When 125 mL of a magnesium nitrate solution is mixed with 175 mL of a 1.50 mol L$^{-1}$ sodium fluoride solution, 0.6231 g of magnesium fluoride (MM = 62.31 g mol$^{-1}$) precipitates - HSC - SSCE Chemistry - Question 34 - 2023 - Paper 1

Next, we calculate the initial moles of sodium fluoride (NaF) after mixing the two solutions:
Initial n(NaF) = 0.175 \text{ L} \times 1.50 \text{ mol L}^{-1} = 0.263 \text{ mol}

The moles of fluoride left after precipitation can be calculated:
n(F^-) = 0.263 \text{ mol} - 2 \times 1.000 \times 10^{-2} \text{ mol} = 0.243 \text{ mol}
The concentration of fluoride after precipitation:
[\text{F}^-] = \frac{0.243 \text{ mol}}{0.300 \text{ L}} = 0.810 \text{ mol L}^{-1}

Given that the solubility product $K_{sp}$ is related to the concentrations of the ions in solution, we define the expression:
K_{sp} = [\text{Mg}^{2+}][\text{F}^-]^2
Substituting the values into the expression:
5.16 \times 10^{-11} = \text{Mg}^{2+}^2
From here, we can isolate [Mg$^{2+}$]:
[\text{Mg}^{2+}] = \frac{5.16 \times 10^{-11}}{(0.810)^2} = 7.90 \times 10^{-11} \text{ mol L}^{-1}