Calculate pH of buffer solution — Henderson-Hasselbalch equation

medium CBSE JEE-MAIN NEET JEE Main 2023 3 min read

Question

A buffer solution is prepared by mixing 0.1 M acetic acid (CH3COOH\text{CH}_3\text{COOH}) with 0.2 M sodium acetate (CH3COONa\text{CH}_3\text{COONa}). If the KaK_a of acetic acid is 1.8×1051.8 \times 10^{-5}, find the pH of the buffer solution.

(JEE Main 2023, similar pattern)


Solution — Step by Step

We have a weak acid (acetic acid) mixed with its conjugate base (acetate ion from sodium acetate). This is an acidic buffer system. The Henderson-Hasselbalch equation applies directly.

For an acidic buffer:

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

This is derived from the equilibrium expression of the weak acid. The “salt” here is the conjugate base — sodium acetate.

pKa=log(Ka)=log(1.8×105)\text{p}K_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) pKa=[log1.8+log105]=[0.25535]=4.74\text{p}K_a = -[\log 1.8 + \log 10^{-5}] = -[0.2553 - 5] = 4.74 pH=4.74+log0.20.1=4.74+log2\text{pH} = 4.74 + \log\frac{0.2}{0.1} = 4.74 + \log 2 pH=4.74+0.301=5.04\text{pH} = 4.74 + 0.301 = \mathbf{5.04}

Why This Works

A buffer resists pH changes because it has a reservoir of both weak acid and conjugate base. When you add a strong acid, the conjugate base neutralises it. When you add a strong base, the weak acid neutralises it. The Henderson-Hasselbalch equation gives the pH based on the ratio of these two components.

Notice that pH depends on the ratio of salt to acid concentrations, not their absolute values. This means diluting a buffer (adding water) does not significantly change its pH — one of the key properties of buffers.

Shortcut: when the salt and acid concentrations are equal, log(1)=0\log(1) = 0, so pH = pKa. This means a buffer is most effective (has maximum buffer capacity) when pH \approx pKa. For acetic acid buffer, this is around pH 4.74.


Alternative Method

Work from the equilibrium expression directly without memorising Henderson-Hasselbalch:

Ka=[H+][CH3COO][CH3COOH]K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} [H+]=Ka×[acid][salt]=1.8×105×0.10.2=9×106[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]} = 1.8 \times 10^{-5} \times \frac{0.1}{0.2} = 9 \times 10^{-6} pH=log(9×106)=6log9=60.954=5.05\text{pH} = -\log(9 \times 10^{-6}) = 6 - \log 9 = 6 - 0.954 = 5.05

The small difference (5.05 vs 5.04) is due to rounding in the log values.


Common Mistake

The most common error: using the Henderson-Hasselbalch equation for a strong acid + strong base system. The equation works ONLY for weak acid/conjugate base (or weak base/conjugate acid) buffers. A mixture of HCl and NaCl is NOT a buffer.

Another trap: forgetting to account for the common ion effect. When sodium acetate dissolves, it fully dissociates to give CH3COO\text{CH}_3\text{COO}^-. This suppresses the ionisation of acetic acid, so you cannot use the simple weak acid pH formula (pH=12(pKalogC)\text{pH} = \frac{1}{2}(\text{p}K_a - \log C)) — you must use the buffer equation instead.

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