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Question
The first ionization constant of H2S is 9.1×10−8. Calculate the concentration of HS− ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2×10−13, calculate the concentration of S2− under both conditions.
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Solution
(i) To calculate [HS−] in absence of HCl:
Let, [HS−]=x M.
H2S⇌H++HS−
The initial concentrations of H2S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.
Ka=[H+][HS−][H2S]
9.1×10−8=x×x0.1−x
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
9.1×10−8=x×x0.1
x=9.54×10−5M=[HS−]
(ii) To calculate [HS−] in presence of HCl:
Let [HS−]=y M.
H2S⇌H++HS−
The initial concentrations of H2S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-y M, y M and y M respectively.
Also, HCl⇌H++Cl−
For HCl [H+]=[Cl−]=0.1M
Ka=[HS−][H+][H2S]
Ka=y(0.1+y)0.1−y
In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.
9.1×10−8=y×0.10.1
y=9.1×10−8M=[HS−]
(iii) To calculate [S2−] in absence of 0.1 M HCl:
HS−⇌H++S2−
[HS−]=9.54×10−5M
Let [S2−]=X M.
[H+]=9.54×10−5 M
Ka=[H+][S2−][HS−]
1.2×10−13=9.54×10−5×X9.54×10−5
X=1.2×10−13M=[S2−]
(iv) To calculate [S2−] in presence of 0.1 M HCl:
Let [S2−]=X′ M.
[HS−]=9.1×10−8 M
[H+]=0.1 M (from HCl)
Ka=[H+][S2−][HS−]
1.2×10−13=0.1×X′9.1×10−8
X′=1.092×10−19=[S2−]
Let, [HS−]=x M.
H2S⇌H++HS−
The initial concentrations of H2S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.
Ka=[H+][HS−][H2S]
9.1×10−8=x×x0.1−x
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
9.1×10−8=x×x0.1
x=9.54×10−5M=[HS−]
(ii) To calculate [HS−] in presence of HCl:
Let [HS−]=y M.
H2S⇌H++HS−
The initial concentrations of H2S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-y M, y M and y M respectively.
Also, HCl⇌H++Cl−
For HCl [H+]=[Cl−]=0.1M
Ka=[HS−][H+][H2S]
Ka=y(0.1+y)0.1−y
In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.
9.1×10−8=y×0.10.1
y=9.1×10−8M=[HS−]
(iii) To calculate [S2−] in absence of 0.1 M HCl:
HS−⇌H++S2−
[HS−]=9.54×10−5M
Let [S2−]=X M.
[H+]=9.54×10−5 M
Ka=[H+][S2−][HS−]
1.2×10−13=9.54×10−5×X9.54×10−5
X=1.2×10−13M=[S2−]
(iv) To calculate [S2−] in presence of 0.1 M HCl:
Let [S2−]=X′ M.
[HS−]=9.1×10−8 M
[H+]=0.1 M (from HCl)
Ka=[H+][S2−][HS−]
1.2×10−13=0.1×X′9.1×10−8
X′=1.092×10−19=[S2−]
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