Question

What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.58 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

pH = pKa + log [PhCOONa]/[PhCOOH]

4.58 = 4.20 + log [PhCOONa]/[PhCOOH]

0.38 = log [PhCOONa]/[PhCOOH]

2.399 = [PhCOONa]/[PhCOOH]

2.399 = A/B ……………….(i)

Dlta T_f = K_f x m

2.0 = 1.86 x m

m = 1.075

weight of 1 L of solution = V x d = 1.01 x 1 = 1.01 kg

weight of PhCOONa = A x 144.10

weight of PhCOOH = B x 122.12

Weight of solute = (A x 144.10 + B x 122.12) g

Wight of solvent = 1.01 - (A x 144.10 + B x 122.12)/1000

m = (A + B)/ {1.01 - (A x 144.10 + B x 122.12)/1000}

1.075 = (A + B)/ {1.01 - (A x 144.10 + B x 122.12)/1000}

1.075 {1.01 - (A x 144.10 + B x 122.12)/1000} = A + B

1.075 {1010 - (A x 144.10 + B x 122.12)} = 1000 (A + B)

1085.75 - 154.9075 A - 131.2575 B = 1000 A + 1000 B

1145.9075 A + 1131.2575 B = 1085.75

1.0554 A + 1.0419 B = 1 ………………………..(ii)

From equation (i) and (ii)

1.0554 (2.399 B) + 1.0419 B = 1

2.5319046 B + 1.0419 B = 1

B = 0.2798 M = [PhCOOH]

A = 2.399 x B = 2.399 x 0.2798 = 0.6544 M = [PhCOOH]