40 millimoles of NaOH are added to 100 mL of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH… · ABC26PH0429
Subject: Physical Chemistry · Chapter: Chemical Equilibrium · Topic: Equilibria – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium
40 millimoles of NaOH are added to 100 mL of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer $\left(\mathrm{K}_{\mathrm{HA}}=1.00 \times 10^{-5}\right)$ is
(a)5.30
(b)5.00
(c)0.30
(d)10.30
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
After base: pH 5.30 = 5 + log[(0.1Y + 0.04)/0.08] → 0.1Y + 0.04 = 0.16, Y = 1.2 M. Initially [A⁻] = [HA] = 1.2 M, so pH = pKa = 5.00 (b).