The solubility of $C a {S O}_{4}$ at $25^{o} C$ is $2.036 g L^{- 1}$ and degree of dissociation of the saturated solution at this temperature is $55$ %. $K_{s p}$ of $C a {S O}_{4}$ is :
(Mol. mass of $C a {S O}_{4} = 136$ )

6.8 \times 10^{- 5}

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3.06 \times 10^{- 5}

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6.12 \times 10^{- 4}

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3.06 \times 10^{- 6}

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Solution

The correct option is A $6.8 \times 10^{- 5}$
The solubility (s) of $C a S O_{4} = 2.036 g / L$
$= \frac{2.036}{136} m o l / L = 0.015 M$
Since degree of dissociation is $55$ % .
$∴ [C a^{2 +}] = [S O_{4}^{2 -}] = 0.015 \times 0.55 M = 8.25 \times 10^{- 3} M$
$∴ K_{s p} = [C a^{2 +}] [S O_{4}^{2 -}] = (8.25 \times 10^{- 3})^{2} \approx 6.8 \times 10^{- 5}$
The value is approximately equal to Option A.

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