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How Buffers Work

How Buffers Work

A mixture of acetic acid and sodium acetate is acidic because the Ka of acetic acid is greater than the Kb of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. The decrease in hydronium ion concentration causes the acetic acid hydrolysis equilibrium to shift to the right, restoring the hydronium ion concentration almost to its original value, and yielding a relatively modest increase in pH:

\({\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)\)

If we add an acid such as hydrochloric acid, the resultant increase in hydronium ion concentration shifts the equilibrium to the left. This effectively converts the added strong acid to a much weaker acid (acetic acid), and the buffer solution thus experiences only a slight decrease in pH.

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, “H subscript 3 O superscript positive sign added, equilibrium position shifts to the left.” 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, “O H subscript negative sign added, equilibrium position shifts to the right.” 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, “C H subscript 3 C O O H,” and the other is labeled, “C H subscript 3 C O O superscript negative sign.” 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, “Buffer solution equimolar in acid and base.” There is an arrow pointing to the right which is labeled, “Add O H superscript negative sign.” 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, “Buffer solution after addition of strong base.” From the middle bars again, there is an arrow that points left. The arrow is labeled, “Add H subscript 3 O superscript positive sign.” 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, “Buffer solution after addition of strong acid.”

This diagram shows the buffer action of these reactions.

A mixture of ammonia and ammonium chloride is basic because the Kb for ammonia is greater than the Ka for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value:

\({\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\)

If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value:

\({\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NH}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\)

The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid.

Example

pH Changes in Buffered and Unbuffered Solutions

Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds.(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

Solution

To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):

Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”
  1. Determine the direction of change. The equilibrium in a mixture of H3O+, \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}},\) and CH3CO2H is:

    \({\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)\)

    The equilibrium constant for CH3CO2H is not given, so we look it up in this appendix: Ka = 1.8 \(×\) 10−5. With [CH3CO2H] = \(\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]\) = 0.10 M and [H3O+] = ~0 M, the reaction shifts to the right to form H3O+.

  2. Determine x and equilibrium concentrations. A table of changes and concentrations follows:

    This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of “[ 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 ].” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, x, x. The fourth column has the following: 0.10, x, 0.10 plus sign x.
  3. Solve for x and the equilibrium concentrations. We find:

    \(x=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}M\)

    and

    \(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\phantom{\rule{0.2em}{0ex}}=0+x=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M\)

    Thus:

    \(\text{pH}=\text{−log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\text{−log}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)\)

    \(=4.74\)

  4. Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = Ka.

(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:

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 “Volume of N a O H solution.” An arrow points right to a second rectangle labeled “Moles of N a O H added.” A second arrow points right to a third rectangle labeled “Additional moles of N a C H subscript 3 C O subscript 2.” Just beneath the first rectangle in the upper left is a rectangle labeled “Volume of buffer solution.” An arrow points right to another rectangle labeled “Initial moles of C H subscript 3 C O subscript 2 H.” This rectangle points to the same third rectangle, which is labeled “ Additional moles of N a C H subscript 3 C O subscript 2.” An arrow points right to a rectangle labeled “ Unreacted moles of C H subscript 3 C O subscript 2 H.” An arrow points from this rectangle to a rectangle below labeled “[ C H subscript 3 C O subscript 2 H ].” An arrow extends below the “Additional moles of N a C H subscript 3 C O subscript 2” rectangle to a rectangle labeled “[ C H subscript 3 C O subscript 2 ].” This rectangle points right to the rectangle labeled “[ C H subscript 3 C O subscript 2 H ].”
  1. Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains:

    \(0.0010\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{0.10\phantom{\rule{0.2em}{0ex}}\text{mol NaOH}}{1\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{mol NaOH}\)

  2. Determine the moles of CH2CO2H. Before reaction, 0.100 L of the buffer solution contains:

    \(0.100\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{0.100\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}}{1\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\)

  3. Solve for the amount of NaCH3CO2 produced. The 1.0 \(×\) 10−4 mol of NaOH neutralizes 1.0 \(×\) 10−4 mol of CH3CO2H, leaving:

    \(\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(0.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)\phantom{\rule{0.2em}{0ex}}=0.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\)

    and producing 1.0 \(×\) 10−4 mol of NaCH3CO2. This makes a total of:

    \(\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)+\left(0.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)\phantom{\rule{0.2em}{0ex}}=1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NaCH}}_{3}{\text{CO}}_{2}\)

  4. Find the molarity of the products. After reaction, CH3CO2H and NaCH3CO2 are contained in 101 mL of the intermediate solution, so:

    \(\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{9.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{mol}}{0.101\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=0.098\phantom{\rule{0.2em}{0ex}}M\)

    \(\left[{\text{NaCH}}_{3}{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}}{0.101\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=0.100\phantom{\rule{0.2em}{0ex}}M\)

    Now we calculate the pH after the intermediate solution, which is 0.098 M in CH3CO2H and 0.100 M in NaCH3CO2, comes to equilibrium. The calculation is very similar to that in part (a) of this example:

    Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

    This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution (see the figure in the previous lesson).
(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 \(×\) 10−5-M solution of HCl). The volume of the final solution is 101 mL.

Solution

This 1.8 \(×\) 10−5-M solution of HCl has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:

\(0.100\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{mol HCl}}{1\phantom{\rule{0.2em}{0ex}}\text{L}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{mol HCl}\)

As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 \(×\) 10−4 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:

\(\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)\phantom{\rule{0.2em}{0ex}}=9.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M\)

The concentration of NaOH is:

\(\cfrac{9.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M\phantom{\rule{0.2em}{0ex}}\text{NaOH}}{0.101\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}M\)

The pOH of this solution is:

\(\text{pOH}=\text{−log}\left[{\text{OH}}^{\text{−}}\right]\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}=3.01\)

The pH is:

\(\text{pH}=14.00\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\text{pOH}=10.99\)

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).

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.

This lesson is part of:

Acid-Base Equilibria

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