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)₃^2-: The concentration of Cu⁺ at equilibrium is A) 5.0 × 10^-14 M. B) 1.2 × 10^-14 M. C) 2.0 × 10^-4 M. D) 1.0 × 10^-4 M. E) none of these

Question

Quizplus · Accepted Answer

First, we can write the balanced chemical equation for the reaction: CuNO₃⁺ + NaCN^- → Cu(CN)₃^2- + NaNO₃⁺ The reaction is an equilibrium reaction, and we are given the initial concentrations of CuNO₃⁺ and NaCN, both at 50.0 mL. We can use an ICE table (initial, change, equilibrium) to solve for the concentration of Cu⁺ at equilibrium. | |CuNO₃⁺ | NaCN | Cu(CN)₃^2- | |--------|-------------------|--------|------------------------| |Initial | 2.0×10^-4 M | 4.0 M | 0 M | |Change | -x | -x/2 | +x | |Equilibrium|2.0×10^-4 M - x | 4.0 M - x/2 | x | At equilibrium, the concentration of Cu(CN)₃^2- is equal to the concentration of CuNO₃⁺ that reacted to form the complex, which is x. To solve for x, we need to use the equilibrium constant expression for the reaction. The equilibrium constant (K) is given as: K = [Cu(CN)₃^2-] / ([Cu⁺][NaCN^-]/2) Substituting in the equilibrium concentrations: K = x / [(2.0×10^-4 - x)(4.0 - x/2)] We are given that K = 6.0×10^-28. Rearranging and solving for x, we get: x = (K[NaCN]/2) / (1 + K[NaCN]/2 + K[NaCN]([CuNO₃⁺]/[Cu⁺])) Substituting in values: x = (6.0×10^-28 × 2.0) / (1 + 6.0×10^-28 × 2.0 + 6.0×10^-28 × (2.0×10^-4 / [Cu⁺])) x ≈ 1.2×10^-14 M Therefore, the concentration of Cu⁺ at equilibrium is approximately 1.2×10^-14 M, which is Choice (B).

A 500-ML Sample of 2