Question 10

1. Calculate initial millimoles:

Millimoles of $Ba^{2+} = 200 \times 0.010 = 2 \text{ mmol}$.

Millimoles of $IO_3^- = 100 \times 0.10 = 10 \text{ mmol}$. Total volume = $300 \text{ mL}$.

2. Precipitation reaction: $Ba^{2+} + 2IO_3^- \rightarrow Ba(IO_3)_2(s)$.

$Ba^{2+}$ is the limiting reagent ($2 \text{ mmol}$ reacts with $4 \text{ mmol}$ $IO_3^-$). Remaining $IO_3^- = 10 - 4 = 6 \text{ mmol}$.

Final concentration of common ion $[IO_3^-] = \frac{6 \text{ mmol}}{300 \text{ mL}} = 0.02 \text{ M}$.

3. Let the solubility of $Ba(IO_3)_2$ in this solution be $s$. $K_{sp} = [Ba^{2+}][IO_3^-]^2 = s(0.02 + 2s)^2$.

Since $s$ is very small, $0.02 + 2s \approx 0.02$. $1.58 \times 10^{-9} = s(0.02)^2 = s(4 \times 10^{-4})$.

$s = \frac{1.58 \times 10^{-9}}{4 \times 10^{-4}} = 0.395 \times 10^{-5} = 3.95 \times 10^{-6} \text{mol dm}^{-3}$.

Therefore, $X = 3.95$.