Multiple Choice
What is the molar solubility of barium fluoride (BaF2) in a 0.15 M NaF solution, given that the solubility product constant (Ksp) for BaF2 is 2.45 × 10^-5?
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
Given that the solubility product constant (Ksp) of PbF2 is 3.6 × 10^{-8}, what is the molar solubility of PbF2 in pure water?
A
2.1 × 10^{-3} mol/L
B
3.6 × 10^{-8} mol/L
C
1.9 × 10^{-3} mol/L
D
1.1 × 10^{-3} mol/L
Verified step by step guidance1
Write the dissociation equation for PbF\_2 in water: \(\mathrm{PbF_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2F^- (aq)}\).
Define the molar solubility of PbF\_2 as \(s\), which represents the concentration of \(\mathrm{Pb^{2+}}\) ions in solution at equilibrium.
Express the equilibrium concentrations of the ions in terms of \(s\): \([\mathrm{Pb^{2+}}] = s\) and \([\mathrm{F^-}] = 2s\) because two fluoride ions are produced per formula unit dissolved.
Write the expression for the solubility product constant \(K_{sp}\) in terms of \(s\): \(K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{F^-}]^2 = s \times (2s)^2 = 4s^3\).
Solve for \(s\) by rearranging the equation: \(s = \sqrt[3]{\frac{K_{sp}}{4}}\). Substitute the given \(K_{sp} = 3.6 \times 10^{-8}\) to find the molar solubility.
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