Q. A monoprotic acid in 1.00 M solution is 0.01 % ionised. The dissociation constant of this acid is
A) $1 \times 10^{- 8}$
B) $1 \times 10^{- 4}$
C) $1 \times 10^{- 6}$
D) $1 \times 10^{- 5}$
Explanation
(a)
For weak electrolytes, according to Ostwald’s dilution law
$\alpha = \sqrt{KV}$
Here, $\alpha = 0.01\% = 0.0001 = 1 \times 10^{- 4}$
$V = \frac{1}{C} = \frac{1}{1.0} = 1$L
$\therefore\ \ \ \ K_{a} = \frac{\alpha^{2}}{V} = \frac{\left( 1 \times 10^{- 4} \right)^{2}}{1} = 1 \times 10^{- 8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.