How many moles of Ag will be precipitated in the above reaction? — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Find [Ag⁺] in saturated AgCl solution.** From the previous part: $K_{sp}(\text{AgCl}) = 10^{-10}$ For the dissolution: $\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$ $$[\text{Ag}^+] = [\text{Cl}^-] = \sqrt{K_{sp}} = \sqrt{10^{-10}} = 10^{-5}\ \text{mol/L}$$ **Step 2: Calculate moles of Ag⁺ in 100 mL of saturated AgCl.** $$n_{\text{Ag}^+} = [\text{Ag}^+] \times V = 10^{-5}\ \text{mol/L} \times 0.100\ \text{L} = 10^{-6}\ \text{mol}$$ **Step 3: Determine how many moles of Ag are precipitated.** The cell reaction (from Q11) has $\log K = 52.88$, so the equilibrium constant $K = 10^{52.88}$ is enormous. The reaction goes essentially to completion. Zinc is added in excess ($6.539 \times 10^{-2}\ \text{g Zn} = 10^{-3}\ \text{mol}$, which is far more than $10^{-6}\ \text{mol Ag}^+$). Therefore, **all** $10^{-6}$ mol of Ag⁺ is precipitated as Ag. $$n_{\text{Ag precipitated}} = 10^{-6}\ \text{mol}$$ $$\boxed{\text{Answer: C — } 10^{-6}\ \text{mol of Ag is precipitated}}$$