Question

The ${{\text{K}}_{{\text{SP}}}}$ of salts ${\text{AB}}$, ${\text{A}}{{\text{B}}_2}$ and ${{\text{A}}_3}{\text{B}}$ are $4.0 \times {10^{ - 8}}$, $3.2 \times {10^{ - 14}}$ and $2.7 \times {10^{ - 15}}$ respectively at temperature ${\text{T}}$. The solubility order of these salts in water at temperature ${\text{T}}$ (in ${\text{mole liter}}{{\text{e}}^{ - 1}}$) is:
A) ${\text{AB}}$ > ${\text{A}}{{\text{B}}_2}$ > ${{\text{A}}_3}{\text{B}}$
B) ${{\text{A}}_3}{\text{B}}$ > ${\text{A}}{{\text{B}}_2}$ > ${\text{AB}}$
C) ${\text{A}}{{\text{B}}_2}$ > ${{\text{A}}_3}{\text{B}}$ > ${\text{AB}}$
D) ${\text{AB}}$ > ${{\text{A}}_3}{\text{B}}$ > ${\text{A}}{{\text{B}}_2}$

Answer 1

Hint: We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.

Complete answer:
We know that the solubility of a salt at any temperature is calculated from its solubility product.
Now, consider a salt ${{\text{A}}_x}{{\text{B}}_y}$. The salt dissociates as follows:
${{\text{A}}_x}{{\text{B}}_y} \rightleftharpoons x{{\text{A}}^{y + }} + y{{\text{B}}^{x - }}$
The solubility product of the salt ${{\text{A}}_x}{{\text{B}}_y}$ is given as follows:
${{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^{y + }}]^x}{[{{\text{B}}^{x - }}]^y}$
Where ${{\text{K}}_{{\text{SP}}}}$ is the solubility product.
We are given three salts ${\text{AB}}$, ${\text{A}}{{\text{B}}_2}$ and ${{\text{A}}_3}{\text{B}}$. We will calculate the solubility and solubility products of the given salts.
Consider salt ${\text{AB}}$ which dissociates as follows:
${\text{AB}} \rightleftharpoons {{\text{A}}^ + } + {{\text{B}}^ - }$
The solubility product of the salt ${\text{AB}}$ is as follows:
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^ + }][{{\text{B}}^ - }]$
For the salt ${\text{AB}}$, $[{{\text{A}}^ + }] = [{{\text{B}}^ - }]$. And $[{{\text{A}}^ + }] = [{{\text{B}}^ - }] = s$, where $s$ is the solubility of the ions. Thus,
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^ + }][{{\text{B}}^ - }] = s \times s = {s^2}$
We are given that the solubility product of salt ${\text{AB}}$ is $4.0 \times {10^{ - 8}}$. Thus,
${s^2} = 4.0 \times {10^{ - 8}}$
$s = 2.0 \times {10^{ - 4}}$
Thus, the solubility of salt ${\text{AB}}$ is $2.0 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}$.
Consider salt ${\text{A}}{{\text{B}}_2}$ which dissociates as follows:
${\text{A}}{{\text{B}}_2} \rightleftharpoons {{\text{A}}^{2 + }} + 2{{\text{B}}^ - }$
The solubility product of the salt ${\text{A}}{{\text{B}}_2}$ is as follows:
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^{2 + }}]{[{{\text{B}}^ - }]^2}$
For the salt ${\text{A}}{{\text{B}}_2}$.
${{\text{K}}_{{\text{SP}}}} = [{{\text{A}}^{2 + }}]{[{{\text{B}}^ - }]^2} = s \times {\left( {2s} \right)^2} = 4{s^3}$
Where $s$ is the solubility of the ions.
We are given that the solubility product of salt ${\text{A}}{{\text{B}}_2}$ is $3.2 \times {10^{ - 14}}$. Thus,
$4{s^3} = 3.2 \times {10^{ - 14}}$
${s^3} = 8 \times {10^{ - 15}}$
$s = 2 \times {10^{ - 5}} = 0.2 \times {10^{ - 4}}$
Thus, the solubility of salt ${\text{A}}{{\text{B}}_2}$ is $0.2 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}$.
Consider salt ${{\text{A}}_3}{\text{B}}$ which dissociates as follows:
${{\text{A}}_3}{\text{B}} \rightleftharpoons 3{{\text{A}}^ + } + {{\text{B}}^{3 - }}$
The solubility product of the salt ${{\text{A}}_3}{\text{B}}$ is as follows:
\[{{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^ + }]^3}[{{\text{B}}^ - }]\]
For the salt ${{\text{A}}_3}{\text{B}}$.
${{\text{K}}_{{\text{SP}}}} = {[{{\text{A}}^ + }]^3}[{{\text{B}}^ - }] = {\left( {3s} \right)^3} \times s = 27{s^4}$
Where $s$ is the solubility of the ions.
We are given that the solubility product of salt ${{\text{A}}_3}{\text{B}}$ is $2.7 \times {10^{ - 15}}$. Thus,
$27{s^4} = 2.7 \times {10^{ - 15}}$
${s^4} = 1 \times {10^{ - 16}}$
$s = 1 \times {10^{ - 4}}$
Thus, the solubility of salt ${{\text{A}}_3}{\text{B}}$ is $1 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}$.
Thus, we have calculated that,
The solubility of salt ${\text{AB}}$ is $2.0 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}$.
The solubility of salt ${\text{A}}{{\text{B}}_2}$ is $0.2 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}$.
The solubility of salt ${{\text{A}}_3}{\text{B}}$ is $1 \times {10^{ - 4}}{\text{ mole litr}}{{\text{e}}^{ - 1}}$.
From this we can write the solubility order as follows:
${\text{AB}}$ > ${{\text{A}}_3}{\text{B}}$ > ${\text{A}}{{\text{B}}_2}$

Thus, the correct answer is option (D) ${\text{AB}}$ > ${{\text{A}}_3}{\text{B}}$ > ${\text{A}}{{\text{B}}_2}$.

Note: We know that solubility is the amount of solute that dissolves in a solvent. The solubility of many solutes increases with the temperature of the solvent. We can see that the units of solubility and molar concentration are the same. Solve for the solubility of each of the given salt carefully to avoid errors.