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Q)

If the solubility of $Ag_2CrO_4$ is S mole/litre. Its solubility product is

(a) $S_2$

(b) $S_3$

(c) $4S_3$

(d) $2S_3$

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A)
Toolbox:
• Dissociation of an electrolyte, $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
• Solubility Product, $K_{sp} = \frac{[A^{y+}]^x [B^{x-}]^y}{[A_x B_y]}$
• In saturated solution, $[A_x B_y]$ = K' = constant

Answer: $4S^3$

Dissociation of $Ag_2CrO_4$
$Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^-$
$[Ag^+]$ = 2s
$[CrO_4^{-2}]$ = s

Solubility Product, $K_{sp} = [Ag^+]^2 [CrO_4^{-2}] = (2S)^2 (S) = 4S^3$

Comment
A)
Solution :
The solubility product is the product of the molar concentration of the ions in the saturated solution raised to appropriate powers
We begin with the equilibrium equation and $Ksp$ expression.
$Ag_2CrO_4 \to 2Ag ^{+} +CrO_4^{2--}$
Given solubility is S,
then $K_{sp} =[2s]^2[s]=4s^3$