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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 54 of 73
Marks: +1, -0
The ionization constant of dimethylamine is 5.4 × 10−410^{-4}. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?
Solution:  
KaK_a = 5.4 × 10−410^{-4}
C = 0.02 M, KaK_a = Cα2C\alpha^2
α2\alpha^2 = KaC\frac{K_a}{C} = 5.4×10−40.02\frac{5.4\times10^{-4}}{0.02} ⇒ α2\alpha^2 = 270 × 10−410^{-4}
α = 0.164
(CH3)2NH+H2O\mathrm{(CH_3)_2NH+H_2O} ⇌ (CH3)2NH2++OH−\mathrm{(CH_3)_2NH_2^+ + OH^-}
Initially :C00At eqili. :C(1−α)CαCα\begin{array}{llll} \text{Initially :} & C & 0 & 0 \\ \text{At eqili. :} & C(1-\alpha) & C\alpha & C\alpha \end{array}
But [OH−][\mathrm{OH}^-] = 0.1 M
KaK_a = Cα × [OH−][\mathrm{OH}^-] ⇒ 5.4 × 10−410^{-4} = Cα×0.1C(1−α)\frac{C\alpha \times 0.1}{C(1-\alpha)}
But α ⋘ 1,
∴ α is neglected in the denominator.
⇒ α = 5.4×10−40.1\frac{5.4\times10^{-4}}{0.1} = 5.40 × 10−310^{-3} = 0.0054
% ionization = 100α = 100 × 0.0054 = 0.54
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