-}\\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 style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{A}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HA}\\left(aq\\right)+{\\text{OH}}^{-}\\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}}^{-}\\right]}{\\left[{\\text{A}}^{-}\\right]}\\end{array}[\/latex]<\/p>\n<p>The pH of the solution is 5.48, so\u00a0pOH = 14.00 \u2212 5.48 = 8.52,\u00a0and\u00a0[OH<sup>\u2212<\/sup>] = 10<sup>\u22128.52<\/sup> = 3.02 [latex]\\times[\/latex] 10<sup>\u22129<\/sup><em>M<\/em><\/p>\n<p>Because of the small size of <em>K<\/em><sub>b<\/sub>, almost all the A<sup>\u2212<\/sup> will be in the form of HA. Therefore,\u00a0[latex]4.8\\times {10}^{-3}=\\frac{x\\left(3.02\\times {10}^{-9}\\right)}{3.9\\times {10}^{-5}-x}[\/latex],\u00a0where\u00a0<em>x<\/em> \u2248 3.9 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em>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 style=\"text-align: center;\">[latex]\\left[{\\text{A}}^{-}\\right]=\\frac{\\left[\\text{HA}\\right]\\left[{\\text{OH}}^{-}\\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>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>buffer capacity: <\/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<p><strong>buffer: <\/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<p><strong>Henderson-Hasselbalch equation: <\/strong>equation used to calculate the pH of buffer solutions<\/p>\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 2e. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/\">https:\/\/openstax.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>: Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction<\/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 2e\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction\"}]","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\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3521","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":17,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3521\/revisions"}],"predecessor-version":[{"id":7861,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3521\/revisions\/7861"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/2988"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3521\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=3521"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=3521"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=3521"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=3521"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}