Question

What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.86 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

Given , pH = 4.86

freezing point = -2.0 ∘C,

density = 1.01 g/mL

consider the freezing point depression . ΔTf = Kf * m

We know pure solvent water freezing point = 0.00 ^oC

So, ΔTf = 0.00 ^oC – (-2.0)

              = 2.0 ^o C

So, molality = ΔTf / Kf

                     = 2.0 ^o C / 1.86 ^oC.m^-1

                     = 1.075 m

We know, Benzoic acid pKa = 4.19

Let's consider Henderson-Hasselbalch equation

pH = pKa + log [C₆H₅COO^- ] / [C₆H₅COOH ]

4.86 = 4.19 + log [C₆H₅COO^- ] / [C₆H₅COOH ]

log [C₆H₅COO^- ] / [C₆H₅COOH ] = 4.86 - 4.19

                                                       = 0.67

[C₆H₅COO^- ] / [C₆H₅COOH ] = 10^0.67

= 4.67

[C₆H₅COO^- ] = 4.67 * [C₆H₅COOH ]

Let [C₆H₅COO^- ] = 4.67 x

total moles = 1.075

1.075 = x moles of acid + 4.67 x moles of * 2

1.075 = x + 5.67 x

1.075 = 6.67 x

x = 0.161 moles

moles of sodium benzoate = 4.67 * 0.161

= 0.751 moles

mass of benzoic acid = 0.161 moles * 122.12 g/mole

= 19.66 g

mass of sodium benzoate = 0.751 moles * 144.11 g/mole

= 108.22 g

total mass of solution = 1000 g + 19.66 g + 108.22 g

                                      = 1127.82 g

Volume of solution =1127.82 g /1.01 g.ml^-1

                               = 1116.65 mL

[C₆H₅COOH] = 0.161 mole / 1.11665L

                             = 0.1441 M