Question

If 12.0 mL of glacial acetic acid (pure HC2H3O2; Ka=1.8×10^−5) is diluted to 1.25 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

a) pH < 1
b) pH = 7
c) pH > 7
d) pH < 7

Answer 1

The pH of the resulting solution can be calculated using the initial concentration of the acetic acid and its dissociation constant. The solution is acidic with a pH of 2.87. The solution contains a conjugate base, acetate ion, which affects the final pH.

Step-by-step explanation:

(a) The initial pH is computed for the acetic acid solution in the usual ICE approach:

[H3O+] = Ka × [CH3CO₂H] = √1.8 × 10-5 × 0.100 = 1.3 × 10-3
pH = -log(1.3 × 10-³) = 2.87

(b) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of