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How buffers work

Covers AS91392

What a buffer is

A buffer is a solution that resists a change in pH when a small amount of acid or base is added, or when it is diluted.
It contains both a weak acid and its conjugate base, in comparable amounts.
Two common ways to make one:
- mix a weak acid with a salt of that acid — ethanoic acid plus sodium ethanoate
- partly neutralise a weak acid with a strong base, so some acid and some conjugate base remain

Why both components are needed

The conjugate base deals with added acid:

$\text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

The weak acid deals with added base:

$\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$

In each case the added ion is converted into something that is already part of the buffer, so it is removed from solution and the pH barely moves.
Both components are essential. A solution of ethanoic acid alone can absorb added base but not added acid, so it is not a buffer.

What actually changes

Adding acid does change the buffer — it converts some $\text{A}^-$ into $\text{HA}$ , so the ratio shifts.
But because $[\text{H}_3\text{O}^+]$ depends on the ratio $[\text{HA}]/[\text{A}^-]$ , and both are present in large amounts, a small addition changes that ratio only slightly.
So the pH changes — just very little.

Buffer capacity

A buffer only works while both components remain in reasonable quantity.
Add enough acid to consume all the conjugate base, and the buffer is exhausted — after that the pH falls sharply.
A buffer with larger amounts of both components has a greater capacity, even at the same pH.
Concentration affects capacity; the ratio affects pH. These are two separate things and questions test the distinction.

Buffers in the body and the lab

Blood is buffered near pH 7.4, mainly by the carbonic acid / hydrogencarbonate pair.
Buffers are used whenever a reaction must be run at a controlled pH — in enzyme work, in electroplating, and in the calibration of pH meters.

Tips and Tricks

Write both equations. A buffer question almost always asks what happens on adding acid and on adding base, and each needs its own equation with the correct species.

Show the added ion being consumed. The point is that $\text{H}_3\text{O}^+$ or $\text{OH}^-$ reacts and is removed, so it does not accumulate. An answer that says "the buffer neutralises it" without showing what reacts with what will not reach Merit.

Use a single arrow in these two equations. The reactions of a strong acid or base with the buffer components go essentially to completion.

Say which component does the work. Added acid is removed by the conjugate base; added base is removed by the weak acid. Getting these the wrong way round is a common slip.

Distinguish capacity from pH. Diluting a buffer barely changes its pH, because the ratio is unchanged — but it reduces its capacity, because there is less of each component to react.

Worked ExampleHow much difference does a buffer make?

A buffer contains $0.100$ mol of ethanoic acid and $0.100$ mol of sodium ethanoate in $1.00$ L of solution. $K_a(\text{CH}_3\text{COOH}) = 1.74 \times 10^{-5}$ .

(a) Calculate the pH of the buffer. (b) Calculate the pH after $0.010$ mol of HCl is added, assuming no volume change. (c) Compare this with the pH change caused by adding $0.010$ mol of HCl to $1.00$ L of pure water.

Part (a) — the initial pH

Rearranging the $K_a$ expression:

$[\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.74 \times 10^{-5} \times \frac{0.100}{0.100}$

The ratio is exactly 1, so

$[\text{H}_3\text{O}^+] = 1.74 \times 10^{-5} \text{ mol L}^{-1}$

$\text{pH} = -\log(1.74 \times 10^{-5}) = 4.76$

Part (b) — after adding acid

Step 1 — Decide what reacts. The added $\text{H}_3\text{O}^+$ is removed by the conjugate base:

$\text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

Step 2 — Adjust the amounts. 0.010 mol of ethanoate is consumed, and 0.010 mol of ethanoic acid is produced:

	Before	Change	After
$\text{CH}_3\text{COOH}$	0.100	+0.010	0.110 mol
$\text{CH}_3\text{COO}^-$	0.100	−0.010	0.090 mol

Step 3 — Recalculate.

$[\text{H}_3\text{O}^+] = 1.74 \times 10^{-5} \times \frac{0.110}{0.090} = 2.127 \times 10^{-5}$

$\text{pH} = -\log(2.127 \times 10^{-5}) = 4.672$

Part (c) — the comparison

Adding 0.010 mol of HCl to 1.00 L of pure water gives $[\text{H}_3\text{O}^+] = 0.010$ mol L⁻¹ directly, since HCl is a strong acid:

$\text{pH} = -\log(0.010) = 2.00$

Pure water starts at pH 7.00, so the pH has fallen by 5.00 units.

	Before	After	Change
buffer	4.76	4.67	0.09
pure water	7.00	2.00	5.00

The same amount of acid caused a pH change more than fifty times smaller in the buffer.

The reason is visible in the working: in water the added $\text{H}_3\text{O}^+$ stays in solution, whereas in the buffer almost all of it is converted to ethanoic acid by the ethanoate ion. What remains is only a small shift in the ratio of the two buffer components — from 1.00 to 1.22 — and since pH depends on the logarithm of that ratio, even that produces only a small change.

How buffers work

Covers AS91392

What a buffer is

A buffer is a solution that resists a change in pH when a small amount of acid or base is added, or when it is diluted.
It contains both a weak acid and its conjugate base, in comparable amounts.
Two common ways to make one:
- mix a weak acid with a salt of that acid — ethanoic acid plus sodium ethanoate
- partly neutralise a weak acid with a strong base, so some acid and some conjugate base remain

Why both components are needed

The conjugate base deals with added acid:

$\text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

The weak acid deals with added base:

$\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$

In each case the added ion is converted into something that is already part of the buffer, so it is removed from solution and the pH barely moves.
Both components are essential. A solution of ethanoic acid alone can absorb added base but not added acid, so it is not a buffer.

What actually changes

Adding acid does change the buffer — it converts some $\text{A}^-$ into $\text{HA}$ , so the ratio shifts.
But because $[\text{H}_3\text{O}^+]$ depends on the ratio $[\text{HA}]/[\text{A}^-]$ , and both are present in large amounts, a small addition changes that ratio only slightly.
So the pH changes — just very little.

Buffer capacity

A buffer only works while both components remain in reasonable quantity.
Add enough acid to consume all the conjugate base, and the buffer is exhausted — after that the pH falls sharply.
A buffer with larger amounts of both components has a greater capacity, even at the same pH.
Concentration affects capacity; the ratio affects pH. These are two separate things and questions test the distinction.

Buffers in the body and the lab

Blood is buffered near pH 7.4, mainly by the carbonic acid / hydrogencarbonate pair.
Buffers are used whenever a reaction must be run at a controlled pH — in enzyme work, in electroplating, and in the calibration of pH meters.

Tips and Tricks

Write both equations. A buffer question almost always asks what happens on adding acid and on adding base, and each needs its own equation with the correct species.

Use a single arrow in these two equations. The reactions of a strong acid or base with the buffer components go essentially to completion.

Say which component does the work. Added acid is removed by the conjugate base; added base is removed by the weak acid. Getting these the wrong way round is a common slip.

Distinguish capacity from pH. Diluting a buffer barely changes its pH, because the ratio is unchanged — but it reduces its capacity, because there is less of each component to react.

Worked ExampleHow much difference does a buffer make?

A buffer contains $0.100$ mol of ethanoic acid and $0.100$ mol of sodium ethanoate in $1.00$ L of solution. $K_a(\text{CH}_3\text{COOH}) = 1.74 \times 10^{-5}$ .

Part (a) — the initial pH

Rearranging the $K_a$ expression:

$[\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.74 \times 10^{-5} \times \frac{0.100}{0.100}$

The ratio is exactly 1, so

$[\text{H}_3\text{O}^+] = 1.74 \times 10^{-5} \text{ mol L}^{-1}$

$\text{pH} = -\log(1.74 \times 10^{-5}) = 4.76$

Part (b) — after adding acid

Step 1 — Decide what reacts. The added $\text{H}_3\text{O}^+$ is removed by the conjugate base:

$\text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

Step 2 — Adjust the amounts. 0.010 mol of ethanoate is consumed, and 0.010 mol of ethanoic acid is produced:

	Before	Change	After
$\text{CH}_3\text{COOH}$	0.100	+0.010	0.110 mol
$\text{CH}_3\text{COO}^-$	0.100	−0.010	0.090 mol

Step 3 — Recalculate.

$[\text{H}_3\text{O}^+] = 1.74 \times 10^{-5} \times \frac{0.110}{0.090} = 2.127 \times 10^{-5}$

$\text{pH} = -\log(2.127 \times 10^{-5}) = 4.672$

Part (c) — the comparison

Adding 0.010 mol of HCl to 1.00 L of pure water gives $[\text{H}_3\text{O}^+] = 0.010$ mol L⁻¹ directly, since HCl is a strong acid:

$\text{pH} = -\log(0.010) = 2.00$

Pure water starts at pH 7.00, so the pH has fallen by 5.00 units.

	Before	After	Change
buffer	4.76	4.67	0.09
pure water	7.00	2.00	5.00

The same amount of acid caused a pH change more than fifty times smaller in the buffer.