Test Index

NCERT Class XI Chemistry Equilibrium Solutions

© examsnet.com
Question : 50 of 73
Marks: +1, -0
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and pKapK_a of bromoacetic acid.
Solution:  
−COOH∣BrCH2\underset{\underset{\mathrm{CH}_2}{\underset{\mathrm{Br}}{\vert}}}{-\mathrm{COOH}} ⇌ −COO−∣BrCH2+H+\underset{\underset{\mathrm{CH}_2}{\underset{\mathrm{Br}}{\vert}}}{-\mathrm{COO}^{-}} + \mathrm{H}^{+}
Initial molar conc.C00Eqm. molar conc.C(1−α)CαCα\begin{array}{lccc} \text{Initial molar conc.} & C & 0 & 0 \\ \text{Eqm. molar conc.} & C(1-\alpha) & C\alpha & C\alpha \end{array}
C = 0.1 M, a = 0.132, [H+][\mathrm{H}^{+}] = Ca = 0.0132 M
pH = – log(0.0132) = 1.8794
KaK_a = Ca2\mathrm{Ca}_2 or KaK_a = 0.1 × (0.132)2(0.132)^2 = 1.74 × 10−310^{-3}
pKapK_a = –log KaK_a = – log(1.74 × 10−310^{-3}) = 2.7595
© examsnet.com
Go to Question:
Prev Question
Next Question