Question 17.SE.1: What is the pH of a solution made by adding 0.30 mol of acet...

Analyze We are asked to determine the pH of a solution of a weak electrolyte (CH_3COOH) and a strong electrolyte (CH_3COONa) that share a common ion, CH_3COO^-.

Plan In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps:

(1) Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution.

(2) Identify the important equilibrium reaction that is the source of H^+ and therefore determines pH.

(3) Tabulate the concentrations of ions involved in the equilibrium.

(4) Use the equilibrium-constant expression to calculate [H^+] and then pH.

Solve

First, because CH_3COOH is a weak electrolyte and CH_3 COONa is a strong electrolyte, the major species in the solution are CH_3COOH (a weak acid), Na^+ (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH_3COO^- (which is the conjugate base of CH_3COOH).

Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH_3COOH:

CH _3 COOH (a q) \rightleftharpoons H ^{+}(a q)+ CH _3 COO ^{-}(a q)

(We have written the equilibrium using H^+(aq) rather than H_3O^+(aq), but both representations of the hydrated hydrogen ion are equally valid.)

Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16:

CH _3 COOH (a q) \rightleftharpoons H ^{+}(a q)+ CH _3 COO ^{-}(a q)

\begin{array}{|l|c|c|c|}\hline \text{Initial}(M) & 0.30 & 0 & 0.30 \\\hline \text{Change}(M) & -x & +x & +x \\\hline \text{Equilibrium}(M) & (0.30-x) & x & (0.30+x) \\\hline\end{array}

The equilibrium concentration of CH_3COO^- (the common ion) is the initial concentration that is due to CH_3COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH_3COOH.

Now we can use the equilibrium expression:

K_a=1.8 \times 10^{-5}=\frac{\left[ H ^{+}\right]\left[ CH _3 COO ^{-}\right]}{\left[ CH _3 COOH \right]}

The dissociation constant for CH_3COOH at 25 °C is from Table 16.2, or Appendix D; addition of CH_3COONa does not change the value of this constant. Substituting the equilibrium concentrations from our table into the equilibrium expression gives:

K_a=1.8 \times 10^{-5}=\frac{x(0.30+x)}{0.30-x}

Because K_a is small, we assume that x is small compared to the original concentrations of CH_3COOH and CH_3COO^- (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving:

K_a=1.8 \times 10^{-5}=\frac{x(0.30)}{0.30}

The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem.

x=1.8 \times 10^{-5}M=\left[ H ^{+}\right]

Finally, we calculate the pH from the equilibrium concentration of H^+(aq):

pH=-\log \left(1.8 \times 10^{-5}\right)=4.74

Comment In Section 16.5, we calculated that a 0.30 M solution of CH_3COOH has a pH of 2.64, corresponding to \left[ H ^{+}\right]=2.3 \times 10^{-3}\,M. Thus, the addition of CH_3COONa has substantially decreased [H^+], as we expect from Le Châtelier’s principle.