Calculate the molar solubility of gold(I) chloride \(\left(K_{s p}=2.0 \times 10^{-13}\right)\) in \(0.10 \mathrm{M} \mathrm{NaCN}\). The complex ion formed is \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) with \(K_{\mathrm{f}}=2 \times 10^{38}\). Ignore any other competing equilibrium systems.

For the solubility of AuCl: \(\mathrm{AuCl} \rightleftharpoons \mathrm{Au}^{+} + \mathrm{Cl}^{-}\) For the formation of the complex ion: \(\mathrm{Au}^{+} + 2\mathrm{CN}^{-} \rightleftharpoons \left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\)

Since NaCN completely dissociates in water, the concentration of cyanide ions is equal to the concentration of NaCN, which is given as 0.10 M.

For the solubility of AuCl, we have the expression: \(K_{sp} = [\mathrm{Au}^{+}][\mathrm{Cl}^{-}]\) For the formation of the complex ion, we have the expression: \(K_{f} = \frac{[\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}]}{[\mathrm{Au}^{+}][\mathrm{CN}^{-}]^2}\)

Let the solubility of AuCl be represented by S. Then, the concentration of Au+ ions and Cl- ions in the solution will also be S, since it forms in a 1:1 ratio from the dissociation of AuCl. Now, the formation of a complex ion \([\mathrm{Au}(\mathrm{CN})_{2}]^-\) reduces the concentration of Au+ ions. Let the concentration of the complex ion be x. Hence, the concentration of Au+ ions left in the solution will be (S - x). From the above equilibrium expressions, we can write: \(K_{sp} = S(S - x)\) ...(1) \(K_{f} = \frac{x}{(S - x)(0.10)^2}\) ...(2)

From equation (2), we can write the expression for x (complex ion concentration) as, \(x = K_{f}(S - x)(0.10)^2\) As \(K_{f}\) is very large, we can consider the concentration of complex ion x to be approximately equal to S. Now we can solve for S: \(x \approx S = K_{f}(S - x)(0.10)^2\) \(S = K_{f}(S - S)(0.10)^2\) \(S = K_{f} \times 0.01 \times S\) Now, we can find the value of S: \(\frac{S}{0.01 K_{f}} = 1\) \(S = \frac{0.01}{K_{f}}\) But we have not taken \(K_{sp}\) into consideration yet. As in the previous analysis, the concentration of complex ion, x, is approximately equal to S. Therefore, we can write equation (1) as: \(K_{sp} = S^2\) Plugging the value of S from above, \(K_{sp} = \left(\frac{0.01}{K_{f}}\right)^2\) Now, we can solve for S: \(S = \sqrt{K_{sp} \times K_{f} \times 0.01^2}\) Plugging the given values of \(K_{sp}\) and \(K_{f}\): \(S = \sqrt{2.0\times 10^{-13} \times 2.0 \times 10^{38} \times 0.0001}\) Finally, we can calculate the solubility S: \(S \approx 2.0 \times 10^{-6}\, M\) The molar solubility of gold(I) chloride in \(0.10\,M\, \mathrm{NaCN}\) is approximately \(2.0 \times 10^{-6}\, M\).