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)`

Options

A. `4xx10^(-2)`
B. `8xx10^(-2)`
C. `2xx10^(-2)`
D. `1.86xx10^(-2)`

Answer

A

Explanation

`DeltaT_"f"=K_"f"xxm`
`m=(DeltaT_"f")/K_"f"=0.372/1.86=0.2`
This means that total molal conc. of all particles is 0.2.
`NaHSO_4rarrNa^++HSO_4^-`
0.08 0.08
`HSO_4^(-)hArrH^++SO_4^(2-)`
(0.08-x) x x
The net particle concentration
0.087 = (0.08 – x) + x + x = 0.2
or x = 0.04
`K_a=([H^+][SO_4^(2-)])/([HSO_4^-])=0.04`