Question 17.9: Calculating the Common-Ion Effect on Acid Ionization (Effect...
Calculating the Common-Ion Effect on Acid Ionization (Effect of a Strong Acid)
The degree of ionization of acetic acid, \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, in a 0.10 \mathrm{M} aqueous solution at 25^{\circ} \mathrm{C} is 0.013. K_{a} at this temperature is 1.7 \times 10^{-5}. Calculate the degree of ionization of \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} in a 0.10 \mathrm{M} solution at 25^{\circ} \mathrm{C} to which sufficient \mathrm{HCl} is added to make it 0.010 \mathrm{M} \mathrm{HCl}. How is the degree of ionization affected?
PROBLEM STRATEGY
This is an acid-ionization problem, but it differs from the simple ionization illustrated by Example 17.2. Here you have a starting concentration of \mathrm{H}_{3} \mathrm{O}^{+} (=0.010 M) from the addition of a strong acid (\mathrm{HCl}). This gives a different type of equation in Step 2, but you solve it in Step 3 by using a similar approximation method. You assume that x is small compared with starting concentrations of acid and \mathrm{H}_{3} \mathrm{O}^{+}, so that the resulting equation is linear, rather than quadratic. The calculation of degree of ionization follows that in Example 17.2.
STEP 1 Starting concentrations are \left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=0.10 \mathrm{M},\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.010 \mathrm{M} (from \mathrm{HCl} ), and \left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}\right]=0. The acetic acid ionizes to give an additional x \mathrm{mol} / \mathrm{L} of \mathrm{H}_{3} \mathrm{O}^{+}and x \mathrm{~mol} / \mathrm{L} of \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}. The table is
STEP 2 You substitute into the equilibrium-constant equation
\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}
obtaining
\frac{(0.010+x) x}{0.10-x}=1.7 \times 10^{-5}
STEP 3 To solve this equation, assume that x is small compared with 0.010 . Then
\begin{gathered} 0.010+x \simeq 0.010 \\ 0.10-x \simeq 0.10 \end{gathered}
The equation becomes
\frac{0.010 x}{0.10} \simeq 1.7 \times 10^{-5}
Solving for x, you get
x=1.7 \times 10^{-5} \times \frac{0.10}{0.010}=1.7 \times 10^{-4}
(Check that x can indeed be neglected in 0.010+x and 0.10-x.) The degree of ionization of \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} is x / 0.10=\mathbf{0 . 0 0 1 7}. This is much smaller than the value for 0.10 M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(0.013), because the addition of \mathrm{HCl} represses the ionization of \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}.