Solveeit Logo
Login

Question

Question: The solubility in water of a sparingly soluble salt \({\text{A}}{{\text{B}}_{\text{2}}}\) is \(1.0 \...

The solubility in water of a sparingly soluble salt AB2{\text{A}}{{\text{B}}_{\text{2}}} is 1.0×10−5 mol L−11.0 \times {10^{ - 5}}{\text{ mol }}{{\text{L}}^{ - 1}}. Its solubility product will be:
A) 4×10−154 \times {10^{ - 15}}
B) 4×10−104 \times {10^{ - 10}}
C) 1×10−151 \times {10^{ - 15}}
D) 1×10−101 \times {10^{ - 10}}

Explanation

Solution

To solve this question we have to first write the dissociation of salt into its two individual ions. Then find the concentration of the ions in terms of molar solubility. Calculate the solubility product using the concentration of ions.

Complete step by step answer:
The maximum amount of any solute that can be dissolved in a solvent at equilibrium is known as the solubility of the solute. The solubility product gives the equilibrium between a solid and its ions in the aqueous solution.
We are given a sparingly soluble salt AB2{\text{A}}{{\text{B}}_{\text{2}}}. The sparingly soluble salt AB2{\text{A}}{{\text{B}}_{\text{2}}} dissociates as follows:
AB2⇌A2++2B−{\text{A}}{{\text{B}}_{\text{2}}} \rightleftharpoons {{\text{A}}^{2 + }} + 2{{\text{B}}^ - }
The solubility in water of a sparingly soluble salt AB2{\text{A}}{{\text{B}}_{\text{2}}} is 1.0×10−5 mol L−11.0 \times {10^{ - 5}}{\text{ mol }}{{\text{L}}^{ - 1}}. Thus,
AB2⇌A2++2B−{\text{A}}{{\text{B}}_{\text{2}}} \rightleftharpoons {{\text{A}}^{2 + }} + 2{{\text{B}}^ - }
Where S is the molar solubility of salt and it is equal to 1.0×10−5 mol L−11.0 \times {10^{ - 5}}{\text{ mol }}{{\text{L}}^{ - 1}}.
We know that the solubility product is the measure of degree of dissociation of the ions in the solution. The solubility product is the product of the concentrations or activities of the ions at equilibrium. Thus, the equation for solubility product is as follows:
Ksp=[Cation][Anion]{{\text{K}}_{{\text{sp}}}} = \left[ {{\text{Cation}}} \right]\left[ {{\text{Anion}}} \right]
Where Ksp{{\text{K}}_{{\text{sp}}}} is the solubility product.
Thus,
Ksp=[A2+][B−]2{{\text{K}}_{{\text{sp}}}} = \left[ {{{\text{A}}^{2 + }}} \right]{\left[ {{{\text{B}}^ - }} \right]^2}
Ksp=(S)×(2S)2{{\text{K}}_{{\text{sp}}}} = \left( {\text{S}} \right) \times {\left( {{\text{2S}}} \right)^2}
Ksp=4S3{{\text{K}}_{{\text{sp}}}} = 4{{\text{S}}^3}
Substitute 1.0×10−5 mol L−11.0 \times {10^{ - 5}}{\text{ mol }}{{\text{L}}^{ - 1}} for the molar solubility and solve for the solubility product. Thus,
Ksp=4(1.0×10−5 mol L−1)3{{\text{K}}_{{\text{sp}}}} = 4{\left( {1.0 \times {{10}^{ - 5}}{\text{ mol }}{{\text{L}}^{ - 1}}} \right)^3}
Ksp=4×10−15{{\text{K}}_{{\text{sp}}}} = 4 \times {10^{ - 15}}
Thus, the solubility product of a sparingly soluble salt AB2{\text{A}}{{\text{B}}_{\text{2}}} is 4×10−154 \times {10^{ - 15}}.

Thus, the correct option is (A) 4×10−154 \times {10^{ - 15}}.

Note: The solubility product is calculated using the concentration of the ions in which the salt has dissociated. Solubility factor depends on various factors such as temperature, pressure and nature of the electrolyte. The concentration of ions is affected by these factors and thus, the solubility product gets affected.