Consider the following statements:
(a) The pH of a mixture containing 400 mL of 0.1 M H₂SO₄ and 400 mL of 0.1 M NaOH will be approximately 1.3.
(b) Ionic product of water is temperature dependent.
(c) A monobasic acid with Ka = 10⁻⁵ has a pH = 5. The degree of dissociation of this acid is 50%.

Solution:- (B)
(a), (b) and (c)
(a) H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
400 × 0.1 = 40
400 × 0.1 = 40
200
∴ [H⁺] = 20/800 = 1/20
⇒ pH = -log(1/20) ≅ 1.3
So (a) is correct
(b) log(Kw₂/Kw₁) = ΔH/2.303R[1/T₁ - 1/T₂]
So, ionic product of water is temp. dependent hence (b) is correct
(c) Ka = 10⁻⁵, pH = 5 ⇒ [H⁺] = 10⁻⁵
Ka = cα²/ (1 - α) ⇒ Ka = [H⁺]α/(1 - α)
10⁻⁵ = 10⁻⁵α/(1 - α)
⇒ 1 - α = α
⇒ α = 1/2 = 50%