Chapter 17.7, Problem 4Q

Silver undergoes similar reactions as those shown for gold. Both metals react with cyanide ion in the presence of oxygen to form soluble complexes, and both are reduced by zinc. The reaction of Ag⁺ with cyanide ion may be viewed as two sequential steps:

(1) $A{g}^{+}(aq)+C{N}^{-}(aq)\rightleftarrows AgCN(s)$

(2) $\begin{array}{l}\frac{\text{AgCN(s) + }C{N}^{-}(aq)\rightleftarrows {{\text{[Ag(CN)}}_{\text{2}}\text{]}}^{-}(aq)}{A{g}^{+}(aq)+2C{N}^{-}(aq)\rightleftarrows {{\text{[Ag(CN)}}_{\text{2}}]}^{-}\left(aq\right)}\\ {\text{K}}_{\text{f}}\text{ = 1}\text{.3 }\times {\text{10}}^{\text{21}}\end{array}$

1. a. Use the solubility product equilibrium constant (Appendix J) of AgCN(s) to determine the equilibrium constant for Step 1.
2. b. Use the equilibrium constants from Step 1 and the overall reaction to determine the equilibrium constant for Step 2.
3. c. Excess AgCN(s) is combined with 1.0 L of 0.0071 M CN⁻ (aq) and allowed to equilibrate. Calculate the equilibrium concentrations of CN⁻ and [Ag(CN)₂]⁻ using the equilibrium constant for Step 2. Assume no change in volume occurs.