The solubility product of AgCl at 298 K is — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Write the dissolution equilibrium for AgCl.** $$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$ **Step 2: Calculate ΔG° for the precipitation reaction (reverse).** From the passage, $\Delta G^\circ_f$ values: $$\Delta G^\circ_f(\text{Ag}^+, aq) = +77\ \text{kJ/mol}$$ $$\Delta G^\circ_f(\text{Cl}^-, aq) = -129\ \text{kJ/mol}$$ $$\Delta G^\circ_f(\text{AgCl}, s) = -109\ \text{kJ/mol}$$ For the **precipitation** reaction $\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$: $$\Delta G^\circ_{\text{ppt}} = \Delta G^\circ_f(\text{AgCl}) - [\Delta G^\circ_f(\text{Ag}^+) + \Delta G^\circ_f(\text{Cl}^-)]$$ $$= -109 - [77 + (-129)] = -109 - (-52) = -57\ \text{kJ/mol}$$ **Step 3: Find ΔG° for dissolution (Ksp reaction).** $$\Delta G^\circ_{\text{dis}} = +57\ \text{kJ/mol}$$ **Step 4: Calculate Ksp.** $$\Delta G^\circ = -RT \ln K_{sp} = -2.303 RT \log K_{sp}$$ $$\log K_{sp} = \frac{-\Delta G^\circ_{\text{dis}}}{2.303 RT} = \frac{-57{,}000}{2.303 \times 8.314 \times 298} = \frac{-57{,}000}{5705} = -9.99 \approx -10$$ $$K_{sp} = 10^{-10}$$ $$\boxed{\text{Answer: B — } K_{sp}(\text{AgCl}) = 10^{-10}}$$