Answer
The pH of this buffer is equal to $3.37$
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 7.2 \times 10^{- 4})$
$pKa = 3.143$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{1}{0.6}$
- 1.667: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{1}{7.2 \times 10^{-4}} = 1389$
- $ \frac{0.6}{7.2 \times 10^{-4}} = 833.3$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.143 + log(\frac{1}{0.6})$
$pH = 3.143 + log(1.667)$
$pH = 3.143 + 0.2218$
$pH = 3.365 \approx 3.37$
