# What mass of sodium benzoate should be added to 140.0 mL of a 0.15 M benzoic acid solution to obtain a buffer with a pH of 4.25?

Jul 12, 2016

$3.3 \setminus g$ of sodium benzoate must be dissolved in the benzoic acid solution to obtain the solution with the desired pH (4.25).

#### Explanation:

The buffer solution is prepared by mixing benzoic acid and sodium benzoate.

Use the Handerson-Hasselbalch equation to determine the number of moles of the salt needed to produce the buffer solution with the desired pH.

The $p {K}_{a}$ values for the common weak acids is provided in tables at the end of most chemistry text books.

The $p {K}_{a}$ value for benzoic acid is 4.20.

n_"Salt" = ??

${n}_{\text{Acid}} = {C}_{M} \times V$

${n}_{\text{Acid}} = 0.15 \setminus \frac{m o l .}{L} \times 0.1400 L$

${n}_{\text{Acid}} = 0.0210 \setminus m o l .$

$- - - - - - - - - - - - - - - -$

Solve for the unknown (${n}_{\text{Salt}}$)

$p H = p {K}_{a} + \log \left(\left({n}_{\text{Salt")/(n_"Acid}}\right)\right)$

$4.25 = 4.20 + \log \left(\left({n}_{\text{Salt")/(n_"Acid}}\right)\right)$

$\log \left(\left({n}_{\text{Salt")/(n_"Acid}}\right)\right) = 4.25 - 4.20$

$\log \left(\left({n}_{\text{Salt")/(n_"Acid}}\right)\right) = 0.05$

$\left({n}_{\text{Salt")/(n_"Acid}}\right) = {10}^{0.05}$

$\left({n}_{\text{Salt")/(n_"Acid}}\right) = 1.1$

${n}_{\text{Salt}} = 0.0210 \times 1.1$

${n}_{\text{Salt}} = 0.0231 \setminus m o l .$

${m}_{\text{Salt}} = 0.0231 \setminus m o l . \times 144.11 \frac{g}{m o l .}$

${m}_{\text{Salt}} \cong 3.3 \setminus g$

$3.3 \setminus g$ of sodium benzoate must be dissolved in the benzoic acid solution to obtain the solution with the desired pH (4.25).