# If a buffer made by dissolving 20.0 g of sodium benzoate (144g/mol) and 10 grams of benzoic acid (122g/mol) in water to a final volume of 500.0 ml, what is the pH of the buffer?

Apr 11, 2016

$p H = 4.43$.

#### Explanation:

To find the pH of the buffered solution in question, we can simply use the Henderson-Hasselbach equation:

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

where, $p {K}_{a}$ for benzoic acid is $4.20$.

and $\left[B e n z o a t e\right] = \frac{n}{V} \mathmr{and} n = \frac{m}{M M} = \frac{20.0 \cancel{g}}{144 \frac{\cancel{g}}{m o l}} = 0.139 m o l$

$\left[B e n z o a t e\right] = \frac{n}{V} = \frac{0.139 m o l}{0.500 L} = 0.278 M$

and $\left[\text{Benzoic acid}\right] = \frac{n}{V} \mathmr{and} n = \frac{m}{M M} = \frac{10.0 \cancel{g}}{122 \frac{\cancel{g}}{m o l}} = 0.0820 m o l$

$\left[B e n z o a t e\right] = \frac{n}{V} = \frac{0.0820 m o l}{0.500 L} = 0.164 M$

Therefore, $p H = 4.20 + \log \left(\frac{0.278}{0.164}\right) = 4.43$

Acid - Base Equilibria | Buffer Solution.