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

💡 Solution & Explanation

The saturated solution of AgCl (Ksp = 10^-10) has an initial Ag^+ concentration of 10^-5 M. In 100 mL (0.1 L) of solution, the total moles of Ag^+ present = 10^-5 mol/L × 0.1 L = 10^-6 moles. Since zinc is added in large excess (10^-3 moles), the huge equilibrium constant ensures that virtually all available Ag^+ is reduced to solid Ag. Thus, 10^-6 moles of Ag will precipitate. Therefore, correct answer is C.