Answer
pH = 1.781
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.
-$[H_3O^+] = [F^-] = x$
-$[HF] = [HF]_{initial} - x = 0.38 - x$
For approximation, we consider: $[HF] = 0.38M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 7.2 \times 10^{- 4}= \frac{x * x}{ 3.8 \times 10^{- 1}}$
$Ka = 7.2 \times 10^{- 4}= \frac{x^2}{ 3.8 \times 10^{- 1}}$
$ 2.736 \times 10^{- 4} = x^2$
$x = 1.654 \times 10^{- 2}$
Percent ionization: $\frac{ 1.654 \times 10^{- 2}}{ 3.8 \times 10^{- 1}} \times 100\% = 4.353\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = x = 1.654 \times 10^{- 2} $
3. Find the pH:
$pH = -log[H_3O^+]$
$pH = -log( 0.01654)$
$pH = 1.781$
