Buffer solutions — how they resist pH change, Henderson-Hasselbalch selection

Question

A buffer solution is prepared by mixing 0.1 M acetic acid ( $K_a = 1.8 \times 10^{-5}$ ) with 0.1 M sodium acetate. Find the pH of the buffer. What happens to the pH when a small amount of HCl is added?

(JEE Main / NEET pattern)

Solution — Step by Step

We have a weak acid (acetic acid, CH₃COOH) and its conjugate base (sodium acetate, CH₃COONa). This is an acidic buffer.

flowchart TD
    A["Need a buffer?"] --> B{"What pH range?"}
    B -->|"pH < 7 (acidic)"| C["Acidic Buffer\nWeak acid + its salt\n(e.g., CH₃COOH + CH₃COONa)"]
    B -->|"pH > 7 (basic)"| D["Basic Buffer\nWeak base + its salt\n(e.g., NH₃ + NH₄Cl)"]
    C --> E["Use: pH = pKa + log([salt]/[acid])"]
    D --> F["Use: pOH = pKb + log([salt]/[base])"]

For an acidic buffer:

\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}

Here, $[\text{salt}] = [\text{CH}_3\text{COONa}] = 0.1$ M and $[\text{acid}] = [\text{CH}_3\text{COOH}] = 0.1$ M.

\text{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74

\text{pH} = 4.74 + \log\frac{0.1}{0.1} = 4.74 + \log 1 = \mathbf{4.74}

When acid and salt concentrations are equal, pH = pKₐ. This is the most efficient buffer point.

When a small amount of HCl (strong acid) is added, the H⁺ ions react with the conjugate base:

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

This consumes the added H⁺ and converts some salt into acid. The ratio $[\text{salt}]/[\text{acid}]$ changes slightly, so pH changes only marginally — not the drastic drop you would see in pure water.

After adding 0.01 mol HCl:

Salt decreases: $0.1 - 0.01 = 0.09$ M
Acid increases: $0.1 + 0.01 = 0.11$ M

\text{pH} = 4.74 + \log\frac{0.09}{0.11} = 4.74 + \log(0.818) = 4.74 - 0.087 = \mathbf{4.65}

pH dropped by only 0.09 units — that is the buffer in action.

Why This Works

A buffer works because it has a reservoir of both the weak acid and its conjugate base. When H⁺ is added, the conjugate base neutralises it. When OH⁻ is added, the weak acid neutralises it. The pH changes very little because the ratio $[\text{salt}]/[\text{acid}]$ inside the logarithm changes slowly — logarithms compress large ratio changes into small pH shifts.

The buffer capacity is maximum when $[\text{salt}] = [\text{acid}]$ (i.e., pH = pKₐ). As the ratio deviates significantly from 1, the buffer becomes weaker. A buffer is effective within the range $\text{pH} = \text{p}K_a \pm 1$ .

Alternative Method — ICE Table Approach

Instead of Henderson-Hasselbalch, set up a full equilibrium ICE table for CH₃COOH dissociation in the presence of common ion CH₃COO⁻. You will arrive at the same answer, but Henderson-Hasselbalch is faster for buffer problems.

To select the right weak acid for a buffer at a given pH, pick one whose pKₐ is close to the desired pH. For a buffer at pH 7, use $\text{H}_2\text{PO}_4^-/\text{HPO}_4^{2-}$ (pKₐ₂ = 7.2). For pH 9-10, use $\text{NH}_4^+/\text{NH}_3$ (pKb = 4.74, so pKₐ = 9.26). This is a common NEET MCQ type.

Common Mistake

Students often confuse which formula to use: Henderson-Hasselbalch for an acidic buffer gives pH, but for a basic buffer you should first find pOH and then convert. Writing $\text{pH} = \text{p}K_b + \log([\text{salt}]/[\text{base}])$ is wrong — that expression gives pOH, not pH. Always check whether your answer makes sense: an acidic buffer should give pH < 7, a basic buffer should give pH > 7.

Question

Solution — Step by Step

Why This Works

Alternative Method — ICE Table Approach

Common Mistake

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