Question

The Ksp of Ag2CrO4 is 1.1*10 ^(-12) at 298 K. The solubility (in mol/l) of Ag2CrO4 in a 0.1M AgNO3 solution is-

Answer 1

Write down the dissociation reaction of Ag2CrO4.Ag2CrO4------>2Ag+ + CrO42-Due to common ion effect,Add the moles of AgNO3 to the solubility const. of Ag+Therefore, net Ag+ ions = 2s+0.1. (Let solubility const. be s)Now,(2s+0.1)^2×s=1.1×10^-12As s s=1.1×10^-10

Answer 2

Answer: $1.1\times 10^{-11}M$

Explanation: The addition of common ion such as chloride results in common ion effect which reduces the solubility of the salt.

The equation for the reaction will be as follows:

$Ag_2CrO_4\leftrightharpoons 2Ag^++Cr_2O_4^{2-}$

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

Thus if solubility of $Ag_2CrO_4$ is s moles/liter, solubility of  $Ag^{+}$ is 2s moles\liter and solubility of $CrO_4^{2-}$ is s moles/liter

Therefore,  

$K_sp=[Ag^+]^2[CO_3^{2-}]$

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

Now , $AgNO_3\rightarrow Ag^++NO_3^-$

1 mole of $AgNO_3$ gives 1 mole of $Ag^+$ and $NO_3^-$

0.1 moles of  $AgNO_3$ will give 0.1 mole of  $Ag^+$ and 0.1 mole of $NO_3^-$

Thus now $[Ag^+]$=(2s+0.1) M and $[NO_3^-]$= s M

$K_{sp}=[2s+0.1]^2[s]$

as 2s<<<0.1, $K_{sp}=[0.1][s]$

$1.1\times 10^{-12}=[0.1][s]$

$s=\frac{1.1\times 10^{-12}}{0.1}=1.1\times 10^{-11}M$