a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\).

To calculate the equilibrium constant for the given reaction, we need to use the given \(K_{\mathrm{sp}}\) value and the overall formation constant. The equation for the equilibrium constant is as follows: \[K = \frac{K_{\mathrm{sp}}}{K_{\mathrm{formation}}}\] In this case, we have: \[K = \frac{1.6 \times 10^{-19}}{1.0 \times 10^{13}}\] Now, we can calculate the equilibrium constant K: \[K = 1.6 \times 10^{-32}\]

To find the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\), we can use the following expression involving the given concentrations and the equilibrium constant that we found in step 1: \[K = \frac{\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}[OH^{-}]^2}{\left[\mathrm{NH}_{3}\right]^4}\] From the problem, we are given that \(\mathrm{OH}^{-}\) has a concentration of \(0.0095M\), and \(\mathrm{NH}_{3}\) has a concentration of \(5.0M\). Using these values, we can solve for the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). We will represent this concentration as \(x\): \[1.6 \times 10^{-32} = \frac{x \cdot (0.0095)^2}{(5.0)^4}\] Solving for \(x\), we obtain the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\): \[x = \frac{1.6 \times 10^{-32} \cdot (5.0)^4}{(0.0095)^2}\] \[x \approx 9.03 \times 10^{-5} M\] Now we can find the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\) using the relationship between the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) and \(\mathrm{Cu}(\mathrm{OH})_{2}\): \[s = \frac{1}{2} [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]\] \[s = \frac{1}{2} (9.03 \times 10^{-5} M)\] \[s \approx 4.52 \times 10^{-5} M\] So, the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\) is approximately \(4.52 \times 10^{-5} M\).