Question 17.SE.2: Calculate the fluoride ion concentration and pH of a solutio...

Analyze We are asked to determine the concentration of F^- and the pH in a solution containing the weak acid HF and the strong acid HCl. In this case the common ion is H^+.

Plan We can again use the four steps outlined in Sample Exercise 17.1.

Solve

Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H^+, and Cl^-. The Cl^-, which is the conjugate base of a strong acid, is merely a spectator ion in any acid–base chemistry. The problem asks for [F^-], which is formed by ionization of HF. Thus, this is the important equilibrium:

HF (a q) \rightleftharpoons H ^{+}(a q)+ F ^{-}(a q)

The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium:

\begin{matrix}\qquad \qquad \qquad \quad \operatorname{HF}(a q) \rightleftharpoons \quad H ^{+}(a q) + F ^{-}(a q)\\ \begin{array}{|c|c|c|c|}\hline \text{Initial (M)}& 0.20 & 0.10 & 0 \\\hline \text{Change (M)}& -x & +x & +x \\\hline \text{Equilibrium}(M) & (0.20-x) & (0.10+x) & x \\\hline\end{array} \end{matrix}

The equilibrium constant for the ionization of HF, from Appendix D, is 6.8 × 10^{-4}. Substituting the equilibrium-constant concentrations into the equilibrium expression gives:

K_a=6.8 \times 10^{-4}=\frac{\left[ H ^{+}\right]\left[ F ^{-}\right]}{[ HF]}=\frac{(0.10+x)(x)}{0.20-x}

If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to:

This F^- concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H+, suppresses the ionization of HF. The concentration of H^+(aq) is:

Thus,

\begin{aligned}&{\left[ H ^{+}\right]=(0.10+x) \,M \simeq 0.10 \,M}\\& pH=1.00\end{aligned}

Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the HCl; the HF makes a negligible contribution by comparison.