Multiple Choice
What is the concentration of Al³⁺ ions in a saturated solution of Al(OH)₃ with a Ksp of 1.3 × 10⁻³³?
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
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
Verified step by step guidance1
Write the dissociation equation for La(IO3)3 in water: \(\mathrm{La(IO_3)_3 (s) \rightleftharpoons La^{3+} (aq) + 3 IO_3^- (aq)}\).
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.
Write the expression for the solubility product constant \(K_{sp}\) in terms of \(s\):
\(K_{sp} = [\mathrm{La^{3+}}][\mathrm{IO_3^-}]^3 = s \times (3s)^3 = s \times 27s^3 = 27s^4\).
Set the expression equal to the given \(K_{sp}\) value and solve for \(s\):
\(7.5 \times 10^{-12} = 27s^4\).
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.
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