Question

A 0.20 M solution of hydrofluoric acid (HF(aq)) has a pH of 3.96 at 25 oC. Calculate the Ka and Percent Ionization of hydrofluoric acid.

Answer 1

Hydrogen fluoride being dissolved in water solvent dissociates according the following scheme:

HF + H₂O = H₃O⁺ + F^-

As one easily note, 1 mole of HF dissociated gives rise to 1 mole H3O+ cations, which concentration affects pH decreasing. However, HF is a quite weak acid and is mostly not dissociated in the water solution. The Ka value can be expressed with the following equation:

$K_a=\frac{[H_{3}O^+][F^{-}]}{[HF]}$

On the other hand:

$pH=-lg[H_{3}O^+]$

and,

$[H_{3}O^+]=10^{-pH}$

Hydroxonium as well as fluoride equilibrium concentrations are the same, while the equilibrium concentration of HF is lower:

$[F^-]=[H_{3}O^+]=10^{-pH}$

$[HF]=C_{HF}-[F^-]=C_{HF}-10^{-pH}$

Now it is time to complete the fragments into one formula:

$K_a=\frac{10^{-2pH}}{C_{HF}-10^{-pH}}=\frac{10^{-2*3.96}}{0.20-10^{-3.96}}=6.00*10^{-8}$

In turn, the percentage of dissociation is a relation between the concentration of the dissociation products and the starting concentration of acid:

$\alpha=\frac{[H_{3}O^+]}{C_{HF}}*100\%=\frac{10^{-pH}}{C_{HF}}*100\%=\frac{10^{-3.96}}{0.20}*100\%=0.055\%$