Question

If $p{K}_b$ for the fluoride ion at ${25}^{\circ }C$ is 10.83
The ionization constant of hydrofluoric acid in water at this temperature will be:
$(Given:{10}^{-10.83}=1.48\times {10}^{-11})$

• A. $1.74\times {10}^{-5}$
• B. $1.52\times {10}^{-3}$
• C. $6.76\times {10}^{-4}$
• D. $8.38\times {10}^{-2}$

Answer 1

The correct option is C $6.76\times {10}^{-4}$

$Given,p{K}_b=10.83-lo{g}_{10}\left(K_b\right)=10.83\Rightarrow K_b={10}^{-10.83}=1.48\times {10}^{-11}$
we know at ${25}^{\circ }C$,
$K_a.K_b=K_w$
$K_a.K_b=1.0\times {10}^{-14}$
$K_a\times 1.479\times {10}^{-11}=1.0\times {10}^{-14}\Rightarrow K_a=6.76\times {10}^{-4}$

hence, the ionization constant$\left(K_a\right)$ of hydrofluoric acid in water is $6.76\times {10}^{-4}$