Question

If $pK_{b}$ for the fluoride ion at $25^\circ $ C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is

• A. $1.74\times 10^{- 5}$
• B. $3.52\times 10^{- 3}$
• C. $6.76\times 10^{- 4}$
• D. $5.38\times 10^{- 2}$

Answer 1

Answer: (C)

$pK_{b}=10.83,$
$\text{log} \, \left(\text{K}\right)_{\text{b}} = - 10.83 - \left(\right. 11 - 0.17 \left.\right)$
On taking antilog
$K_{b}=1.479\times 10^{- 11}$
$K_{a}.K_{b}=K_{w}$
or $K_{a}\times 1.479\times 10^{- 11}=1.0\times 10^{- 14}$
$K_{a}=6.76\times 10^{- 4}$