20mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

NaOH + CH3COOH  CH3COONa + H2O 0.1 M, 20 ml 0.1 M, 50 ml Millimole = 0.1 × 20 = 2, 0.1 × 50 = 5 L.R. = NaOH So 5 – 2 = (3) (2) So Resultant solution is Acidic buffer solution So pH = pKa + log       acid salt pH = 4.76 + log 2 pH = 4.76 + (–0.18) = 4.58 = 458 ×10–2