CaSO4 with Ksp-4-9- 10-5 is found to have 4- 10-7 M in Ca2- and 2-5- 10-8 M in SO42- The equilibrium state of this solution with respect to CaSO4 is -

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Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

CaSO4 with Ks...

Question

$(C a S O_{4})$ with $K_{s p} = 4.9 \times 10^{- 5}$ is found to have $4 \times 10^{- 7} M$ in $C a^{2 +}$ and $2.5 \times 10^{- 8} M$ in $S O_{4}^{2 -}$ .
The equilibrium state of this solution with respect to $(C a S O_{4})$ is :

Unsaturated solution

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Saturated solution

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Supersaturated solution

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Cannot be determined

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Solution

The correct option is A Unsaturated solution
The equilibrium established will be:
$C a S O_{4} (s) ⇌ C a^{2 +} (a q .) + S O_{4}^{2 -} (a q .)$

Finding the ion product $Q_{s p}$
$Q_{s p} = (4 \times 10^{- 7}) \times (2.5 \times 10^{- 8}) Q_{s p} = 10^{- 14} < K_{s p} = 4.9 \times 10^{- 5}$

Since, $\frac{Q s p}{K_{s p}} < 1$
$\Rightarrow$ The solution is unsaturated in $C a S O_{4} .$

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