Answer
The pH of this buffer is equal to $6.77$.
Work Step by Step
1. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.25}{0.1}$
- 2.5: It is.
2. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.25}{ 4.3 \times 10^{-7}} = 0.5814\times 10^{6}$
- $ \frac{0.1}{ 4.3 \times 10^{-7}} = 0.2326\times 10^{6}$
3. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 6.37 + log(\frac{0.25}{0.1})$
$pH = 6.37 + 0.3979$
$pH = 6.77$
