Answer
The $pH$ of this buffer is about $10.3$.
Work Step by Step
1. Calculate the pKa value for $HC{O_3}^-$
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 11})$
$pKa = 10.25$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.125}{0.105}$
- 1.19: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.125}{5.6 \times 10^{-11}} = 2.232\times 10^{9}$
- $ \frac{0.105}{5.6 \times 10^{-11}} = 1.875\times 10^{9}$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 10.25 + log(\frac{0.125}{0.105})$
$pH = 10.25 + 0.07572$
$pH = 10.33$
