Question

If the solubility of Ag2CrO4 is S mole/litre. Its solubility product is

Answer 1

4s3 as ag2cro4- 2ag+ + cro4^2-

S 2s s

Ksp= 2s×2s×s

=4s3

Answer 2

Answer: The solubility product of $Ag_2CrO_4$ is $4s^3$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the silver chromate is given as:

$Ag_2CrO_4\leftrightharpoons 2Ag^++CrO_4^{2-}$

We are given:

Solubility of $Ag_2CrO_4$ = S mol/L

By stoichiometry of the reaction:

1 mole of $Ag_2CrO_4$ gives 2 moles of $Ag^{+}$ and 1 mole of $CrO_4^{2-}$.

When the solubility of $Ag_2CrO_4$ is S moles/liter, then the solubility of $Ag^{+}$ will be 2S moles\liter and solubility of $CrO_4^{2-}$ will be S moles/liter.

Expression for the equilibrium constant of $Ag_2CrO_4$ will be:

$K_{sp}=[Ag^+]^2[CrO_4^{2-}]$

$K_{sp}=[2s]^2[s]=4s^3$

Hence, the solubility product of $Ag_2CrO_4$ is $4s^3$