A solution contains 2.2 × 10^-3 M in Cu2+ and 0.33 M in LiCN. If the formation constant (Kf) for the complex ion Cu(CN)4^2- is 1.0 × 10^25, how much copper ion (Cu2+) remains at equilibrium?

• A. 0.33 M
• B. 1.1 × 10^-3 M
• C. 2.2 × 10^-3 M
• D. 2.2 × 10^-28 M

Answer: (D)

Verified step by step guidance

1. Identify the initial concentrations of the ions involved: Cu2+ is 2.2 × 10^-3 M and LiCN is 0.33 M.
2. Write the balanced equation for the formation of the complex ion: Cu2+ + 4CN- ⇌ Cu(CN)4^2-.
3. Use the formation constant (Kf) expression for the complex ion: Kf = [Cu(CN)4^2-] / ([Cu2+][CN-]^4).
4. Assume that x is the change in concentration of Cu2+ that forms the complex ion. At equilibrium, [Cu2+] = (2.2 × 10^-3 - x) M and [CN-] = (0.33 - 4x) M.
5. Substitute the equilibrium concentrations into the Kf expression and solve for x, which represents the amount of Cu2+ that reacts. The remaining Cu2+ concentration will be (2.2 × 10^-3 - x) M.