The correct option is C 6.76×10−4
pKb=10.83
F−+H2O⇌HF+OH−
Kb=[HF][OH−][F−] ...(1)
We know,
Kw=[H3O+][OH−]=10−14 ...(2)
Dissociation of HF in water is represented by the equation.
HF+H2O⇌H3O++F−
Ka=[H3O+][F−][HF]
Dividing (2) by (1),
KwKb=[H3O+][OH−][F−][HF][OH−]
KwKb=Ka
Taking log on both sides,
logKa=logKw−logKb=−pKw+pKb
=−14+10.83=−3.17 =−4+0.83 =¯4.83
K=antilog ¯4.83=6.76×10−4