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.)

Express your answers using two significant figures separated by a comma.

answer is not 0.11 and 0.44

answer is not 0.11 and 0.45

answer is not 0.11 and 0.46

Textbook says Tf is 1.86

Answer 1

Freezing point depression = 0oC - (-2^oC) = 2^oC

Tf = i *Kf *m

or, 2 = 2 * 1.86 oC /gm/mol * m

or, m = 0.538 m

0.538 moles in 1000gm water.

1000gm water = 1000gm * 1.01 g/mL = 1010mL

Molarity of the buffer in the solution = 0.538 moles *1000mL /1010 mL = 0.533 M

------------------------

pH = pKa = log [benzoate/benzoic acid]

pKa of benzoic acid = 4.20

or, 4.78 = 4.20 + log [benzoate/benzoic acid]

or, [benzoate/benzoic acid] = 3.8

decimal fraction of benzoate = 3.8/1+3.8 = 0.79

decimal fraction of benzoic acid = 1/1+3.8 =0.21

Molarity of benzoate = 0.53M *0.79 = 0.42

Molarity of benzoic acid = 0.21 * 0.533 M = 0.11