Given:

Ionization constant of acetic acid = 1.74 × 10^–5

0.05 M solution

Ionization of CH₃COOH

CH₃COOH ⇒ CH₃COO^– + H⁺

By applying the formula,

K_a =

K_a =

At equilibrium [CH₃COO^– ]= [H⁺]

Hence,

K_a=

As K_a = 1.74 × 10^–5 (given)

[CH₃COOH] = 0.05 M

∴ 1.74 × 10^–5 =

⇒[H⁺]² = 1.74 × 10^–5 × 5 × 10^-2M

⇒[H⁺] =

⇒[H⁺] = 9.33 × 10^-4 M

As, [CH₃COO^–] = [H⁺]

Thus, the concentration of acetate ion is 9.33 × 10^-4M

To calculate pH, we apply the formula,

pH = -log[H⁺]

⇒pH = -log(9.33 × 10^-4)

⇒pH = -log 9.33 – log 10^-4

⇒pH = - log 9.33 – (-4) log 10

⇒pH = 4 - log 9.33

⇒pH = 4 – 0.9699

⇒pH = 3.03

Thus, its pH is 3.03.