Which of the following will produce a buffer solution when mixed in equal volumes? (i) 0.1 mol dm^-3 NH4OH and

NCERT Exemplar · Q11

Q. Which of the following will produce a buffer solution when mixed in equal volumes?

(i) 0.1 mol dm^-3 NH4OH and 0.1 mol dm^-3 HCl

(ii) 0.05 mol dm^-3 NH4OH and 0.1 mol dm^-3 HCl

(iii) 0.1 mol dm^-3 NH4OH and 0.05 mol dm^-3 HCl

(iv) 0.1 mol dm^-3 CH4COONa and 0.1 mol dm^-3 NaOH

Andhra Pradesh BieapMCQ· 1mImportance★★★★★est

72% · 112/155 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Buffer Solution pH

What is a Buffer Solution?

Imagine you're making lemonade. If you add a few drops of lemon juice to a glass of water, the pH drops sharply — it becomes very acidic. But if you add the same few drops to a glass of already acidic lemonade, the pH barely changes. Why? Because lemonade contains a buffer — a mixture that resists pH change when small amounts of acid or base are added.

A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid). It "soaks up" added H⁺ or OH⁻ ions without letting the pH swing wildly.

Note

The key is that both components must be present in significant amounts. A weak acid alone won't buffer — you need its conjugate base partner too.

The Intuition: A Chemical Sponge

Think of a buffer as a two-way sponge:

If you add acid (H⁺): The conjugate base in the buffer grabs the extra H⁺, turning into the weak acid. The H⁺ is "absorbed" — pH barely drops.
If you add base (OH⁻): The weak acid donates an H⁺ to neutralise the OH⁻, turning into the conjugate base. The OH⁻ is "absorbed" — pH barely rises.

The buffer works best when the amounts of weak acid and conjugate base are roughly equal. That's when the sponge is most "spongy" — it can absorb shocks in either direction.

The Precise Statement: The Henderson–Hasselbalch Equation

For a buffer made from a weak acid $HA$ and its conjugate base $A^-$ , the pH is given by:

\text{pH} = \text{p}K_a + \log_{10} \left( \frac{[A^-]}{[HA]} \right)

Where:

$\text{p}K_a = -\log_{10} K_a$ (a measure of the weak acid's strength — lower pKa = stronger acid)
$[A^-]$ = concentration of the conjugate base
$[HA]$ = concentration of the weak acid

This equation tells you exactly how the pH depends on the ratio of base to acid, not their absolute amounts.

Tip

When $[A^-] = [HA]$ , the ratio is 1, $\log(1) = 0$ , so $\text{pH} = \text{p}K_a$ . This is the buffer's optimal pH — it resists change most strongly here.

Why This Works: A Quick Derivation

Start from the weak acid equilibrium:

HA \rightleftharpoons H^+ + A^-

The acid dissociation constant is:

K_a = \frac{[H^+][A^-]}{[HA]}

Take negative logs of both sides:

-\log K_a = -\log [H^+] - \log \frac{[A^-]}{[HA]}

Which gives:

\text{p}K_a = \text{pH} - \log \frac{[A^-]}{[HA]}

Rearrange:

\text{pH} = \text{p}K_a + \log \frac{[A^-]}{[HA]}

That's it. The derivation is just algebra on the definition of $K_a$ .

Watch out

The Henderson–Hasselbalch equation assumes that the concentrations $[HA]$ and $[A^-]$ are the initial concentrations you mixed. It works well when both are much larger than $[H^+]$ or $[OH^-]$ from dissociation — which is true for a properly made buffer.

Example: Making an Acetate Buffer

You mix 0.1 M acetic acid ( $\text{p}K_a = 4.76$ ) with 0.1 M sodium acetate. What's the pH?

$\text{pH} = 4.76 + \log \frac{0.1}{0.1} = 4.76 + \log 1 = 4.76$ …

Start your 14-day free trial to unlock the full solution →

A buffer requires a weak acid/base and its conjugate in comparable amounts. Mixing 0.1 M NH₄OH with 0.05 M HCl (equal volumes) leaves half the NH₄OH unreacted and produces an equal amount of NH₄Cl — a perfect buffer. The correct option is (iii).

A buffer solution resists pH change. The classic recipe is a weak acid and its salt (conjugate base) or a weak base and its salt (conjugate acid), both in roughly equal concentrations. Here we have NH₄OH (ammonium hydroxide, a weak base) and HCl (a strong acid). When they react, NH₄OH + HCl → NH₄Cl + H₂O. The product NH₄Cl is the salt of the weak base — it provides the conjugate acid NH₄⁺. So a buffer forms if, after reaction, we have significant amounts of both NH₄OH (weak base) and NH₄⁺ (its conjugate acid) left in solution.

Let’s check each option. We mix equal volumes, so the number of moles of each reactant is simply its concentration multiplied by the same volume V. We can compare moles directly using the given concentrations.

Option (i): 0.1 M NH₄OH and 0.1 M HCl

Moles of NH₄OH = 0.1V, moles of HCl = 0.1V. They react 1:1, so both are completely consumed. Only NH₄Cl remains — that’s just a salt solution, not a buffer. No weak base left. ✗
Option (ii): 0.05 M NH₄OH and 0.1 M HCl

Moles of NH₄OH = 0.05V, moles of HCl = 0.1V. HCl is in excess. All NH₄OH is used up, and leftover HCl (0.05V moles) makes the solution strongly acidic. No buffer. ✗
Option (iii): 0.1 M NH₄OH and 0.05 M HCl

Moles of NH₄OH = 0.1V, moles of HCl = 0.05V. HCl is the limiting reagent. It reacts completely, consuming 0.05V moles of NH₄OH and producing 0.05V moles of NH₄Cl. That leaves 0.05V moles of NH₄OH unreacted. So after mixing, we have 0.05V moles of NH₄OH (weak base) and 0.05V moles of NH₄⁺ from NH₄Cl (conjugate acid) — equal amounts in the same total volume. That’s a textbook buffer solution. ✓ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$Q. Observe the following data given below. Which set of buffer solution has highest pH? Buffer solution Volume (in mL) of 1M solution of weak acid Volume (in mL) of 0.1 M sodium salt of weak acid I 4.0 4.0 II 4.0 40.0 III 40.0 4.0 IV 0.1 10.0 (A) I (B) II (C) III (D) IV$
›Reveal solutionSolution
Converting each buffer's volumes+concentrations into actual moles of acid and salt shows buffer IV has the largest salt-to-acid mole ratio, and by Henderson–Hasselbalch that means the highest pH.

Concept and Intuition

A buffer's pH is governed by the Henderson–Hasselbalch equation, $\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{n_{\text{salt}}}{n_{\text{acid}}}$ (using moles, since both components share the same final solution volume once mixed). Because all four buffers use the same weak acid (same p $K_a$ ), comparing pH reduces entirely to comparing the salt/acid mole ratio — and one must be careful that the acid solution is 1 M while the salt solution is only 0.1 M, so volumes alone are not directly comparable; the tenfold concentration difference must be folded in.

Step-by-Step Solution

Moles of acid $= V_{\text{acid}}(\text{mL})\times1\ \text{M}/1000$ ; moles of salt $=V_{\text{salt}}(\text{mL})\times0.1\ \text{M}/1000$ .

I: acid $=4.0\times1/1000=0.0040$ mol; salt $=4.0\times0.1/1000=0.00040$ mol. Ratio $=0.00040/0.0040=0.1$ .

II: acid $=4.0\times1/1000=0.0040$ mol; salt $=40.0\times0.1/1000=0.0040$ mol. Ratio $=0.0040/0.0040=1$ .

III: acid $=40.0\times1/1000=0.040$ mol; salt $=4.0\times0.1/1000=0.00040$ mol. Ratio $=0.00040/0.040=0.01$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\mathrm{HCN}$
›Reveal solutionSolution

For an $\text{HCN}/\text{NaCN}$ buffer, $\dfrac{[\text{HCN}]}{[\text{CN}^-]}=10^{\,p K_a-pH}$ . The printed options are inconsistent with the stated data (the clean result is $1\times10^{-1}$ mol), so the official exam key (C) $9\times10^{-3}$ is reported.

Using Henderson–Hasselbalch with $p K_a=-\log(10^{-10})=10$ and $pH=9$ :
$pH=pK_a+\log\frac{[\text{CN}^-]}{[\text{HCN}]}\;\Longrightarrow\;\log\frac{[\text{CN}^-]}{[\text{HCN}]}=9-10=-1,$
so $\dfrac{[\text{CN}^-]}{[\text{HCN}]}=10^{-1}$ , i.e. $[\text{HCN}]=10\,[\text{CN}^-]=10\times0.01=0.1\ \text{mol}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$K_a$
›Reveal solutionSolution
Partial neutralisation of formic acid by NaOH creates an acid/conjugate-base buffer; the Henderson-Hasselbalch equation gives $pH=4.74$ .

Concept and Intuition

When a weak acid is only partially neutralised by a strong base, the resulting solution contains both the leftover weak acid and its conjugate base (salt) — a buffer. Its pH is found from the Henderson-Hasselbalch equation, using the mole ratio of salt to acid (volume cancels out since both are in the same final solution).

Step-by-Step Solution

Moles of $NaOH = 0.5\,L \times 0.5\,M = 0.25$ mol.

Moles of $HCOOH$ (formic acid) $= 0.5\,L\times0.55\,M = 0.275$ mol.

$NaOH$ is the limiting reagent — it fully reacts: $HCOOH + NaOH \rightarrow HCOONa + H_2O$ , consuming $0.25$ mol of each and producing $0.25$ mol $HCOONa$ .

Leftover $HCOOH = 0.275 - 0.25 = 0.025$ mol — this plus the $0.25$ mol $HCOONa$ forms an acidic buffer in the combined $1\,L$ solution.

$pK_a = -\log(K_a) = -\log(1.8\times10^{-4}) = 4 - \log(1.8) = 4 - 0.26 = 3.74$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\mathrm{L^{-1}}$
›Reveal solutionSolution
Mixing a weak acid with less than the stoichiometric amount of strong base creates an acid/conjugate-base buffer; the Henderson–Hasselbalch equation converts the given pH directly into the leftover-acid-to-salt ratio.

Concept and Intuition

Adding NaOH to HCOOH converts some HCOOH into its conjugate base HCOO⁻ (as HCOONa). If there's more acid than base, the base is fully consumed, leaving a mixture of unreacted HCOOH and newly formed HCOONa — a buffer whose pH is governed by the Henderson–Hasselbalch equation.

Step-by-Step Solution

Moles of NaOH $=0.5\text{L}\times0.5\text{M}=0.25$ mol.

Moles of HCOOH initially $=0.5\text{L}\times x\text{M}=0.5x$ mol.

NaOH (limiting) converts 0.25 mol HCOOH → 0.25 mol HCOONa, leaving $(0.5x-0.25)$ mol HCOOH. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{NH_4OH}$
›Reveal solutionSolution
This tests the Henderson–Hasselbalch equation for a basic buffer; the answer is $pK_b =$ 4.8, option (D).

Concept and Intuition

For a basic buffer of a weak base and its salt (conjugate acid), $pOH = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]}$ . We're given the mixed volumes/concentrations and the resulting pH, so we work backward through pOH to find $pK_b$ .

Step-by-Step Solution

Moles of $\mathrm{NH_4OH}$ = $0.030\ \text{L} \times 0.2\ \text{M} = 0.006$ mol.

Moles of $\mathrm{NH_4Cl}$ = $0.030\ \text{L} \times 2\ \text{M} = 0.06$ mol.

Ratio $\dfrac{[\text{salt}]}{[\text{base}]} = \dfrac{0.06}{0.006} = 10$ (volumes are equal and cancel, so mole ratio = concentration ratio). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$pK_a$
›Reveal solutionSolution
Partial neutralization of acetic acid by NaOH creates an acetic acid/acetate buffer; the Henderson–Hasselbalch equation gives $\mathrm{pH}=4.2$ .

Concept and Intuition

When a weak acid is partially neutralized by a strong base, and some weak acid remains along with the salt (conjugate base) formed, the resulting solution is a buffer. Its pH is found using the Henderson–Hasselbalch equation, $\mathrm{pH}=pK_a+\log\dfrac{[\text{conjugate base}]}{[\text{weak acid}]}$ , using the moles (or concentrations, since they share the same total volume) of each species remaining after the acid-base reaction goes to completion.

Step-by-Step Solution

Moles of $\mathrm{CH_3COOH}$ initially: $0.100\ \text{L} \times 0.5\ \text{M} = 0.05\ \text{mol}$

Moles of NaOH: $0.050\ \text{L} \times 0.2\ \text{M} = 0.01\ \text{mol}$

Reaction: $\mathrm{CH_3COOH + NaOH \to CH_3COONa + H_2O}$ . NaOH (0.01 mol) is the limiting reagent, consuming 0.01 mol of acetic acid and producing 0.01 mol of acetate (CH $_3$ COO $^-$ ).

Remaining acetic acid: $0.05-0.01=0.04$ mol. Acetate formed: $0.01$ mol. This is a buffer (both species present in the same total volume, so the mole ratio equals the concentration ratio). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\mathrm{HCN\,/\,CN^-}$
›Reveal solutionSolution
Human blood pH is buffered by the carbonic acid–bicarbonate system, $\mathrm{H_2CO_3/HCO_3^-}$ .

Concept and Intuition

A buffer resists pH change by containing both a weak acid and its conjugate base in appreciable amounts, so that added $\mathrm{H^+}$ or $\mathrm{OH^-}$ is absorbed by one of the pair. In blood, dissolved $\mathrm{CO_2}$ forms carbonic acid ( $\mathrm{H_2CO_3}$ ), which is in equilibrium with bicarbonate ion ( $\mathrm{HCO_3^-}$ ); the lungs (removing $\mathrm{CO_2}$ ) and kidneys (regulating $\mathrm{HCO_3^-}$ excretion) work with this equilibrium to keep blood pH tightly in the 7.26–7.42 range (physiological pH ≈ 7.4, with the exam's range corresponding to this buffer's typical stated span).

Step-by-Step Solution

Identify the four candidate buffer systems given: HCN/CN⁻, H₂CO₃/HCO₃⁻, CH₃COOH/CH₃COO⁻, H₃PO₄/H₂PO₄⁻.

Recall the standard NCERT/physiological chemistry fact: the primary buffer of blood plasma is the carbonic acid–bicarbonate system. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$CH_3COOH$
›Reveal solutionSolution
Maximum buffer capacity occurs at half-neutralization (pH = pKa); for 8 mmol of acetic acid, that needs 4 mmol NaOH = 20 mL.

Concept and Intuition

A buffer resists pH change best when the concentrations of the weak acid and its conjugate base are equal — this is the point of maximum buffer capacity, corresponding to exactly half the acid being converted to its conjugate base (half-equivalence point, pH = pKa there).

Step-by-Step Solution

Total millimoles of $CH_3COOH$ = $40\ \text{mL} \times 0.2\ M = 8$ mmol.

Maximum buffer capacity requires converting exactly half of this acid to acetate: $8/2 = 4$ mmol of NaOH needed. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$Q. Which of the following will make a basic buffer solution? (A) 100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH (B) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH (C) 50 mL of 0.1 M KOH + 25 mL of 0.1 M CH3COOH (D) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH$
›Reveal solutionSolution
A basic buffer requires a weak base left over along with its salt (conjugate acid) — only option (D), where excess NH4OH remains after partial neutralization by HCl, satisfies this.

Concept and Intuition

A buffer solution resists pH changes because it contains a conjugate acid–base pair in significant, comparable amounts. A basic buffer specifically needs a weak base plus a salt of that weak base (its conjugate acid), such as NH4OH + NH4Cl. Mixing a strong acid with a strong base, or a weak acid with a strong base in exact stoichiometric amounts, just gives a simple salt solution (or excess strong electrolyte) — not a buffer.

Step-by-Step Solution

Compute millimoles in each option using $mmol = mL\times M$ .

(A): CH3COOH $=10$ mmol, NaOH $=10$ mmol — exact neutralization, product is only CH3COONa (a hydrolyzing salt solution, not a buffer).

(B): HCl $=10$ mmol, NaOH $=10$ mmol — exact neutralization to NaCl, a neutral salt, no buffer.

(C): KOH $=5$ mmol, CH3COOH $=2.5$ mmol — KOH is in excess; the leftover strong base dominates pH rather than forming a genuine weak-acid/weak-base buffer pair. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$K_a$
›Reveal solutionSolution
This is an acidic buffer (HA + its salt NaA); Henderson–Hasselbalch gives pH = 5.3.

Concept and Intuition

Mixing a weak acid with its conjugate base (as the sodium salt) creates a buffer solution. The Henderson–Hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\text{conjugate base}]}{[\text{acid}]}$ , lets us find the pH directly from the mole ratio, since both species share the same final (mixed) volume.

Step-by-Step Solution

$\mathrm{p}K_a = -\log(10^{-5}) = 5$ .

Moles of HA $= 0.1\ \mathrm{L} \times 0.1\ \mathrm{mol/L} = 0.01\ \mathrm{mol}$ ; moles of $\mathrm{A^-}$ (from NaA) $= 0.1\ \mathrm{L}\times 0.2\ \mathrm{mol/L} = 0.02\ \mathrm{mol}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$K_a$
›Reveal solutionSolution
NaOH exactly neutralizes the HCl present, leaving the weak acid HA (0.01 M) to establish its own equilibrium, giving pH = 5.7.

Concept and Intuition

When a strong base is added to a mixture of a strong acid and a weak acid, it reacts with the fully-dissociated $H^+$ from the strong acid first (since those protons are freely available), before touching the largely-undissociated weak acid. If the base exactly matches the strong acid's moles, the weak acid is left completely unreacted, simplifying the problem to a standard weak-acid pH calculation.

Step-by-Step Solution

Moles of HCl $= 0.1\ M \times 0.5\ L = 0.05\ mol = 50$ mmol (fully dissociated, gives 50 mmol H⁺).

Moles of HA $= 0.02\ M \times 0.5\ L = 0.01\ mol = 10$ mmol (weak acid, mostly undissociated).

Moles of NaOH $= \dfrac{2\ g}{40\ g/mol} = 0.05\ mol = 50$ mmol.

NaOH neutralizes the strong acid's H⁺ first: $50$ mmol NaOH reacts exactly with $50$ mmol HCl, leaving 0 mmol NaOH and 0 mmol HCl remaining, while HA (10 mmol) is untouched. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$25^0C$
›Reveal solutionSolution
The pH of a salt of a weak acid and weak base doesn't depend on its concentration, so the 2 M solution has the same pH as the 1 M solution: 7.005.

Concept and Intuition

For a salt like ammonium acetate, both the cation ( $NH_4^+$ ) and the anion ( $CH_3COO^-$ ) hydrolyse. The equilibrium $[H^+] = \sqrt{\dfrac{K_w K_a}{K_b}}$ (derived from the hydrolysis equilibria) has no concentration term in it at all — the hydrolysis constant expressions conveniently cancel the salt concentration. So changing the concentration of such a salt does not change its pH (to the approximation used in this standard formula).

Step-by-Step Solution

Standard formula: $pH = 7 + \dfrac12(pK_a - pK_b)$ .

Verify with the given data: $pH = 7 + \dfrac12(4.76 - 4.75) = 7 + 0.005 = 7.005$ — this exactly matches the given 1 M value, confirming the formula (and its concentration-independence) applies here.

Since the formula contains no concentration term, the pH of the 2 M solution is computed the same way and gives the same result. …
Start your 14-day free trial to see the full solution →

Buffer solution	Volume (in mL) of 1M solution of weak acid	Volume (in mL) of 0.1 M sodium salt of weak acid
I	4.0	4.0
II	4.0	40.0
III	40.0	4.0
IV	0.1	10.0
(A) I
(B) II
(C) III
(D) IV

←Previous questionQ10 — Acidity of BF3 can be explained on the basis of which of the following…→Next questionQ12 — In which of the following solvents is silver chloride most soluble? (i…

Q.Which of the following will produce a buffer solution when mixed in equal volumes?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days