Question

The solubility of lithium sodium hexafluoroaluminate ${\text{L}}{{\text{i}}_{\text{3}}}{\text{N}}{{\text{a}}_{\text{3}}}{{\text{(Al}}{{\text{F}}_{\text{6}}}{\text{)}}_{\text{2}}}$ is a mol litre-1. Its solubility product is equal to-
A)${{\text{s}}^8}$
 B) ${\text{12}}{{\text{s}}^3}$
C) $18{{\text{s}}^3}$
 D) $2916{{\text{s}}^8}$

Answer 1

Hint: The solubility product is the product of concentration of ions in water. To find the solubility product, dissociate the lithium sodium hexafluoroaluminate into ions and then multiply the concentration of the ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation.
Formula used- ${{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}$

Complete step by step answer:
Given, solubility of lithium sodium hexafluoroaluminate${\text{L}}{{\text{i}}_{\text{3}}}{\text{N}}{{\text{a}}_{\text{3}}}{{\text{(Al}}{{\text{F}}_{\text{6}}}{\text{)}}_{\text{2}}}$ is in mol litre-1. We have to find its solubility product. Solubility Product is the product of concentration of ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium. It is denoted by ${{\text{K}}_{{\text{sp}}}}$ . Now the given salt will dissociate into following ions in water-
\[\;\;\;{\text{L}}{{\text{i}}_{\text{3}}}{\text{N}}{{\text{a}}_{\text{3}}}{{\text{(Al}}{{\text{F}}_{\text{6}}}{\text{)}}_{\text{2}}} \rightleftarrows {\text{3L}}{{\text{i}}^ + } + 3{\text{N}}{{\text{a}}^ + } + 2{\left( {{\text{Al}}{{\text{F}}_6}} \right)_2}^{3 - }\]
Let solubility of salt be’s’ mol litre-1. Then concentration of ions will be $\left[ {{\text{L}}{{\text{i}}^ + }} \right] = \left[ {{\text{N}}{{\text{a}}^ + }} \right] = 3{\text{s}}$ and$\left[ {{\text{Al}}{{\text{F}}_6}^{3 - }} \right] = 2{\text{s}}$ . Now ${{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}$
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {{\text{L}}{{\text{i}}^ + }} \right]^3}{\left[ {{\text{N}}{{\text{a}}^ + }} \right]^3}{\left[ {{\text{Al}}{{\text{F}}_6}^{3 - }} \right]^2} = {\left( {3{\text{s}}} \right)^3}{\left( {3{\text{s}}} \right)^3}{\left( {2{\text{s}}} \right)^2}$ $ = 27{{\text{s}}^3} \times 27{{\text{s}}^3} \times 4{{\text{s}}^2}$
Since on multiplication the base (s) is same so the powers of the base (s) will add.
$ \Rightarrow $ ${{\text{K}}_{{\text{sp}}}} = 2916{{\text{s}}^8}$

Hence the correct answer is ‘D’.

Note:
The solubility of a substance is the total amount of the solute that can be dissolved in the solvent at equilibrium while solubility product constant (${{\text{K}}_{{\text{sp}}}}$ ) is an equilibrium constant which represents the level at which the a solute dissolves in a solution or we can say it provides insight into the equilibrium between the solute and its constituent ions that are dissociated across the solution.