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

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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?

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

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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. ✓ …

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

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