# How do you calculate PKa using the Henderson-Hasselbalch equation?

## PH=6.98. Mix 25ml of 0.1M $N a {H}_{2} P {O}_{4}$ and 25ml of 0.1M $N {a}_{2} H P {O}_{4}$ and dilute with water in a 250ml Volumetric flask.

May 16, 2016

$p {K}_{a} = 6.98 - \log \left(\frac{0.05 M}{0.05 M}\right) = 6.98$

#### Explanation:

The Henderson-Hasselbalch equation is:

$p H = p {K}_{a} + \log \left(\frac{\left[B a s e\right]}{\left[A c i d\right]}\right)$

Therefore, $p {K}_{a} = p H - \log \left(\frac{\left[B a s e\right]}{\left[A c i d\right]}\right)$

$p {K}_{a} = 6.98 - \log \left(\frac{0.05 M}{0.05 M}\right) = 6.98$

Note that the new concentrations are divided by $2$ because the total volume doubles ${V}_{s o \ln .} = 25 m L + 25 m L = 50 m L$

Notice that n this case the $p H = p {K}_{a}$ because the $\left[N {a}_{2} H P {O}_{4}\right] = \left[N a {H}_{2} P {O}_{4}\right]$ and the $\log$ of the ratio is equal to zero.

Here is a video that explains more this topic (topic starts at minute 9:02):
Acid - Base Equilibria | Buffer Solution.