Question

Acetic acid (ethanoic acid) is an “organic” acid with the chemical formula CH3COOH (may also be written as HC2H3O2). Acetic acid has a distinctive sour taste and pungent smell; vinegar is roughly 3-9% acetic acid by volume. Although it is classified as a weak acid, concentrated acetic acid is corrosive and can attack the skin.

(a) A particular brand of vinegar has a H3O+ concentration of 0.015 M. What is the pH and pOH of the solution?
(b) The Ka of acetic acid is 1.80 × 10-5. What is its pKa?
(c) What is the pH of a 0.100 M solution of acetic acid?
(d) What is the acetate ion (CH3COO-) concentration of the solution in (c)?
(e) What is the percent ionization of the acetic acid in (c)?
(f) Wouldyouexpecttrichloroaceticacid(CCl3COOH)tohavealowerorhigherpKavaluethanacetic
acid? Expla

Answer 1

Therefore, the percent ionization of acetic acid in the solution is approximately 4.2%.

Step-by-step explanation:

(a) To find the pH and pOH of the solution, we need to use the equation pH + pOH = 14. Since the concentration of H3O+ is 0.015 M, we can use the equation pH = -log[H3O+].

Calculating pH:

pH = -log(0.015) ≈ 1.82

To find the pOH, we can subtract the pH from 14:

pOH = 14 - pH ≈ 12.18

Therefore, the pH of the solution is approximately 1.82 and the pOH is approximately 12.18.

(b) The pKa is a measure of the acidity of a substance and is given by the equation pKa = -log(Ka). The given Ka of acetic acid is 1.80 × 10^-5.

Calculating pKa:

pKa = -log(1.80 × 10^-5) ≈ 4.74

Therefore, the pKa of acetic acid is approximately 4.74.

(c) To find the pH of a 0.100 M solution of acetic acid, we can use the equation for the ionization of acetic acid, which is given by the equation CH3COOH ⇌ H3O+ + CH3COO-.

Using the Ka value of acetic acid (1.80 × 10^-5), we can set up an equilibrium expression:

Ka = [H3O+][CH3COO-] / [CH3COOH]

Since the initial concentration of acetic acid is 0.100 M and we assume that x is the concentration of H3O+ and CH3COO- formed, we can set up the equation:

1.80 × 10^-5 = x^2 / (0.100 - x)

Solving for x using the quadratic equation gives us x ≈ 0.0042 M.

Therefore, the concentration of H3O+ is approximately 0.0042 M, and the pH of the solution is given by pH = -log(0.0042) ≈ 2.38.

(d) The concentration of the acetate ion (CH3COO-) can be assumed to be equal to the concentration of H3O+ since acetic acid is a weak acid and undergoes partial ionization. Therefore, the concentration of the acetate ion in the solution is approximately 0.0042 M.

(e) The percent ionization can be calculated by dividing the concentration of ionized acetic acid (H3O+) by the initial concentration of acetic acid and multiplying by 100.

Percent ionization = (0.0042 M / 0.100 M) × 100 ≈ 4.2%

Therefore, the percent ionization of acetic acid in the solution is approximately 4.2%.

(f) Trichloroacetic acid (CCl3COOH) would have a lower pKa value than acetic acid. This is because the electronegative chlorine atoms in trichloroacetic acid pull electron density away from the carboxyl group, making the dissociation of H+ easier and increasing the acidity. Therefore, trichloroacetic acid would have a lower pKa value than acetic acid.