A 20.0 mL sample of 0.120 mol L⁻¹ lactic acid, a weak monobasic acid (Ka = 1.4 × 10⁻⁴), is titrated…

(a) Weak acid alone. $\dfrac{[\text{H}^+]^2}{0.120 - [\text{H}^+]} = 1.4 \times 10^{-4}$ gives $[\text{H}^+] = 4.029 \times 10^{-3}\ \text{mol L}^{-1}$, so $\text{pH} = 2.39$. (The short form $\sqrt{K_a C}$ gives the same pH to two decimals, since only about $3\%$ of the acid ionises.)

(b) A buffer. Acid: $0.120 \times 20.0 = 2.40\ \text{mmol}$. Base added: $0.0800 \times 10.0 = 0.80\ \text{mmol}$. So $0.80\ \text{mmol}$ of lactate forms and $1.60\ \text{mmol}$ of acid is left. With $\text{p}K_a = 3.854$:

(c) Equivalence. $V = \dfrac{2.40\ \text{mmol}}{0.0800\ \text{mol L}^{-1}} = 30.0\ \text{mL}$.

(d) The salt alone. $2.40\ \text{mmol}$ of lactate in $50.0\ \text{mL}$ is $0.0480\ \text{mol L}^{-1}$. $K_b = \dfrac{1.0 \times 10^{-14}}{1.4 \times 10^{-4}} = 7.143 \times 10^{-11}$, so $[\text{OH}^-] = \sqrt{K_b C} = \sqrt{3.429 \times 10^{-12}} = 1.852 \times 10^{-6}\ \text{mol L}^{-1}$. Then $\text{pOH} = 5.73$ and $\text{pH} = 8.27$: basic, as expected for the salt of a weak acid and a strong base.

(e) Past equivalence. Base added: $3.20\ \text{mmol}$, so $0.80\ \text{mmol}$ of $\text{OH}^-$ is left over in $60.0\ \text{mL}$: $[\text{OH}^-] = 0.0133\ \text{mol L}^{-1}$, $\text{pOH} = 1.88$ and $\text{pH} = 12.12$. The lactate adds almost nothing next to this.