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 \(0.10 M\)

First, we write the balanced chemical equation for the dissolution of CuCl in water: \[ \mathrm{CuCl}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q) + \mathrm{Cl}^{-}(a q) \] Now, we can write the expression for the solubility product, \(K_{sp}\), for this reaction: \[ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}] \] Since the stoichiometry of the reaction is 1:1, we can let the solubility of CuCl in water be \(s\) and then we have: \[ K_{sp} = (s)(s) = s^2 \] Using the given value for \(K_{sp}\), we can solve for the solubility of CuCl: \[ s^2 = 1.2 \times 10^{-6} \] \[ s = \sqrt{1.2 \times 10^{-6}} \]

To find the solubility of copper(I) chloride in pure water, we just need to find the value of \(s\): \[ s = \sqrt{1.2 \times 10^{-6}} \approx 1.1 \times 10^{-3}\, M \] So the solubility of copper(I) chloride in pure water is approximately \(1.1 \times 10^{-3}\, M\).

Now, we need to consider the complex ion formation as well as the dissolution of CuCl in water. The balanced chemical equation for the complex ion formation is: \[ \mathrm{Cu}^{+}(a q) + 2\, \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \] We are given that the concentration of chloride ions, [\(\mathrm{Cl}^{-}\)], is 0.10 M. Let \(x\) be the solubility of CuCl in 0.10 M chloride solution. Since the complex formation involves 1 Cu+ ion and 2 Cl- ions, we can write the concentrations of these ions as: \[[\mathrm{Cu}^{+}] = x \] \[[\mathrm{Cl}^{-}] = 0.10 + 2x \] Now, we can write the expression for the equilibrium constant, \(K\) for the complex ion formation: \[ K = \frac{[\mathrm{CuCl}_{2}^{-}]}{[\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]^2 } \] Since the stoichiometry is 1:1, we can let the concentration of CuCl2- be \(x\), and then we get: \[ K = \frac{x}{x(0.10 + 2x)^2 } \]

Using the given value of \(K\), we can solve the equation and get the value of \(x\), which represents the solubility of CuCl in 0.10 M chloride solution: \[\frac{x}{x(0.10 + 2x)^2 } = 8.7 \times 10^{4} \] \[ 0.10 + 2x = \sqrt[3]{\frac{1}{8.7 \times 10^{4}}} \] \[ x = \frac{1}{2}\left (\sqrt[3]{\frac{1}{8.7 \times 10^{4}}} - 0.10\right) \approx 1.9 \times 10^{-4}\, M \] So the solubility of copper(I) chloride in 0.10 M chloride solution is approximately \(1.9 \times 10^{-4}\, M\).