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

Question

If pKb for the fluoride ion at 25° C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is (A) 1.74× 10- 5 (B) 3.52× 10- 3 (C) 6.76× 10- 4 (D) 5.38× 10- 2

Tardigrade Admin · Accepted Answer

pKb=10.83, textlog ( textK) textb = - 10.83 - (. 11 - 0.17 .) On taking antilog Kb=1.479× 10- 11 Ka.Kb=Kw or Ka× 1.479× 10- 11=1.0× 10- 14 Ka=6.76× 10- 4

Q. If $p K_{b}$ for the fluoride ion at $2 5^{\circ}$ C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is

$1.74 \times 1 0^{- 5}$

$3.52 \times 1 0^{- 3}$

$6.76 \times 1 0^{- 4}$

$5.38 \times 1 0^{- 2}$

$p K_{b} = 10.83,$
$log (K)_{b} = - 10.83 - (11 - 0.17)$
On taking antilog
$K_{b} = 1.479 \times 1 0^{- 11}$
$K_{a} . K_{b} = K_{w}$
or $K_{a} \times 1.479 \times 1 0^{- 11} = 1.0 \times 1 0^{- 14}$
$K_{a} = 6.76 \times 1 0^{- 4}$

Q. If pKb​ for the fluoride ion at 25∘ C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is

1.74×10−5

3.52×10−3

6.76×10−4

5.38×10−2