Question

The solubility of $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$ in water at ${25}^{\circ }\text{C}$ is $1\times {10}^{-4}$ mole/litre. What is the solubility in $0.01\text{M}$ $\text{N}{\text{a}}_2\text{C}{\text{o}}_3$ solution? Assume no hydrolysis of ${\text{Co}}_{3-}^2$ ion.
A)$6\times \text{1}{\text{0}}^{-6}$ mole/litre
B)$4\times {10}^{-5}$ mole/litre
C)${10}^{-5}$ mole/litre
D)$2\times {10}^{-5}$ mole/litre

Answer 1

Hint: We can use the formula of ${\text{K}}_{\text{sp}}$ Solubility product which is the product of concentrations of ions with the power raised to the number of ions.
${\text{K}}_{\text{sp}}$$=\text{product of concentration of ions}$

Complete step by step answer:
Given that the solubility ‘s’ of $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$ in water at ${25}^{\circ }\text{C}$ is $1\times {10}^{-4}$ mole/litre.
Let the solubility be‘s’ when $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$dissociates into following ions in water, then-
$\text{A}{\text{g}}_2\text{C}{\text{o}}_3\rightleftharpoons 2\text{A}{\text{g}}^{+}+{\text{Co}}_{3-}^2$ then its solubility product is given by-
${\text{K}}_{\text{sp}}$$=\text{product of concentration of ions}$
$\Rightarrow {\text{K}}_{\text{sp}}={\left[2\text{A}{\text{g}}^{+}\right]}^2\left[{\text{Co}}_{3-}^2\right]$
 then on putting the values we get-
$\Rightarrow {\text{K}}_{\text{sp}}={\left[2\text{s}\right]}^2\left[\text{s}\right]=4{\text{s}}^3$ --- (i)
We have to find the solubility of $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$ solubility in $0.01\text{M}$ $\text{N}{\text{a}}_2\text{C}{\text{o}}_3$ solution.
Now let the solubility of $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$ in $0.01\text{M}$ $\text{N}{\text{a}}_2\text{C}{\text{o}}_3$ solution be ‘a’. Since there is no hydrolysis of${\text{Co}}_{3-}^2$ ion so its concentration will be $0.01$.Now when $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$dissociates into following ions in$\text{N}{\text{a}}_2\text{C}{\text{o}}_3$, then-
$\text{A}{\text{g}}_2\text{C}{\text{o}}_3\rightleftharpoons 2\text{A}{\text{g}}^{+}+{\text{Co}}_{3-}^2$ then its solubility product is given by-
$\Rightarrow {\text{K}}_{\text{sp}}={\left[2\text{A}{\text{g}}^{+}\right]}^2\left[{\text{Co}}_{3-}^2\right]$
On putting the value of eq. (i), we get-
$\Rightarrow {\text{K}}_{\text{sp}}={\left[2\text{a}\right]}^2\left[0.01\right]=4{\text{s}}^3$
And we know that s =$1\times {10}^{-4}$(given) , then putting the value of s in above equation we get-
$\Rightarrow 4{\text{a}}^2\times {10}^{-2}=4\times {\left({10}^{-4}\right)}^3$
On solving this equation, we get-
$$\begin{gathered} \Rightarrow 4{\text{a}}^2\times {10}^{-2}=4\times {10}^{-12} \\ \Rightarrow {\text{a}}^2={\displaystyle \frac{{10}^{-12}}{{10}^{-2}}} \end{gathered}$$
We know that ${\displaystyle \frac{{\text{x}}^a}{{\text{x}}^b}}={\text{x}}^{\text{a - b}}$ so using this in the above equation, we get-
$\Rightarrow {\text{a}}^2={10}^{-12+2}={10}^{-10}$
Now we will remove the square-root to get the value of a-
$\Rightarrow \text{a = }\sqrt{{10}^{-10}}=\sqrt{{10}^{-5}\times {10}^{-5}}={10}^{-5}$
Hence the solubility of $\text{A}{\text{g}}_2\text{C}{\text{o}}_3$ in $0.01\text{M}$ $\text{N}{\text{a}}_2\text{C}{\text{o}}_3$ solution is ${10}^{-5}$ .

So the correct option is ‘C’.

Note:
There is a difference between solubility and solubility products. Solubility is the concentration of ions of solute in a solvent. Solubility product is the product of the solubility of the solute.