1. **State the problem:** We have a 1.0 L buffer solution with 0.250 M benzoic acid (C6H5COOH) and 0.250 M sodium benzoate (C6H5COONa). We add 0.010 mol of Ca(OH)2 and want to find the new pH. 2. **Relevant formula:** The pH of a buffer solution is given by the Henderson-Hasselbalch equation: $$\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right)$$ where $\mathrm{p}K_a = -\log K_a$. 3. **Calculate initial pH:** $$\mathrm{p}K_a = -\log(6.3 \times 10^{-5}) = 4.20$$ Initial concentrations: $$[\mathrm{HA}] = 0.250\,M, \quad [\mathrm{A}^-] = 0.250\,M$$ Initial pH: $$\mathrm{pH} = 4.20 + \log \left( \frac{0.250}{0.250} \right) = 4.20 + \log(1) = 4.20$$ 4. **Effect of adding Ca(OH)2:** Ca(OH)2 dissociates to give 2 OH⁻ ions per formula unit: $$\mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^-$$ Moles of OH⁻ added: $$0.010 \times 2 = 0.020\,\mathrm{mol}$$ 5. **Reaction with benzoic acid:** OH⁻ reacts with benzoic acid (HA) to form A⁻: $$\mathrm{HA} + \mathrm{OH}^- \rightarrow \mathrm{A}^- + \mathrm{H}_2\mathrm{O}$$ Moles before addition: $$n_{HA} = 0.250 \times 1.0 = 0.250\,\mathrm{mol}$$ $$n_{A^-} = 0.250 \times 1.0 = 0.250\,\mathrm{mol}$$ After reaction: $$n_{HA,new} = 0.250 - 0.020 = 0.230\,\mathrm{mol}$$ $$n_{A^-,new} = 0.250 + 0.020 = 0.270\,\mathrm{mol}$$ 6. **New concentrations:** Volume is 1.0 L, so concentrations equal moles: $$[\mathrm{HA}] = 0.230\,M, \quad [\mathrm{A}^-] = 0.270\,M$$ 7. **Calculate new pH:** $$\mathrm{pH} = 4.20 + \log \left( \frac{0.270}{0.230} \right) = 4.20 + \log(1.174) = 4.20 + 0.07 = 4.27$$ **Final answer:** $$\boxed{4.27}$$