Question

What is the molar solubility of ${Ag}_2{CO}_3(K_{sp}=4\times {10}^{-13})$ in 0.1M ${Na}_2{CO}_3$ solution?

• A. ${10}^{-6}$
• B. ${10}^{-7}$
• C. $2\times {10}^{-6}$
• D. $2\times {10}^{-7}$

Answer 1

The correct option is A ${10}^{-6}$

$K_{sp}\left({Ag}_{2}C{O}_3\right)=4\times {10}^{-13}$
conc. of ${Ag}_{2}C{O}_3$ solution - 0.M
Firstly, write the equation for ${Ag}_{2}C{O}_3$

	${Ag}_{2}C{O}_3$	$\rightleftharpoons$	$2{Ag}^{+}$	$+$	${CO}_3^{2-}$
initial concentration			$0$		$0.1$
final concentration			$2\alpha$		$(0.1+\alpha )$

Let $\alpha$ be the amount dissociated in moles,
The initial conc. of ${CO}_3^{2-}$ in the solution is from ${Na}_2{CO}_3$ solution.
Now, $K_{sp}=\left[cation\right]\left[anion\right]$ at equilibrium
$K_{sp}\left({Ag}_{2}C{O}_3\right)={\left[{Ag}^{+}\right]}^2\left[{CO}_3^{2-}\right]$
Substituting equilibrium concentrations, we get
$4\times {10}^{-13}={\left(2\alpha \right)}^2(0.1+\alpha )Since\alpha <<0.1,(0.1+\alpha )\approx \left(0.1\right)4\times {10}^{-13}=4{\alpha }^2\times 0.1\frac{{10}^{-13}}{0.1}={\alpha }^2\alpha =\sqrt{{10}^{-12}}\alpha ={10}^{-6}M$