Answer
The pH of this buffer is equal to $6.5$
Work Step by Step
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 2 \times 10^{- 7})$
$pKa = 6.699$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.1}{0.15}$
- 0.6667: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.1}{2 \times 10^{-7}} = 5\times 10^{5}$
- $ \frac{0.15}{2 \times 10^{-7}} = 7.5\times 10^{5}$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 6.699 + log(\frac{0.1}{0.15})$
$pH = 6.699 + log(0.6667)$
$pH = 6.699 + (-0.176)$
$pH = 6.523$
