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

`2.85, 1.76%`

Explanation

`{:(,HOCl,+,H_(2)O,hArr,H_(3)O^(+),+,ClO^(-)),("Initial conc.",0.08,,,,,,),("At. Eqm.",0.08-x,,,,x,,x" (x=Amount of HOCl dissociated)"),(,~= 0.08,,,,,,):}`
`K_(a)=(x^(2))/(0.08) or x=sqrt((2.5xx10^(-5))(0.08))=1.41xx10^(-3)`
`pH = - log [H_(3)O^(+)]=-log (1.41xx10^(-3))=2.85`
Alternatively, directly `pH = (1)/(2) [pK_(a) - log C]`
% dissociation = `("Amount dissociated")/("Amount taken")xx100=(1.41xx10^(-3))/(0.08) xx100=1.76%`