Calculate the degree og ionisation of `0.04 M HOCl` solution having ionisation constant `1.25 xx 10^(-4)` ?

Options

A. `0.025`
B. `0.5`
C. `0.25`
D. `0.055`

Answer

D

Explanation

`{:(HOCl,hArr,OH^(+),+,Cl^(+),),(C,,,,,),(C (a - alpha),,C alpha,,C alpha,):}`
`K_(a) = (c alpha^(2))/(1 - alpha) = (0.04 xx alpha^(2))/(1 - alpha) " " [{:("As",,alpha lt lt lt 1),(because,,1-alpha=1):}]`
`1.25 xx 10^(-4) = 0.04 alpha^(2)`
`alpha^(2) = (1.25 xx 10^(-4))/(0.04) = (125 xx 10^(-4))/(4)`
`alpha = (11)/(2) xx 10^(-2) = (0.11)/(2) = 0.055`