Question

0.1 Mole of CH₃COOH and 0.05 moles of sodium acetate are mixed and the solution is made up to two litres. On adding 0.01 moles of HCl to this buffer, the change in pH of the buffer is (neglecting dilution effects, K_a of acetic acid = 1.8 × 10^–5) –

[Note → log 4/11 = –0.4393; log 2 = 0.30]

• A. 0.44
• B. 0.08
• C. 0.05
• D. 0.1393

Answer 1

Answer: (D)

0.1393

Answer 2

pH₁ = pK_a + log $\frac{0.05}{0.10}$ = pK_a – log 2

pH₂ = pK_a + log $\frac{0.04}{0.11}$ = pK_a – log $\frac{11}{4}$

∆pH = – log $\left( \frac{11}{4} \right)$ + log 2 = log 2 + log $\frac{4}{11}$

= 0.30 – 0.4393 = 0.1393 = 0.14