Answer
Percent ionization $ = 3.742\%$
$pH = 2.029$
Work Step by Step
** You can find the ka values on the Appendix II
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [F^-] = x$
-$[HF] = [HF]_{initial} - x = 0.25 - x$
For approximation, we consider: $[HF] = 0.25M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 3.5 \times 10^{- 4}= \frac{x * x}{ 0.25}$
$Ka = 3.5 \times 10^{- 4}= \frac{x^2}{ 0.25}$
$ 8.75 \times 10^{- 5} = x^2$
$x = 9.354 \times 10^{- 3}$
Percent ionization: $\frac{ 9.354 \times 10^{- 3}}{ 0.25} \times 100\% = 3.742\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [F^-] = x = 9.354 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HF] \approx 0.25M$
3 . Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 9.354 \times 10^{- 3})$
$pH = 2.029$
