Question

If $p{K}_b$ for fluoride ion at ${25}^{o}C$ 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. $6.75\times {10}^{-4}$
• D. $5.38\times {10}^{-2}$

Answer 1

The correct option is C $6.75\times {10}^{-4}$

$K_a\times K_b=K_w$
$\therefore K_a=K_w/K_b=\frac{{10}^{-14}}{1.48\times {10}^{-11}}=6.75\times {10}^{-4}$