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. 203Buffer Solutions
8 MCQs3 revision cards9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 24 Sep 2026
Buffer Solutions, explained for NEET
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 6, page 203):
- 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.
Can you answer these Buffer Solutions 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?
Show answer and why every option is right or wrong
Answer: B. An acidic buffer requires a weak acid and its conjugate base (salt with strong base). CH₃COOH is a weak acid and CH₃COONa provides the conjugate base CH₃COO⁻ (NCERT Class 11 Chemistry Chapter 6, page 203).
Why A is wrong: A: HCl is a strong acid. A strong acid + its salt does not form a buffer — there is no equilibrium reservoir to absorb added H⁺.
Why C is wrong: C: NaOH is a strong base and NaCl is a neutral salt. No weak acid or weak base is present, so no buffer action occurs.
Why D is wrong: D: HNO₃ is a strong acid. Like option A, a strong acid + its salt cannot resist pH change upon acid/base addition.
MCQ 2Easy RecallPractice
Which of the following is a basic buffer?
Show answer and why every option is right or wrong
Answer: B. A basic buffer requires a weak base and its salt with a strong acid. NH₄OH is a weak base and NH₄Cl provides the conjugate acid NH₄⁺ (NCERT Class 11 Chemistry Chapter 6, page 203).
Why A is wrong: A: This is an acidic buffer (weak acid + conjugate base), not a basic buffer.
Why C is wrong: C: HCl is a strong acid and KCl is its salt. Strong acid + salt does not form any buffer.
Why D is wrong: D: This is simply two portions of a strong base mixed together. No weak base–conjugate acid pair exists, so no buffer action.
MCQ 3Easy RecallPractice
For an acidic buffer, when [salt] = [acid], what is the pH of the solution?
Show answer and why every option is right or wrong
Answer: A. From the Henderson-Hasselbalch equation: pH = pKₐ + log([salt]/[acid]). When [salt] = [acid], log(1) = 0, so pH = pKₐ (NCERT Class 11 Chemistry Chapter 6, page 203).
Why B is wrong: B: pH = 7 is for pure water at 25 °C. A buffer's pH at equal concentrations equals pKₐ, which is generally not 7 for a weak acid.
Why C is wrong: C: pK_b applies to the basic-buffer version (pOH = pK_b + log([salt]/[base])). For an acidic buffer the relevant constant is pKₐ.
Why D is wrong: D: 14 − pKₐ equals pK_b of the conjugate base, not the buffer pH.
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?
Show answer and why every option is right or wrong
Answer: C. pH = pKₐ + log([NaA]/[HA]) = 4.75 + log(0.10/0.10) = 4.75 + 0 = 4.75 (NCERT Class 11 Chemistry Chapter 6, page 203).
Why A is wrong: A: pH = 7.00 would imply a neutral solution. The Henderson-Hasselbalch equation gives pH = pKₐ when concentrations are equal, and pKₐ = 4.75 ≠ 7.
Why B is wrong: B: 9.25 = 14 − 4.75 = pK_b of the conjugate base. This would be the pOH answer for a basic-buffer analogue, not the pH of this acidic buffer.
Why D is wrong: D: 3.75 has no basis in the Henderson-Hasselbalch equation for equal concentrations. Likely an arithmetic error subtracting 1 from pKₐ.
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:
Show answer and why every option is right or wrong
Answer: A. pH = pKₐ + log([salt]/[acid]) = 4.76 + log(0.02/0.20) = 4.76 + log(0.1) = 4.76 + (−1) = 3.76 (NCERT Class 11 Chemistry Chapter 6, page 203).
Why B is wrong: B: pH = pKₐ = 4.76 is only true when [salt] = [acid]. Here [salt]/[acid] = 0.1, so log = −1 and pH is lower.
Why C is wrong: C: 5.76 = 4.76 + 1. This would result from log(10) = +1, meaning [salt]/[acid] = 10. Here the ratio is 0.1, giving log = −1, not +1. This is the sign-confusion distractor.
Why D is wrong: D: 2.76 = 4.76 − 2. This would require log(0.01), i.e., [salt]/[acid] = 0.01, not 0.1.
MCQ 6Direct ApplicationPractice
When a small amount of HCl is added to a CH₃COOH / CH₃COONa buffer, which species neutralises the added H⁺?
Show answer and why every option is right or wrong
Answer: C. The conjugate base CH₃COO⁻ (from CH₃COONa) reacts with added H⁺: CH₃COO⁻ + H⁺ → CH₃COOH. This consumes the added acid and converts conjugate base to weak acid, causing only a slight ratio change (NCERT Class 11 Chemistry Chapter 6, page 203).
Why A is wrong: A: CH₃COOH is the weak acid component. It does not neutralise added H⁺ — it would need to accept a proton it already has. CH₃COOH neutralises added OH⁻, not H⁺.
Why B is wrong: B: Na⁺ is a spectator ion from the strong base NaOH. It does not participate in any proton-transfer equilibrium.
Why D is wrong: D: While water has some buffering capacity via autoionisation, in a buffer solution the conjugate base CH₃COO⁻ is present at far higher concentration and is the primary neutraliser.
MCQ 7Concept TrapPractice
Two acidic buffers are prepared:
Buffer I: 1.0 M CH₃COOH + 1.0 M CH₃COONa
Buffer II: 0.01 M CH₃COOH + 0.01 M CH₃COONa
Both use the same weak acid (same pKₐ). Which statement is correct?
Show answer and why every option is right or wrong
Answer: B. Both buffers have [salt]/[acid] = 1, so pH = pKₐ + log(1) = pKₐ for both — same pH. Buffer capacity depends on total concentrations: Buffer I (1.0 M each) can neutralise more added acid or base before the ratio shifts drastically, so it has higher buffer capacity (NCERT Class 11 Chemistry Chapter 6, page 203).
Why A is wrong: A: The Henderson-Hasselbalch equation depends on the RATIO [salt]/[acid], not absolute concentrations. Both ratios equal 1, so both have the same pH.
Why C is wrong: C: Buffer capacity increases with higher total concentrations of the acid-salt pair. Buffer II at 0.01 M has lower capacity than Buffer I at 1.0 M.
Why D is wrong: D: The pH values are identical (both have ratio = 1). Only the buffer capacities differ.
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:
Show answer and why every option is right or wrong
Answer: D. pH = pKₐ + log([salt]/[acid]) = 5.00 + log(1.0/0.10) = 5.00 + log(10) = 5.00 + 1.00 = 6.00 (NCERT Class 11 Chemistry Chapter 6, page 203).
Why A is wrong: A: pH = pKₐ = 5.00 only when [salt] = [acid]. Here [salt] is 10× higher, so the log term adds +1.
Why B is wrong: B: 4.00 = 5.00 − 1. This would result from log(0.1), i.e., [salt]/[acid] = 0.1. The actual ratio is 10, not 0.1 — this is the inversion-of-ratio distractor.
Why C is wrong: C: pH = 7.00 has no basis. It would require log([salt]/[acid]) = 2.00, meaning [salt]/[acid] = 100, not 10.
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Buffer Solutions: quick recall before you leave
How do you solve a Buffer Solutions question? A 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.)
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What to remember before solving Buffer Solutions questions
Which Buffer Solutions formulas do you need for NEET?
Henderson-Hasselbalch (buffer)
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)
More in Equilibrium: 8 exam traps and mistakes · 4 formulas · 4 question patterns from its other lessons.
Buffer Solutions questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys.
NEET 2022
12.57
25.57
33.57
44.57
NTA Answer: Option 2(final)
Sources
NCERT refs: Class 11 Chemistry Chapter 6, p.203
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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