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

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-
$\Rightarrow 4{{\text{a}}^2} \times {10^{ - 2}} = 4 \times {10^{ - 12}} \\\ \Rightarrow {{\text{a}}^2} = \dfrac{{{{10}^{ - 12}}}}{{{{10}^{ - 2}}}} \\\$
We know that $\dfrac{{{{\text{x}}^a}}}{{{{\text{x}}^b}}} = {{\text{x}}^{{\text{a - b}}}}$ 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.