Ncert Exercises & Solutions

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

{CaSO}_{4 (s)} \leftrightarrow {Ca}^{2 +}_{(aq)} + {SO}_{4}^{2 -}_{(aq)}

K_{sp} = [{Ca}^{2 +}] [{SO}_{4}^{2 -}]

Let the solubility of {CaSO}_{4} be s .

Then, K_{sp} = s^{2}

9.1 \times 10^{- 6} = s^{2}

s = 3.02 \times 10^{- 3} mol / L

Molecular mass of {CaSO}_{4} = 136 g / mol

Solubility of {CaSO}_{4} in gram / L

= 3.02 \times 10^{- 3} \times 136

= 0.41 g / L

This means that we need 1 L of water to dissolve 0.41 g of {CaSO}_{4}

Therefore, to dissolve 1 g of {CaSO}_{4} we require = \frac{1}{0.41} L = 2.44 L of water .