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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???

Ayush Chaturvedi , 8 Years ago
Grade 12
anser 2 Answers
Soumika Hallder

Last Activity: 7 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

Last Activity: 4 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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