If the salts $M_{2} X, Q Y_{2}$ and $P Z_{3}$ have the same solubilities, their $K_{s p}$ values are related as:

K_{s p} (M_{2} X) = K_{s p} (Q Y_{2}) < K_{s p} (P Z_{3})

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K_{s p} (M_{2} X) > K_{s p} (Q Y_{2}) = K_{s p} (P Z_{3})

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K_{s p} (M_{2} X) = K_{s p} (Q Y_{2}) = K_{s p} (P Z_{3})

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K_{s p} (M_{2} X) > K_{s p} (Q Y_{2}) > K_{s p} (P Z_{3})

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Solution

The correct option is A $K_{s p} (M_{2} X) = K_{s p} (Q Y_{2}) < K_{s p} (P Z_{3})$
$M_{2} X ⇌ 2 M^{+} + X^{2 -}$ $K_{s p} = [2 S]^{2} [S] = 4 S^{3}$
$Q Y_{2} ⇌ Q^{2 +} + 2 X^{-}$ $K_{s p} = [S] [2 S]^{2} = 4 S^{3}$
$P Z_{3} ⇌ P^{3 +} + 3 X^{-}$ $K_{s p} = [S] [3 S]^{3} = 27 S^{4}$
So $K_{s p} (M_{2} X) = K_{s p} (Q Y_{2}) < K_{s p} (P Z_{3})$

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