Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is

Equilibrium

Solution:

Ionisation constant, $K_{a}=\frac{a^{2} \times c}{1-a}$
$\alpha$= degree of dissociation $= 0.001 \%$
$=\frac{0.001}{100}=1\times10^{-5}$
$C$= concentration $= 0.1 \,M$
when $\alpha$ is very small
$K=\alpha^{2} \times C$
$=(1 \times 10^{-5})^{2} \times 0.1=1\times 10^{-11}$