Answer
$pH = 7.423$
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 6.3 \times 10^{- 8})$
$pKa = 7.201$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{2.5 \times 10^{-3}}{1.5 \times 10^{-3}} = 1.667$ : It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{2.5 \times 10^{-3}}{ 6.3 \times 10^{-8}} = 3.968\times 10^{4}$
- $ \frac{1.5 \times 10^{-3}}{ 6.3 \times 10^{-8}} = 2.381\times 10^{4}$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 7.201 + log(\frac{2.5 \times 10^{-3}}{1.5 \times 10^{-3}})$
$pH = 7.201 + 0.2218$
$pH = 7.423$
