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Question: A solution of 0.1 M \[C{l^ - }\], 0.1 M \(B{r^ - }\) and 0.1 M \({I^ - }\) solid \(AgN{O_3}\) is gra...

A solution of 0.1 M Cl−C{l^ - }, 0.1 M Br−B{r^ - } and 0.1 M I−{I^ - } solid AgNO3AgN{O_3} is gradually added to this solution. Assuming that the addition of AgNO3AgN{O_3} does not change the volume.
Given: Ksp{K_{sp}}(AgCl) =1.7×10−10= 1.7 \times {10^{ - 10}}, Ksp{K_{sp}}(AgBr)=5×10−13= 5 \times {10^{ - 13}}, Ksp{K_{sp}}(AgI)=8.5×10−17= 8.5 \times {10^{ - 17}}. What will be the conc. Of both ions when the third ion starts precipitating.
A. Br−=2.9×10−4;I−=5×10−8B{r^ - } = 2.9 \times {10^{ - 4}};{I^ - } = 5 \times {10^{ - 8}}
B. Br−=5×10−8;I−=2.9×10−4B{r^ - } = 5 \times {10^{ - 8}};{I^ - } = 2.9 \times {10^{ - 4}}
C. Br−=5.8×10−4;I−=10×10−8B{r^ - } = 5.8 \times {10^{ - 4}};{I^ - } = 10 \times {10^{ - 8}}
D. Br−=10×10−4;I−=5.8×10−8B{r^ - } = 10 \times {10^{ - 4}};{I^ - } = 5.8 \times {10^{ - 8}}

Explanation

Solution

The solubility product (Ksp{K_{sp}}) of the compound is the product of the concentration of both the ions formed after the dissociation of compound. The concentration of the ionic compound remains the same in its constituent ion at equilibrium.

Complete step by step answer:
The concentration of Cl−C{l^ - } ion is 0.1 M, the concentration of Br−B{r^ - } ion is 0.1 M, the concentration of I−{I^ - } ion is 0.1 M
Ksp{K_{sp}}(AgCl) =1.7×10−10= 1.7 \times {10^{ - 10}}, Ksp{K_{sp}}(AgBr).., Ksp{K_{sp}}(AgI)=8.5×10−17= 8.5 \times {10^{ - 17}}
As, the Ksp{K_{sp}} value of AgCl is maximum therefore it will precipitate at last.
The Ksp{K_{sp}}of AgCl is given as shown below.
Ksp=[Ag+][Cl−]{K_{sp}} = [A{g^ + }][C{l^ - }]
Substitute the given values in the above equation.
⇒1.7×10−10=[Ag+][0.1M]\Rightarrow 1.7 \times {10^{ - 10}} = [A{g^ + }][0.1M]
⇒[Ag+]=1.70×10−9M\Rightarrow [A{g^ + }] = 1.70 \times {10^{ - 9}}M
When AgCl will start to precipitate, then the concentration of [Ag+][A{g^ + }]is.
The Ksp{K_{sp}}of AgBr is given as shown below.
Ksp=[Ag+][Br−]{K_{sp}} = [A{g^ + }][B{r^ - }]
Substitute the given values in the above equation.
⇒5.00×10−13=[1.70×10−9][Br−]\Rightarrow 5.00 \times {10^{ - 13}} = [1.70 \times {10^{ - 9}}][B{r^ - }]
⇒[Br−]=2.94×10−4M\Rightarrow [B{r^ - }] = 2.94 \times {10^{ - 4}}M
When AgCl will start to precipitate, then the concentration of [Br−][B{r^ - }]is 2.94×10−4M2.94 \times {10^{ - 4}}M

The Ksp{K_{sp}}of AgI is given as shown below.
Ksp=[Ag+][I−]{K_{sp}} = [A{g^ + }][{I^ - }]
Substitute the given values in the above equation.
⇒8.50×10−17=[1.70×10−9][I−]\Rightarrow 8.50 \times {10^{ - 17}} = [1.70 \times {10^{ - 9}}][{I^ - }]
⇒[I−]=5.00×10−8M\Rightarrow [{I^ - }] = 5.00 \times {10^{ - 8}}M
When AgCl will start to precipitate, then the concentration of [I−][{I^ - }]is 5.00×10−8M5.00 \times {10^{ - 8}}M.

So, the correct answer is Option A.

Note: The concentration of all the anions are the same and on adding silver nitrate, the compound starts to precipitate as the halogen anion reacts with the silver cation to form a white precipitate.