# Question #c1b5b

Nov 21, 2016

The pH of a buffer solution depends on the ratio of concentrations of an acid and its conjugate base, using the Henderson-Hasselbalch equation.
$p H = p {K}_{a} - {\log}_{10} \left(\frac{\left[H A\right]}{\left[{A}^{-}\right]}\right)$

#### Explanation:

In this example, the acid is $N {H}_{4}^{+}$, which has a $p {K}_{a} = 9.24$

Total volume of solution = 30mL + 15 mL = 45 mL = 0.45 L

Acid concentration: $\frac{0.1 M \times 0.015 L}{0.045 L} = 0.033 M$

Base concentration: $\frac{0.2 M \times 0.030 L}{0.045 L} = 0.133 M$

$p H = p {K}_{a} - {\log}_{10} \left(\frac{\left[H A\right]}{\left[{A}^{-}\right]}\right) = 9.24 - {\log}_{10} \left(\frac{0.033}{0.133}\right) = 9.84$

Sanity Check:
If the acid and base concentrations were equal, the $p H$ would be equal to the $p {K}_{a}$, but here the base concentration is 4X the acid concentration, so the solution is a little bit more basic and the $p H$ is a little bit higher than $p {K}_{a}$.