BUFFER SOLUTIONS

17.6. BUFFER SOLUTIONS

A buffer solution is a solution of a weak acid and its conjugate base or a weak base and its conjugate acid. The main property of a buffer solution is its resistance to changes in its pH despite the addition of small quantities of strong acid or strong base. The student must know the following three things about buffer solutions:

1. How they are prepared.

2. What they do.

3. How they do it.

A buffer solution may be prepared by the addition of a weak acid to a salt of that acid or addition of a weak base to a salt of that base. For example, a solution of acetic acid and sodium acetate is a buffer solution. The weak acid (HC 2 H 3 O 2 ) and its conjugate base (C 2 H 3 O 2 − , from the sodium acetate) constitute a buffer solution. There are other ways to make such a combination of weak acid plus conjugate base (Problem 17.26).

A buffer solution resists change in its acidity. For example, a certain solution of acetic acid and sodium acetate has a pH of 4.0. When a small quantity of NaOH is added, the pH goes up to 4.2. If that quantity of NaOH had been added to the same volume of an unbuffered solution of HCl at pH 4, the pH would have gone up to a value as high as 12 or 13.

The buffer solution works on the basis of Le Chˆatelier’s principle. Consider the equation for the reaction of acetic acid with water:

The solution of HC 2 H 3 O 2 and C 2 H 3 O 2 − in H 2 O results in the relative quantities of each of the species in the equation as shown under the equation. If H 3 O + is added to the equilibrium system, the equilibrium shifts left to

ACID-BASE THEORY

[ CHAP. 17

in pH. If OH − is added to the solution, it reacts with H 3 O + present. But the removal of that H 3 O + is a stress, which causes this equilibrium to shift to the right, replacing much of the H 3 O + removed by the OH − . The pH does not rise nearly as much in the buffered solution as it would have in an unbuffered solution. Calculations may be made as to how much the pH is changed by the addition of a strong acid or base. You should first determine how much of each conjugate would be present if that strong acid or base reacted completely with the weak acid or conjugate base originally present. Use the results as the initial values of concentrations for the equilibrium calculations.

EXAMPLE 17.14.

A solution containing 0.250 mol HC 2 H 3 O 2 and 0.150 mol NaC 2 H 3 O 2 in 1.00 L is treated with 0.010 mol NaOH. Assume that there is no change in the volume of the solution. (a) What was the original pH of the

solution? (b) What is the pH of the solution after the addition of the NaOH? K a = 1.8 × 10 −5 .

Ans. (a)

0.150 + x Assuming x to be negligible when added to or subtracted from larger quantities, we find

Equilibrium

0.250 − x

( 0.150)x K a = −5

0.250 = 1.8 × 10 x = 3.0 × 10 −5

The assumption was valid.

pH = 4.52

(b) We assume that the 0.010 mol of NaOH reacted with 0.010 mol of HC 2 H 3 O 2 to produce 0.010 mol more of C 2 H 3 O 2 − (Sec. 10.3). That gives us

The pH has risen from 4.52 to 4.57 by the addition of 0.010 mol NaOH. (That much NaOH would have raised the pH of 1.0 L of an unbuffered solution of HCl originally at the same pH to a final pH value of 12.00.)