Determine the pH of 0.30 M acetic acid with Ka 1.8×10^-5
To determine the pH of the acetic acid solution, we need to calculate the concentration of the hydronium ions (H3O+) in the solution first.
Given that acetic acid (CH3COOH) is a weak acid, we can assume that it only partially dissociates in water:
CH3COOH + H2O ↔ CH3COO- + H3O+
From the dissociation reaction above, we know that the initial concentration of acetic acid (CH3COOH) is 0.30 M, and the concentration of hydronium ions (H3O+) is equal to the concentration of the dissociated acetic acid, which we assume to be x.
The Ka expression for this reaction is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Substitute the given values and x for [H3O+] into the Ka expression:
1.8×10^-5 = (x)(x) / 0.30
Solving for x:
x^2 = (1.8×10^-5)(0.30)
x = √((1.8×10^-5)(0.30)) ≈ 0.001897 M
Therefore, the concentration of hydronium ions in the solution is approximately 0.001897 M.
To determine the pH of the solution, we need to calculate the negative logarithm of the hydronium ion concentration:
pH = -log[H3O+]
pH = -log(0.001897)
pH ≈ 2.72
The pH of the 0.30 M acetic acid solution is approximately 2.72.