Assuming no change in volume, calculate the minimum mass of NaCl necessary to dissolve 0.01 mole of — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

We want to dissolve $0.01 \text{ moles}$ of AgCl in $100 \text{ L}$, so the required concentration is $S = 10^{-4} \text{ M}$. The dissolution reaction with excess $\text{Cl}^-$ is $\text{AgCl}(s) + \text{Cl}^- \rightleftharpoons \text{AgCl}_2^-$. The equilibrium constant $K_{eq} = K_{sp} \times K_f = (2.0 \times 10^{-10})(2.5 \times 10^5) = 5.0 \times 10^{-5}$. $K_{eq} = \frac{[\text{AgCl}_2^-]}{[\text{Cl}^-]} \implies 5.0 \times 10^{-5} = \frac{10^{-4}}{[\text{Cl}^-]} \implies [\text{Cl}^-] = \frac{10^{-4}}{5 \times 10^{-5}} = 2 \text{ M}$. The total $\text{Cl}^-$ added must be roughly $2 \text{ M}$. In $100 \text{ L}$, this is $200 \text{ moles}$ of NaCl. Mass $= 200 \times 58.5 \text{ g/mol} = 11700 \text{ g} = 11.7 \text{ kg}$.