Multiple Choice
Calculate the molar solubility of barium fluoride (BaF2) in pure water, given that its solubility product constant (Ksp) is 2.45 x 10^-5.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
Calculate the molar solubility of magnesium fluoride (MgF2) in a solution that is 0.350 M in NaF. For magnesium fluoride, Ksp = 5.16 x 10^-11.
A
1.47 x 10^-10 M
B
2.35 x 10^-4 M
C
4.21 x 10^-10 M
D
2.05 x 10^-5 M
Verified step by step guidance1
Start by writing the balanced dissolution equation for magnesium fluoride (MgF2): \( \text{MgF}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{F}^- (aq) \).
Express the solubility product constant (Ksp) for MgF2 in terms of the concentrations of the ions: \( K_{sp} = [\text{Mg}^{2+}][\text{F}^-]^2 \).
Let the molar solubility of MgF2 be \( s \). In the presence of 0.350 M NaF, the fluoride ion concentration becomes \( [\text{F}^-] = 0.350 + 2s \).
Substitute the expressions for the ion concentrations into the Ksp expression: \( K_{sp} = s(0.350 + 2s)^2 \).
Assume \( s \) is small compared to 0.350 M, so \( (0.350 + 2s) \approx 0.350 \). Simplify the expression to \( K_{sp} = s(0.350)^2 \) and solve for \( s \).
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