Titrating a weak acid with a strong base gives a different curve from the strong–strong case. Between the start and the equivalence point the flask holds a buffer, so the Henderson–Hasselbalch equation does most of the work.
Key idea
Start: weak-acid pH. Before equivalence: a buffer, $\text{pH} = \text{p}K_a + \log(n_{\mathrm{A^-}}/n_\mathrm{HA})$; at the halfway point $\text{pH} = \text{p}K_a$. At equivalence: only $\mathrm{A^-}$, a weak base, so the pH is above 7. After: excess $\mathrm{OH^-}$ sets the pH.
Why it works
Each portion of NaOH turns some HA into $\mathrm{A^-}$. Halfway, half is converted, so $n_{\mathrm{A^-}} = n_\mathrm{HA}$ and the log term is zero. That is how $\text{p}K_a$ values are measured. At equivalence every HA has become $\mathrm{A^-}$, which takes protons from water and makes $\mathrm{OH^-}$, so the pH is basic. The jump at equivalence is smaller than for a strong acid, because the curve starts higher.
Worked examples
Worked example: Acetic acid with NaOH
$25.0\ \text{mL}$ of $0.100\ \text{M}$ acetic acid ($K_a = 1.8 \times 10^{-5}$) is titrated with $0.100\ \text{M}$ NaOH. Find the pH at $0$, $12.5$, $20.0$ and $25.0\ \text{mL}$.
- 0 mL: weak acid, $x = \sqrt{(1.8 \times 10^{-5})(0.100)} = 1.34 \times 10^{-3}$; pH 2.87. (Solving exactly gives 2.88, the value on the curve above; the small-x shortcut is off by 0.01.)
- 12.5 mL: half neutralized, pH $= \text{p}K_a = 4.74$.
- 20.0 mL: HA $2.50 - 2.00 = 0.50\ \text{mmol}$, $\mathrm{A^-}$ $2.00\ \text{mmol}$; $\text{pH} = 4.74 + \log 4.0 = 5.34$.
- 25.0 mL: $2.50\ \text{mmol}$ $\mathrm{A^-}$ in $50.0\ \text{mL}$ = $0.0500\ \text{M}$. $K_b = 5.6 \times 10^{-10}$, $$[\mathrm{OH^-}] = \sqrt{(5.6 \times 10^{-10})(0.0500)} = 5.3 \times 10^{-6}$$; pOH 5.28, pH 8.72.
Answer: 2.87, 4.74, 5.34, 8.72.
Worked example: pKa from a curve
In a titration of an unknown weak acid, equivalence comes at $18.0\ \text{mL}$ of base, and the pH at $9.0\ \text{mL}$ is 3.75. What is $K_a$?
- $9.0\ \text{mL}$ is halfway to equivalence, so $\text{p}K_a = 3.75$.
- $K_a = 10^{-3.75} = 1.8 \times 10^{-4}$ (formic acid has this value).
Answer: $K_a = 1.8 \times 10^{-4}$.
After equivalence, the excess hydroxide sets the pH just as in the strong-acid case, so the two curves become the same from there on.
Common mistake
Thinking the equivalence point is always at pH 7. It is at 7 only when both acid and base are strong. With a weak acid, the conjugate base left at equivalence makes the solution basic; with a weak base titrated by a strong acid, the conjugate acid makes it acidic. Check: ask what species is in the flask at equivalence.
You can run this titration on the bench in the Mixing Lab.