Answer
$pH = 2.14$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [F^-] = x$
-$[HF] = [HF]_{initial} - x = 0.15 - x$
For approximation, we consider: $[HF] = 0.15M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 3.5 \times 10^{- 4}= \frac{x * x}{ 0.15}$
$Ka = 3.5 \times 10^{- 4}= \frac{x^2}{ 0.15}$
$ 5.25 \times 10^{- 5} = x^2$
$x = 7.246 \times 10^{- 3}$
Percent dissociation: $\frac{ 7.246 \times 10^{- 3}}{ 0.15} \times 100\% = 4.83\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [F^-] = x = 7.246 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HF] \approx 0.15M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 7.246 \times 10^{- 3})$
$pH = 2.14$
