Buffer Solutions

Add one drop of strong acid to pure water and the pH crashes several units in an instant. Add that same drop to blood plasma, and the pH barely twitches. The plasma is buffered: it holds a large reserve of a weak acid together with its conjugate base, ready to soak up any small intruder before it can swing the pH. This idea — a buffer solution that resists pH change on small additions of acid or base, or on dilution — sits inside Unit VI, Equilibrium, of CBSE Class 11 Chemistry, right after ionisation constants, pH and $\text{p}K_a$ . The same reservoir logic explains why some salts, dissolved alone in water, quietly turn the solution acidic or basic through hydrolysis. This article builds the Henderson-Hasselbalch equation from the common-ion effect, works a real buffer-pH calculation, and separates true buffers from salts that merely look similar.

The core insight

A buffer resists pH change because it holds a large, comparable reservoir of a weak acid (or base) and its conjugate salt, so added $\text{H}^+$ or $\text{OH}^-$ is consumed by the reservoir instead of accumulating freely, and the resulting pH is fixed by $\text{pH} = \text{p}K_a + \log_{10}([\text{salt}]/[\text{acid}])$ .

The core idea

A buffer solution is a mixture that keeps its pH nearly constant when small amounts of acid or base are added, or when it is diluted. Two kinds appear in the syllabus:

Acidic buffer — a weak acid plus its salt with a strong base, e.g. $\text{CH}_3\text{COOH}$ + $\text{CH}_3\text{COONa}$ , giving pH below 7.
Basic buffer — a weak base plus its salt with a strong acid, e.g. $\text{NH}_4\text{OH}$ + $\text{NH}_4\text{Cl}$ , giving pH above 7.

Why it resists change. In an acetate buffer the solution holds a large pool of un-ionised $\text{CH}_3\text{COOH}$ and a large pool of $\text{CH}_3\text{COO}^-$ (from the fully dissociated salt). A small dose of strong acid is mopped up by the excess $\text{CH}_3\text{COO}^-$ :

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

A small dose of strong base is mopped up by the excess acid:

\text{CH}_3\text{COOH} + \text{OH}^- \to \text{CH}_3\text{COO}^- + \text{H}_2\text{O}

Either way, only the ratio $[\text{CH}_3\text{COO}^-]/[\text{CH}_3\text{COOH}]$ shifts slightly, and since pH depends on the logarithm of that ratio, the pH itself barely moves.

The governing equation. For a weak acid HA and its salt,

\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{salt}]}{[\text{acid}]}

where: $\text{p}K_a = -\log_{10}K_a$ is the acid's dissociation constant on a log scale, $[\text{salt}]$ is the concentration of the fully dissociated conjugate-base salt, and $[\text{acid}]$ is the concentration of the remaining un-ionised weak acid. A parallel relation, $\text{pOH} = \text{p}K_b + \log_{10}([\text{salt}]/[\text{base}])$ , governs basic buffers. When $[\text{salt}] = [\text{acid}]$ the log term vanishes and $\text{pH} = \text{p}K_a$ — the point of maximum buffering capacity, where the reservoir is most evenly split between acid and base forms and can absorb the largest addition of either without much pH drift.

Salt hydrolysis (qualitative). A salt's ions can react with water even with no separate acid or base added, shifting pH away from 7:

Strong acid + strong base (NaCl): neither ion hydrolyses; solution stays neutral.
Weak acid + strong base ( $\text{CH}_3\text{COONa}$ ): $\text{CH}_3\text{COO}^-$ hydrolyses, releasing $\text{OH}^-$ ; solution is basic.
Strong acid + weak base ( $\text{NH}_4\text{Cl}$ ): $\text{NH}_4^+$ hydrolyses, releasing $\text{H}^+$ ; solution is acidic.
Weak acid + weak base ( $\text{CH}_3\text{COONH}_4$ ): both ions hydrolyse; the net pH depends on which of $K_a$ , $K_b$ is larger.

Figure

Two beakers compared, plain water has a large pH drop arrow after one drop of acid while a buffer beaker with weak acid and conjugate base reservoirs has only a tiny pH drop arrow after the same drop

Derivation AHenderson-Hasselbalch equation from the acid equilibrium

The goal is to show, starting only from the equilibrium expression for a weak acid HA in water, why pH comes out as $\text{p}K_a$ plus a logarithm of the salt-to-acid concentration ratio — the single relation that lets you read off buffer pH directly from composition without solving a quadratic.

Step 1. Write the equilibrium. A weak acid dissociates as $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ , with $K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ .

Step 2. Solve for $[\text{H}^+]$ . Rearranging gives $[\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]}$ . In a buffer, $[\text{HA}]$ is essentially the stated concentration of the weak acid (its own ionisation is suppressed by the common ion $\text{A}^-$ from the added salt), and $[\text{A}^-]$ is essentially the stated concentration of the salt, since the salt is a strong electrolyte and dissociates completely.

Step 3. Take negative logarithms of both sides. $-\log_{10}[\text{H}^+] = -\log_{10}K_a - \log_{10}\dfrac{[\text{acid}]}{[\text{salt}]}$ , which is $\text{pH} = \text{p}K_a - \log_{10}\dfrac{[\text{acid}]}{[\text{salt}]}$ .

Step 4. Flip the fraction to remove the leading minus sign, using $-\log_{10}(x/y) = \log_{10}(y/x)$ : $\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{salt}]}{[\text{acid}]}$ . This is the Henderson-Hasselbalch equation. It shows immediately that pH depends only on the ratio of salt to acid, not on the absolute concentrations — so mild dilution barely changes pH, because both concentrations scale down together and their ratio stays fixed.

Worked example

Problem: Calculate the pH of a buffer solution containing $0.10\ \text{mol L}^{-1}$ acetic acid ( $\text{CH}_3\text{COOH}$ ) and $0.15\ \text{mol L}^{-1}$ sodium acetate ( $\text{CH}_3\text{COONa}$ ), given $K_a(\text{CH}_3\text{COOH}) = 1.8\times10^{-5}$ .

Step 1. Find $\text{p}K_a$ . $\text{p}K_a = -\log_{10}(1.8\times10^{-5}) = 4.74$ .

Step 2. Identify $[\text{salt}]$ and $[\text{acid}]$ . $\text{CH}_3\text{COONa}$ is a strong electrolyte and dissociates fully, so $[\text{salt}] = 0.15\ \text{mol L}^{-1}$ . The acid stays essentially at $[\text{acid}] = 0.10\ \text{mol L}^{-1}$ , since the common-ion $\text{CH}_3\text{COO}^-$ from the salt suppresses the acid's own ionisation.

Step 3. Apply Henderson-Hasselbalch. $\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{salt}]}{[\text{acid}]} = 4.74 + \log_{10}(1.5) = 4.74 + 0.18 = 4.92$ .

Answer: $\text{pH} \approx 4.92$ , close to $\text{p}K_a = 4.74$ because salt and acid concentrations are comparable — this buffer sits near its point of maximum buffering capacity.

Competitive stretch

If instead $1\ \text{mL}$ of $1\ \text{M}$ HCl were added to $1\ \text{L}$ of this buffer, the added $\text{H}^+$ ( $0.001\ \text{mol}$ ) would react with $\text{CH}_3\text{COO}^-$ , giving new concentrations $[\text{acid}] \approx 0.101\ \text{mol L}^{-1}$ and $[\text{salt}] \approx 0.149\ \text{mol L}^{-1}$ . Recomputing, $\text{pH} = 4.74 + \log_{10}(0.149/0.101) = 4.74 + 0.169 \approx 4.91$ — a shift of only about $0.01$ pH unit, versus a drop of several units the same acid would cause in pure water.

Common traps

Believing a buffer keeps pH exactly constant — it resists large swings from small additions; the pH still shifts slightly, just far less than in an unbuffered solution.
Treating any weak acid plus any salt as a buffer — the salt must be the conjugate salt of that same weak acid (a common-ion pair), not an unrelated salt.
Using pre-mixing concentrations without accounting for full dissociation of the salt — the salt is assumed to dissociate completely, so its stated concentration is used directly as $[\text{salt}]$ .
Assuming all salts give neutral solutions — only strong-acid/strong-base salts are reliably neutral; weak-acid/weak-base combinations hydrolyse and shift pH away from 7.
Forgetting that buffering capacity is finite — enough strong acid or base can still overwhelm the reservoir once one of the two species (acid or salt) is almost used up, at which point the pH begins to change sharply, just as in an unbuffered solution.

Ask the tutor: Want to try designing a buffer that holds a target pH, or see how the acetate buffer above responds to added strong base instead of acid?

Why does adding the salt of a weak acid make the buffer work, instead of just using the weak acid alone?

A weak acid by itself has very little $\text{A}^-$ around, since it barely ionises, so it cannot neutralise much added base. Adding its salt injects a large, ready-made pool of $\text{A}^-$ without needing further ionisation. Now both a proton donor (HA) and a proton acceptor ( $\text{A}^-$ ) exist in comparable, substantial amounts, so the mixture can absorb either added acid or added base — a lone weak acid can only absorb base, not acid.

If I mix ammonium chloride and sodium acetate, is that a buffer?

No. $\text{NH}_4\text{Cl}$ is the salt of a weak base with a strong acid, and $\text{CH}_3\text{COONa}$ is the salt of a weak acid with a strong base — they belong to two different conjugate pairs. Mixing them causes some mutual hydrolysis and shifts pH somewhat, but there is no single weak acid or base paired with its own conjugate salt in large, comparable amounts, so it does not follow the Henderson-Hasselbalch behaviour of a true buffer.

Does diluting a buffer change its pH?

Only very slightly, and that is the point of the ratio form of the equation. Diluting scales $[\text{salt}]$ and $[\text{acid}]$ down by the same factor, so their ratio — and hence $\log_{10}([\text{salt}]/[\text{acid}])$ — stays almost unchanged, and pH stays close to $\text{p}K_a + \log_{10}(\text{ratio})$ . This resistance to dilution is a second, separate sense in which buffers "resist change," beyond resisting added acid or base.

Buffer Solutions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads