Answer
The $pH$ of this solution is equal to $4.73$.
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.3 \times 10^{- 5})$
$pKa = 4.886$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.06}{0.085}$
- 0.7058: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.06}{1.3 \times 10^{-5}} = 4615$
- $ \frac{0.085}{1.3 \times 10^{-5}} = 6537$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.886 + log(\frac{0.06}{0.085})$
$pH = 4.886 + -0.1512$
$pH = 4.735$
