Question

You will be using a mixture of 5:1:4 volume ratio of butanol, glacial acetic acid (pure acetic acid), and water in this lab. Assume that butanol is the same as water in terms of a pH calculation. The density of acetic acid is 1.05 g/ml. The molecular weight of acetic acid is 60.05 g/mol. The Ka of acetic acid HAc is 1.8*10^-5. You can make the assumption that [H+] = [Ac-] << [HAc] here. What is the pH of this solution?

Answer 1

First we calculate the concentration of acetic acid in solution from the volume ratio, for a total volume of 10 mL:

M = 1 mL AA / 10 mL sol * (1000 mL sol / 1 L sol) * (1.05 g AA / 1 mL) * (1 mol AA / 60.05 g) = 1.75 M

For the dissociation reaction:

HA + H2O = A- + H3O +

It has to that the expression of Ka is:

Ka = [A-] * [H3O +] / [HA]

Substituting:

Ka = X ^ 2 / 1.75 = 1.8x10 ^ -5

Clearing:

X = [H3O +] = √1.8x10 ^ -5 / 1.75 = 3.21x10 ^ -3 M

We calculate pH = - Log [H3O +]

pH = - Log (3.21x10 ^ -3) = 2.49

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