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

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)

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best_answer

We have,
The solubility product of calcium sulphate is 9.1\times 10^{-6}.
given mass of calcium sulphate = 1g

Ionization of calcium sulphate;

CaSO_4\rightleftharpoons Ca^{2+}+SO_{4}^{2-}
Therefore, 
K_s_p = [Ca^{2+}][SO_{4}^{2-}]

Let the solubility of calcium sulphate be s.
Then, 
\\K_{sp} = s^2\\ 9.1\times 10^{-6}=s^2\\ s = \sqrt{9.1\times 10^{-6}}
     =3.02\times 10^{-3} mol/L

Thus,
mass/ (mol. wt)\times volume =3.02\times 10^{-3} Molarity
mass = 3.02\times 136\times 10^{-3} = 0.41 g

So, that to dissolve 1 g of calcium sulphate we need = 1/0.41 L = 2.44 L of water.

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manish

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