Calculating the pH of a 0.1 M Solution of Acetic Acid

 

Calculating the pH of a 0.1 M Solution of Acetic Acid

Acetic acid (CH3COOH) is a weak acid that partially dissociates in water, forming hydrogen ions (H+) and acetate ions (CH3COO-). The dissociation of acetic acid can be represented by the equation:

CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

The dissociation constant (Ka) for acetic acid is given as 1.8 x 10^-5. This constant represents the equilibrium between the reactants and products of the dissociation reaction.

To calculate the pH of a 0.1 M solution of acetic acid, we need to consider the equilibrium concentrations of the hydrogen ions and the acetate ions in the solution.

Let’s assume that x mol/L of CH3COOH dissociates to form x mol/L of H+ and x mol/L of CH3COO-. At equilibrium, the concentrations will be 0.1 M – x for CH3COOH and x each for H+ and CH3COO-.

Using the expression for the dissociation constant (Ka) of acetic acid:

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

Substitute the equilibrium concentrations into the equation:

1.8 x 10^-5 = x * x / (0.1 – x)

Since the dissociation of acetic acid is relatively small, we can assume that x is much smaller than 0.1. This allows us to approximate 0.1 – x as 0.1.

1.8 x 10^-5 = x^2 / 0.1

Rearranging the equation:

x^2 = 1.8 x 10^-6

x = sqrt(1.8 x 10^-6) ≈ 4.24 x 10^-4

Since x represents the concentration of H+, which corresponds to the concentration of hydronium ions (H3O+), we can consider [H3O+] ≈ 4.24 x 10^-4 M in the solution.

To calculate the pH, we use the formula:

pH = -log[H3O+]

pH = -log(4.24 x 10^-4)

pH ≈ -log(4.24) + log(10^-4)

pH ≈ -(-0.373) – 4

pH ≈ 4.37

Therefore, the pH of a 0.1 M solution of acetic acid is approximately 4.37.