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Question

The pKb for fluoride ion at 25C is 10.83. The ionisation constant of hydrofluoric acid at this temperature is:
antilog ¯4.83=6.76×104

A
1.74×105
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B
3.52×103
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C
6.76×104
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D
5.38×102
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Solution

The correct option is C 6.76×104
pKb=10.83
F+H2OHF+OH
Kb=[HF][OH][F] ...(1)
We know,
Kw=[H3O+][OH]=1014 ...(2)
Dissociation of HF in water is represented by the equation.
HF+H2OH3O++F
Ka=[H3O+][F][HF]
Dividing (2) by (1),
KwKb=[H3O+][OH][F][HF][OH]
KwKb=Ka
Taking log on both sides,
logKa=logKwlogKb=pKw+pKb
=14+10.83=3.17 =4+0.83 =¯4.83
K=antilog ¯4.83=6.76×104

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