Question

The concentration of $\left[A{g}^{+}\right]$ ion in a saturated solution of sparingly soluble salt $A{g}_{2}Cr{O}_4$ at equilibrium is $2.0\times {10}^{-4}$. Calculate the value of $K_{sp}$ for $A{g}_{2}Cr{O}_4$

• A. $4\times {10}^{-12}$
• B. $2\times {10}^{-6}$
• C. $2\times {10}^{-6}$
• D. $2\times {10}^{-11}$

Answer 1

The correct option is A $4\times {10}^{-12}$

Given, $\left[A{g}^{+}\right]=2.0\times {10}^{-4}$
The equilibrium between the undissolved solid $A{g}_{2}Cr{O}_4\left(s\right)$ and the ions in a saturated solution can be represented by the solution
$A{g}_{2}Cr{O}_4\left(s\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+Cr{O}_4^{2-}\left(aq\right)t=t_{eq}c-s2ss$

$K_{sp}=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}]$
Also, $\left[Cr{O}_4^{2-}\right]=\frac{1}{2}\left[A{g}^{+}\right]$
$\left[A{g}^{+}\right]=2s=2.0\times {10}^{-4}Ms=1.0\times {10}^{-4}M$

$\therefore \left[Cr{O}_4^{2-}\right]=s=1\times {10}^{-4}M$
So, $K_{sp}=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}]K_{sp}=(2.0\times {10}^{-4}{)}^2\times (1.0\times {10}^{-4})$
$K_{sp}=4\times {10}^{-12}$