# 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 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):

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:

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:

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:

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 $$×$$ 10−5M solution of HCl). The volume of the final solution is 101 mL.

### Solution

This 1.8 $$×$$ 10−5M 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.