COMEDK 2025 Afternoon Solved Paper

Concept:The equilibrium constant for the base hydrolysis of $\mathrm{F}^{-}$ is the base dissociation constant $\mathrm{K}_{\mathrm{b}}$, which is related to the acid dissociation constant of $\mathrm{HF}$ via $\mathrm{K}_{\mathrm{b}} = \frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}$. The pH of a weak acid is approximated using $[\mathrm{H}^{+}] \approx \sqrt{\mathrm{K}_{\mathrm{a}}C}$.Explanation:First, find $\mathrm{K}_{\mathrm{b}}$ for the reaction $\mathrm{F}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HF} + \mathrm{OH}^{-}$.Given $\mathrm{K}_{\mathrm{a}} = 3.5 \times 10^{-4}$ for $\mathrm{HF}$ and $\mathrm{K}_{\mathrm{w}} = 1.0 \times 10^{-14}$.$\mathrm{K}_{\mathrm{b}} = \frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} = 2.86 \times 10^{-11}$.Next, calculate the pH of the $0.1\,\mathrm{M}$ $\mathrm{HF}$ solution.Using the approximation $[\mathrm{H}^{+}] \approx \sqrt{\mathrm{K}_{\mathrm{a}}C}$:$[\mathrm{H}^{+}] \approx \sqrt{(3.5 \times 10^{-4})(0.1)} = \sqrt{3.5 \times 10^{-5}} = 5.9 \times 10^{-3}\,\mathrm{M}$.$\mathrm{pH} = -\log(5.9 \times 10^{-3}) \approx 2.23$.Among the given options, only option C has the correct $\mathrm{K}_{\mathrm{eq}} = 2.86 \times 10^{-11}$, even though its pH value ($3.23$) does not match the calculated pH. The correct pH is approximately $2.23$.Answer:Option C (with the correct equilibrium constant).