Q. The ionisation constant of acetic acid is $1.8 \times 10^{- 5}$ at what concentration it will be dissociated to 2%?
A) 0.025 M
B) 0.045 M
C) 0.240 M
D) 0.082 M
Explanation
(b)
$CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$
$C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$
$C(1 - \alpha)\ \ \ \ \ \ \ \ \ \ \ \ \ C\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ C\alpha$
$K = \frac{C\alpha \bullet C\alpha}{C(1 - \alpha)} = \frac{C\alpha^{2}}{(1 - \alpha)}$
If $1 > > > C\$then $K = C\alpha^{2}$
$C = \frac{\ 1.8 \times 10^{- 5}}{(0.02)^{2}} = 0.045\ M$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.