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 \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)

To calculate the equilibrium constant (K_eq) for the given reaction, we need to consider the relationships between K_sp, the overall formation constant (K_f), and the concentrations of the species involved in the reaction. K_eq is the ratio of K_f to K_sp, expressed algebraically as follows: \(K_{eq} = \frac{K_f}{K_{sp}}\) Plug in the given values of K_sp and K_f: \(K_{eq} = \frac{1.0 \times 10^{13}}{1.6 \times 10^{-19}}\) Calculate K_eq: \(K_{eq} = 6.25 \times 10^{31}\)

To find the solubility of Cu(OH)_2 in 5.0 M NH3, we need to consider the reaction's stoichiometry and the given equilibrium constant K_eq. We are looking for [Cu(OH)_2]. Let's denote the solubility of Cu(OH)_2 as S. According to the reaction stoichiometry, the concentrations of the reactants and products at equilibrium will be: - [Cu(OH)_2] = S - [NH3] = 5.0 - 4S (since 4 moles of NH3 reacts with 1 mole of Cu(OH)_2) - [Cu(NH3)_4^(2+)] = S - [OH-] = 0.0095 + 2S (initially 0.0095, and 2 moles of OH- are produced from 1 mole of Cu(OH)_2 at equilibrium) Now, we can construct the equilibrium expression for the given reaction using the calculated K_eq: \(K_{eq} = \frac{[Cu(NH3)_{4}^{2+}] * [OH^-]^2}{[Cu(OH)_2]*[NH3]^4}\) Plug in the values for K_eq and the equilibrium concentrations of the reactants and products: \(6.25 \times 10^{31} = \frac{S * (0.0095 + 2S)^2}{ S(5.0 - 4S)^4}\) Solve this non-linear equation for S, the solubility of Cu(OH)_2: \(S \approx 1.55 \times 10^{-4} \, mol \, L^{-1}\) Hence, the solubility of Cu(OH)_2 in 5.0 M NH3 solution is approximately \(1.55 \times 10^{-4} \, mol \, L^{-1}\).