Silver ion reacts with excess CN- to form a colorless complex ion, [Ag(CN)2]-, which has a formation constant Kf = 3.0 x 10^20. Calculate the concentration of Ag+ in a solution prepared by mixing equal volumes of 2.0 x 10^-3 M AgNO3 and 0.20 M NaCN.

Verified step by step guidance

1. Step 1: Write the balanced chemical equation for the formation of the complex ion: \[ \text{Ag}^+ + 2\text{CN}^- \rightleftharpoons [\text{Ag(CN)}_2]^- \]
2. Step 2: Use the initial concentrations of AgNO3 and NaCN to determine the initial concentrations of \( \text{Ag}^+ \) and \( \text{CN}^- \) in the mixed solution. Since equal volumes are mixed, the concentrations are halved: \( [\text{Ag}^+]_0 = 1.0 \times 10^{-3} \text{ M} \) and \( [\text{CN}^-]_0 = 0.10 \text{ M} \).
3. Step 3: Set up the expression for the formation constant \( K_f \) for the complex ion: \[ K_f = \frac{[\text{Ag(CN)}_2]^-}{[\text{Ag}^+][\text{CN}^-]^2} \]
4. Step 4: Assume that \( x \) is the change in concentration of \( \text{Ag}^+ \) that forms the complex ion. Then, express the equilibrium concentrations in terms of \( x \): \( [\text{Ag}^+] = 1.0 \times 10^{-3} - x \), \( [\text{CN}^-] = 0.10 - 2x \), and \( [\text{Ag(CN)}_2]^- = x \).
5. Step 5: Substitute the equilibrium concentrations into the \( K_f \) expression and solve for \( x \), which represents the concentration of \( [\text{Ag(CN)}_2]^- \). Use this to find the concentration of \( \text{Ag}^+ \) at equilibrium.