What is the minimum volume of pure water (in Liters) required to completely dissolve exactly of soli — Ionic Equilibrium Chemistry Question
Question
What is the minimum volume of pure water (in Liters) required to completely dissolve exactly $1 \text{ g}$ of solid calcium sulphate ($CaSO_4$) at $298 \text{ K}$? (Given $K_{sp}$ for $CaSO_4 = 9 \times 10^{-6}$ and Molar Mass $= 136 \text{ g/mol}$). If the volume is $V$ Liters, calculate the value of $V$ accurate to two decimal places.
💡 Solution & Explanation
Let the molar solubility of $CaSO_4$ in pure water be $S$. $K_{sp} = S^2 \Rightarrow S = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3} \text{ mol/L}$. The number of moles in $1 \text{ g}$ of $CaSO_4$ is $n = \frac{\text{mass}}{\text{molar mass}} = \frac{1}{136} \text{ moles}$. Since Molarity $S = \frac{n}{V}$, the required volume $V = \frac{n}{S} = \frac{1/136}{3 \times 10^{-3}} = \frac{1}{136 \times 3 \times 10^{-3}} = \frac{1}{0.408} \approx 2.45 \text{ L}$.