Question

A very expensive method of making vinegar (pH=3.00) is to prepare a solution of acetic acid in the laboratory. At 25 degrees C, pure acetic acid is a liquid with a density of 1.049 g ml^-1. The pKa of acetic acid is 4.76. Calculate the volume of acetic acid that must be added to water to make 500.0 mL of vinegar. Show work.

Answer 1

Given data:

pH of vinegar (a solution of acetic acid) = 3

The density of pure acetic acid = 1.049 g/mL

pKa of acetic acid = 4.76

The molar mass of acetic acid = 60.052 g/mol

The concentration of acetic acid = (mass of acetic acid/molar mass of acetic acid)*(1000/volume (in mL))

= (1.049/60.052)*(1000/1), i.e. 17.4682 mol/L

According to Henderson-Hasselbulch equation

pH = pKa + Log([A^-]/[HA])

3 = 4.76 + Log([A^-]/[HA])

i.e. -Log([A^-]/[HA]) = 4.76-3

i.e. Log([HA]/[A^-]) = 1.76

i.e. [HA]/[A^-] = 57.544

At pH = 3, [H⁺] = [A^-] = 10^-3 mol/L

Therefore, [HA] = 57.544*10^-3 mol/L

i.e. The concentration of vinegar = 57.544*10^-3 mol/L

Formula: M₁V₁ = M₂V₂, where M₁ and M₂ are the molar concentrations of pure acetic acid and vinegar, respectively; V₁ and V₂ are the volumes (in L) of pure acetic acid and vinegar, respectively.

Here, we want to make 500 mL, i.e. 0.5 L of vinegar.

i.e. 17.4682*V1 = 57.544*10^-3*0.5

Therefore, V1 = 0.00165 L, i.e. 1.65 mL

Hence, to make 500mL of vinegar, 3.3/2, i.e. 1.65 mL of acetic acid must be added to water.