The freezing point of `0.08 molal NaHSO_(4)` is `-0.345^(@)C`. Calculate the percentage of `HSO_(4)+O` ions that transfers a proton to water. Assume `100%` ionization of `NaHSO_(4)` and `K_(t)` for `H_(2)O = 1.86 K molality^(-1)`.

Explanation

`NaHSO_(4)+H_(2)O(aq)toNa^(+)+H_(3)O^(+)+SO_(4)^(2-)`
`t=0" " 1" "0" "0" "0`
`"at eq. " " "1-alpha" "alpha" "alpha" "alpha`
`i=(1+2 alpha)/1`
`DeltaT_(f)=ixxK_(f)xxm`
`0.345=(1+2alpha)xx1.86xx0.08`
So `alpha=0.6591`
% dissociation `=65.92%`