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Q.Define the buffer solution.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2017Subjective· 1mImportance★★★★★
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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 HAHA and its conjugate base A−A^-, the pH is given by:

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

Where:

  • pKa=−log⁡10Ka\text{p}K_a = -\log_{10} K_a (a measure of the weak acid's strength — lower pKa = stronger acid)
  • [A−][A^-] = concentration of the conjugate base
  • [HA][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][A^-] = [HA], the ratio is 1, log⁡(1)=0\log(1) = 0, so pH=pKa\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⇌H++A−HA \rightleftharpoons H^+ + A^-

The acid dissociation constant is:

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

Take negative logs of both sides:

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

Which gives:

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

Rearrange:

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

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

Watch out

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

Example: Making an Acetate Buffer

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

pH=4.76+log⁡0.10.1=4.76+log⁡1=4.76\text{pH} = 4.76 + \log \frac{0.1}{0.1} = 4.76 + \log 1 = 4.76 …

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