Answer
The pH of the $HF$ pure solution is equal to $1.58$
Work Step by Step
1. We get these concentrations at equilibrium:
-$[H_3O^+] = [F^-] = x$
-$[HF] = [HF]_{initial} - x = 1 - x$
For approximation, we consider: $[HF] = 1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 6.8 \times 10^{- 4}= \frac{x * x}{ 1}$
$Ka = 6.8 \times 10^{- 4}= \frac{x^2}{ 1}$
$ 6.8 \times 10^{- 4} = x^2$
$x = 2.608 \times 10^{- 2}$
Percent dissociation: $\frac{ 2.608 \times 10^{- 2}}{ 1} \times 100\% = 2.608\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [F^-] = x = 2.608 \times 10^{- 2}M $
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.02608)$
$pH = 1.584$