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 \(5.0M\) \(\mathrm{NH}_{3} .\) In \(5.0M\) \(\mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\) .

Step by step solution

01

Find the equilibrium constant K for the reaction

: According to the given information, we have the Ksp value for Cu(OH)₂ and the overall formation constant for Cu(NH₃)₄²⁺ as: \(K_{sp}(\mathrm{Cu(OH)_2}) = 1.6 \times 10^{-19}\) \(β_{4}(\mathrm{Cu(NH_3)_4^{2+}}) = 1.0 \times 10^{13}\) For the following reaction, $$\mathrm{Cu(OH)_2}(s)+4 \mathrm{NH_3}(aq) \rightleftharpoons \mathrm{Cu(NH_3)_4^{2+}}(aq)+2 \mathrm{OH^-}(aq)$$ the equilibrium constant K can be found using Ksp and β₄ as follows: $$K = \frac{1}{K_{sp}(\mathrm{Cu(OH)_2})} \times β_{4}(\mathrm{Cu(NH_3)_4^{2+}})$$

02

Calculate the equilibrium constant K

: Plug the values of Ksp and β₄ into the formula, we get: $$K = \frac{1}{1.6 \times 10^{-19}} \times 1.0 \times 10^{13}$$ $$K = 6.25 \times 10^{31}$$ Now that we have the equilibrium constant K, we can proceed to calculate the solubility of Cu(OH)₂ in 5.0M NH₃.

03

Write down the expression for K in terms of concentrations

: The equilibrium constant K for the reaction can be expressed as: $$K = \frac{[\mathrm{Cu(NH_3)_4^{2+}}][\mathrm{OH^-}]^2}{[\mathrm{NH_3}]^4}$$

04

Set up the solubility expression

: From the equilibrium expression, the solubility of Cu(OH)₂, denoted by S, is related to the equilibrium concentrations as follows: $$[\mathrm{Cu(NH_3)_4^{2+}}] = S$$ $$[\mathrm{OH^-}] = S + 0.0095$$ $$[\mathrm{NH_3}] = 5.0 - 4S$$ Now we can rewrite the K expression in terms of S: $$K = \frac{S(S + 0.0095)^2}{(5.0 - 4S)^4}$$

05

Solve for S

: Insert the value of K and solve for S: $$6.25 \times 10^{31} = \frac{S(S + 0.0095)^2}{(5.0 - 4S)^4}$$ Approximating 5.0 − 4S to 5 since S is relatively small, we get: $$6.25 \times 10^{31} = \frac{S(S + 0.0095)^2}{5^4}$$ $$S(S + 0.0095)^2 = 6.25 \times 10^{31} \times 625$$ Now, solve for S: $$S ≈ 7.5 \times 10^{-6} \,\mathrm{mol/L}$$ So, the solubility of Cu(OH)₂ in a 5.0M NH₃ solution is approximately 7.5 x 10⁻⁶ mol/L.

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