Calculate the `pH` of `0.08` solution of `HOCI` (hydrochlorous acid). The ionisation constant of the acid is `2.5 xx 10^(-5)`. Determine the percent dissociation of `HOCI`.

Answer

1.7

Explanation

`HOCl(aq) + H_2O(l) hArr H_3O^(+)(aq) + ClO^(-)(aq)`
If c is the initial concentration and `alpha` is the degree of dissociation , then
`{:("Initial conc.",c,0,0),("At equi.",c(1-alpha),c alpha, c alpha):}`
`K_a=([H_3O^+][ClO^-])/([HOCl])`
`((calpha)xx(calpha))/(c(1-alpha))approx calpha^2`
`2.5xx10^(-5)=0.08 alpha^2`
or `alpha=((2.5xx10^(-5))/0.08)^(1/2)=1.77xx10^(-2)`
`therefore [H_3O^+]=calpha = 0.08xx1.77xx10^(-2)`
`=1.41xx10^(-3)`
`pH=-log [H_3O^+]`
`=-log (1.41xx10^(-3))=2.85`
Percent dissociation =`"[HOCl ] dissociated"/"[HOCl] undissociated"xx100`
[HOCl] dissociation =`calpha =0.08xx1.77xx10^(-2)`
`=1.41xx10^(-3)`
[HOCl] undissociated = `c (1-alpha)approx c = 0.08`
Per cent dissociation =`(1.41xx10^(-3))/0.08xx100`
=1.76%