What will be the value of pH of 0.01 mol dm^-3 CH₃COOH (K_a = 1.74 × 10^-5 )?

Ionization of CH₃COOH occurs as :

CH₃COOH + H₂O ⇋ H₃O⁺ + CH₃COO^-

Initial concentration: 0.01 0 0

Equilibrium concentration: 0.01 – x x x

ionization constant =

since , x>>0.01 hence, 0.01 – x ~ 0.01 .

x²/0.01 = 1.74 × 10^-5 (given)

X² = 1.74 × 10^-5 x 0.01

X = 4.2 x 10^-4

Or, pH = -log (4.2 x 10^-4 ) = 3.4