Calculate the `pH` of `0.08M` solution of hypochlorous acid, `HOCl`. The ionization constant of the acid is `2.5xx10^(-5)`. Determine the percent dissociation of `HOCl`.

Explanation

`underset(O^(-)(aq))(HOCl)(aq)+H_(2)O(l) hArr H_(3)O^(+)(aq)+CI`
Initial concentration (M)
`0.08 " " 0 " " 0`
Change to reach
equilibrium concentration (M)
`-x" "+x " "+x`
equilbrium concentration (M)
`0.08-x" "x" "x`
`K_(a)=[{H_(2)O^(+)][CIO^(-)]//HOCl]}`
`x^(2)//(0.08-x)`
As ` x lt lt 0.08` therefore, `0.08-x=0.08`
`x^(2)//0.08=2.5xx10^(-5)`
`x^(2)=2.0xx10^(-6)`, thus `x=1.41xx10^(-3)`
`[H^(+)]=1.41xx10^(-3)M`
Therefore,
Percent dissociation
`={[HOCI]_("dissociated")xx100//[HOCI]_("initial")}`
`=1.41xx10^(-3)//0.08=1.76%`
`pH=-log(1.41xx10^(-3))=2.85`