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Question: The solubility product of AgI at \[25^\circ C\] is \[1.0 \times {10^{ - 16}}\] \[mo{l^2}{L^{ - 2}}\]...

The solubility product of AgI at 25∘C25^\circ C is 1.0×10−161.0 \times {10^{ - 16}} mol2L−2mo{l^2}{L^{ - 2}} . The solubility of AgI in 10−4{10^{ - 4}} N solution of KI at 25∘C25^\circ C is approximately (in mol/L):
A. 1.0×10−161.0 \times {10^{ - 16}}
B. 1.0×10−121.0 \times {10^{ - 12}}
C. 1.0×10−101.0 \times {10^{ - 10}}
D. 1.0×10−81.0 \times {10^{ - 8}}

Explanation

Solution

To Solve this question, we must first find the solubility of AgI from the solubility product given to us. After that, we must find the product of the solubility of AgI and the concentration of KI to find the solubility of AgI in KI.

Complete step by step answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Solubility can be understood as the maximum amount of solute that can be dissolved in the given solvent at equilibrium. It can be regarded as the maximum degree to which the given compound can dissolve itself in the given solvent. Solubility product constant can be understood as the equilibrium constant for the dissolution of a given solute into a given solvent.
In the given question, AgI is dissociated into its constituent ions. This reaction can be given as:
AgI⇌Ag++I−AgI \rightleftharpoons A{g^ + } + {I^ - }
Hence, the solubility product for this compound can be given as:
Since AgI is a binary electrolyte, the solubility product can be given as
Ksp=S2{K_{sp}} = {S^2}
Where Ksp{K_{sp}} is the solubility product and S is the solubility of AgI.
Hence, substituting the given values, we get
1.0×10−16=S21.0 \times {10^{ - 16}} = {S^2}
Hence, S = 1.0  x  10−16=10−8\sqrt {1.0\,\,x\,\,{{10}^{ - 16}}} = {10^{ - 8}} mol/L
The solubility of AgI in KI can be given by the product of the solubility of AgI and the concentration of KI:
Solubility of AgI in KI = 10−8×10−4{10^{ - 8}} \times {10^{ - 4}} = 1.0×10−121.0 \times {10^{ - 12}} mol/L

Hence, Option B is the correct option.

Note:
The value of the solubility product varies on the basis of the temperature of the system and the surroundings. Higher the temperature of the system, higher is the solubility product of the solute.