Q. An acid solution of 0.005 M has a pH of 5. The degree of ionisation of acid is
A) $0.1 \times 10^{- 2}$
B) $0.2 \times 10^{- 2}$
C) $0.5 \times 10^{- 4}$
D) $0.6 \times 10^{- 6}$
Explanation
(b)
Acid, [$H^{+}$]=$10^{- pH} = 10^{- 5}$
$\alpha = \frac{actual\ ionisation}{molar\ concentration} = \frac{10^{- 5}}{0.005} = 0.2 \times 10^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.