Answer
The pH of this solution is equal to $4.14$;
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 4})$
$pKa = 3.745$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.25}{0.1}$
- 2.5: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.25}{1.8 \times 10^{-4}} = 1389$
- $ \frac{0.1}{1.8 \times 10^{-4}} = 555.6$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.745 + log(\frac{0.25}{0.1})$
$pH = 3.745 + 0.3979$
$pH = 4.143$
