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Question: The conductivity of a saturated solution of \({\text{A}}{{\text{g}}_{\text{3}}}{\text{P}}{{\text{O}}...

The conductivity of a saturated solution of Ag3PO4{\text{A}}{{\text{g}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}} is 9×10−6Sm - 19 \times {10^{ - 6}}{\text{S}}{{\text{m}}^{{\text{ - 1}}}} and its equivalent conductivity is 1.50×10−4Sm2eq - 11.50 \times {10^{ - 4}}{\text{S}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}. The Ksp{{\text{K}}_{{\text{sp}}}} of Ag3PO4{\text{A}}{{\text{g}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}} is:
A.4.32×10−184.32 \times {10^{ - 18}}
B.1.8×10−91.8 \times {10^{ - 9}}
C.8.64×10−138.64 \times {10^{ - 13}}
D.None of these

Explanation

Solution

To answer this question, you should recall the concept of solubility and solubility products. The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution.
Formula used:
λeq=k×1000N{\lambda _{eq}} = \dfrac{{{\text{k}} \times 1000}}{{\text{N}}}
where λeq{\lambda _{eq}} is equivalent conductivity

Complete step by step solution:
When a salt is dissolved in a solvent the strong forces of attraction of solute must be overcome by the interactions between ions and the solvent. The solvation enthalpy of ions is always negative which means that energy is released during this process. The nature of the solvent determines the amount of energy released during solvation that is solvation enthalpy. Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution.
We are given in the question that: k=9×10−6Sm - 1{\text{k}} = 9 \times {10^{ - 6}}{\text{S}}{{\text{m}}^{{\text{ - 1}}}}and λeq=1.5×10−4Sm2eq - 1{\lambda _{eq}} = 1.5 \times {10^{ - 4}}{\text{S}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}
Substituting these values in the equation we have, Normality:
N=k×1000λeq=9×10−6×10−2×10001.4×10−4×104=6×10−5{\text{N}} = \dfrac{{{\text{k}} \times 1000}}{{{\lambda _{eq}}}} = \dfrac{{9 \times {{10}^{ - 6}} \times {{10}^{ - 2}} \times 1000}}{{1.4 \times {{10}^{ - 4}} \times {{10}^4}}} = 6 \times {10^{ - 5}}.
The reaction for the given reactant can be written as:
Ag3PO4→3Ag++PO4−A{g_3}P{O_4} \to 3A{g^ + } + P{O_4}^ -
So, 3 electrons are transferred.
Solubility can be calculated:
S=Nn=6×10−63=2×10−5mol/L  {\text{S}} = \dfrac{{\text{N}}}{{\text{n}}} = \dfrac{{6 \times {{10}^{ - 6}}}}{3} = 2 \times {10^{ - 5}}{\text{mol/L}}\;.
The solubility product of the above reaction can be written as:
Ksp=(3S)3.S=27S4{{\text{K}}_{{\text{sp}}}} = \left( {3{\text{S}}} \right)3.{\text{S}} = 27{{\text{S}}^4} .
Now substituting the calculated values of solubility:
Ksp=27×(2×10−5)4=4.32×10−18{{\text{K}}_{{\text{sp}}}} = 27 \times {(2 \times {10^{ - 5}})^4} = 4.32 \times {10^{ - 18}}

Hence, the correct option is A.

Note:
You should know about the difference between solubility and solubility product constant. The solubility of a substance in a solvent is the total amount of the solute that can be dissolved in the solvent at equilibrium. On the other hand, the solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.