Question

Three sparingly soluble salts have same solubility products are given below:
$\mathrm{I}.A_{2}X$ $\mathrm{I}\mathrm{I}.AX$ $\mathrm{I}\mathrm{I}\mathrm{I}.A{X}_3$
Their solubility in a standard solution will be such that:
A: $\mathrm{I}\mathrm{I}\mathrm{I}>\mathrm{I}\mathrm{I}>\mathrm{I}$
B: $\mathrm{I}\mathrm{I}\mathrm{I}>\mathrm{I}>\mathrm{I}\mathrm{I}$
C: $\mathrm{I}\mathrm{I}>\mathrm{I}\mathrm{I}\mathrm{I}>\mathrm{I}$
D: $\mathrm{I}\mathrm{I}>\mathrm{I}>\mathrm{I}\mathrm{I}\mathrm{I}$

Answer 1

Hint: 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, each concentration raised to the power equal to the number of ions produced on dissociation. It is represented by$K_{sp}$.
Formula used: $A_x{B}_y\stackrel{}{\leftrightarrows }x{A}^{+}+y{B}^{-}$
$K_{sp}={\left[A^{y+}\right]}^x{\left[B^{x-}\right]}^y$
Where $K_{sp}$is solubility product,$\left[A^{y+}\right]$is concentration of$A^{y+}$ and$\left[B^{x-}\right]$is concentration of$B^{x-}$.

Complete answer:
Sparingly soluble salts are those salts whose solubility is very low. The 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. Given salts have the same solubility product. Let their solubility product$K_{sp}$ be $m$.
For salt $A_{2}X$ :
Solubility product $K_{sp}=m$ (stated above)
Let solubility of salt be $s_1$
At equilibrium $A_{2}X\stackrel{}{\leftrightarrows }2A^{+}+X^{-}$
$\left[A^{+}\right]=\left[X^{2-}\right]=s_1$ , $x=2$ and $y=1$ (according to the formula stated above)
Applying the above formula $K_{sp}={\left[A^{y+}\right]}^x{\left[B^{x-}\right]}^y$ (stated above)
$m={\left[A^{+}\right]}^2\left[X^{2-}\right]$
$m=s_1^2\times s_1$
$m=s_1^3$
$s_1={\left(m\right)}^{{\displaystyle \frac{1}{3}}}$
For salt$AX$ :
Solubility product $K_{sp}=m$ (stated above)
Let solubility of salt be $s_2$
At equilibrium $AX\stackrel{}{\leftrightarrows }{A}^{+}+X^{-}$
$\left[A^{+}\right]=\left[X^{-}\right]=s_2$ , $x=1$ and $y=1$ (according to the formula stated above)
Applying the above formula $K_{sp}={\left[A^{y+}\right]}^x{\left[B^{x-}\right]}^y$ (stated above)
$m=\left[A^{+}\right]\left[X^{-}\right]$
$m=s_2\times s_2$
$m=s_2^2$
$s_2={\left(m\right)}^{{\displaystyle \frac{1}{2}}}$
For salt $A{X}_3$ :
Solubility product $K_{sp}=m$ (stated above)
Let solubility of salt be$s_3$
At equilibrium $A{X}_3\stackrel{}{\leftrightarrows }{A}^{+}+3X^{-}$
$\left[A^{3+}\right]=\left[X^{-}\right]=s_3$ , $x=1$ and $y=3$ (according to the formula stated above)
Applying the above formula $K_{sp}={\left[A^{y+}\right]}^x{\left[B^{x-}\right]}^y$ (stated above)
$m=\left[A^{3+}\right]{\left[X^{-}\right]}^3$
$m=s_3\times s_3^3$
$m=s_3^4$
$s_3={\left(m\right)}^{{\displaystyle \frac{1}{4}}}$
Now$s_1={\left(m\right)}^{{\displaystyle \frac{1}{3}}}$,$s_2={\left(m\right)}^{{\displaystyle \frac{1}{2}}}$,$s_3={\left(m\right)}^{{\displaystyle \frac{1}{4}}}$
From these values we can clearly see that$s_2>s_1>s_3$ .$s_1$Is solubility of$A_{2}X$,$s_1$is solubility of $AX$and $s_3$is solubility of $A{X}_3$. In the question $A_{2}X$ is denoted as $\mathrm{I}$, $AX$is denoted as $\mathrm{I}\mathrm{I}$and $A{X}_3$ is stated as $\mathrm{I}\mathrm{I}\mathrm{I}$. This means $\mathrm{I}\mathrm{I}>\mathrm{I}>\mathrm{I}\mathrm{I}\mathrm{I}$ (as $s_2>s_1>s_3$and $s_1$Is solubility of$A_{2}X$,$s_1$is solubility of $AX$and $s_3$is solubility of $A{X}_3$).

So, correct answer is option D that is$\mathrm{I}\mathrm{I}>\mathrm{I}>\mathrm{I}\mathrm{I}\mathrm{I}$.


Note:
If the solution of a weak electrolyte, a strong electrolyte having a common ion is added, the ionization of the weak electrolyte is further suppressed. For example, addition of $C{H}_{3}COONa$ to $C{H}_{3}COOH$ solution suppresses the ionization of $C{H}_{3}COOH$.