Answer
$pH = 4.745$
Work Step by Step
1. Calculate the pKa value for the acid:
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
2. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.745 + log(\frac{0.25}{0.25})$
$pH = 4.745 + log(1)$
$pH = 4.745 + 0$
$pH = 4.745$
