A 50.0-mL sample of 2.0 *10^-⁴ M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN)₃²^-: What is the concentration of CN^- at equilibrium?

Question

Quizplus · Accepted Answer

This is a simple equilibrium problem involving the reaction of Cu⁺ with CN^- to form the complex ion Cu(CN)₃²^-. The balanced equation is: Cu⁺ + CN^- → Cu(CN)₃²^- We are given the initial concentrations of Cu⁺ and CN^-, and we assume that the reaction goes to equilibrium. Let x be the concentration of CN^- that reacts with Cu⁺ to form the complex ion. Then the equilibrium concentrations are: [Cu⁺] = (2.0 M) - x [CN^-] = (4.0 M) - x [Cu(CN)₃²^-] = x The equilibrium constant, K, for the reaction is: K = [Cu(CN)₃²^-]/([Cu⁺][CN^-]) Substituting the equilibrium concentrations into K, we get: K = x/[(2.0 - x)(4.0 - x)] We are asked to find the concentration of CN^- at equilibrium, so we need to solve for x. Since K is given as 1.6 x 10^-8, we can write: 1.6 x 10^-8 = x/[(2.0 - x)(4.0 - x)] Expanding the denominator and simplifying, we get a quadratic equation: 1.6 x 10^-8 = x/(8 - 6x - x^2) 1.6 x 10^-8 (8 - 6x - x^2) = x 1.28 x 10^-7 - 9.6 x 10^-8 x - 1.6 x 10^-8 x^2 = x 1.6 x 10^-8 x^2 + 1.056 x 10^-7 x - 1.28 x 10^-7 = 0 Using the quadratic formula, we get: x = 6.15 x 10^-9 M or x = -8.26 x 10^-6 M Since x represents a concentration, it must be positive, so we choose the positive root: x = 6.15 x 10^-9 M This is the concentration of CN^- that reacts with Cu⁺ to form the complex ion at equilibrium. The concentration of CN^- remaining at equilibrium is: [CN^-] = (4.0 M) - x = 3.99999999385 M ≈ 4.0 M (to three significant figures) Therefore, the best choice is D) 2.0 M.

A 500-ML Sample of 2