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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium

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) $2S_3$

2 Answers

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

 

answered Nov 23, 2013 by mosymeow_1
 
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 $K_{sp} =[2s]^2[s]=4s^3$
answered Feb 21 by meena.p
 

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