Acetic acid + sodium acetate. Henderson-Hasselbalch equation.

Buffer Solution — How It Resists pH Change

Question

A buffer solution is prepared by mixing 0.1 mol of acetic acid (CH₃COOH) and 0.1 mol of sodium acetate (CH₃COONa) in 1 litre of water. Calculate the pH of this buffer. What happens to the pH when 0.01 mol of HCl is added?

Given: $K_a$ of acetic acid = $1.8 \times 10^{-5}$ , so $pK_a = 4.74$

Solution — Step by Step

The Henderson-Hasselbalch equation is the only tool you need for buffer pH:

pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}

Here, salt = CH₃COONa (gives CH₃COO⁻), acid = CH₃COOH. Both are 0.1 mol/L.

pH = 4.74 + \log\frac{0.1}{0.1} = 4.74 + \log(1) = 4.74 + 0 = \mathbf{4.74}

When we add 0.01 mol HCl, the H⁺ ions don’t just sit there — they react with the acetate ions (CH₃COO⁻), the basic component of the buffer:

\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}

This is why buffers resist pH change: the added acid is consumed by the conjugate base rather than free H⁺ accumulating in solution.

Before adding HCl:

CH₃COOH = 0.1 mol
CH₃COO⁻ = 0.1 mol

After 0.01 mol H⁺ reacts with 0.01 mol CH₃COO⁻:

CH₃COOH = 0.1 + 0.01 = 0.11 mol
CH₃COO⁻ = 0.1 − 0.01 = 0.09 mol

pH = 4.74 + \log\frac{0.09}{0.11} = 4.74 + \log(0.818)

\log(0.818) \approx -0.087

pH = 4.74 - 0.087 \approx \mathbf{4.65}

The pH dropped by only 0.09 units despite adding a strong acid. Without the buffer, 0.01 mol HCl in 1 L would give pH = 2. That’s the buffer action in numbers.

Why This Works

A buffer works because it contains both a weak acid and its conjugate base in significant concentrations. The acid neutralises any added base (OH⁻), and the conjugate base neutralises any added acid (H⁺). Neither component gets fully used up — they shift concentrations slightly, which only nudges the log ratio.

The Henderson-Hasselbalch equation captures this beautifully. pH depends on a ratio, not absolute concentrations. So even after adding HCl, the ratio shifts from 1:1 to 0.09:0.11 — a small change in a log scale.

Buffer capacity is maximum when [Salt] = [Acid], i.e., when $pH = pK_a$ . In this problem, the initial condition is exactly the maximum-capacity point. This is why the pH barely budges.

Alternative Method — Using ICE Without Henderson-Hasselbalch

If you distrust the formula (or it slips your mind during JEE), work from $K_a$ directly.

At equilibrium, CH₃COOH partially dissociates. But in the presence of CH₃COO⁻ (common ion), dissociation is suppressed. Set up:

K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}

1.8 \times 10^{-5} = \frac{(0.1 + x)(x)}{(0.1 - x)}

Since $x \ll 0.1$ (buffer suppresses dissociation):

1.8 \times 10^{-5} \approx \frac{0.1 \cdot x}{0.1} = x

So $[\text{H}^+] = 1.8 \times 10^{-5}$ , giving $pH = -\log(1.8 \times 10^{-5}) = 4.74$ . ✓

Same answer, more steps. Henderson-Hasselbalch is the shortcut for this reason — use it in JEE Main where time is the real constraint.

Common Mistake

Writing the ratio upside down. Many students write $\frac{[\text{Acid}]}{[\text{Salt}]}$ in the Henderson-Hasselbalch equation. The correct form is always $\frac{[\text{Salt}]}{[\text{Acid}]}$ , i.e., conjugate base over weak acid. This mistake flips the log term and gives $pH = 4.74 - 0 = 4.74$ here (same answer, by coincidence, when ratio is 1:1) — but kills you when concentrations differ. Verify: more salt than acid → ratio > 1 → log is positive → pH > pKₐ. That makes chemical sense because more base pulls the buffer alkaline.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using ICE Without Henderson-Hasselbalch

Common Mistake

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