Answer
The pH of this buffer is equal to $4.95$.
Work Step by Step
1. Calculate the pKa value for $CH_3COOH$.
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.241}{0.15}$
- 1.607: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.241}{1.8 \times 10^{-5}} = 1.339\times 10^{4}$
- $ \frac{0.15}{1.8 \times 10^{-5}} = 8333$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.745 + log(\frac{0.241}{0.15})$
** We calculated the concentration of the base in 85a.
$pH = 4.745 + 0.2059$
$pH = 4.95$
