Question

The molar solubility of a silver sulphate is $1.5 \times {10^{ - 2}}$ mol ${L^{ - 1}}$. The solubility product of the salt will be:
A.$2.25 \times {10^{ - 4}}$
B.$1.35 \times {10^{ - 5}}$
C.$1.7 \times {10^{ - 6}}$
D.$3.0 \times {10^{ - 3}}$

Answer 1

Hint:Solubility product of a salt is basically the product of the solubilities or concentration of the products (ionic species) of the reaction, where the concentration of each ion is raised to the power respective to their stoichiometric coefficients in a balanced chemical equation.

Complete step by step answer:
Molar solubility determines the number of moles of solute that can be dissolved per liter of a solution, before the solution becomes saturated.
The balanced chemical equation for the dissociation of silver sulphate $(A{g_2}S{O_4})$ is,
$A{g_2}S{O_4}(s) \rightleftharpoons 2A{g^ + }(aq) + SO_4^{2 - }(aq)$
Let the solubility of $A{g_2}S{O_4}$ be ‘s’. So, the solubilities of $2$ moles of $A{g^ + }$ and one mole of $SO_4^{2 - }$ at equilibrium will be $‘2s’$ and ‘s’ respectively.
So, the solubility product for the above reaction is,
${K_{sp}} = {[A{g^ + }]^2}[SO_4^{2 - }]$
$ \Rightarrow {K_{sp}} = {(2s)^2}(s)$
$ \Rightarrow {K_{sp}} = (4{s^2})(s)$
$ \Rightarrow {K_{sp}} = 4{s^3}$
As given in the question, the molar solubility of $A{g_2}S{O_4}$ is $1.5 \times {10^{ - 2}}$ mol ${L^{ - 1}}$, that is,
$s = 1.5 \times {10^{ - 2}}$ mol ${L^{ - 1}}$
So, ${K_{sp}} = 4 \times {(1.5 \times {10^{ - 2}})^3}$
$ \Rightarrow {K_{sp}} = 4 \times 3.375 \times {10^{ - 6}}$
$ \Rightarrow {K_{sp}} = 13.5 \times {10^{ - 6}}$
$ \Rightarrow {K_{sp}} = 1.35 \times {10^{ - 5}}$
Thus, the solubility product of $A{g_2}S{O_4}$ is $1.35 \times {10^{ - 5}}$.

Hence option B is correct.

Note:
The solubility product denoted by ${K_{sp}}$ is the equilibrium constant for the dissolution of a solute in solid state into an aqueous solution. While writing the equation of solubility product of a salt, we do not write the concentration of the solid constituent of the reaction (because the concentration of pure solid species remains constant), only the species in aqueous state is included in the formula.
The equilibrium constant for the dissociation of $A{g_2}S{O_4}$ can be written as,
$K = \dfrac{{{{[A{g^ + }]}^2}[SO_4^{2 - }]}}{{[A{g_2}SO_4^{2 - }]}}$
$ \Rightarrow K[A{g_2}SO_4^{2 - }] = {[A{g^ + }]^2}[SO_4^{2 - }]$
Where $K[A{g_2}SO_4^{2 - }] = $ constant, known as the solubility product constant.
Therefore, $K[A{g_2}SO_4^{2 - }] = {K_{sp}} = {[A{g^ + }]^2}[SO_4^{2 - }]$