An aqueous solution contains 0.010 M Br^- and 0.010 M I^-.If Ag⁺ is added until AgBr(s)just begins to precipitate,what are the concentrations of Ag⁺ and I^-? (K_sp of AgBr = 5.4 × 10^-13,K_sp of AgI = 8.5 × 10^-17)

Question

Quizplus · Accepted Answer

1. Write the equation for the precipitation of AgBr: Ag⁺ + Br^- → AgBr(s) + I^- 2. Write the equilibrium constant expression for the reaction: K = [Ag⁺][Br^-]/[AgBr] 3. Set up an ICE table for the reaction: Initial: [Ag⁺] = 0, [Br^-] = 0.010 M, [AgBr] = 0 Change: [Ag⁺] = -x, [Br^-] = -x, [AgBr] = +x Equilibrium: [Ag⁺] = 0.010 - x, [Br^-] = 0.010 - x, [AgBr] = x 4. Substitute the equilibrium concentrations into the equilibrium constant expression and solve for x: K = [Ag⁺][Br^-]/[AgBr] 5.4 × 10^-13 = (0.010 - x)(0.010 - x)/x Solving for x gives x = 5.4 × 10^-11 M 5. Substitute x into the equilibrium concentrations to find the final concentrations: [Ag⁺] = 0.010 - x = 5.4 × 10^-13 M [Br^-] = 0.010 - x = 5.4 × 10^-13 M 6. Use the law of mass action to find the concentration of I^- at equilibrium: K = [Ag⁺][I^-]/[AgI] 8.5 × 10^-17= (5.4 × 10^-11)([I^-]/x) Solving for [I^-] gives [I^-] = 1.6 × 10^-6 M Therefore, the correct answer is (C) [Ag⁺] = 5.4 × 10^-11 M, [I^-] = 1.6 × 10^-6 M.

An Aqueous Solution Contains 0