The freezing point of a 0.08 molal solution of `NaHSO^(4)` is `-0.372^(@)C`. Calculate the dissociation constant for the reaction.

`K_(f)` for water =`1.86 K m^(-1)`

Explanation

`NaHSO_(4)` dissociates as:
`NaHSO_(4) rarr Na^(o+) +HSO_(4)^(ө)`
Since the concentration of `NaHSO_(4)` is `0.08 m`
`[Na^(o+)]=0.08 m`,`[HSO_(4)^ө]=0.08 m`
Now, `HSO_(4)^(ө)` also dissociates as

If alpha is the degree of dissociation, then after dissociation at equilibrium

`[H^(o+)]=0.08alpha`,`SO_(4)^(2-)]=0.08alpha`
Total concentration of all ions (i.e.,`Nao+,H^(o+)`, `HSO_(4)^ө`, and `SO_(4)^(2-)`)
`=0.08+0.08(1-alpha)+0.08alpha+0.08alpha`
`=0.16+0.08alpha`
`:.`Van't Hoff factor , `i="Observed moles of solute"/"Normal moles of solute"`
=`(0.16+0.08alpha)/(0.08)=2+alpha`
Now, `DeltaT_(f)=iK_(f)xxm`
`=ixx1.86xx0.88=0.1488i`
or `0.372=0.1488 i`
`:.i=(0.372/0.1488)=2.5`
Thus, `2+alpha =2.5` or `alpha=0.5`
Dissociation constant for the reaction is
`K=([H^(o+)][SO_(4)^(2-)])/[[HSO_(4)^(ө)]`
`[H^(o+)]=0.08xx0.5=0.04`
`[HSO_(4)^(ө)]=0.08 xx (1-0.5)=0.04`
`[SO_(4)^(2-)]=0.08xx(0.5)=0.04`
`:.K=((0.04)xx(0.04))/(0.04) =4xx10^(-2)`