If `pK_(b)` for fluoride ion at `25^(@)C` is `10.83`, the ionisation constant of hydrofluoric acid in water at this temperature is

Options

A. `1.74 xx 10^(-5)`
B. `3.52 xx 10^(-3)`
C. `6.75 xx 10^(-4)`
D. ` 5.38 xx 10^(-2)`

Answer

C

Explanation

`pK_(b)=10.83 `
`F^(-)+H_(2)O hArr HF+OH^(-)`
`K_(b)==([HF][OH^(-)])/([F^(-)])" " ….(i)`
We know
`K_(w)=[H_(3)O^(+)][OH^(-)]=10^(-14)" "...(ii)`
Dissociation of HF in water is represented by the equation.
`HF+H_(2)O hArr H_(3)O^(+)+F^(-)`
`K_(a)=([H_(3)O^(+)][F^(-)])/([HF])`
Dividing (ii) by (i)
`(K_(w))/(K_(b))=([H_(3)O^(+)][OH^(-)][F^(-)])/([HF][OH^(-)])`
`(K_(w))/(K_(b))=K_(a)`
Taking log on both sides
`log K_(a)=log K_(w)-log K_9b)=-pK_(w)+K_(b)`
`=-14 +10.83=-3.17=-4+0.83`
`bar(4).83`
`K=` antilog `bar(4).83=6.76xx10^(-4)`