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Question: An aqueous solution contains \(0.10\;M\) \[{H_2}S\] and \(0.20\;M\) \(HCl\). If the equilibrium cons...

An aqueous solution contains 0.10  M0.10\;M H2S{H_2}S and 0.20  M0.20\;M HClHCl. If the equilibrium constants for the formation of HS−H{S^ - } ions from H2S{H_2}S is 1.0×10−71.0 \times {10^{ - 7}} and that of S2−{S^{2 - }} from HS−H{S^ - } ions is 1.2 × 10−131.2\, \times \,{10^{ - 13}} then the concentration of S2−{S^{2 - }} ions in aqueous solution is:
A.5 × 10−195\, \times \,{10^{ - 19}}
B.5 × 10−85\, \times \,{10^{ - 8}}
C.3 × 10−203\, \times \,{10^{ - 20}}
D.6 × 10−216\, \times \,{10^{ - 21}}

Explanation

Solution

For this question we will first check the reaction for decomposition ofH2S{H_2}S. H2S{H_2}S Decomposes to form HS−H{S^ - } andS−{S^ - }. Further HS−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+{H^ + } ion and HS−H{S^ - } ion respectively.
The reaction is as follows:
H2S  ⇌H+(aq) + HS−(aq){H_2}S\,\, \rightleftharpoons {H^ + }\left( {aq} \right)\, + \,H{S^ - }\left( {aq} \right) ………………………………………. (1)

Another equation is for decomposition of HS-
HS− ⇌ H+ + S2−H{S^ - }\, \rightleftharpoons \,{H^ + }\, + \,{S^{2 - }} ……………………………………………………………. (2)
Equilibrium equation for 1 st reaction:
K1 =   [H+][HS−][H2S]{K_1}\, = \,\,\dfrac{{\,[{H^ + }][H{S^ - }]}}{{[{H_2}S]}}
Equilibrium reaction for 2 nd reaction:
K2 = [H+][S−] [HS−]{K_2}\, = \,\dfrac{{[{H^ + }][{S^ - }]}}{{\,[H{S^ - }]}}
Adding equation 1 and 2 we get,
H2S (aq) ⇌ 2H+(aq) + S2− (aq) K = K1 × K2 K = [H+][HS−][H2S]   × [H+][S2−][H2S]  {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 × 1.2 × 10−13 = 1.2 × 10−20  = \,{10^{ - 7}}\, \times \,1.2\, \times \,{10^{ - 13}} \\\ = \,1.2\, \times \,{10^{ - 20}} \\\
At time t=0t = 0,
Concentration of H2S{H_2}S is 0.10M and concentration of HClHCl is 0.20  M0.20\;M
Substituting the values in the final equilibrium expression, we get
K = [0.2]2[x][0.1] 1.2 × 10−20 = 0.2 × 0.2 × x x = 3 X 10−20   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 × 10−203\, \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.
Kc =  [product][reactants]{{\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.
Kp  =   [partial pressure of product][partial pressure of reactants]{{\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 ofHS−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.