Question

If solubility product of a sparingly soluble salt, $A{g}_{2}S{O}_3$ is $8\times {10}^{-14}$ at ${25}^{o}C$ . Calculate the concentration of $A{g}^{+}\left(aq\right)$ in its saturated solution at equilibrium.
Given : $\sqrt[3]{320}=2.71$

• A. $4\times {10}^{-14}$
• B. $2.82\times {10}^{-7}$
• C. $1.41\times {10}^{-7}$
• D. $5.42\times {10}^{-5}$

Answer 1

The correct option is D $5.42\times {10}^{-5}$

Given $K_{sp}=6.6\times {10}^{-6}$
Lets consider solid $A{g}_{2}S{O}_3$ is in contact with its saturated aqueous solution.The equilibrium is established between undissolved $A{g}_{2}S{O}_3$ and its ions. The equilibrium reaction is given by :
$A{g}_{2}S{O}_3\left(s\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+S{O}_3^{2-}\left(aq\right)t=0c00t=t_{eq}c-s2ss$

$K_{sp}=\left[A{g}^{+}{]}^2\right[S{O}_3^{2-}]$
$K_{sp}=(2s{)}^2\times s=4s^3$
$K_{sp}=4s^3=8\times {10}^{-14}$
$s^3=20\times {10}^{-15}$
$s=2.71\times {10}^{-5}$
$\left[A{g}^{+}\right]\left(aq\right)=2s=2\times (2.71\times {10}^{-5})\left[A{g}^{+}\right]\left(aq\right)=5.42\times {10}^{-5}$