What would be the PH of a 0.20M CH3COOH solution if no salt were present ?
The pH of a solution of acetic acid (CH3COOH) can be calculated using its dissociation constant (Ka) and the initial concentration of the acid. Acetic acid is a weak acid, and it only partially dissociates in water:
CH3COOH ⇌ CH3COO- + H+
The dissociation constant (Ka) for acetic acid at 25°C is approximately 1.8 × 10^-5. The expression for the Ka of acetic acid is:
Ka = [CH3COO-][H+] / [CH3COOH]
For a 0.20 M solution of acetic acid, we can make the simplifying assumption that the concentration of the acetic acid at equilibrium will be approximately equal to the initial concentration (0.20 M) because the acid only partially dissociates. We will represent the concentration of H+ and CH3COO- at equilibrium as "x". The Ka expression becomes:
1.8 × 10^-5 = (x)(x) / (0.20 - x)
Since acetic acid is a weak acid and only dissociates to a small extent, we can assume that "x" will be much smaller than 0.20 M, thus we can simplify this to:
1.8 × 10^-5 ≈ x^2 / 0.20
Now, solve for "x":
x^2 = 1.8 × 10^-5 * 0.20
x^2 = 3.6 × 10^-6
Now take the square root of both sides to find the concentration of H+:
x = √(3.6 × 10^-6)
x ≈ 1.9 × 10^-3
The concentration of H+ ions (x) is approximately 1.9 × 10^-3 M. The pH is the negative logarithm (base 10) of the hydrogen ion concentration:
pH = -log[H+]
pH = -log(1.9 × 10^-3)
pH ≈ 2.72
Therefore, the pH of a 0.20 M solution of CH3COOH, with no additional salt present, would be approximately 2.72. Keep in mind that this is an approximation, as the initial assumption that the change in concentration of acetic acid (0.20 - x) is approximately equal to the initial concentration (0.20) may not be precisely accurate for more exact calculations. However, this approximation is reasonable for most practical purposes, particularly at this concentration of a weak acid.