A solution contains a mixture of Ag+ (0.10 M) and Hg2+ (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodine ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated ? Ksp:AgI = 8.5 × 10−17, HgI2 = 2.5 × 10−26
A solution contains a mixture of Ag+ (0.10 M) and Hg2+ (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodine ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated ? Ksp:AgI = 8.5 × 10−17, HgI2 = 2.5 × 10−26
The correct option is A
The [I−] needed for precipitation of Ag+ and Hg2+2 are derived as:
For Agl:
[Ag+][l−] = Ksp(Agl)
(0.1)[l−] = 8.5 × 10−17
[l−] = 8.5 × 10−16 M ............(1)
For Hg2I2: [Hg2+2][I−]2 = Ksp(Hg2I2
(0.1)[I−]2 = 2.5 × 10−26
[I−] = 5 × 10−13 M .............(2)
Since [I−] required for precipitation of AgI is less and thus AgI begins to precipitate first. Also, it will continue upto addition of [I−] = 5 × 10−13 M when Hg2I2 begins to precipitate and thus, maximum [I−] for AgI precipitation = 5 × 10−13 M
Now, at this concentration of I−,
[Ag+]left in solution is [Ag+]left[I−] = Ksp(AgI) = Ksp(Ag+)
[Ag+]left = 8.5 × 10−175.0 × 10−13 = 1.7 × 10−4 M
0.1 M Ag+ will be left = 1.7 × 10−4 M Ag+ in solution
100 M Ag+ will be left = 0.17% M Ag+
% of Ag precipitated =99.83%.
The [I−] needed for precipitation of Ag+ and Hg2+2 are derived as:
For Agl:
[Ag+][l−] = Ksp(Agl)
(0.1)[l−] = 8.5 × 10−17
[l−] = 8.5 × 10−16 M ............(1)
For Hg2I2: [Hg2+2][I−]2 = Ksp(Hg2I2
(0.1)[I−]2 = 2.5 × 10−26
[I−] = 5 × 10−13 M .............(2)
Since [I−] required for precipitation of AgI is less and thus AgI begins to precipitate first. Also, it will continue upto addition of [I−] = 5 × 10−13 M when Hg2I2 begins to precipitate and thus, maximum [I−] for AgI precipitation = 5 × 10−13 M
Now, at this concentration of I−,
[Ag+]left in solution is [Ag+]left[I−] = Ksp(AgI) = Ksp(Ag+)
[Ag+]left = 8.5 × 10−175.0 × 10−13 = 1.7 × 10−4 M
0.1 M Ag+ will be left = 1.7 × 10−4 M Ag+ in solution
100 M Ag+ will be left = 0.17% M Ag+
% of Ag precipitated =99.83%.
