Buffer Solution Chemistry Math Question

Question

How many grams of benzoic acid would you add to 25 mL of 0.45M sodium benzoate in order to prepare a buffer solution at pH=3.8?

K_a of benzoic acid is 6.3E-5

Answer 1

When dealing with this type of problem, you should really be familiar with the Henderson Hasselbalch equation and how to use it. pH = pKa + log [salt]/[acid].

You are given the Ka, so you can find the pKa. pKa = -log Ka = -log 6.3x10^-5 = 4.20

You are also given the pH = 3.8. So, all we have to do is solve for [salt]/[acid]

3.8 = 4.20 + log [salt]/[acid]

log [salt]/[acid] = -0.4

[salt]/[acid] = 0.398

We know that we have 25 ml (0.025 L) of 0.45 M (0.45 moles/L) of sodium benzoate (the salt) , so we can find the moles of sodium benzoate from 0.025 L x 0.45 moles/L = 0.01125 moles

We want the ratio of sodium benzoate / benzoic acid = 0.398, so we can find the moles of benzoic acid needed.

0.01125/x = 0.398 and x = 0.0283 moles of benzoic acid

grams benzoic acid = 0.0283 moles x 122 g/mole = 3.4 g benzoic acid is needed (assuming no change in volume)