JEE Main 2026 Chemistry question
20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (pKa value of acetic acid is 4.75)
Answer: (B)
Solution
Total acid = 2.84 mmol. NaOH added = 1.42 mmol, exactly half-neutralising the acid, giving equal moles of acid and salt: pH = pKa = 4.75.