Question 17.10: Calculating the Common-Ion Effect on Acid Ionization (Effect...

STEP 1 Consider the equilibrium

\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}(a q)

Initially, 1 \mathrm{~L} of solution contains 0.10 \mathrm{~mol} of acetic acid. Sodium acetate is a strong electrolyte, so 1 \mathrm{~L} of solution contains 0.20 \mathrm{~mol} of acetate ion. When the acetic acid ionizes, it gives x mol of hydronium ion and x mol of acetate ion. This is summarized in the following table:

STEP 2 The equilibrium-constant equation is

\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\right]}{\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]}=K_{a}

Substituting into this equation gives

\frac{x(0.20+x)}{0.10-x}=1.7 \times 10^{-5}

STEP 3 To solve the equation, assume that x is small compared with 0.10 and 0.20 . Then

\begin{aligned} & 0.20+x \simeq 0.20 \\ & 0.10-x \simeq 0.10 \end{aligned}

The equilibrium equation becomes

\frac{x(0.20)}{0.10} \simeq 1.7 \times 10^{-5}

Hence,

x \simeq 1.7 \times 10^{-5} \times \frac{0.10}{0.20}=8.5 \times 10^{-6}

(Note that x is indeed much smaller than 0.10 or 0.20 .) The hydronium-ion concentration is 8.5 \times 10^{-6} \mathrm{M}, and

\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left(8.5 \times 10^{-6}\right)=\mathbf{5 . 0 7}

For comparison, the \mathrm{pH} of 0.10  \mathrm{M} acetic acid is 2.88 .