Given that the solubility product constant (Ksp) of La(IO3)3 is 7.5 × 10^{-12}, what is the molar solubility of La(IO3)3 in pure water?

• A. 1.2 × 10^{-3} mol/L
• B. 1.3 × 10^{-3} mol/L
• C. 1.2 × 10^{-3} mol/L
• D. 1.2 × 10^{-4} mol/L

Answer: (A)

Verified step by step guidance

1. Write the dissociation equation for La(IO3)3 in water: \(\mathrm{La(IO_3)_3 (s) \rightleftharpoons La^{3+} (aq) + 3 IO_3^- (aq)}\).
2. Define the molar solubility as \(s\), which represents the concentration of \(\mathrm{La^{3+}}\) ions in solution at equilibrium. Then, the concentration of \(\mathrm{IO_3^-}\) ions will be \$3s$ because of the 1:3 stoichiometric ratio.
3. Write the expression for the solubility product constant \(K_{sp}\) in terms of \(s\): \(K_{sp} = [\mathrm{La^{3+}}][\mathrm{IO_3^-}]^3 = s \times (3s)^3 = s \times 27s^3 = 27s^4\).
4. Set the expression equal to the given \(K_{sp}\) value and solve for \(s\): \$7.5 \times 10^{-12} = 27s^4$.
5. Isolate \(s\) by dividing both sides by 27 and then take the fourth root: \(s = \sqrt[4]{\frac{7.5 \times 10^{-12}}{27}}\). This \(s\) is the molar solubility of La(IO3)3 in pure water.