The weak-acid calculation gives $[\mathrm{H_3O^+}]$. Dividing by the starting concentration tells you what share of the acid molecules ionized, and that share depends on how dilute the solution is.
Key idea
$\text{percent ionization} = \dfrac{[\mathrm{H_3O^+}]_\text{eq}}{C} \times 100\%$. When the small-x approximation holds, it is about $\sqrt{K_a/C} \times 100\%$, so diluting a weak acid makes a larger fraction of it ionize (even though $[\mathrm{H_3O^+}]$ goes down).
Why it works
Ionization makes two particles from one: $\mathrm{HA \rightleftharpoons H^+ + A^-}$ (water left out). Dilute an equilibrium mixture tenfold: the top of $Q = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$ falls by $10 \times 10 = 100$ and the bottom by 10, so $Q$ drops to $\tfrac{1}{10}K$. The reaction shifts forward, ionizing more. Algebraically, $x/C \approx \sqrt{K_aC}/C = \sqrt{K_a/C}$: a hundredfold dilution gives ten times the percent ionization.
Worked examples
Worked example: Concentrated and dilute acetic acid
Find the percent ionization of acetic acid ($K_a = 1.8 \times 10^{-5}$) at $0.10\ \text{M}$ and at $0.0010\ \text{M}$.
- At $0.10\ \text{M}$: $x = 1.34 \times 10^{-3}$, so $\dfrac{1.34 \times 10^{-3}}{0.10} = 1.3\%$.
- At $0.0010\ \text{M}$ the shortcut gives $\sqrt{1.8 \times 10^{-8}} = 1.34 \times 10^{-4}$: 13%, over the 5% limit. Solve $x^2 + (1.8 \times 10^{-5})x - 1.8 \times 10^{-8} = 0$: $x = 1.25 \times 10^{-4}\ \text{M}$.
- Percent ionization $= \dfrac{1.25 \times 10^{-4}}{0.0010} = 12.5\%$.
Answer: 1.3% and 12.5%: a hundredfold dilution raised it almost tenfold.
Worked example: From percent to Ka
A $0.20\ \text{M}$ weak acid is 0.50% ionized. Find $K_a$.
- $x = 0.0050 \times 0.20 = 1.0 \times 10^{-3}\ \text{M}$.
- $K_a = \dfrac{(1.0 \times 10^{-3})^2}{0.20 - 0.0010} = 5.0 \times 10^{-6}$.
Answer: $K_a = 5.0 \times 10^{-6}$.
Worked example: Dilution and pH
Diluting $0.10\ \text{M}$ acetic acid (pH 2.87) a hundredfold gives $0.0010\ \text{M}$. What is the new pH, and why did it rise by less than 2 units?
- From the first example, $[\mathrm{H_3O^+}] = 1.25 \times 10^{-4}\ \text{M}$, so $\text{pH} = 3.90$.
- A strong acid diluted a hundredfold would rise by exactly 2 pH units. Here the pH rose by only 1.03, because the larger fraction that ionizes makes up part of the loss.
Answer: $\text{pH} = 3.90$.
Percent ionization is also a quick way to compare acids at the same concentration: a strong acid is 100% ionized, while at $0.10\ \text{M}$ HF is about 8% ionized and acetic acid about 1.3%.
Common mistake
Thinking dilution lowers the percent ionization because the solution is "weaker." $[\mathrm{H_3O^+}]$ does fall, but the fraction ionized rises. Check with $\sqrt{K_a/C}$: smaller $C$ gives a larger fraction.