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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 45 of 73
Marks: +1, -0
The first ionization constant of H2S\mathrm{H_2S} is 9.1 × 10−810^{-8}. Calculate the concentration of HS−\mathrm{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\mathrm{H_2S} is 1.2 × 10−1310^{-13}, calculate the concentration of S2−\mathrm{S}^{2-} under both conditions.
Solution:  
H2S\mathrm{H_2S} ⇌ HS−+H+\mathrm{HS}^{-}+\mathrm{H}^{+}
InitiallyC00At equilibriumC−CαCαCα\begin{array}{lccc} \text{Initially} & C & 0 & 0 \\ \text{At equilibrium} & C - C\alpha & C\alpha & C\alpha \end{array}
KaK_a = [HS−][H+][H2S]\frac{[\mathrm{HS}^{-}][\mathrm{H}^{+}]}{[\mathrm{H_2S}]} = Cα21−α\frac{C\alpha^2}{1-\alpha} = Cα2C\alpha^2 (Since 1 ⋙ α)
and [Hs−][\mathrm{Hs}^{-}] = Cα = KaC\sqrt{K_a C}
[Hs−][\mathrm{Hs}^{-}] = 9.1×10−8\sqrt{9.1 \times 10^{-8}}}×0.1} = 9.54 × 10−5 M10^{-5}\,\mathrm{M}
∴ [Hs−][\mathrm{Hs}^{-}] = 9.54 × 10−5 M10^{-5}\,\mathrm{M}
In 0.1 M HCl,
H2S0.1 M→(0.1−x)\underset{0.1\,\mathrm{M} \rightarrow (0.1 - x)}{\mathrm{H_2S}} ⇌ HS−0→x+H+0→x\underset{0 \rightarrow x}{\mathrm{HS}^{-}} + \underset{0 \rightarrow x}{\mathrm{H}^{+}}
HCl0.1 M\underset{0.1\,\mathrm{M}}{\mathrm{HCl}} → Cl−0.1 M+H+0.1 M\underset{0.1\,\mathrm{M}}{\mathrm{Cl}^{-}} + \underset{0.1\,\mathrm{M}}{\mathrm{H}^{+}}
KaK_a = [HS−][H+][H2S]\frac{[\mathrm{HS}^{-}][\mathrm{H}^{+}]}{[\mathrm{H_2S}]} = 9.1 × 10−810^{-8} ⇒ KaK_a = 0.1×[HS−]0.1\frac{0.1 \times [\mathrm{HS}^{-}]}{0.1} = 9.1 × 10810^{8}
⇒ [HS−][\mathrm{HS}^{-}] = 9.1 × 10−8 M10^{-8}\,\mathrm{M}
∴ The concentration of HS−\mathrm{HS}^{-} has decreased in 0.1 M HCl.
To calculate the concentration of S2−\mathrm{S}^{2-} ion:
HS−\mathrm{HS}^{-} → H++S2−\mathrm{H}^{+}+\mathrm{S}^{2-}
H2S\mathrm{H_2S} ⇌Ka1\xrightleftharpoons{K_{a_1}} H++HS−\mathrm{H}^{+}+\mathrm{HS}^{-}
HS−\mathrm{HS}^{-} ⇌Ka2\xrightleftharpoons{K_{a_2}} H++S2−\mathrm{H}^{+}+\mathrm{S}^{2-}
H2S\mathrm{H_2S} ⇌Ka\xrightleftharpoons{K_a} 2H++S2−2\mathrm{H}^{+}+\mathrm{S}^{2-}
Overall dissociation constant of H2S\mathrm{H_2S}
KaK_a = Ka1×Ka2K_{a_1} \times K_{a_2} = 9.1 × 10−810^{-8} × 1.2 × 10−1310^{-13} = 1.092 × 10−2010^{-20}
H2S0.1 M→0.1−x\underset{0.1\,\mathrm{M} \rightarrow 0.1 - x}{\mathrm{H_2S}} ⇌ 2H+0→2x+S2−0→x\underset{0 \rightarrow 2x}{2\mathrm{H}^{+}} + \underset{0 \rightarrow x}{\mathrm{S}^{2-}}
KaK_a = [H+]2[S2−][H2S]\frac{[\mathrm{H}^{+}]^2[\mathrm{S}^{2-}]}{[\mathrm{H_2S}]} ⇒ 1.092 × 10−2010^{-20} = (2x)2x0.1\frac{(2x)^2 x}{0.1} ⇒ x = 6.5 × 10−8 M10^{-8}\,\mathrm{M}
In presence of 0.1 M HCl, KaK_a = [H+]2[S2−][H2S]\frac{[\mathrm{H}^{+}]^2[\mathrm{S}^{2-}]}{[\mathrm{H_2S}]}
1.092 × 10−2010^{-20} = (0.1)2[S2−]0.1\frac{(0.1)^2[\mathrm{S}^{2-}]}{0.1} = 1.092 × 10−19 M10^{-19}\,\mathrm{M}
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