Buffer Solutions

8 MCQs2 revision cards9-step worked example

Source: NCERT EquilibriumPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

A buffer solution resists pH change when small amounts of acid or base are added. This is the property NEET tests — not equilibrium shifts, not Ksp, not Kp/Kc conversion, but the mechanism by which a weak acid–conjugate base (or weak base–conjugate acid) pair absorbs added H⁺ or OH⁻ without significant pH change.

Two types of buffer:

Acidic buffer — weak acid + its salt with a strong base (e.g., CH₃COOH + CH₃COONa). pH < 7.

Basic buffer — weak base + its salt with a strong acid (e.g., NH₄OH + NH₄Cl). pH > 7.

The Henderson-Hasselbalch equation (NCERT Class 11 Chemistry Chapter 7, page 42):

Acidic buffer: pH = pKₐ + log₁₀([salt]/[acid])
Basic buffer: pOH = pK_b + log₁₀([salt]/[base])

When [salt] = [acid], the log term vanishes and pH = pKₐ. This is the maximum buffer capacity point — a common NEET conceptual question.

How the buffer works: Add a small amount of strong acid → extra H⁺ reacts with the conjugate base (salt component), converting it to the weak acid. The ratio [salt]/[acid] shifts only slightly, so pH barely changes. Add a small amount of strong base → extra OH⁻ reacts with the weak acid, converting it to the conjugate base. Again, the ratio shifts only slightly.

Buffer capacity is finite. Once the weak acid or conjugate base component is consumed, the buffer fails and pH changes sharply. NEET may ask which buffer has higher capacity — it is the one with higher total concentrations of the acid-salt pair.

Watch-out: A solution of a strong acid + its salt (e.g., HCl + NaCl) is NOT a buffer. The weak-acid/weak-base component is essential. Identifying what qualifies as a buffer is a frequent recall question.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following pairs constitutes an acidic buffer solution?

MCQ 2Easy RecallPractice

Which of the following is a basic buffer?

MCQ 3Easy RecallPractice

For an acidic buffer, when [salt] = [acid], what is the pH of the solution?

MCQ 4Direct ApplicationPractice

The pKₐ of a weak acid HA is 4.75. A buffer is prepared with 0.10 M HA and 0.10 M NaA. What is the pH of this buffer?

MCQ 5Direct ApplicationPractice

A buffer is made from 0.20 M CH₃COOH (pKₐ = 4.76) and 0.02 M CH₃COONa. The pH of this buffer is:

MCQ 6Direct ApplicationPractice

When a small amount of HCl is added to a CH₃COOH / CH₃COONa buffer, which species neutralises the added H⁺?

MCQ 7Concept TrapPractice

Two acidic buffers are prepared:

MCQ 8CalculationPractice

The pKₐ of a weak acid HA is 5.00. A buffer is prepared with 0.10 M HA and 1.0 M NaA. The pH of this buffer is:

Quick recall before you leave

Worked Example

1
Given
- Weak base: NH₄OH, concentration = 0.20 M - Salt (conjugate acid): NH₄Cl, concentration = 0.20 M - pK_b = 4.74 - Temperature = 25 °C (so pH + pOH = 14)
2
Required
pH of the basic buffer.
3
Concept
This is a basic buffer (weak base + salt with strong acid). Use the Henderson-Hasselbalch equation for a basic buffer: pOH = pK_b + log₁₀([salt]/[base]). Then convert: pH = 14 − pOH.
4
Formula
pOH = pK_b + log₁₀([salt]/[base]) pH = 14 − pOH
5
Substitution
pOH = 4.74 + log₁₀(0.20/0.20)
6
Calculation
[salt]/[base] = 0.20/0.20 = 1 log₁₀(1) = 0 pOH = 4.74 + 0 = 4.74 Note on exact values: the ratio 1 and its logarithm 0 are exact mathematical values and do not affect significant-figure counting. pH = 14.00 − 4.74 = 9.26
7
Final answer
pH = 9.26
8
Common trap
The common error is forgetting to convert pOH to pH. A student who stops at pOH = 4.74 and reports it as the pH would give an acidic value for what is clearly a basic buffer — a quick sanity check (basic buffer → pH > 7) catches this.
9
Similar NEET-style question
Calculate the pH of a buffer made from 0.50 M NH₄OH (pK_b = 4.74) and 0.05 M NH₄Cl at 25 °C. (Answer: pOH = 4.74 + log(0.05/0.50) = 4.74 − 1 = 3.74; pH = 14 − 3.74 = 10.26.) ---

Before solving, remember these

Formula

Henderson-Hasselbalch (buffers)

pH = pKa + log([salt]/[acid]) for acidic buffer (weak acid + conjugate base). Buffer resists pH change on small acid/base addition.

-- NCERT Class 11 Chemistry, Ch. 6, p. 42

Formulas

Henderson-Hasselbalch (buffer)

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

pH of acidic buffer in terms of conjugate base/acid concentrations. For basic buffer: pOH = pKb + log10([salt]/[base]).

Symbol	Quantity	SI Unit
pKa	-log Ka	-
[salt]	conjugate base conc	mol/L
[acid]	weak acid conc	mol/L

Valid when

Buffer (weak acid + conjugate base)
Concentrations not too dilute
Approximate (assumes negligible dissociation)

Ka, Kb, Kw relationship

K_{a} K_{b} = K_{w}

Stronger acid → weaker conjugate base, and vice versa. pKa + pKb = 14.

Symbol	Quantity	SI Unit
Ka	acid dissociation	-
Kb	base dissociation	-
Kw	water 10^-14	-

Valid when

Conjugate acid-base pair
25°C

Equilibrium constant K_p and K_c

K_{p} = K_{c} (R T)^{Δ n}

Convert between pressure-based and concentration-based equilibrium constants. T in K; R = 0.0821 L atm/mol/K (when K_p in atm).

Symbol	Quantity	SI Unit
K_p	pressure constant	-
K_c	concentration constant	-
Δn	mole change	-

Valid when

Gas-phase equilibrium
Same temperature

Solubility product

K_{sp} = [M^{+}]^{a} [X^{-}]^{b}

Equilibrium constant for sparingly soluble salt. Q < K_sp: dissolves; Q > K_sp: precipitates.

Symbol	Quantity	SI Unit
K_sp	solubility product	-

Valid when

Sparingly soluble salt
Saturated solution

pH and pOH

pH = - lo g [H^{+}]

Logarithmic acidity scale. Pure water at 25°C: pH = 7 = pOH.

Symbol	Quantity	SI Unit
[H+]	hydrogen ion conc	mol/L
[OH-]	hydroxide conc	mol/L

Valid when

Aqueous solution
Use Kw = 10^-14 at 25°C

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Catalyst does NOT shift equilibrium

Category: Overthinking

Student claims catalyst shifts equilibrium toward products. Catalyst speeds up forward AND reverse equally; equilibrium position unchanged.

When it triggers

Question lists catalyst addition among options for shifting equilibrium.

How to avoid

Catalyst lowers activation energy of BOTH forward and reverse reactions equally. Time to reach equilibrium decreases; equilibrium position is unchanged.

Trap

Δn sign in K_p ↔ K_c conversion

Category: Sign Convention

Δn = (mol gas product) - (mol gas reactant). Sign matters; K_p = K_c (RT)^Δn.

When it triggers

Convert K_p ↔ K_c for gas-phase reaction.

How to avoid

Count moles of gas only (ignore solids/liquids). Δn = product - reactant. If Δn = 0, K_p = K_c. If Δn = +1, K_p = K_c × RT.

Trap

K_sp stoichiometric powers

Category: Similar Terms

For salt M_aX_b: K_sp = [M⁺]^a [X⁻]^b. Student uses [M⁺][X⁻] regardless of stoichiometry.

When it triggers

K_sp problem with non-1:1 salt (e.g. CaF₂, Mg(OH)₂, Ag₂CrO₄).

How to avoid

Write dissolution: M_aX_b → aM + bX. Then K_sp = [M]^a · [X]^b. For CaF₂ ↔ Ca + 2F: K_sp = s · (2s)² = 4s³.

Mistake

Believes catalyst shifts equilibrium toward products.

Root cause: concept gap

Correction

Catalyst speeds up forward AND reverse equally. Equilibrium position is unchanged. Time to reach equilibrium decreases.

Mistake

Counts solid/liquid moles in Δn for K_p ↔ K_c conversion.

Root cause: concept gap

Correction

Δn = (moles GAS product) - (moles GAS reactant). Solids and liquids excluded.

Mistake

Uses K_sp = [M+][X-] regardless of salt stoichiometry.

Root cause: formula misuse

Correction

For M_aX_b: K_sp = [M]^a · [X]^b. CaF₂: K_sp = s · (2s)² = 4s³.

Mistake

Predicts wrong direction of shift on pressure change.

Root cause: concept gap

Correction

Increase P → shift toward FEWER moles of gas. Decrease P → shift toward MORE moles. Reactions with same number of moles of gas: P doesn't shift equilibrium.

Mistake

Treats weak acid as fully dissociated when computing pH.

Root cause: concept gap

Correction

Weak acid: only fraction ionises. [H+] ≈ √(Ka · C). For 0.1 M acetic acid (Ka ~10⁻⁵): [H+] ~10⁻³ (pH ~3), not 1 (pH=1).

Past Year Questions

15 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Among the following, choose the ones with equal number of atoms. A. 212 g of Na CO (s) [molar mass = 106 g] 2 3 B. 248 g of Na O(s) [molar mass = 62 g] 2 C. 240 g of NaOH(s) [molar mass = 40 g] D. 12 g of H (g) [molar mass = 2 g] 2 E. 220 g of CO (g) [molar mass = 44 g] 2 Choose the correct answer from the options given below :

1B, D, and E only

2A, B, and C only

3A, B, and D only

4B, C, and D only

NTA Answer: Option 3(final)

NEET 2025

Sugar ‘X’ A. is found in honey B. is a keto sugar C. exists in  and  – anomeric forms. D. Is laevorotatory. ‘X’ is :

1Sucrose

2D-Glucose

3D-Fructose

4Maltose

NTA Answer: Option 3(final)

NEET 2025

For the reaction A(g)  2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given : R = 0.0831 L atm mol–1 K–1] K for the reaction at 1000 K is P

10.021

283.1

32.077 × 105

40.033

NTA Answer: Option 4(final)

NEET 2024Revised key

C [A] = [B] = [C] = 2 × 10–3 M. Then, which of the following is correct?

1Reaction is at equilibrium.

2Reaction has a tendency to go in forward direction.

3Reaction has a tendency to go in backward direction.

4Reaction has gone to completion in forward direction.

NTA Answer: Option 3(revised_final)

NEET 2023

The relation between n , (n = the number of permissible values of magnetic quantum number (m)) for a m m given value of azimuthal quantum number (l), is

1l = 2n + 1

2n = 2l2 + 1 m m n –1

3n = l + 2

4l  m m 2

NTA Answer: Option 4(final)

NEET 2023

Amongst the given options which of the following molecules/ ion acts as a Lewis acid?

1H O

2BF 2 3

3OH–

4NH 3

NTA Answer: Option 2(final)

NEET 2023

Which complex compound is most stable?

1CoNH  NO    3 3 3 3

2CoCl en NO  2 2 3

3CoNH   SO   3 6 2 4 3

4CoNH  H OBrNO   3 4 2  3 2

NTA Answer: Option 2(final)

NEET 2022

Choose the correct statement:

1Both diamond and graphite are used as dry lubricants.

2Diamond and graphite have two dimensional network.

3Diamond is covalent and graphite is ionic.

4Diamond is sp3 hybridised and graphite is sp2 hybridized.

NTA Answer: Option 4(final)

NEET 2022

Which one is not correct mathematical equation for Dalton’s Law of partial pressure? Here p = total pressure of gaseous mixture

1p =p, where  = mole fraction of ith gas in gaseous mixture i i i i p = pressure of ith gas in pure state i

2p = p + p + p 1 2 3 RT RT RT

3p=n +n +n 1 V 2 V 3 V

4p = p, where p = partial pressure of ith gas i i i  = mole fraction of ith gas in gaseous mixture i

NTA Answer: Option 1(final)

NEET 2022

A 10.0 L flask contains 64 g of oxygen at 27ºC. (Assume O 2 gas is behaving ideally). The pressure inside the flask in bar is (Given R = 0.0831 L bar K–1 mol–1)

14.9

22.5

3498.6

449.8

NTA Answer: Option 1(final)

NEET 2022

- 19 - 3 O 2 ( g ) 2 O 3 ( g ) for the above reaction at 298 K, K is found to be 3.0 × 10–59. If the concentration of O at equilibrium is C 2 0.040 M then concentration of O in M is 3

11.2 × 1021

24.38 × 10–32

31.9 × 10–63

42.4 × 1031

NTA Answer: Option 2(final)

NEET 2022

Find the emf of the cell in which the following reaction takes place at 298 K - 20 - N (i s ) + 2 A g + ( 0 .0 0 1 M ) → N 2i + ( 0 .0 0 1 M ) + 2 A g ( s )  G iv e n th a t E c e ll = 1 0 .5 V , 2 .3 0 3 F R T = 0 .0 5 9 a t 2 9 8 K 

11.05 V

21.0385 V

31.385 V

40.9615 V Answer (NA)

NTA Answer: Option 5(final)

NEET 2021

BF is planar and electron deficient compound. 3 Hybridization and number of electrons around the central atom, respectively are : sp2 sp3

1and 8

2and 4 sp3 sp2

3and 6

4and 6

NTA Answer: Option 4(final)

NEET 2021

Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is :

1Beryllium chloride

2Calcium chloride

3Strontium chloride

4Magnesium chloride

NTA Answer: Option 1(final)

NEET 2021

Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O and 2 g H confined in 2 2 a total volume of one litre at 0°C is : mol–1K–1, [Given R = 0.082 L atm T = 273 K]

126.02

22.518

32.602

425.18

NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Convert between K_p and K_c using K_p = K_c (RT)^Δn. Identify Δn carefully.

Multi StepMedium

Common distractors

wrong delta n sign

Counts moles incorrectly

Predict equilibrium shift on changing concentration, P, T, or adding catalyst.

InterpretationMedium

Common distractors

treats catalyst as shifting

Believes catalyst shifts equilibrium

Compute pH of weak acid/base solution using Ka or Kb. Approximation: [H+] ≈ √(Ka·C).

Multi StepMedium

Common distractors

ignores weak vs strong

Treats weak acid as fully dissociated

Compute solubility from K_sp or vice versa. Apply common-ion effect when relevant.

Multi StepMedium

Common distractors

forgets stoichiometric coefficient power

Uses [M+]^1 [X-]^1 for M(X)_2 salt

Sources

NCERT refs: Class 11 Chemistry Chapter 7, p.42

Test yourself on this topic with real past-paper questions:

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Buffer Solutions

Lesson

Practice MCQs

Quick recall before you leave

Henderson-Hasselbalch equation (acidic buffer)

Henderson-Hasselbalch equation (basic buffer)

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Correction

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources