Question

What is the molar solubility of $A{g_2}C{O_3}$ (${K_{sp}} = 4 \times {10^{ - 13}}$) in 0.1 M $N{a_2}C{O_3}$ solution?
A. ${10^{ - 6}}$
B. ${10^{ - 7}}$
C. $2 \times {10^{ - 6}}$
D. $2 \times {10^{ - 7}}$

Answer 1

Hint: The molar solubility is related with the solubility product and defined as the maximum moles of solute which can dissolve in the one liter solution. The molar solubility product is the ratio of products concentration present in equilibrium.

Complete step by step answer:
Given,
The ${K_{sp}}$ of $A{g_2}C{O_3}$is $4 \times {10^{ - 13}}$
The molarity of $N{a_2}C{O_3}$ solution is 0.1 M.
Silver carbonate is an ionic compound which on dissolving in water dissociates into silver cation and carbonate anion.
The dissociation reaction of $A{g_2}C{O_3}$ is shown below.
$A{g_2}C{O_3} \rightleftharpoons 2A{g^ + } + CO_3^{2 - }$
The initial concentration of $A{g^ + }$ is 0 and $CO_3^{2 - }$ is 0.1.
The final concentration of $A{g^ + }$ is $2\alpha$ and $CO_3^{2 - }$ is $(0.1 + \alpha )$.
$\alpha$ denote the dissociated moles in the solution.
The initial concentration of $CO_3^{2 - }$ is obtained from the concentration $N{a_2}C{O_3}$.
The ${K_{sp}}$ of the solution is known as the solubility product constant which is an equilibrium constant for the solid compound which dissolves in the solution.
In solubility product constant, the concentration of the ionic product is multiplied and if any coefficient is present in front of the product it is raised with the coefficient power.
The ${K_{sp}}$ of the $A{g_2}C{O_3}$solution is given as shown below.
${K_{sp}} = {[A{g^ + }]^2}[CO_3^{2 - }]$
Substitute the values of concentration in the above equation.
$\Rightarrow 4 \times {10^{ - 13}} = {[2\alpha ]^2}[0.1 + \alpha ]$
As, $\alpha$ is less than 0.1 ($\alpha < < 0.1$) and $(0.1 + \alpha ) \approx (0.1)$
Then, the equation is written as shown below.
$\Rightarrow 4 \times {10^{ - 13}} = 4{\alpha ^2} \times 0.1$
$\Rightarrow {\alpha ^2} = \dfrac{{{{10}^{ - 13}}}}{{0.1}}$
$\Rightarrow \alpha = \sqrt {{{10}^{ - 12}}}$
$\Rightarrow \alpha = {10^{ - 6}}$
Thus, the molar solubility of $A{g_2}C{O_3}$ (${K_{sp}} = 4 \times {10^{ - 13}}$) in 0.1 M $N{a_2}C{O_3}$ solution is ${10^{ - 6}}$.
So, the correct answer is “Option A”.

Note:
In the solubility product constant the concentration of the reactant compound does not take part. Make sure to determine the initial and final concentration of the ionic product formed during the reaction.