Question

# Solubility product constants of (${{K}_{sp}}$) of salts of the types $MX$, $M{{X}_{2}}$ and ${{M}_{3}}X$ at temperature T are $4.0\times {{10}^{-8}}$, $3.2\times {{10}^{-14}}$ and $2.7\times {{10}^{-15}}$ respectively. Solubility (mol $d{{m}^{-3}}$) of the salts at temperature T are in the order:(A)- $MX>M{{X}_{2}}>{{M}_{3}}X$ (B)- ${{M}_{3}}X>M{{X}_{2}}>MX$ (C)- $M{{X}_{2}}>{{M}_{3}}X>MX$ (D)- $MX>{{M}_{3}}X>M{{X}_{2}}$

Hint: Solubility product of a salt at any temperature is the product of the molar concentration of its ions where each concentration term is raised to the number of ions produced on dissociation of one molecule of the salt in saturated solution.
Solubility of a salt at any given temperature can be calculated from its solubility product. If salt of the type${{M}_{x}}{{N}_{y}}$ dissociates in solution.
${{M}_{x}}{{N}_{y}}\rightleftarrows x{{M}^{y-}}+y{{N}^{x-}}$
Then, it solubility product, ${{K}_{sp}}$ will be given as:${{K}_{sp}}={{\left[ {{M}^{y+}} \right]}^{x}}{{\left[ {{N}^{x-}} \right]}^{y}}$

To determine the order of solubility of the salts $MX$, $M{{X}_{2}}$ and ${{M}_{3}}X$, let us first calculate the solubility of these salts one by one and then compare the values.
Solubility of salt of the type $MX$ is calculated in the following steps:
Consider the dissociation of the salt $MX$ in solution as:
\begin{align} & MX\rightleftarrows {{M}^{+}}+{{X}^{-}} \\ & s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s\,\,\,\,\,\,\,\,\,\,s \\ \end{align}
We can write the solubility product constant for $MX$ as: ${{K}_{sp}}=\left[ {{M}^{+}} \right]\left[ {{X}^{-}} \right]$.
Now the concentration of ${{M}^{+}}$ and ${{X}^{-}}$ (in mol $d{{m}^{-3}}$) ions produced on dissociation is equal to the amount of salt that is dissolve. For salt of the type $MX$, $\left[ {{M}^{+}} \right]=\left[ {{X}^{-}} \right]$. 
Let $\left[ {{M}^{+}} \right]=\left[ {{X}^{-}} \right]=s$. Here, $s$ represents the solubility of the ions produced on dissociation of $MX$ in solution. Then, the expression for ${{K}_{sp}}$ becomes:
${{K}_{sp}}=\left[ {{M}^{+}} \right]\left[ {{X}^{-}} \right]=s\times s={{s}^{2}}$
Given solubility product of the salt $MX$, ${{K}_{sp}}=4.0\times {{10}^{-8}}$.
Substituting the value of ${{K}_{sp}}=4.0\times {{10}^{-8}}$ in the above equation of ${{K}_{sp}}$ and solving for $s$, we get
\begin{align} & {{K}_{sp}}={{s}^{2}}=4.0\times {{10}^{-8}} \\ & \Rightarrow {{s}^{2}}=4.0\times {{10}^{-8}} \\ \end{align}
Taking the square on both sides, we get
\begin{align} & \sqrt{{{s}^{2}}}=\sqrt{4.0\times {{10}^{-8}}} \\ & \sqrt{{{s}^{2}}}=\sqrt{{{(2.0\times {{10}^{-4}})}^{2}}} \\ & s=2.0\times {{10}^{-4}} \\ \end{align}
Therefore, the solubility of salt of the type $MX$ is $2.0\times {{10}^{-4}}mol\,d{{m}^{-3}}$

Similarly, the solubility of salt of the type $M{{X}_{2}}$ can be calculated from its solubility product constant.
\begin{align} & M{{X}_{2}}\rightleftarrows {{M}^{2+}}+2{{X}^{-}} \\ & s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s\,\,\,\,\,\,\,\,\,\,\,\,2s \\ \end{align}
Here, solubility product, ${{K}_{sp}}$ will be written as : ${{K}_{sp}}=\left[ {{M}^{2+}} \right]{{\left[ {{X}^{-}} \right]}^{2}}=s\times {{(2s)}^{2}}=4{{s}^{3}}$
Given, solubility product of the salt $M{{X}_{2}}$, ${{K}_{sp}}=3.2\times {{10}^{-14}}$.
Taking the value of ${{K}_{sp}}$ to be $3.2\times {{10}^{-14}}$ in the above equation, we obtain
\begin{align} & {{K}_{sp}}=4{{s}^{3}}=3.2\times {{10}^{-14}} \\ & \Rightarrow 4{{s}^{3}}=3.2\times {{10}^{-14}} \\ \end{align}
Dividing both sides by 4, we get
\begin{align} & \dfrac{4{{s}^{3}}}{4}=\dfrac{3.2\times {{10}^{-14}}{{s}^{3}}}{4} \\ & {{s}^{3}}=0.8\times {{10}^{-14}}=8\times {{10}^{-15}} \\ \end{align}
Taking cube root on both sides and then calculating the value of $s$ as follows:
\begin{align} & \sqrt[3]{{{s}^{3}}}=\sqrt[3]{8\times {{10}^{-15}}} \\ & \sqrt[3]{{{s}^{3}}}=\sqrt[3]{{{(2\times {{10}^{-5}})}^{3}}} \\ & s=2\times {{10}^{-5}} \\ \end{align}
The solubility of salt of the type $M{{X}_{2}}$ comes out to be $2.0\times {{10}^{-5}}mol\,d{{m}^{-3}}$ or $0.2\times {{10}^{-4}}mol\,d{{m}^{-3}}$.

Now, let us finally calculate the solubility of salt of the type ${{M}_{3}}X$.
\begin{align} & {{M}_{3}}X\rightleftarrows 3{{M}^{+}}+{{X}^{3-}} \\ & s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3s\,\,\,\,\,\,\,\,\,\,s \\ \end{align}
Solubility constant will be given as: ${{K}_{sp}}={{\left[ {{M}^{+}} \right]}^{3}}\left[ {{X}^{-}} \right]={{(3s)}^{3}}\times s=27{{s}^{4}}$
Given value of the ${{K}_{sp}}$ for ${{M}_{3}}X$ is $2.7\times {{10}^{-15}}$.
Substituting ${{K}_{sp}}=2.7\times {{10}^{-15}}$ and solving, we get
\begin{align} & {{K}_{sp}}=27{{s}^{4}}=2.7\times {{10}^{-15}} \\ & \Rightarrow 27{{s}^{4}}=2.7\times {{10}^{-15}} \\ \end{align}
On dividing both sides by 27, we can simply as:
\begin{align} & {{s}^{4}}=0.1\times {{10}^{-15}}=1\times {{10}^{-16}} \\ & \\ \end{align}
Taking fourth root on both sides and then solving for $s$, we obtain
\begin{align} & \sqrt[4]{{{s}^{4}}}=\sqrt[4]{1\times {{10}^{16}}} \\ & \sqrt[4]{{{s}^{4}}}=\sqrt[4]{{{(1\times {{10}^{-4}})}^{4}}} \\ & s=1\times {{10}^{-4}} \\ & \\ \end{align}
Therefore, solubility of salt of the type ${{M}_{3}}X$ is $1\times {{10}^{-4}}mol\,d{{m}^{-3}}$.
Salt
Solubility ($mold{{m}^{-3}}$)
$MX$
$2.0\times {{10}^{-4}}mol\,d{{m}^{-3}}$
$M{{X}_{2}}$
$0.2\times {{10}^{-4}}mol\,d{{m}^{-3}}$
${{M}_{3}}X$
$1\times {{10}^{-4}}mol\,d{{m}^{-3}}$
On comparing the values of solubility of the salts, it is clear that the order of solubility is:
$MX>{{M}_{3}}X>M{{X}_{2}}$

Hence, the correct option is (D).

Note: It is to be noted that the unit of solubility is the same as that of the molar concentration of ions in solution. Do not make any mistake while calculating. Solve for the solubility of each salt step by step to avoid errors in calculation.