Question

What should the $ {H^ + } $ concentration be in solution that is $ 0.25\;M $ in $ C{o^{2 + }} $ to prevent the precipitation of $ CoS $ when the solution is saturated with $ {{\text{H}}_2}{\text{S}} $ . A saturated solution of $ {{\text{H}}_2}{\text{S}} $ is $ 0.1\;{\text{M}} $ :-
 $ {{\text{K}}_{sp}}\left( {CoS} \right) = \;5 \times {10^{ - 22}} $ , $ {{\text{K}}_1}\left( {{{\text{H}}_2}{\text{S}}} \right) = \;1.1 \times {10^{ - 8}} $ , $ {{\text{K}}_2}\left( {{{\text{H}}_2}{\text{S}}} \right) = \;1.1 \times {10^{ - 13}} $
(A) $ 0.523 $
(B) $ 0.234 $
(C) $ 0.725 $
(D) $ 0.924 $

Answer 1

Hint: We are required to find the concentration of $ {H^ + } $ ions so that the precipitation of Cobalt sulphide $ CoS $ can be prevented. We will determine the concentration of $ {H^ + } $ ions by observing the reactions that are taking place here. We will calculate equilibrium constant for each reaction and by using we will find the concentration of each ion.

Complete answer:
Let’s write the values that are given to us
concentration of Cobalt in the solution, $ C{o^{2 + }} = \;0.25\;{\text{M}} $
concentration of Hydrogen sulphide, $ {{\text{H}}_2}{\text{S}}\; = \;0.1\;{\text{M}} $
Solubility product of Cobalt sulphide, $ {{\text{K}}_{sp}}\left( {CoS} \right) = \;5 \times {10^{ - 22}} $
Equilibrium constant for first reaction of Hydrogen sulphide, $ {{\text{K}}_1}\left( {{{\text{H}}_2}{\text{S}}} \right) = \;1.1 \times {10^{ - 8}} $
Equilibrium constant for second reaction of Hydrogen sulphide, $ {{\text{K}}_2}\left( {{{\text{H}}_2}{\text{S}}} \right) = \;1.1 \times {10^{ - 13}} $
Let’s see the reactions:
first reaction: $ {{\text{H}}_2}{\text{S}}\; \rightleftharpoons \;{{\text{H}}^ + }\; + \;H{S^ - } $
second reaction: $ H{S^ - }\; \rightleftharpoons \;{{\text{H}}^ + }\; + \;{{\text{S}}^ - } $
adding both first and second reactions we get,
 $ {{\text{H}}_2}{\text{S}}\; \rightleftharpoons \;2{{\text{H}}^ + }\; + \;{{\text{S}}^ - } $
calculating Equilibrium constant $ {\text{K}} $ for the above reaction
 $ {\text{K}}\; = \;{{\text{K}}_1}\; \times \;{{\text{K}}_2} $
substituting given values in the above equation, we get
 $ {\text{K}}\; = \;1.1\; \times \;{10^{ - 8}}\;\; \times \;1.1\; \times \;{10^{ - 13}} $
 $ {\text{K}}\; = \;1.1\; \times \;{10^{ - 21}} $
Equilibrium constant for the combined reaction is obtained, now writing the formula of equilibrium constant for the combined reaction
 $ {\text{K}}\; = \dfrac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}\left[ {{{\text{S}}^ - }} \right]}}{{\left[ {{{\text{H}}_2}{\text{S}}} \right]}} $ ……… (first)
Let's take the above equation as the first equation.
From the formula of the Solubility product of Cobalt sulphide, we will calculate the concentration of sulphide ion as it is unknown.
 $ {{\text{K}}_{sp}}\;\left( {CoS} \right) = \left[ {C{o^{2 + }}} \right]\left[ {{{\text{S}}^ - }} \right] $
substituting given values in the above formula, we get
 $ \left[ {{{\text{S}}^ - }} \right] = \dfrac{{5\; \times \;{{10}^{ - 22}}}}{{0.25}} $
 $ \left[ {{{\text{S}}^ - }} \right] = \;20\; \times \;{10^{ - 22}} $
putting the obtained value of concentration of sulphide ion and equilibrium constant in first equation, we have
 $ 1.1\; \times \;{10^{ - 21}}\; = \dfrac{{\;20\; \times \;{{10}^{ - 22\;}}{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{0.1}} $
now calculating concentration of $ {H^ + } $ ions
 $ {\left[ {{{\text{H}}^ + }} \right]^2}\; = \dfrac{{\;1.1\; \times \;{{10}^{ - 21}}\; \times \;0.1}}{{20\; \times \;{{10}^{ - 22\;}}}} $
 $ \left[ {{{\text{H}}^ + }} \right]\;\; = \;0.234 $
The concentration of $ {H^ + } $ ions is $ \;0.234 $ .
So, option (B) is the correct answer.

Note:
Carefully observe the reactions to calculate the equilibrium constant of the reaction, when two reactions are combined the equilibrium constant of the combined reaction is given by product of the equilibrium constant of the first and second reaction.