Question

The solubility products of three sparingly soluble salt ${{M}_{2}}X$, $MX$ and $M{{X}_{3}}$ are identical. What will be the order of their solubilities?
A. $M{{X}_{3}}>{{M}_{2}}X>MX$
B. $M{{X}_{3}}>MX>{{M}_{2}}X$
C. $MX>{{M}_{2}}X>M{{X}_{3}}$
D. $MX>M{{X}_{3}}>{{M}_{2}}X$

Answer 1

Hint: You should know that; solubility product of an electrolyte at a particular temperature equals to the product of the molar concentration of its ions in a saturated solution, each concentration raised to the power equal to the number of ions produced on dissociation.

Complete step by step solution:
Let us first know about what does sparingly soluble solute mean. So, sparingly soluble salts are those whose solubility is very low. And we know that, solubility of a substance is defined as the amount of substance which is soluble in $100mL$ of water. The solubility product of an electrolyte at a particular temperature is defined as the product of the molar concentration of its ions in a saturated solution. As per the given question, the salts have equal solubility. So, let us consider the solubility product (which is represented by ${{K}_{sp}}$) be ‘m’.
The formula used for solubility product in a reaction given below is:
${{M}_{x}}{{X}_{y}}\rightleftharpoons x{{M}^{+}}+y{{X}^{-}}$
${{K}_{sp}}={{[{{M}^{y+}}]}^{x}}{{[{{X}^{x-}}]}^{y}}$, where
${{K}_{sp}}$ is the soluble product,
$[{{M}^{y+}}]$ is the concentration of ${{M}^{y+}}$ and
$[{{X}^{x-}}]$ is the concentration of ${{X}^{x-}}$.${{M}_{2}}X$
Now let us find out the solubility product of the given salts.
For salt ${{M}_{2}}X$
The solubility product is m as stated above. And let solubility of the salt be ${{s}_{1}}$.
At equilibrium, the reaction will be:
${{M}_{2}}X\rightleftharpoons 2{{M}^{+}}+{{X}^{-}}$
So, $[{{M}^{+}}]=[{{X}^{2-}}]={{s}_{1}}$ as $x=2\text{ and }y=1$ (applying the formula).
Therefore, ${{K}_{sp}}={{[{{M}^{y+}}]}^{x}}{{[{{X}^{x-}}]}^{y}}$ as stated above we will get:
$m={{[{{M}^{+}}]}^{2}}[{{X}^{2-}}]$
Then, $m={{s}_{1}}^{2}\times {{s}_{1}}={{s}_{1}}^{3}$
Thus, ${{s}_{1}}={{(m)}^{\dfrac{1}{3}}}$
Similarly, for salt $MX$ whose reaction will be:
$MX\rightleftharpoons {{M}^{+}}+{{X}^{-}}$ where, $x=1\text{ and }y=1$.
Then, the value of m will be considering the solubility of salt as ${{s}_{2}}$:
$m=[{{M}^{+}}][{{X}^{-}}]$
So, $m={{s}_{2}}\times {{s}_{2}}=s_{2}^{2}$
Thus, ${{s}_{2}}={{(m)}^{\dfrac{1}{2}}}$.
For salt $M{{X}_{3}}$, whose solubility be considered as ${{s}_{3}}$ and the reaction at equilibrium will be:
$M{{X}_{3}}\rightleftharpoons {{M}^{+}}+3{{X}^{-}}$, where $x=1\text{ and }y=3$
So, $m=[{{M}^{3+}}]{{[{{X}^{-}}]}^{3}}$
Then, $m={{s}_{3}}\times {{s}_{3}}^{3}={{s}_{3}}^{4}$
Thus, ${{s}_{3}}={{(m)}^{\dfrac{1}{4}}}$
We can see that, ${{s}_{2}}>{{s}_{1}}>{{s}_{3}}$.
So, $MX>{{M}_{2}}X>M{{X}_{3}}$.

Hence, the correct option is C.

Note: If in a solution of weak electrolyte, when a strong electrolyte is added (having a common ion) then the ionization of that weak electrolyte is suppressed.