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The solubility product of MgF(2) is 7.4x...

The solubility product of `MgF_(2)` is `7.4xx10^(-11)`. Calculate the solubility of `MgF_(2)` in `0.1`M NaF solution

A

`7.4xx10^(-9)`

B

`3.7xx10^(-9)`

C

`3.7xx10^(-11)`

D

`7.4xx10^(-11)`

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The correct Answer is:
To calculate the solubility of \( MgF_2 \) in a \( 0.1 \, M \) NaF solution, we can follow these steps: ### Step 1: Write the dissociation equation for \( MgF_2 \) The dissociation of \( MgF_2 \) in water can be represented as: \[ MgF_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2F^{-} (aq) \] ### Step 2: Define the solubility product expression The solubility product \( K_{sp} \) for \( MgF_2 \) is given by: \[ K_{sp} = [Mg^{2+}][F^{-}]^2 \] Given \( K_{sp} = 7.4 \times 10^{-11} \). ### Step 3: Set up the equilibrium concentrations Let the solubility of \( MgF_2 \) in the \( 0.1 \, M \) NaF solution be \( S \). The concentration of \( Mg^{2+} \) will be \( S \) and the concentration of \( F^{-} \) will be \( 2S \) from the dissociation of \( MgF_2 \) plus \( 0.1 \, M \) from the NaF solution: \[ [F^{-}] = 2S + 0.1 \] ### Step 4: Substitute into the \( K_{sp} \) expression Substituting the equilibrium concentrations into the \( K_{sp} \) expression, we get: \[ K_{sp} = [Mg^{2+}][F^{-}]^2 = S(2S + 0.1)^2 \] Setting this equal to the given \( K_{sp} \): \[ 7.4 \times 10^{-11} = S(2S + 0.1)^2 \] ### Step 5: Expand and simplify the equation Expanding \( (2S + 0.1)^2 \): \[ (2S + 0.1)^2 = 4S^2 + 0.4S + 0.01 \] Thus, the equation becomes: \[ 7.4 \times 10^{-11} = S(4S^2 + 0.4S + 0.01) \] This simplifies to: \[ 7.4 \times 10^{-11} = 4S^3 + 0.4S^2 + 0.01S \] ### Step 6: Solve the cubic equation This is a cubic equation in \( S \). However, since \( S \) is expected to be small, we can neglect the higher order terms. Assuming \( S \) is small compared to \( 0.1 \): \[ 7.4 \times 10^{-11} \approx 0.01S \] Solving for \( S \): \[ S \approx \frac{7.4 \times 10^{-11}}{0.01} = 7.4 \times 10^{-9} \, M \] ### Conclusion The solubility of \( MgF_2 \) in \( 0.1 \, M \) NaF solution is approximately: \[ \boxed{7.4 \times 10^{-9} \, M} \]
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