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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 AgIAgI at 25∘C25^\circ C is 1.0×10−16mol2L−21.0 \times {10^{ - 16}}mo{l^2}{L^{ - 2}} . The solubility of AgIAgI in 10−4N{10^{ - 4}}N solution of KIKI at 25∘C25^\circ C is approximately (molL−1)(mol{L^{ - 1}}):
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

In the given two solutions, the iodide ion is the common ion. AgIAgI is a sparingly soluble salt whereas KIKI is a strong electrolyte or salt that has a high degree of dissolution. Such salts have different extents of solubility.

Complete step by step answer:
The solubility of ionic solids in water varies greatly. There are some salts that are highly soluble in water or even hygroscopic and some salts are less or slightly soluble. Now if we go on adding such salt to a solvent at a particular temperature then there comes a stage beyond which salt cannot dissolve further in the solution and thus the solution becomes saturated with the salt. The remaining salt is in undissolved form. Consider a salt ABAB added to a solvent then the two equilibria existing in the solution are
AB⇌       AB⇌        A+BAB \rightleftharpoons \,\,\,\,\,\,\,AB \rightleftharpoons \,\,\,\,\,\,\,\,A + B
(solid) (unionized) (ions)
Now applying law of mass action we get-
[A+][B−]=K[AB]=KK′=Ks[{A^ + }][{B^ - }] = K[AB] = K{K'} = {K_s}
Where Ks{K_s} is the solubility product. It is defined as the product of concentration of ions in a saturated solution of an electrolyte at a given temperature.
Now in case of AgIAgI -
AgI⇌Ag++I−AgI \rightleftharpoons A{g^ + } + {I^ - }
Ksp=[Ag+][I−]{K_{sp}} = [A{g^+}][{I^ - }]
Salts like AgIAgI , BaSO4BaS{O_4} are sparingly soluble electrolytes. A saturated solution of sparingly soluble electrolyte contains a very small amount of dissolved electrolyte. It is assumed that the whole of the dissolved electrolyte is present in the form of ions. If we consider ′S′'S' as the solubility of the salt in molL−1mol{L^{ - 1}} then,
AgI⇌Ag++I−AgI \rightleftharpoons A{g^ + } + {I^ - }
Let SS be the solubility of AgIAgI in molL−1mol{L^{ - 1}} .
Then, Ksp=[Ag+][I−]=S×S=S2{K_{sp}} = [A{g^ + }][{I^ - }] = S \times S = {S^2}
Solubility of two electrolytes having common ions when they are dissolved in the same solution is called simultaneous solubility.
AgIAgI and KIKI solutions have I−{I^ - } ions as common in them.
Now let the solubility of AgIAgI be SS . Then,

AgIAgIAgAgII
SSSSSS

So, [Ag+]=[I−]=S[A{g^ + }] = [{I^ - }] = S
KIKI is a strong electrolyte. It dissociates completely. The concentration of KIKI solution is 10−4N{10^{ - 4}}N so it shall provide a concentration of I−{I^ - } ion =10−4N = {10^{ - 4}}N .
Now the total concentration of I−{I^ - } ions is =(10−4+S) = ({10^{ - 4}} + S)
It is given that solubility product of AgIAgI is Ksp=1.0×10−16mol2L−2{K_{sp}} = 1.0 \times {10^{ - 16}}mo{l^2}{L^{ - 2}}
So, Ksp=[Ag+][I−]{K_{sp}} = [A{g^ + }][{I^ - }]
=S(10−4+S)= S({10^{ - 4}} + S)
Ksp=S2+10−4×S{K_{sp}} = {S^2} + {10^{ - 4}} \times S
Neglecting S2{S^2} since solubility is a small quantity and after squaring it becomes smaller.
Therefore, Ksp=10−4×S{K_{sp}} = {10^{ - 4}} \times S
1.0×10−16=10−4×S1.0 \times {10^{ - 16}} = {10^{ - 4}} \times S
S=1.0×10−1610−4=1.0×10−12molL−1S = {{1.0 \times {{10}^{ - 16}}}}{{{{10}^{ - 4}}}} = 1.0 \times {10^{ - 12}}mol{L^{ - 1}}
The correct option is B.

Note:
Ionic product is the product of the concentration of ions each raised to their stoichiometric coefficients.
When Ksp>Kip{K_{sp}} > {K_{ip}} , then the solution is unsaturated in which more solute can be dissolved.( Kip{K_{ip}} is the ionic product and Ksp{K_{sp}} is the solubility product).
When Ksp<Kip{K_{sp}} < {K_{ip}} , the solution becomes supersaturated and precipitation takes place.
When Ksp=Kip{K_{sp}} = {K_{ip}} , the solution is saturated in which no more solute can be dissolved.