Calculate the `pH` of `0.001MHOCl` having `25%` dissocation .Also calculate dissocation constant of the acid ,Take `log2=0.3`

Explanation

`{:(,HOCl hArr ,H^(+),OCl^(-)),(t=0,a,0,0),(t=eq,a-a alpha,a alpha,a alpha):}`
so `[H^(+)]=aalpha=10^(-3)xx(25)/(100)=2.5xx10^(-4)`
So` pH=3.6`
Now `K_(a)=(aalpha)/(a(1-alpha))=(aalpha^(2))/(1-alpha)=(1)/(12)xx10^(-3)`