The pKa of the given buffer solution is 4.

Problem Solution

The buffer consists of acetic acid (weak acid, HA) at 0.04 M and sodium acetate (conjugate base, A⁻) at 0.4 M, with pH 5. Apply the Henderson-Hasselbalch equation:

pH = pK_a + log₁₀([A^–]/[HA])

Substitute values: 5 = pK_a + log₁₀(0.4/0.04). The ratio 0.4/0.04 = 10, so log₁₀(10) = 1. Thus, 5 = pK_a + 1, giving pK_a = 4.

This assumes initial concentrations approximate equilibrium values, valid for typical buffer problems in exams like CSIR NET.

Article Introduction

The pKa of buffer solution pH 5 with 0.4 M sodium acetate and 0.04 M acetic acid is a key CSIR NET Life Sciences question testing buffer chemistry fundamentals. This acetate buffer example demonstrates precise pKa calculation using the Henderson-Hasselbalch equation, essential for biochemistry and enzyme kinetics topics.

Henderson-Hasselbalch Equation

Buffers resist pH changes via weak acid (HA) and conjugate base (A⁻) equilibrium. The equation is pH = pK_a + log₁₀([A^–]/[HA]), derived from K_a = [H⁺][A^–]/[HA].

Here, [A⁻] = 0.4 M (sodium acetate), [HA] = 0.04 M (acetic acid), pH = 5. Ratio = 10, log = 1, so pKa = 5 – 1 = 4 (integer).

Step-by-Step Calculation

• Compute ratio: 0.4/0.04 = 10
• Logarithm: log₁₀(10) = 1
• Rearrange: pKa = pH – log(ratio) = 5 – 1 = 4

No options provided; direct integer answer is 4, matching acetic acid’s typical pKa range (4.74-4.76) adjusted for this ratio.

Exam Relevance

For CSIR NET, recognize buffers maintain pH near pKa ±1 for max capacity. This pH 5 setup (pKa 4) falls within effective range. Practice similar problems for molecular biology, biotech sections.