Question

What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.78 and has a freezing point of -2.0 ∘C? (Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.)

Answer 1

Using the Literature value of benzoic acid Ka of 6.3 X 10^-5 from
pKa = -log(6.3 X 10^-5) = 4.20
& Using the henderson - hasselbalch equation
pH = pKa + log [A-] / [HA]
4.78 = 4.20 + log [A-] / [HA]
4.78 - 4.20 = log [A-] / [HA]
0.58 = log [A-] / [HA]

[A-] / [HA] = antilog(0.58) = 3.80

3.80 = [A-] / [HA]

which means that the benzoate concentration is 3.80 times that of the acid concentration
[HA] = X Molar
[A-] = 3.80X Molar

Using the freezing point depression
dT = Kf (molality) ....dT is always positive as it is the change in temperature
2.0 ^oC = (1.86^oC / molal) (molality)

molality = 2.0 / 1.86
total molality = 1.075 molal = 1.075 total moles dissociated in with 1000 grams of water

1.075 total moles = moles Acid & twice moles of sodium benzoate dissociated

1.075 total moles = X moles Acid + twice moles of 3.80X

1.075 total moles = X moles Acid + 7.6X

1.075 total moles = 8.6 X moles HA

moles of HA in with 1000 grams of water = (1.075/8.6) = 0.125 moles acid so
3.80X = (3.80 x 0.125) = 0.475 moles of sodium benzoate in with 1000 grams of water

use molar masses to find grams

Molar mass of Sodium benozate = 144.11 g/mol

(0.475 moles of sodium benzoate) x (144.11 g/mole) = 68.45grams of sodium benzoate

Molar mass of benzoic acid = 122.12 g/mol
(0.125 moles of benzoic acid) x (122.12 g/mole) = 15.26 grams of benzoic acid
in with 1000 grams of water ,
in the
"total molality = 1.075 molal = 1.075 total moles dissociated in with 1000 grams of water"

68.45 g sodium benzoate + 15.26 g benzoic acid + 1000 g water = 1083.71 grams of solution

so use density to find the volume of solution: (Density = (Mass / Volume)
(1083.71grams of solution) / (1.01 grams / ml) = 1072.98 ml

1L = 1000 mL
Therefore, volume of solution = 1.072 litres

find molarity of Benzoic acid
(0.125 moles of benzoic acid) / 1.072 L = 0.116 Molar Benzoic acid
which in your 2 sig fig problem rounds off to
0.11 M benzoic acid,

find molarity of sodium benzoate:
(0.475 moles of sodium benzoate) / 1.072 L = 0.443 Molar
which rounds to 2 sig figs as
0.44 M sodium benzoate