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

`{:(HOCl(aq)+H_(2)O(l),hArr,H_(3)O^(+)(aq),+,CiO^(-)(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^(+)][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
`={[HOCl]_("dissociated")//[HOCl]_("initial")}xx100`
`=1.41xx10^(-3)xx10^(2)//0.08=1.76%`
`pH=-log(1.41xx10^(-3)=2.85`