Question

A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is

• A. 1 X 10^-8
• B. 1 X 10^-4
• C. 1 X 10^-6
• D. 1 X 10^-5

Answer 1

The correct option is A

1 X 10^-8

$K=\frac{{\alpha }^{2}C}{1-\alpha };\alpha =\frac{0.01}{100}$
$\therefore K={\alpha }^{2}C={\left[\frac{0.01}{100}\right]}^2\times 1=1\times {10}^{-8}$