Thank you for reporting, we will resolve it shortly
Q.
A monoprotic acid in a $0.1 \,M$ solution ionizes to $0.001 \%$. If its ionization constant is $\frac{10^{-x}}{100}$, the value of $x$ is __
Equilibrium
Solution:
For monoprotic acid,
$K _{ a }=\frac{ a ^{2} C }{(1- a )} \approx a ^{2} C$
where $\alpha$ is very small.
$\alpha \frac{0.001}{100}=10^{-5}$
$K _{ a } =10^{-5} \times 10^{-5} \times 0.1 $
$=1 \times 10^{-11} $
Since $K _{ a }=\frac{10^{- x }}{100}=10^{-11}$
$10^{- x }=100 \times 10^{-11}$
Or $x =9$