The $K_{sp}$ of $CaF_2$ is $1.7 \times 10^{-10}$. What is the volume in $mL$ of a saturated solution containing $10 \ mg$ of $CaF_2$? (Molecular mass of $Ca = 40, F = 19$).
(D) $1$. Calculate the molar mass of $CaF_2$: $M = 40 + 2 \times 19 = 78 \ g/mol$.
$2$. The solubility expression for $CaF_2$ is $K_{sp} = 4s^3$.
$3$. $s = (K_{sp} / 4)^{1/3} = (1.7 \times 10^{-10} / 4)^{1/3} = (0.425 \times 10^{-10})^{1/3} = (42.5 \times 10^{-12})^{1/3} \approx 3.49 \times 10^{-4} \ mol/L$.
$4$. Moles of $CaF_2$ in $10 \ mg$ $(0.01 \ g)$ = $0.01 / 78 \approx 1.282 \times 10^{-4} \ mol$.
$5$. Volume of saturated solution = $\text{moles} / \text{solubility} = (1.282 \times 10^{-4}) / (3.49 \times 10^{-4}) \approx 0.367 \ L = 367 \ mL$.
$2$. The solubility expression for $CaF_2$ is $K_{sp} = 4s^3$.
$3$. $s = (K_{sp} / 4)^{1/3} = (1.7 \times 10^{-10} / 4)^{1/3} = (0.425 \times 10^{-10})^{1/3} = (42.5 \times 10^{-12})^{1/3} \approx 3.49 \times 10^{-4} \ mol/L$.
$4$. Moles of $CaF_2$ in $10 \ mg$ $(0.01 \ g)$ = $0.01 / 78 \approx 1.282 \times 10^{-4} \ mol$.
$5$. Volume of saturated solution = $\text{moles} / \text{solubility} = (1.282 \times 10^{-4}) / (3.49 \times 10^{-4}) \approx 0.367 \ L = 367 \ mL$.
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