Answer
The pH of this acetic acid buffer is equal to $4.92$.
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.74$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.075}{0.05}$
- 1.5: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.075}{1.8 \times 10^{-5}} = 4277$
- $ \frac{0.05}{1.8 \times 10^{-5}} = 3888$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.74 + log(\frac{0.075}{0.05})$
$pH = 4.74 + log(1.5)$
$pH = 4.74 + 0.18$
$pH = 4.92$
