Question

If $p{K}_b$ for fluoride ion at ${25}^{\circ }C$ is $10.83$, then the ionisation constant of hydrofluoric acid in water at this temperature is:

• A. $3.52\times {10}^{-3}$
• B. $6.75\times {10}^{-4}$
• C. $5.38\times {10}^{-2}$
• D. $1.74\times {10}^{-5}$

Answer 1

Answer: (B)

$6.75\times {10}^{-4}$

It is given that $p{K}_b=10.83$ for fluoride.

Hence, the base dissociation constant $=K_b={10}^{-10.83}=1.48\times {10}^{-11}$

The relationship between the dissociation constants for hydrofluoric acid and fluoride ions is as given below:

$K_a\times K_b=K_w$,

where $K_w$ is the ionic product of water, $K_a$ is the dissociation constant of an acid and $K_b$ is the dissociation constant of a base..

Substituting values in the above expression, we get
$\therefore K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.48\times {10}^{-11}}=6.75\times {10}^{-4}$

Hence, the ionisation constant of hydrofluoric acid in water is $6.75\times {10}^{-4}$.