Question

If $K_{sp}$ of $A{g}_{2}C{O}_3$ is, $8\times {10}^{-12}$, the molar solubility of$A{g}_{2}C{O}_3$ in $0.1MAgN{O}_3$ is:

• A. $8\times {10}^{-12}M$
• B. $8\times {10}^{-10}M$
• C. $8\times {10}^{-11}M$
• D. $8\times {10}^{-13}M$

Answer 1

The correct option is B

$8\times {10}^{-10}M$

Solubility product of silver carbonate $A{g}_{2}C{O}_{3}is{K}_{sp}=8\times {10}^{-12}$

The $A{g}_{2}C{O}_3$ is in the $0.1A{g}_{2}N{O}_3$ solution.

The solute like silver carbonate dissolves in the solution and forms corrresponding silver and carbonate ions.
The equilibrium reaction between the ions and solute is as shown below.

$A{g}_{2}C{O}_3\left(s\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+C{O}_3^{2-}$

Before S 0 0
After - 2S+0.1 S

The $A{g}_{2}C{O}_3$ is dissolved in the 0.1 M $A{g}_{2}N{O}_3$ solution. Both salts have the common $A{g}^{+}$. Thus, the silver ion from both the salt contributes towards the molar solubility product for the$A{g}_{2}C{O}_3$ is written as follows.

$K_{sp}={\left[A{g}^{+}\right]}^2\left[C{O}_3^{2-}\right]$

Let's substitute the values in the above equation we have,
$8\times {10}^{-12}={(0.1+2S)}^2\left(S\right)$

The value of 'S' is very small compared to 0.1, hence,
$0.1+2S\approx 0.1$.

The equation now becomes,
$8\times {10}^{-12}={\left(0.1\right)}^2\left(S\right)$

$S=\frac{8\times {10}^{-12}}{{\left(0.1\right)}^2}=8\times {10}^{-10}M$

Therefore, molar solubility of a silver carbonate in presence of silver nitrate is equal to $8\times {10}^{-10}M$

Hence, (B) is the correct option.