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`.

Explanation

`HOCl(aq) + H_(2)O (l) hArr H_(3)O^(+)(aq) + ClO^(-)(aq)`
`{:("Initial concentration (M)",,),(0.08,0,0):}`
Change to reach
`{:("equilibrium concentration",,),([M],,),(-x,+x,+x),("equilibrium concentration (M)",,),(0-08-x,x,x):}`
`K_(a) = {[H_(3)O^(+)][ClO^(-)]//[HOCl]}`
`= x^(2)//(0.08-x)`
As `x lt lt 0.08`, therefore `0.08 - x = 0.08`
`x^(2)//0.08 = 2.5 xx 10^(-5)`
`x^(2) = 2.0 xx 10^(-6)`, thus `x = 1.41 xx 10^(-3)`
`[H^(+)] = 1.41 xx 10^(-3) M`.
Therefore,
Percent dissociation
`={[HOCl]_("dissociated")//[HOCl]_("Initial")] xx 100`
`=1.41 xx 10^(-3) xx 10^(2)//0.08 = 1.76 %`
`pH = -"log"(1.41 xx 10^(-3)) = 2.85`.