25.00 mL of ethanoic acid of concentration 0.1000 mol/L is titrated with sodium hydroxide of…

Step 1. Start. $x^2 + K_a x - K_a c = 0$ gives $[\text{H}^+] = 1.33 \times 10^{-3}\ \text{mol/L}$. $\text{pH} = 2.88$.

Step 2. Half-way. Acid and ethanoate are in equal amounts, so $\text{pH} = \text{p}K_a = 4.745$, which is $4.75$ to two places.

Step 3. Equivalence. The total volume is $50.00\ \text{mL}$, so $[\text{CH}_3\text{COO}^-] = 0.0500\ \text{mol/L}$. $K_b = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$. $[\text{OH}^-] = \sqrt{5.56 \times 10^{-10} \times 0.0500} = 5.27 \times 10^{-6}\ \text{mol/L}$. $\text{pOH} = 5.28$ and $\text{pH} = 8.72$.

Step 4. After 30.00 mL. Excess base: $\dfrac{0.1000 \times 5.00\ \text{mL}}{55.00\ \text{mL}} = 9.09 \times 10^{-3}\ \text{mol/L}$. $\text{pOH} = 2.04$, so $\text{pH} = 11.96$.