The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

The balanced chemical equation for the dissolution of copper(I) chloride in water is:\[ \mathrm{CuCl}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q)+\mathrm{Cl}^{-}(a q) \] The \(K_{sp}\) expression for this reaction is: \[ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}] \]

Let the solubility of copper(I) chloride in water be denoted as \(x\) mol/L. Then, based on the balanced chemical equation, the concentrations of \(\mathrm{Cu}^{+}\) and \(\mathrm{Cl}^{-}\) will also be \(x\) mol/L. Plugging in the values into the \(K_{sp}\) expression: \[ 1.2 \times 10^{-6} = (x)(x) \] Solving for \(x\), we get: \[ x = \sqrt{1.2 \times 10^{-6}} \approx 1.1 \times 10^{-3}\,\text{mol/L} \] Thus, the solubility of copper(I) chloride in pure water is approximately \(1.1 \times 10^{-3}\,\text{mol/L}\).

The balanced chemical equation for the formation of the complex ion CuCl2- is: \[ \mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \] And its equilibrium constant expression is: \[ K = \frac{[\mathrm{CuCl}_{2}^{-}]}{[\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]^2} \]

Since the NaCl solution is 0.10 M, it means that the initial concentration of \(\mathrm{Cl}^{-}\) is \(0.10\) mol/L. To find the solubility of copper(I) chloride in the NaCl solution, we first assume the solubility to be \(y\) mol/L, and the concentration of \(\mathrm{Cu}^{+}\) to be \(y\) mol/L. Considering the formation of the complex ion, \[ [\mathrm{CuCl}_{2}^{-}] = K \times [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]^2 = (8.7 \times 10^4) \times (y)(0.10 + y)^2 \] Now, we apply the solubility product constant (\(K_{sp}\)) to find solubility: \[ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-} + 2[\mathrm{CuCl}_{2}^{-}]] \] Plugging in the values: \(1.2 \times 10^{-6} = y(0.10 + y + 2(8.7 \times 10^4)(y)(0.10 + y)^2)\) Since \(y\) is very small compared to 0.10, we can assume that \(y << 0.10\) and simplify the equation: \(1.2 \times 10^{-6} \approx y(0.10)\) Solving for y, we get: \[ y \approx \frac{1.2 \times 10^{-6}}{0.10} = 1.2 \times 10^{-5}\,\text{mol/L} \] Thus, the solubility of copper(I) chloride in 0.10 M NaCl solution is approximately \(1.2 \times 10^{-5}\,\text{mol/L}\).