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) to Na^(+) + HSO_(4)^(-)`
Since the concentration of `NaHSO_(4)` is 0.08 m ,
`[Na^(+)] = 0.08m , HSO_(4)^(-) = 0.08 m`
Now `HSO_(4)^(-) ` also dissociates as :
`HSO_(4)^(-) hArr H^(+) + SO_(4)^(2-)`
If `alpha` is the degree of dissociaiton , then after dissociation at equilibrum ,
`[HSO_(4)^(-)] = 0.08 (1-alpha)`
`[H^(+)] = 0.08 alpha,[SO_(4)^(2-)] = 0.08 alpha`
Total concentration of all ions (i.e., `Na^(+) ,H^(+) , HSO_(4)^(-) ` and `SO_(4)^(2-)`).
` = 0.08 + 0.08 (1-alpha) + 0.08 alpha + 0.08 alpha`
` = 0.16 + 0.08 alpha`
`therefore ` Van.t Hoff factor , `i=(0.16 + 0.08 alpha)/(0.08) = 2+alpha`
Now `, Delta T_(f) = iK_(f) xx m`
` = i xx 1.86 xx 0.08 =0.1488 i`
and `Delta T_(f) = 0 - (- 0.372) = 0.372`
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 reaciton is
`K = ([H^(+)][SO_(4)^(2-)])/(HSO_(4)^(-))`
`[H^(+)] = 0.08 xx 0.5 = .01, [HSO_(4)^(-)] = 0.08 (1-0.5) = .04`
`[SO_(4)^(2-)] = 0.08 xx 0.5= 0.04`
`K = ((0.04) xx (0.04))/((0.04))= 4.0 xx 10^(-2)`