Question

At 298 K, $ K_{\text{sp}} $ of Cu(OH)₂ is $ 2.2 \times 10^{-20} $. What is its solubility in $ 10^{-7} \, \text{M} \, \text{CuCl}_2 $ solution?

• A. \( 7.4 \times 10^{-10} \)
• B. \( 5.5 \times 10^{-18} \)
• C. \( 1.1 \times 10^{-5} \)
• D. \( 7.4 \times 10^{-9} \)

Hint:

For solubility problems, remember to write out the equilibrium expression for \( K_{\text{sp}} \) and substitute the known concentrations.

Answer 1

The Correct Option is A

We are given the solubility product of Cu(OH)₂, \( K_{\text{sp}} = 2.2 \times 10^{-20} \), and the concentration of Cu²⁺ in solution from CuCl₂, which is \( 10^{-7} \, \text{M} \). The solubility of Cu(OH)₂ in this solution is determined by calculating the concentration of hydroxide ions from the given \( K_{\text{sp}} \). For the reaction: \[ \text{Cu(OH)}_2 (s) \rightleftharpoons \text{Cu}^{2+} (aq) + 2 \text{OH}^- (aq) \] At equilibrium, the concentration of Cu²⁺ is \( 10^{-7} \, \text{M} \), and the concentration of OH⁻ is derived using \( K_{\text{sp}} \). The correct solubility can be calculated by applying the relation: \[ K_{\text{sp}} = [\text{Cu}^{2+}] [\text{OH}^-]^2 \] Solving for the concentration of OH⁻ gives us the correct answer. The result matches \( 7.4 \times 10^{-10} \).