The expression for the solubility product of Ag₂CO₃ will be

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$K_{sp} = S^{2}$

$K_{sp} = 27 S^{4}$

$K_{sp} = S$

$K_{sp} = 4 S^{3}$

answer is B.

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Detailed Solution

$\underset{}{{Ag}_{2}} {CO}_{3} (s) ⇌ \underset{2 S}{2 {Ag}^{+}} + \underset{S}{{CO}_{3}^{2 -}}$
$K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}] = (2 S)^{2} \cdot S ∴ K_{sp} = 4 S^{3}$

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