Question

An aqueous solution contains $0.10\;M$ \[{H_2}S\] and $0.20\;M$$HCl$. If the equilibrium constants for the formation of $H{S^ - }$ ions from ${H_2}S$ is $1.0 \times {10^{ - 7}}$ and that of ${S^{2 - }}$ from $H{S^ - }$ ions is $1.2\, \times \,{10^{ - 13}}$ then the concentration of ${S^{2 - }}$ ions in aqueous solution is:
A.$5\, \times \,{10^{ - 19}}$
B.$5\, \times \,{10^{ - 8}}$
C.$3\, \times \,{10^{ - 20}}$
D.$6\, \times \,{10^{ - 21}}$

Answer 1

Hint: For this question we will first check the reaction for decomposition of${H_2}S$. ${H_2}S$ Decomposes to form $H{S^ - }$ and${S^ - }$. Further $H{S^ - }$ decomposes to form Hydrogen ion and sulphide ion. From this equation we will check the equilibrium constant and find the answer and substitute the values to get the answer.

Complete step by step answer:
Hydrogen Sulphide decomposes to form ${H^ + }$ ion and $H{S^ - }$ ion respectively.
The reaction is as follows:
            ${H_2}S\,\, \rightleftharpoons {H^ + }\left( {aq} \right)\, + \,H{S^ - }\left( {aq} \right)$ ………………………………………. (1)

Another equation is for decomposition of HS-
                                          $H{S^ - }\, \rightleftharpoons \,{H^ + }\, + \,{S^{2 - }}$ ……………………………………………………………. (2)
Equilibrium equation for 1 st reaction:
                                       ${K_1}\, = \,\,\dfrac{{\,[{H^ + }][H{S^ - }]}}{{[{H_2}S]}}$
Equilibrium reaction for 2 nd reaction:
                                        ${K_2}\, = \,\dfrac{{[{H^ + }][{S^ - }]}}{{\,[H{S^ - }]}}$
Adding equation 1 and 2 we get,
$
  {H_2}S\,(aq)\, \rightleftharpoons \,2{H^ + }(aq)\, + \,{S^{2 - }}\,(aq) \\
  K\, = \,{K_1}\, \times \,{K_2} \\
  K\, = \,\dfrac{{[{H^ + }][H{S^ - }]}}{{[{H_2}S]\,\,}}\, \times \,\dfrac{{[{H^ + }][{S^{2 - }}]}}{{[{H_2}S]}} \\
 $
On putting the values we get,
$
   = \,{10^{ - 7}}\, \times \,1.2\, \times \,{10^{ - 13}} \\
   = \,1.2\, \times \,{10^{ - 20}} \\
$
At time $t = 0$,
Concentration of ${H_2}S$ is 0.10M and concentration of $HCl$ is $0.20\;M$
Substituting the values in the final equilibrium expression, we get
$
  K\, = \,\dfrac{{{{[0.2]}^2}[x]}}{{[0.1]}} \\
  1.2\, \times \,{10^{ - 20}}\, = \,0.2\, \times \,0.2\, \times \,x \\
  x\, = \,3\,X\,{10^{ - 20}}\, \\
 $

So, the correct answer is $3\, \times \,{10^{ - 20}}$ which matches with option C.

Additional Information:
The equilibrium constant expression is the ratio of the concentrations of a reaction at equilibrium. Each equilibrium constant expression has a constant value known as K, the equilibrium constant.
${{\text{K}}_{\text{c}}}{\text{ = }}\,\dfrac{{{\text{[product]}}}}{{{\text{[reactants]}}}}$
 Equilibrium constant can also be represented by using partial pressure where instead of Kc we use Kp.
${{\text{K}}_{\text{p}}}\,{\text{ = }}\,\dfrac{{\,{\text{[partial}}\,{\text{pressure}}\,{\text{of}}\,{\text{product]}}}}{{{\text{[partial}}\,{\text{pressure}}\,{\text{of}}\,{\text{reactants]}}}}$

Note:
While solving this question don’t forget to consider the decomposition of$H{S^ - }$. Also it should be noted that the concentration of solid should not be mentioned in the equilibrium expression. Only the concentration of liquids or gases are considered.