What is the pH of a 0.05 M solution of acetic acid knowing that it was 2% ionized

Acetic acid = HAc

.........HAc ==> H^+ + Ac^-
I........0.05....0......0
C........-x......x......x
E......0.05-x...x......x
You know x = 0.05 x 0.02 = ? = (H^+).

Then pH = -log(H^+) = ?

2.8

2.8

4.1

To determine the pH of a solution of acetic acid, you need to consider its dissociation in water. Acetic acid (CH3COOH) ionizes partially in water to form acetate ions (CH3COO-) and hydrogen ions (H+). The equation for the dissociation of acetic acid is as follows:

CH3COOH ⇌ CH3COO- + H+

The extent of ionization, also known as the ionization constant (Ka), indicates the percentage of the acid that has dissociated. In this case, you mentioned that acetic acid is 2% ionized. This means that 2% of the initial acetic acid has dissociated into acetate ions and hydrogen ions.

To calculate the pH, we need to determine the concentration of the hydrogen ions. Since acetic acid is weak, we can assume that the concentration of acetate ions is equal to the concentration of hydrogen ions. Let's call this concentration x.

Now let's set up an equilibrium expression for the ionization of acetic acid:

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

Since the concentration of CH3COO- is equal to x, and the concentration of CH3COOH is the initial concentration minus the concentration of CH3COO-, we can write:

Ka = [x][x] / [0.05 - x]

Since acetic acid is 2% ionized, we can say that [x] = 0.02 * 0.05 (2% of the initial concentration). Therefore:

Ka = (0.02 * 0.05)(0.02 * 0.05) / [0.05 - (0.02 * 0.05)]

To find the value of Ka, we need the specific value of the equilibrium constant at this temperature. Once you have the value of Ka, you can use it to calculate the concentration of hydrogen ions (H+) and determine the pH using the formula:

pH = -log[H+]

Please note that without the specific value of the ionization constant (Ka), it is not possible to calculate the pH accurately. The value of Ka can vary with temperature and specific experimental conditions.