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If pKb for fluoride ion at 25 degree celsius is 10.83,the ionisation constant of hydrofluoric acid in water at this temperature is???

If pKb for fluoride ion at 25 degree celsius is 10.83,the ionisation constant of hydrofluoric acid in water at this temperature is???

Grade:12

2 Answers

Soumika Hallder
11 Points
6 years ago
pKb = 10.83     kb = 10^-10.83 = 1.48*10^-11
HF + H2O =  F- + H3O+
Ka * Kb = Kw
Ka = Kw/Kb = 10^-14/1.48*10^-11 = 6.76 * 10^-4
ankit singh
askIITians Faculty 614 Points
3 years ago
Kb = 10.83     kb = 10^-10.83 = 1.48*10^-11
HF + H2O =  F- + H3O+
Ka * Kb = Kw
Ka = Kw/Kb = 10^-14/1.48*10^-11 = 6.76 * 10^-4

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