Question

If the solubility product of a sparingly soluble salt $M{{X}_{2}}$ at ${{25}^{\circ }}C$ is $1.0\text{ x 1}{{\text{0}}^{-11}}$, the solubility of salt in $moles\text{ }{{L}^{-1}}$ at this temperature will be:
(a)- $2.46\text{ x 1}{{\text{0}}^{14}}$
(b)- $\text{1}\text{.35 x 1}{{\text{0}}^{-4}}$
(c)- $2.60\text{ x 1}{{\text{0}}^{-7}}$
(d)- $\text{1}\text{.20 x 1}{{\text{0}}^{-10}}$

Answer 1

Hint: In equilibrium, the given salt will form cations and anions. The solubility of the cation will be s and the solubility of the anion will be 2s. So, the formula used will be ${{K}_{sp}}=4{{s}^{3}}$.

Complete step by step answer:
First, let us study the solubility product:
If a sparingly soluble salt $AB$ is stirred with water, only a small amount of it goes into solution while most of the salt remains undissolved. But whenever the little amount of salt dissolves, it gets completely dissociated into ions. In other words, there exists a dynamic equilibrium between the undissolved solid salt and the ions which it furnishes in solution when a sparingly soluble salt is added to water.

Thus, the equilibrium can be represented as:
$AB(s)\rightleftharpoons {{A}^{+}}(aq)+{{B}^{-}}(aq)$
We can also write according to the law of chemical equilibrium,
 $K=\dfrac{[{{A}^{+}}][{{B}^{-}}]}{[AB]}$

Since the concentration of the undissociated solid remains constant, we may write,
$[{{A}^{+}}][{{B}^{-}}]=K\text{ x }\!\![\!\!\text{ AB }\!\!]\!\!\text{ =}{{\text{K}}_{sp}}$
Where ${{K}_{sp}}$ is the solubility product is equal to the product of $[{{A}^{+}}][{{B}^{-}}]$.
So, here the sparingly soluble salt is given $M{{X}_{2}}$. The equilibrium reaction will be,
$M{{X}_{2}}\rightleftharpoons {{M}^{+}}+2{{X}^{-}}$

Let s represents the solubility of both the ions, so
The solubility of $[{{M}^{+}}]=s$
The solubility of $[{{X}^{-}}]=2s$ (because 2 X atoms are present in one molecule of $M{{X}_{2}}$)
The formula of ${{K}_{sp}}$ for this reaction will be:
${{K}_{sp}}=(s)\ \text{x }{{(2s)}^{2}}$
${{K}_{sp}}=4{{s}^{3}}$

Given the solubility product of $M{{X}_{2}}$ is $1.0\text{ x 1}{{\text{0}}^{-11}}$, hence
$1.0\text{ x 1}{{\text{0}}^{-11}}=4{{s}^{3}}$
${{s}^{3}}=\dfrac{1.0\text{ x 1}{{\text{0}}^{-11}}}{4}$
$s=\sqrt[3]{\dfrac{1.0\text{ x 1}{{\text{0}}^{-11}}}{4}}$
$s=1.36\ \text{x 1}{{\text{0}}^{-4}}$
So, the solubility of the salt is $1.36\text{ x 1}{{\text{0}}^{-4}}$.
So, the correct answer is “Option B”.

Note: The concentration of the ions must be raised to the power equal to the number of ions produced in the solution. The solubility product is different from the ionic product as the solubility product is only restricted to saturated solutions but the ionic product is for both saturated and unsaturated solutions.