To determine the solubility of Cu(IO₃)₂ given its solubility product (Ksp) of 1.08 × 10⁻⁷, we need to follow a systematic approach. Let's break it down step-by-step. ### Step 1: Write the Dissociation Equation When Cu(IO₃)₂ dissolves in water, it dissociates into its constituent ions: \[ \text{Cu(IO}_3\text{)}_2 (s) \rightleftharpoons \text{Cu}^{2+} (aq) + 2 \text{IO}_3^{-} (aq) \] ### Step 2: Express the Solubility Product (Ksp) The solubility product expression for this dissociation can be written as: \[ K_{sp} = [\text{Cu}^{2+}][\text{IO}_3^{-}]^2 \] Let \( s \) be the solubility of Cu(IO₃)₂ in mol/dm³. When 1 mole of Cu(IO₃)₂ dissolves, it produces 1 mole of Cu²⁺ and 2 moles of IO₃⁻. Therefore, we can express the concentrations of the ions in terms of \( s \): - \([\text{Cu}^{2+}] = s\) - \([\text{IO}_3^{-}] = 2s\) ### Step 3: Substitute into the Ksp Expression Now, substituting these expressions into the Ksp equation gives: \[ K_{sp} = [\text{Cu}^{2+}][\text{IO}_3^{-}]^2 = s(2s)^2 \] This simplifies to: \[ K_{sp} = s(4s^2) = 4s^3 \] ### Step 4: Set Up the Equation Now we can set this equal to the given Ksp value: \[ 4s^3 = 1.08 \times 10^{-7} \] ### Step 5: Solve for s To find \( s \), we first divide both sides by 4: \[ s^3 = \frac{1.08 \times 10^{-7}}{4} = 2.7 \times 10^{-8} \] Next, we take the cube root of both sides: \[ s = \sqrt[3]{2.7 \times 10^{-8}} \] Calculating this gives: \[ s \approx 3.0 \times 10^{-3} \, \text{mol/dm}^3 \] ### Conclusion: Correct Answer Thus, the solubility of Cu(IO₃)₂ is approximately **3.0 × 10⁻³ mol/dm³**, which corresponds to option **C**. ### Explanation of Other Options - **Option A (2.7 × 10⁻⁸ mol/dm³)**: This value is too low and does not account for the stoichiometry of the dissociation reaction. - **Option B (9.0 × 10⁻⁸ mol/dm³)**: This value is also incorrect as it does not reflect the correct calculation of the solubility product. - **Option D (9.0 × 10⁻³ mol/dm³)**: This value is too high and does not match the calculated solubility. ### Revision Summary - The solubility product (Ksp) is used to find the solubility of sparingly soluble salts. - Write the dissociation equation and express ion concentrations in terms of solubility (s). - Substitute into the Ksp expression and solve for s. - Always check the stoichiometry of the dissociation to ensure correct calculations.