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).

Asked by Pragya Singh | 1 year ago |  153

##### Solution :-

$$CaSO_{4(s)}\leftrightarrow Ca^{2+}_{(aq)} + SO_{4}^{2-}(aq)$$

Ksp = $$[Ca^{2+}][SO_{4}^{2-}]$$

Assuming the solubility of calcium sulphate is x. So,

Ksp = x2

$$\;9.1\times 10^{-6} = x^{2}$$

$$9.1×10−6=x^2$$

$$∴ \;x = 3.02\times 10^{-3} mol/L$$

$$∴x=3.02×10−3mol/L$$

Now, molecular mass os calcium sulphate is 136g/mol. Solubility in calcium sulphate in g/mol is

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

$$=3.02×10−3×136$$

$$= 0.41 g/L$$

i.e. 1 litre H2O will be required to dissolve 0.41g of calcium sulphate.Thus, minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K is

=$$\dfrac{1}{0.41} L = 2.44 L$$

Answered by Abhisek | 1 year ago

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