14.6 Buffers

How Buffers Work

A mixture of acetic acid and sodium acetate is acidic because the K_a of acetic acid is greater than the K_b 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:

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.

Figure 14.18 This diagram shows the buffer action of these reactions.

A mixture of ammonia and ammonium chloride is basic because the K_b for ammonia is greater than the K_a 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:

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:

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 14.20

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):

1. Step 1.
   Determine the direction of change. The equilibrium in a mixture of H₃O⁺, ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}},$ and CH₃CO₂H is:
   

   $${\text{CH}}_3{\text{CO}}_2\text{H}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}(aq)$$

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   The equilibrium constant for CH₃CO₂H is not given, so we look it up in Appendix H: K_a = 1.8 $\times$ 10⁻⁵. With [CH₃CO₂H] = $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ = 0.10 M and [H₃O⁺] = ~0 M, the reaction shifts to the right to form H₃O⁺.
2. Step 2. Determine x and equilibrium concentrations. A table of changes and concentrations follows:
   
3. Step 3.
   Solve for x and the equilibrium concentrations. We find:
   

   $$x=1.8\times {10}^{\mathrm{-5}}M$$

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   and
   

   $$[{\text{H}}_3{\text{O}}^{\text{+}}]=0+x=1.8\times {10}^{\mathrm{-5}}M$$

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   Thus:
   

   $$\text{pH}=\text{−log}[{\text{H}}_3{\text{O}}^{\text{+}}]=\text{−log}(1.8\times {10}^{\mathrm{-5}})$$

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   $$=4.74$$

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

(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:

1. Step 1.
   Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains:
   

   $$0.0010\sout{\text{L}}\times \left(\frac{0.10\text{mol NaOH}}{1\sout{\text{L}}}\right)=1.0\times {10}^{\mathrm{-4}}\text{mol NaOH}$$

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

   $$0.100\sout{\text{L}}\times \left(\frac{0.100\text{mol}{\text{CH}}_3{\text{CO}}_2\text{H}}{1\sout{\text{L}}}\right)=1.00\times {10}^{\mathrm{-2}}\text{mol}{\text{CH}}_3{\text{CO}}_2\text{H}$$

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3. Step 3.
   Solve for the amount of NaCH₃CO₂ produced. The 1.0 $\times$ 10⁻⁴ mol of NaOH neutralizes 1.0 $\times$ 10⁻⁴ mol of CH₃CO₂H, leaving:
   

   $$(1.0\times {10}^{\mathrm{-2}})-(0.01\times {10}^{\mathrm{-2}})=0.99\times {10}^{\mathrm{-2}}\text{mol}{\text{CH}}_3{\text{CO}}_2\text{H}$$

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   and producing 1.0 $\times$ 10⁻⁴ mol of NaCH₃CO₂. This makes a total of:
   

   $$(1.0\times {10}^{\mathrm{-2}})+(0.01\times {10}^{\mathrm{-2}})=1.01\times {10}^{\mathrm{-2}}\text{mol}{\text{NaCH}}_3{\text{CO}}_2$$

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

   $$[{\text{CH}}_3{\text{CO}}_2\text{H}]=\frac{9.9\times {10}^{\mathrm{-3}}\text{mol}}{0.101\text{L}}=0.098M$$

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   $$[{\text{NaCH}}_3{\text{CO}}_2]=\frac{1.01\times {10}^{\mathrm{-2}}\text{mol}}{0.101\text{L}}=0.100M$$

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   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:
   
   This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution (Figure 14.17).

(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 $\times$ 10⁻⁵-M solution of HCl). The volume of the final solution is 101 mL.

Solution

This 1.8 $\times$ 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:

$$0.100\text{L}\times \left(\frac{1.8\times {10}^{\mathrm{-5}}\text{mol HCl}}{1\text{L}}\right)=1.8\times {10}^{\mathrm{-6}}\text{mol HCl}$$

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As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 $\times$ 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:

$$(1.0\times {10}^{\mathrm{-4}})-(1.8\times {10}^{\mathrm{-6}})=9.8\times {10}^{\mathrm{-5}}M$$

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The concentration of NaOH is:

$$\frac{9.8\times {10}^{\mathrm{-5}}M\text{NaOH}}{0.101\text{L}}=9.7\times {10}^{\mathrm{-4}}M$$

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The pOH of this solution is:

$$\text{pOH}=\text{−log}\left[{\text{OH}}^{\text{−}}\right]=\text{−log}\left(9.7\times {10}^{\mathrm{-4}}\right)=3.01$$

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The pH is:

$$\text{pH}=14.00-\text{pOH}=10.99$$

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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).

Check Your Learning

Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 $\times$ 10⁻⁵ M HCl solution from 4.74 to 3.00.

Answer:

Initial pH of 1.8 $\times$ 10⁻⁵ M HCl; pH = −log[H₃O⁺] = −log[1.8 $\times$ 10⁻⁵] = 4.74
Moles of H₃O⁺ in 100 mL 1.8 $\times$ 10⁻⁵ M HCl; 1.8 $\times$ 10⁻⁵ moles/L $\times$ 0.100 L = 1.8 $\times$ 10⁻⁶
Moles of H₃O⁺ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L $\times$ 0.0010 L = 1.0 $\times$ 10⁻⁴ moles; final pH after addition of 1.0 mL of 0.10 M HCl:

$$\text{pH}=\text{−log}[{\text{H}}_3{\text{O}}^{\text{+}}]=\text{−log}\left(\frac{\text{total moles}{\text{H}}_3{\text{O}}^{\text{+}}}{\text{total volume}}\right)=\text{−log}\left(\frac{1.0\times {10}^{\mathrm{-4}}\text{mol}+1.8\times {10}^{\mathrm{-6}}\text{mol}}{101\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)}\right)=3.00$$

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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.

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 14.19). 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.

Figure 14.19 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)

The buffer capacity 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 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M 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.

Selection of Suitable Buffer Mixtures

There are two useful rules of thumb for selecting buffer mixtures:

1. 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 14.20 shows 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.
   

   Figure 14.20 The graph, an illustration of buffering action, shows change of pH as an increasing amount of a 0.10-M NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH₃CO₂H] = 0.10 M and $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]=0.10M.$
2. 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.

Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H₂CO₃, and the bicarbonate ion, ${\text{HCO}}_3{}^{\text{−}}.$ When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:

An added hydroxide ion is removed by the reaction:

The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H₃O⁺ is converted to H₂CO₃ and OH^- is converted to HCO₃^-). 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.

The Henderson-Hasselbalch Equation

The ionization-constant expression for a solution of a weak acid can be written as:

Rearranging to solve for [H₃O⁺], we get:

Taking the negative logarithm of both sides of this equation, we arrive at:

which can be written as

where pK_a is the negative of the common logarithm of the ionization constant of the weak acid (pK_a = −log K_a). 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 Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.

Link to Learning

View information on the buffer system encountered in natural waters.