Question

What is the minimum volume of water required to dissolve 1g of calcium sulfate at 298 K? (For calcium sulfate, K_sp is 9.1 × 10^–6).

Answer 1

CaSO4 ⇌ Ca^2+ + SO4^2-, Ksp = 4s^2 = 9.1 * 10^{-6} 

\( s = \sqrt{\frac{9.1 \times 10^{-6}}{4}} = 1.51 \times 10^{-3} \) M

Volume

Moles CaSO₄ = \( \frac{1}{136.14} = 7.34 \times 10^{-3} \)

\( [\ce{CaSO4}] = 2s = 3.02 \times 10^{-3} \) M

V = \( \frac{7.34 \times 10^{-3}}{3.02 \times 10^{-3}} = 2.43 \) L = 2435 mL