Question

If pKb for fluoride at 25°c is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is?

(a) 3.52×10^-3

(b) 6.75×10^-4

(c) 5.38×10^-2

(d) 1.74×10^-5

The question was asked in an international level competition.

Query is from Ionization of Water, Weak acids and Weak Bases in chapter Water Basics of Biochemistry

Answer 1

The correct answer is (b) 6.75×10^-4

To explain: Kw = Ka × Kb

Ka = Kw / Kb

Ka = 10^-14/-log (10.83) = 6.75 × 10^-4.