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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 45
Total: 73
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.
Solution:  
H2S ⇌ HS−+H+
InitiallyC00
At‌equilibriumC−CαCαCα

Ka =
[HS−][H+]
[H2S]
 =
Cα2
(1−α)
 = Cα2 (Since 1 ⋙ α)
and [Hs−] = Cα = √KaC
[Hs−] = √9.1×10−8}×0.1} = 9.54 × 10−5M
∴ [Hs−] = 9.54 × 10−5M
In 0.1 M HCl,
H2S
0.1M→(0.1−x)
 ⇌
HS−
0→x
+
H+
0→x
HC
l
0.1M
 →
Cl−
0.1M
+
H+
0.1M
Ka =
[HS−][H+]
[H2S]
 = 9.1 × 10−8 ⇒ Ka =
0.1×[HS−]
0.1
 = 9.1 × 108
⇒ [HS−] = 9.1 × 10−8M
∴ The concentration of HS– has decreased in 0.1 M HCl.
To calculate the concentration of S2– ion:
HS− → H++S2−
H2S
Ka1
⇌
 H++HS−
HS−
Ka2
⇌
 H++S2−
H2S
Ka
⇌
 2H++S2−
Overall dissociation constant of H2S
Ka = Ka1×Ka2 = 9.1 × 10−8 × 1.2 × 10−13 = 1.092 × 10−20
H2S
0.1M→0.1−x
 ⇌
2H+
0→2x
+
S2−
0→x
Ka =
[H+]2[S2−]
[H2S]
 ⇒ 1.092 × 10−20 =
(2x)2x
0.1
 ⇒ x = 6.5 × 10−8M
In presence of 0.1 M HCl, Ka =
[H+]2[S2−]
[H2S]
1.092 × 10−20 =
(0.1)2[S2−]
(0.1)
 = 1.092 × 10−19M
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