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MCQ Solution: Ionic Equilibrium

A solution contains a mixture of Ag⁺ (0.1M) and Hg₂²⁺ (0.1M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated?
K_sp (AgI) = 8.5 $\times$ 10^–17

K_sp (Hg₂I₂) = 2.5 $\times$ 10^–26

1) 90.83%
2) 97.83%
3) 89.93%
4) 99.83%

Solution:

Let us first calculate [I^–] to precipitate AgI and Hg₂I₂

K_sp[AgI] = [Ag⁺] [I^–]

8.5 $\times$ 10^–17 = (0.1) [I^–]

[I^–] to precipitate as AgI = (8.5 $\times$ 10^-17)

K_sp(Hg₂²⁺) = [Hg₂I₂][I^–] = 8.5 $\times$ 10^–16 M

2.5 $\times$ 10^–26 = 0.1 [I^–]²

[I^–] to precipitate Hg₂I₂ = 5.0 $\times$ 10^–13 M

[I^–] to precipitate AgI is smaller. Therefore, Ag I will start precipitating first. On further addition of I^– more AgI will precipitate and when [I^–] ³ 5.0 ´ 10^–13 J, Mg₂I₂ will start precipitating. The maximum concentration of Ag⁺ at this stage will thus be calculated as:

K_sp(AgI) = [Ag⁺] [I^–]

8.5 $\times$ 10^–17 = [Ag⁺] (5.0 ´ 10^–13)

or, [Ag⁺] = 1.7 $\times$ 10^–4 M

Percentage of Ag⁺ remained precipitated = [(1.7 $\times$ 10^–4 M)/0/1] $\times$ 100 = 0.17%

Thus percentage of Ag⁺ precipitated = 99.83%