1. Understand the Concepts

* Acetic Acid (CH3COOH): A weak acid, meaning it doesn't fully ionize in water.

* pH: A measure of the acidity or alkalinity of a solution. A lower pH indicates a stronger acid.

* Ka: The acid dissociation constant, which indicates the strength of an acid. For acetic acid, Ka = 1.8 x 10^-5.

2. Calculate the Molarity of the Solution

* 3% acetic acid means 3 g of acetic acid per 100 mL of solution.

* Convert grams to moles:

* Molar mass of acetic acid (CH3COOH) = 60.05 g/mol

* Moles of acetic acid = (3 g) / (60.05 g/mol) = 0.05 mol

* Convert mL to L: 100 mL = 0.1 L

* Calculate Molarity (M):

* Molarity = (moles of solute) / (volume of solution in liters)

* Molarity = (0.05 mol) / (0.1 L) = 0.5 M

3. Set Up the Equilibrium Expression

Acetic acid ionizes in water according to the following equilibrium:

CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)

* Initial Concentrations:

* [CH3COOH] = 0.5 M

* [H+] = 0 (initially)

* [CH3COO-] = 0 (initially)

* Change in Concentrations:

* -x (for CH3COOH)

* +x (for H+)

* +x (for CH3COO-)

* Equilibrium Concentrations:

* [CH3COOH] = 0.5 - x

* [H+] = x

* [CH3COO-] = x

4. Use the Ka Expression to Solve for [H+]

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

1.8 x 10^-5 = (x)(x) / (0.5 - x)

5. Simplify the Equation (Assumption)

Since Ka is small, we can assume that x is much smaller than 0.5. This allows us to simplify the equation:

1.8 x 10^-5 ≈ x^2 / 0.5

6. Solve for x (which is equal to [H+])

* x^2 = 9 x 10^-6

* x = √(9 x 10^-6) = 3 x 10^-3 M

7. Calculate the pH

pH = -log[H+]

pH = -log(3 x 10^-3)

pH ≈ 2.52

Therefore, the pH of a 3% acetic acid solution is approximately 2.52.