Question

If a saturated solution is prepared by dissolving $A{g}_{2}C{O}_3$ in water has $\left[A{g}^{+}\right]=2.56\times {10}^{-4}$. What is the value of $K_{sp}$ for $A{g}_{2}C{O}_3$?

• A. $2.56\times {10}^{-4}$
• B. $2.56\times {10}^{-8}$
• C. $8.39\times {10}^{-12}$
• D. $8.39\times {10}^{-10}$

Answer 1

The correct option is C $8.39\times {10}^{-12}$

Given, $\left[A{g}^{+}\right]=2.56\times {10}^{-4}$
In saturated solution of $A{g}_{2}C{O}_3\left(s\right)$
the dissociation equation is given by
$A{g}_{2}C{O}_3\left(s\right)+H_{2}O\rightleftharpoons 2A{g}^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$K_{sp}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]$
Also, $\left[C{O}_3^{2-}\right]=\frac{1}{2}\left[A{g}^{+}\right]$
$\left[A{g}^{+}\right]=2.56\times {10}^{-4}M$
$\therefore \left[C{O}_3^{2-}\right]=1.28\times {10}^{-4}M$
So $K_{sp}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]=(2.56\times {10}^{-4}{)}^2\times (1.28\times {10}^{-4})$
$K_{sp}=8.39\times {10}^{-12}$