If the freezing point of `0.1 M HA(aq)` solution is `-0.2046^(@)C` then `pH` of solution is
`(` If `K_(f)` water `=1.86mol^(-1)kg^(-1))`

Options

A. 1
B. 2
C. 1.3
D. 1.7

Answer

B

Explanation

`DeltaT_(f)=iK_(f)m`
`i=(DeltaT_(f))/(K_(f)m)=(0.2046)/(1.86 xx 0.1)=(204.6)/(186)`
`{:(,HA, hArr ,H^(+),+,A^(-)),("Initial",0.1,,,,),("At eqm.",0.1-x,,x,,x):}`
`i=(0.1 -x+2x)/(0.1)=(0.1+x)/(0.1)=1+10x ....(ii)`
Eqn. (i) and (ii), `x=10^(-2)`
`:. pH=-log[H^(+)]`
`=-log 10^(-2)=2`