If pKb for fluoride ion at 250C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is

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1.74 ×10-5

3.74 ×10-3

6.75 ×10-4

5.38 ×10-2

answer is C.

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Detailed Solution

HF + H2O ⇌ F- + H3O+Ka × Kb = Kw pKa + pKb = pKwpKa = 14-10.83 ⇒ pKa = 3.17Ka = 6.75 × 10-4

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