Question

Consider the reaction \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) (a) Calculate \(K\) given that for \(\mathrm{Cu}(\mathrm{OH})_{2} K_{\mathrm{sp}}=2 \times 10^{-19}\) and for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} K_{\mathrm{f}}=2 \times 10^{12}\) (b) Determine the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) (in mol/L) in \(4.5 \mathrm{M} \mathrm{NH}_{3}\)

Short answer

Answer: The equilibrium constant, \(K\), for the reaction is \(4 \times 10^{-7}\). The solubility of copper hydroxide in 4.5 M ammonia is approximately \(1.8 \times 10^{-8} \mathrm{M}\).

Step-by-step solution

Write the equilibrium expressions for both constants

Write the equilibrium expressions for both \(K_{sp}\) and \(K_f\). For \(K_{sp}\) of \(\mathrm{Cu}(\mathrm{OH})_{2}\), the expression will be: \(K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{OH^{-}}]^2\) For \(K_f\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), the expression will be: \(K_f = \frac{[\mathrm{Cu}(NH_3)_4^{2+}]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}\)

Calculate the equilibrium constant, \(K\)

We can find the equilibrium constant, \(K\), by multiplying \(K_{sp}\) and \(K_f\). This gives: \(K = K_{sp} \times K_f = (2 \times 10^{-19})(2 \times 10^{12}) = 4 \times 10^{-7}\)

Write the expression for the reaction quotient, \(Q\)

The reaction quotient, \(Q\), is similar to the equilibrium constant, but it relates to the concentrations of the reactants and products before the system reaches equilibrium. For this reaction, the \(Q\) expression will be: \(Q = \frac{[\mathrm{Cu}(NH_3)_4^{2+}][\mathrm{OH^{-}}]^2}{[\mathrm{Cu}(\mathrm{OH})_{2}][\mathrm{NH_3}]^4}\)

Determine the solubility of copper hydroxide in 4.5 M ammonia

Let \(s\) be the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in the ammonia solution. When \([\mathrm{NH}_{3}] = 4.5 M\), we can set the reaction quotient, \(Q\), equal to the equilibrium constant, \(K\), substituting the concentrations with the solubility \(s\) to find the solubility: \(4 \times 10^{-7} = \frac{(s)(2s)^2}{(s)(4.5)^4}\) Solve for \(s\): \(s(4 \times 10^{-7}) = (2s)^3 (4.5)^{-4}\) \(s = \frac{4 \times 10^{-7} (4.5)^{-1}}{4}\) \(s = 1.8 \times 10^{-8} \mathrm{M}\) The solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in 4.5 M \(\mathrm{NH}_{3}\) is approximately \(1.8 \times 10^{-8} \mathrm{M}\).