Ionic EquilibriumHard

Question

If pKb for fluoride ion at 25°C is 10.3, the ionization constant of hydrofluoric acid in water at this temperature is (log 2 = 0.3)

Options

A.2 × 10–4
B.2 × 10–3
C.2 × 10–5
D.5 × 10–11

Solution

$P^{K_{b}} = 10.3 = - \log K_{b} \Rightarrow K_{b} = 5 \times 10^{- 11}$

$\therefore K_{a(HF)} = \frac{K_{w}}{K_{b}\left( F^{-} \right)} = \frac{10^{- 14}}{5 \times 10^{- 11}} = 2 \times 10^{- 4}$

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