106pKb = 10.83
So, pKa = 3.17
-log(Ka) = 3.17
==> Ka = 10-3.17 = (C)
If pKb for fluoride ion at 25oC is 10.83,the ionisation constant of hydrofluoric acid in water at this temperature would be
A. 1.74 X 10-5
B. 3.52 X 10-3
C. 6.75 X 10-4
D. 5.38 X 10-2
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