Question

The solubility of \[{A_2}{B_3}\] is \[s\,mol\,{L^{ - 1}}\]. Its solubility product is 108. The value of ′n′ will be:

Answer 1

Hint: The solubility product, \[{K_{sp}}\], applies in situations where salts do not fully dissolve in a solvent. The solvent is normally water.

Complete answer:
The solubility product, \[{K_{sp}}\], applies in situations where salts do not fully dissolve in a solvent the solvent is normally water. A substance solubility product is the mathematical product of its ion concentrations which is dissolved raised to the power of the concentrations stoichiometric coefficients.
The given reaction is,
\[{A_2}{B_3} \rightleftharpoons 2{A_3}^ + + 3{B_2}^ - \]
Now after putting the values in the formula for solubility we get,
\[{K_{sp}} = 108{S^n}\]
Let S be the solubility of A and B,
So after putting the respective values of the reactants and products in the formula we get:
\[108{S^n} = {(2S)^2}{(3S)^3}\]
After putting the values of the square, we get:
\[108{S^n} = 4{S^2} \times 27{S^3}\]
Now on multiplying the RHS we get,
\[108{S^n} = 108{S^5}\]
On simplifying we get,
n = 5

So the value of n is 5.

Additional Information The smaller the solubility product of a substance, the lower is the solubility. The solubility product is a heterogeneous equilibrium constant, a specific form of the equilibrium constant. Solubility products vary with temperature, so the temperature at which a solubility product was taken must always be written.

Note: When a slightly soluble ionic compound is added to water, some of it dissolves to make a solution, establishing an equilibrium between the pure solid and a solution of its ions.