Buffer Solutions

ChemistryEquilibriumFor JEE aspirants

A buffer solution resists changes in pH when small amounts of acid or base are added. Every buffer contains a conjugate acid-base pair: an acidic buffer combines a weak acid with its conjugate base (like $C H_{3} C O O H$ + $C H_{3} C O O N a$ ), and a basic buffer combines a weak base with its conjugate acid (like $N H_{4} O H$ + $N H_{4} C l$ ). The Henderson-Hasselbalch equation $pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$ predicts buffer pH within one unit of $p K_{a}$ . Titration curves for strong-strong, weak-strong, and polyprotic combinations reveal buffer regions, equivalence points, and guide indicator choice.

Key Formulas - Quick Reference

Henderson-Hasselbalch (acidic buffer): $pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$
Henderson-Hasselbalch (basic buffer): $pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$
At half-equivalence: $pH = p K_{a}$ (for WA + SB titration).
Buffer range: $p K_{a} - 1 \leq pH \leq p K_{a} + 1$ .
Maximum buffer capacity: when $[salt] = [acid]$ .
Buffer capacity: $β = \frac{2.303 [ acid ] [ salt ]}{[ acid ] + [ salt ]}$ .
pH at equivalence (WA + SB): $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ .
pH at equivalence (SA + WB): $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ .
Amphiprotic species: $pH = \frac{1}{2} (p K_{a 1} + p K_{a 2})$ .

1. What is a Buffer Solution?

A buffer is a solution whose pH changes only slightly when a small amount of strong acid or strong base is added, or when it is diluted. It does not mean the pH never changes - only that the change is far smaller than would occur in an unbuffered solution.

Blood plasma is a natural buffer maintained near pH 7.4 by the $H_{2} C O_{3}$ / $H C O_{3}^{-}$ system. Cellular buffers, ocean chemistry, and countless laboratory procedures rely on buffer solutions for stable pH control.

2. Types of Buffers

Acidic buffer: weak acid + salt of that acid with a strong base. Example: $C H_{3} C O O H$ + $C H_{3} C O O N a$ ; phthalic acid + potassium hydrogen phthalate; boric acid + borax. pH lies below 7.
Basic buffer: weak base + salt of that base with a strong acid. Example: $N H_{4} O H$ + $N H_{4} C l$ ; glycine + glycine hydrochloride. pH lies above 7.
Salt buffer: aqueous solution of a salt of a weak acid and a weak base, such as $C H_{3} C O O N H_{4}$ .

3. Mechanism of Acidic Buffer Action

Consider the classic acidic buffer: acetic acid + sodium acetate ( $C H_{3} C O O H$ + $C H_{3} C O O N a$ ).

Two equilibria set up in solution:

C H_{3} C O O H ⇌ C H_{3} C O O^{-} + H^{+}

(weak acid, feebly ionised)

C H_{3} C O O N a ⟶ C H_{3} C O O^{-} + N a^{+}

(salt, fully ionised)

The common ion $C H_{3} C O O^{-}$ from the salt suppresses ionisation of the acid, so the buffer contains a large reservoir of both molecular acid and acetate ion — the two species that stand ready to absorb any incoming stress.

Attack 1: Add H⁺ (drop of HCl)

Acetate ion mops up the added proton to form molecular acetic acid, which is only feebly ionised:

$C H_{3} C O O^{-} + H^{+} ⟶ C H_{3} C O O H$

Result: free $H^{+}$ concentration barely changes.

Attack 2: Add OH⁻ (drop of NaOH)

Molecular acetic acid mops up the added hydroxide to form more acetate ion and water:

$C H_{3} C O O H + O H^{-} ⟶ C H_{3} C O O^{-} + H_{2} O$

Result: free $O H^{-}$ concentration barely changes.

Figure 1: Buffer action - the conjugate base neutralises added

H^{+}

, and the weak acid neutralises added

O H^{-}

, so overall pH barely changes.

4. Mechanism of Basic Buffer Action

Consider the classic basic buffer: ammonium hydroxide + ammonium chloride ( $N H_{4} O H$ + $N H_{4} C l$ ).

Two equilibria set up in solution:

N H_{4} O H ⇌ N H_{4}^{+} + O H^{-}

(weak base, feebly ionised)

N H_{4} C l ⟶ N H_{4}^{+} + C l^{-}

(salt, fully ionised)

The common ion $N H_{4}^{+}$ from the salt suppresses ionisation of the base, so the buffer contains a large reservoir of both molecular base and ammonium ion, ready to absorb any incoming stress.

Attack 1: Add H⁺

Molecular $N H_{4} O H$ mops up the added proton to form ammonium ion and water:

$N H_{4} O H + H^{+} ⟶ N H_{4}^{+} + H_{2} O$

Result: free $H^{+}$ concentration barely changes.

Attack 2: Add OH⁻

Ammonium ion (from the salt) mops up the added hydroxide to form more molecular $N H_{4} O H$ , which is only feebly ionised:

$N H_{4}^{+} + O H^{-} ⟶ N H_{4} O H$

Result: free $O H^{-}$ concentration barely changes.

5. Henderson-Hasselbalch Equation (Acidic Buffer)

For $H A ⇌ H^{+} + A^{-}$ :

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} \Rightarrow [H^{+}] = K_{a} \cdot \frac{[ H A ]}{[ A ^{-} ]}

Because the salt is completely ionised and the acid is weak (its dissociation is further suppressed by common ion effect), we can take $[H A] \approx [acid]_{0}$ and $[A^{-}] \approx [salt]_{0}$ . Taking $- lo g$ of both sides:

pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

6. Henderson-Hasselbalch Equation (Basic Buffer)

By identical reasoning for a weak base:

pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}

Convert to pH using $pH = 14 - pOH$ at 25 °C.

7. Preparing a Buffer of Target pH

Pick a weak acid (or base) whose $p K_{a}$ (or $p K_{b}$ ) is within one unit of the target pH.
Compute the required salt-to-acid ratio using the Henderson-Hasselbalch equation.
Mix the calculated amounts in water.

Example: to make a pH 4.74 buffer, use acetic acid ( $p K_{a} = 4.74$ ) with equal concentration of sodium acetate.

8. Buffer Capacity - Definition and Derivation JEE Advanced

Buffer capacity $β$ is the number of moles of strong acid or strong base that must be added to one litre of buffer to change its pH by exactly one unit:

β = \frac{d n}{d pH}

For a buffer of a weak acid $H A$ (concentration $a$ ) and its salt (concentration $b$ ), adding $x$ moles of strong base per litre converts $x$ moles of acid into salt. The Henderson-Hasselbalch equation gives:

pH = p K_{a} + lo g \frac{b + x}{a - x}

Differentiating and using $\frac{d}{d x} ln \frac{b + x}{a - x} = \frac{1}{b + x} + \frac{1}{a - x} = \frac{a + b}{( a - x ) ( b + x )}$ :

β = \frac{2.303 ( a - x ) ( b + x )}{a + b}

For small $x$ (buffer barely used), $β \approx \frac{2.303 ab}{a + b}$ .

9. Maximum Buffer Capacity

Setting $\frac{d β}{d b} = 0$ (for fixed total concentration $a + b$ ) gives $a = b$ . So the buffer capacity is maximum when the concentrations of the weak acid and its salt are equal, i.e., $pH = p K_{a}$ .

Example: for two 1 L buffers of acetic acid + sodium acetate at $p K_{a} = 4.74$ :

0.1 M acid + 0.1 M salt: $β = 2.303 \times \frac{0.1 \times 0.1}{0.2} = 0.115$ .
0.2 M acid + 0.2 M salt: $β = 2.303 \times \frac{0.2 \times 0.2}{0.4} = 0.230$ .

The more concentrated buffer has twice the capacity, as expected.

10. Buffer Range: pKa ± 1

The Henderson-Hasselbalch equation shows that $pH = p K_{a} + lo g ([salt] / [acid])$ . The ratio $[salt] / [acid]$ can practically vary between 0.1 and 10 without the buffer failing, giving a useful pH range of $p K_{a} \pm 1$ . Outside this window, one component is essentially depleted and the buffer no longer resists pH change.

11. Physiological Buffers - Blood Bicarbonate

Human blood is buffered near pH 7.4 by the $H_{2} C O_{3}$ / $H C O_{3}^{-}$ system:

pH = p K_{a} + lo g \frac{[ H C O _{3}^{-} ]}{[ H _{2} C O _{3} ]}, p K_{a} \approx 6.1

The normal ratio of $[H C O_{3}^{-}] : [H_{2} C O_{3}]$ in blood is about 20:1, giving $pH = 6.1 + lo g 20 = 6.1 + 1.3 = 7.4$ . Respiratory (rapid) and renal (slow) mechanisms adjust the two components to keep the ratio (and thus the pH) constant.

12. Acid-Base Titrations and Indicator Choice

Every titration produces a characteristic pH curve whose shape is determined by the strengths of the acid and base involved. Four standard combinations cover almost every case seen in JEE and NEET:

Titration Types at a Glance

Strong acid + Strong base (HCl + NaOH)

pH stays low, jumps sharply from ~3 to ~11 at equivalence, then rises slowly. Equivalence pH = 7 (neutral salt).

Weak acid + Strong base (CH₃COOH + NaOH)

pH rises gradually through a buffer region where $pH = p K_{a}$ at half-equivalence, then jumps to a basic equivalence pH (~8-9) because acetate hydrolyses.

Strong acid + Weak base (HCl + NH₃)

Mirror image of the previous case: pH falls through a buffer region, then jumps to an acidic equivalence pH (~5) because $N H_{4}^{+}$ hydrolyses.

Weak acid + Weak base

No sharp pH jump. Direct titration is unreliable and typically avoided.

Polyprotic and diprotic titrations

Polyprotic acids (e.g. $H_{3} P O_{4}$ ): one inflection per proton removed. Between adjacent equivalence points, the solution contains an amphiprotic species whose pH is the mean of the flanking $p K_{a}$ values: $pH = \frac{1}{2} (p K_{a, n} + p K_{a, n + 1})$ .
Diprotic bases (e.g. $N a_{2} C O_{3}$ ): two inflections, one for each proton added.

Full numerical walkthroughs of pH at every stage are worked out in Solved Example 4 (weak acid vs strong base) and Solved Example 7 (triprotic

H_{3} P O_{4}

titration).

13. Choosing Indicators for Each Titration Type

Titration type	pH at equivalence	Suitable indicator
Strong acid vs strong base	7	Methyl orange or phenolphthalein
Weak acid vs strong base	~8-9 (basic)	Phenolphthalein
Strong acid vs weak base	~5-6 (acidic)	Methyl orange
Weak acid vs weak base	~7	No sharp jump; indicators unreliable
Polyprotic (each step)	Varies	Match indicator to each equivalence pH

Solved Example 1

Calculate the amount of

(N H_{4})_{2} S O_{4}

(in g) needed to add to 500 mL of 0.2 M

N H_{3}

to give a buffer of pH 9.3.

p K_{b}

for

N H_{3} = 4.7

Solution:

$pOH = 14 - 9.3 = 4.7 = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$ , so $lo g \frac{[ salt ]}{0.2} = 0$ , giving $[salt] = 0.2$ M.

Since $(N H_{4})_{2} S O_{4}$ provides 2 $N H_{4}^{+}$ per formula unit, required $(N H_{4})_{2} S O_{4}$ = 0.1 M.

Moles = 0.1 × 0.5 = 0.05; mass = 0.05 × 132 = 6.6 g.

Solved Example 2

Blood pH is 7.4, maintained by

H_{2} C O_{3}

N a H C O_{3}

. What volume of 5 M

N a H C O_{3}

must be mixed with 10 mL of 2.5 M

H_{2} C O_{3}

to maintain pH 7.4? (

p K_{a}

for

H_{2} C O_{3} = 6.7

lo g 2 = 0.3

Solution:

$pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$ ; $7.4 = 6.7 + lo g \frac{[ salt ]}{[ acid ]}$ , so ratio = $1 0^{0.7} = 5$ .

Equating moles: $5 \times V = 5 \times 2.5 \times 10 \times 1 0^{- 3} \Rightarrow V = 25$ mL.

Solved Example 3

Calculate the buffer capacity of 1 L of (a) 0.1 M

C H_{3} C O O H

+ 0.1 M

C H_{3} C O O N a

and (b) 0.2 M

C H_{3} C O O H

+ 0.2 M

C H_{3} C O O N a

. Which is the better buffer?

Solution:

$β = \frac{2.303 ab}{a + b}$ .

(a) $β = \frac{2.303 \times 0.1 \times 0.1}{0.2} = 0.115$ .
(b) $β = \frac{2.303 \times 0.2 \times 0.2}{0.4} = 0.230$ .

The 0.2 M buffer has twice the capacity, so it is the better buffer.

Solved Example 4

Full walkthrough: titrate 50 mL of 0.1 M

C H_{3} C O O H

(

p K_{a} = 4.74

) with 0.1 M NaOH. Compute pH at 0, 10, 25, 40, 50, 60 mL added.

Solution:

0 mL: pure weak acid. $pH = \frac{1}{2} (p K_{a} - lo g C) = \frac{1}{2} (4.74 + 1) = 2.87$ .

10 mL added: 1 mmol NaOH added, 5 - 1 = 4 mmol acid remains, 1 mmol salt formed. Buffer region: $pH = 4.74 + lo g (1/4) = 4.74 - 0.60 = 4.14$ .

25 mL added (half-eq): $[salt] = [acid]$ , so $pH = p K_{a} = 4.74$ .

40 mL added: 4 mmol NaOH, 1 mmol acid left, 4 mmol salt. $pH = 4.74 + lo g 4 = 5.34$ .

50 mL added (equivalence): pure sodium acetate solution at $C = 5/100 = 0.05$ M. $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C) = \frac{1}{2} (14 + 4.74 - 1.30) = 8.72$ .

60 mL added: excess NaOH = 1 mmol in 110 mL; $[O H^{-}] = 1/110 = 9.1 \times 1 0^{- 3}$ ; $pOH = 2.04$ ; $pH = 11.96$ .

Solved Example 5

Prepare a buffer of pH 4.0 using acetic acid (

K_{a} = 1.8 \times 1 0^{- 5}

) and sodium acetate. How much sodium acetate is needed in 1 L of 0.1 M acetic acid?

Solution:

$p K_{a} = 4.74$ ; $pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$ ; $4.0 = 4.74 + lo g \frac{[ salt ]}{0.1}$ .

$lo g \frac{[ salt ]}{0.1} = - 0.74 \Rightarrow [salt] = 0.1 \times 1 0^{- 0.74} \approx 0.0182$ M.

Moles of sodium acetate needed in 1 L = 0.0182; mass = 0.0182 × 82 ≈ 1.49 g.

Solved Example 6

Calculate the pH of the amphiprotic species

N a H_{2} P O_{4}

solution, given

p K_{a 1} = 3

p K_{a 2} = 7

for

H_{3} P O_{4}

Solution:

For an amphiprotic species like $H_{2} P O_{4}^{-}$ , the pH is essentially independent of concentration and equals the mean of the two flanking $p K_{a}$ values:

$pH = \frac{p K _{a 1} + p K _{a 2}}{2} = \frac{3 + 7}{2} = 5$ .

Similarly $N a H C O_{3}$ solution: $pH = \frac{1}{2} (p K_{a 1} + p K_{a 2})$ of $H_{2} C O_{3}$ ≈ 8.35.

Solved Example 7

Full triprotic walkthrough: titrate 50 mL of 0.1 M

H_{3} P O_{4}

with 0.1 M NaOH (

p K_{a 1} = 3

p K_{a 2} = 7

p K_{a 3} = 11

). Find pH at 0, 25, 50, 75, 100, 125, 150 mL added.

Solution:

0 mL: pure weak acid stage 1. $pH = \frac{1}{2} (3 + 1) = 2$ .

25 mL: half of stage 1 buffer $H_{3} P O_{4}$ / $H_{2} P O_{4}^{-}$ . $pH = p K_{a 1} = 3$ .

50 mL (Eq 1): pure $N a H_{2} P O_{4}$ amphiprotic. $pH = \frac{1}{2} (3 + 7) = 5$ .

75 mL: half of stage 2 buffer $H_{2} P O_{4}^{-}$ / $H P O_{4}^{2 -}$ . $pH = p K_{a 2} = 7$ .

100 mL (Eq 2): pure $N a_{2} H P O_{4}$ amphiprotic. $pH = \frac{1}{2} (7 + 11) = 9$ .

125 mL: half of stage 3 buffer $H P O_{4}^{2 -}$ / $P O_{4}^{3 -}$ . $pH = p K_{a 3} = 11$ .

150 mL (Eq 3): pure $N a_{3} P O_{4}$ . $pH = \frac{1}{2} (p K_{w} + p K_{a 3} + lo g C) \approx \frac{1}{2} (14 + 11 - 2.4) = 11.3$ .

Solved Example 8

To 1 L of a buffer containing 0.10 mol each of

N H_{3}

and

N H_{4} C l

, add (a) 0.02 mol HCl, (b) 0.02 mol NaOH. Find the change in pH.

p K_{b} (N H_{3}) = 4.74

Solution:

Initial pH: $pOH = p K_{b} + lo g (0.10/0.10) = 4.74$ , so $pH = 9.26$ .

(a) Add 0.02 mol HCl: converts 0.02 mol $N H_{3}$ into 0.02 mol $N H_{4}^{+}$ . New ratio: salt/base = 0.12/0.08. $pOH = 4.74 + lo g (0.12/0.08) = 4.74 + 0.176 = 4.92$ ; $pH = 9.08$ . Change: $- 0.18$ .

(b) Add 0.02 mol NaOH: converts 0.02 mol $N H_{4}^{+}$ into 0.02 mol $N H_{3}$ . New ratio: salt/base = 0.08/0.12. $pOH = 4.74 + lo g (0.08/0.12) = 4.74 - 0.176 = 4.56$ ; $pH = 9.44$ . Change: $+ 0.18$ .

Either way the pH shift is small - the mark of a good buffer.

Common Mistakes to Avoid

Watch out

Applying the Henderson-Hasselbalch equation when one component is nearly depleted. It only holds when both acid and salt are in appreciable amounts (roughly within a factor of 10 of each other).
Confusing $[salt]$ with $[conjugate acid]$ in the basic buffer form. In $pOH = p K_{b} + lo g ([salt] / [base])$ , the salt is the conjugate acid of the base.
Using $p K_{a}$ for the weak acid in a basic buffer or $p K_{b}$ for the weak base in an acidic buffer. Match the constant to the species in the equation.
Picking phenolphthalein for a strong-acid + weak-base titration, or methyl orange for a weak-acid + strong-base titration. The indicator's range must overlap the equivalence pH.
Treating pH at equivalence as always 7. Only true for strong-strong titrations; weak-strong equivalence points are shifted.
Forgetting that a catalyst does not shift equilibrium; this applies to buffer equilibrium too.
Ignoring salt hydrolysis at the equivalence point of a weak-strong titration. The equivalence solution is a salt of a weak acid (or weak base), so its pH is not neutral.
Assuming buffer capacity is a fixed property independent of composition. Capacity depends on both the total concentration and the ratio $[acid] / [salt]$ ; it peaks at $pH = p K_{a}$ .

Frequently Asked Questions

Q1. What exactly does a buffer solution do?

A buffer resists changes in pH when small amounts of strong acid or strong base are added to it. It contains a weak conjugate acid-base pair: the acid mops up any added $O H^{-}$ and the base mops up any added $H^{+}$ , so the free $H^{+}$ (and hence pH) barely changes.

Q2. What is the Henderson-Hasselbalch equation?

It is $pH = p K_{a} + lo g ([salt] / [acid])$ for an acidic buffer, or $pOH = p K_{b} + lo g ([salt] / [base])$ for a basic buffer. Derived from the weak acid dissociation equilibrium after using the common-ion approximation.

Q3. Why is buffer capacity maximum at pH = pKa?

Buffer capacity depends on the product $[acid] [salt] / ([acid] + [salt])$ . For a fixed total concentration, this product is maximum when the two are equal, which by the Henderson-Hasselbalch equation corresponds to $pH = p K_{a}$ .

Q4. What is the useful pH range of a buffer?

About $p K_{a} \pm 1$ . Within this range the salt-to-acid ratio stays between 0.1 and 10, so both components remain in usable amounts. Outside this window one component is nearly depleted and the buffer no longer resists pH change.

Q5. Why is the equivalence point of a weak-acid titration not at pH 7?

At equivalence, all the weak acid has been converted into its conjugate base (a salt). This salt is basic because the conjugate base hydrolyses in water to regenerate small amounts of $O H^{-}$ . The resulting pH is above 7. Similarly, a strong-acid + weak-base titration gives an acidic equivalence pH.

Q6. How do I choose an indicator for a titration?

Pick an indicator whose colour-change range overlaps the sharp vertical jump in the titration curve near the equivalence point. Phenolphthalein (8.3-10) works for strong-strong and weak-acid + strong-base titrations. Methyl orange (3.1-4.4) works for strong-strong and strong-acid + weak-base.

Q7. Why is there no sharp equivalence point in a weak-acid + weak-base titration?

Both partners hydrolyse, and the pH change near equivalence is gradual rather than a sudden jump. No indicator gives a clean colour change, so this type of titration is generally not done directly - it is unreliable for endpoint detection.

Q8. How does the blood bicarbonate buffer maintain pH?

Blood plasma contains $H_{2} C O_{3}$ (from dissolved $C O_{2}$ ) and $H C O_{3}^{-}$ in a ratio of about 20:1, giving pH 7.4 via Henderson-Hasselbalch with $p K_{a} \approx 6.1$ . Any excess $H^{+}$ (from metabolic acids) is neutralised by bicarbonate; excess $O H^{-}$ is neutralised by $H_{2} C O_{3}$ . Respiration adjusts $C O_{2}$ (and hence $H_{2} C O_{3}$ ), and kidneys adjust $H C O_{3}^{-}$ over longer time scales.

Q9. What is the difference between an acidic buffer, a basic buffer, and a salt buffer?

An acidic buffer is a weak acid + salt with strong base (pH below 7). A basic buffer is a weak base + salt with strong acid (pH above 7). A salt buffer is a solution of a salt of a weak acid and a weak base (like $C H_{3} C O O N H_{4}$ ), which is inherently buffered because both ions can absorb small perturbations.

Previous year questions on Buffer Solutions

9 questions from past papers, each with a step-by-step solution.

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