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The pKb for fluoride ion at 25^@C is 10....

The `pK_b` for fluoride ion at `25^@C` is 10.83, the ionization constant of hydrofluoric acid at this temperature is

A

`1.74 xx 10^(-5)`

B

`3.52 xx 10^(-3)`

C

`6.76 xx 10^(-4)`

D

`5.38 xx 10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`F^(-) to H_(2)O Leftrightarrow HF+OH^(-)`
`K_(b)=([HF][OH^(-)])/([F^(-)]) ………..(i)`
`K_(w)=[H_(3)O^+] [OH^(-)]=10^(-14) ……..(ii)`
Dissociation of HF in water is represented by the equation `HF+H_(2)O Leftrightarrow H_(3)O^+ +F^(-)`
`K_(a)=([H_(3)O^+] [F^(-)])/([HF]) ............(iii)`
Multiplying eqn (i) and (iii),
`K_(b). K_(a)=[H_(3)O^+] [OH^-]=K_(w). K_w/K_b=K_b`
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