Buffer Calculations: pH After Adding Acid or Base

The whole point of a buffer is that its pH barely changes when you add a little acid or alkali. The best way to really believe that, and to answer the exam questions that test it, is to calculate it. This guide gives you a method that always works and then walks through examples, including one where the buffer fails.

You’ll need the Henderson–Hasselbalch equation:

pH = pKa + log(moles A⁻ ÷ moles HA)

The key idea

A buffer contains a weak acid (HA) and its conjugate base (A⁻). Each one neutralises a different intruder:

• Added strong acid (H⁺) is removed by the conjugate base: A⁻ + H⁺ → HA
• Added strong base (OH⁻) is removed by the weak acid: HA + OH⁻ → A⁻ + H₂O

These reactions go essentially to completion. So adding acid converts some A⁻ into HA, and adding base converts some HA into A⁻. The ratio changes a little, so the pH changes a little, but nowhere near as much as it would in pure water.

The method

1. Work out the starting moles of HA and A⁻.
2. Work out the moles of strong acid or base added.
3. Do the neutralisation as a simple before/after table. The added H⁺ or OH⁻ is used up completely.
4. Plug the new moles into Henderson–Hasselbalch.

Always use moles, not concentrations, in steps 1–3. The volume changes when you add something, but because HA and A⁻ share the same final volume, it cancels in the ratio.

Example 1: the starting buffer

A buffer is made from 0.100 mol ethanoic acid and 0.100 mol sodium ethanoate in 1.00 dm³ of solution. pKa = 4.76.

pH = 4.76 + log(0.100 ÷ 0.100) = 4.76

Example 2: adding strong acid

0.010 mol of HCl is added to the buffer in example 1. Find the new pH.

The H⁺ reacts with ethanoate:

	CH₃COO⁻	H⁺	CH₃COOH
Before	0.100	0.010	0.100
Change	−0.010	−0.010	+0.010
After	0.090	0	0.110

pH = 4.76 + log(0.090 ÷ 0.110) = 4.76 + log(0.818) = 4.76 − 0.087 = 4.67

Change: −0.09 pH units.

Example 3: the same acid added to water

0.010 mol of HCl is added to 1.00 dm³ of pure water.

[H⁺] = 0.010 mol/dm³ → pH = 2.00

Change: from 7.00 to 2.00, a drop of 5 pH units.

Compare that with the buffer’s change of 0.09. The same amount of acid moves the pH of water about 55 times further. That’s what a buffer does.

Example 4: adding strong base

0.010 mol of NaOH is added to the original buffer from example 1.

The OH⁻ reacts with ethanoic acid:

	CH₃COOH	OH⁻	CH₃COO⁻
Before	0.100	0.010	0.100
Change	−0.010	−0.010	+0.010
After	0.090	0	0.110

pH = 4.76 + log(0.110 ÷ 0.090) = 4.76 + 0.087 = 4.85

Change: +0.09. The buffer is symmetrical because it started with equal amounts.

Example 5: working with volumes

A buffer is made by mixing 100 cm³ of 0.200 mol/dm³ ammonia with 100 cm³ of 0.150 mol/dm³ ammonium chloride. pKa(NH₄⁺) = 9.25. Find the pH before and after adding 10.0 cm³ of 0.500 mol/dm³ HCl.

Before:

• Moles NH₃ = 0.200 × 0.100 = 0.0200 mol
• Moles NH₄⁺ = 0.150 × 0.100 = 0.0150 mol
• pH = 9.25 + log(0.0200 ÷ 0.0150) = 9.25 + 0.125 = 9.37

Adding HCl:

• Moles H⁺ = 0.500 × 0.0100 = 0.00500 mol
• The base (NH₃) reacts: NH₃ + H⁺ → NH₄⁺
• NH₃ after = 0.0200 − 0.0050 = 0.0150 mol
• NH₄⁺ after = 0.0150 + 0.0050 = 0.0200 mol

pH = 9.25 + log(0.0150 ÷ 0.0200) = 9.25 − 0.125 = 9.13

Change: −0.25. Bigger than in example 2, because proportionally more acid was added relative to the buffer.

Example 6: overwhelming a buffer

The buffer from example 1 (0.100 mol each of CH₃COOH and CH₃COO⁻ in 1.00 dm³) has 0.150 mol of HCl added.

There’s only 0.100 mol of ethanoate to react with 0.150 mol of H⁺:

	CH₃COO⁻	H⁺	CH₃COOH
Before	0.100	0.150	0.100
Change	−0.100	−0.100	+0.100
After	0	0.050	0.200

All the conjugate base is gone, and 0.050 mol of strong acid is left over. The buffer has been destroyed. The pH is now controlled by the excess strong acid (the weak ethanoic acid contributes almost nothing in comparison):

[H⁺] ≈ 0.050 mol/dm³ → pH ≈ 1.30

Henderson–Hasselbalch can’t be used here, because one of the buffer components has run out. The amount of acid or base a buffer can absorb before this happens is its buffer capacity. See buffer capacity and buffer range.

Example 7: adding water

The buffer from example 1 is diluted to 10.0 dm³ with water. What happens to the pH?

Moles of HA and A⁻ are unchanged, so the ratio is unchanged:

pH = 4.76 + log(0.100 ÷ 0.100) = 4.76

Diluting a buffer doesn’t change its pH (within the limits of the approximation). It does reduce its capacity, because there’s less acid and base per cm³ to absorb additions.

Example 8: a target pH after addition

A biochemist has 500 cm³ of phosphate buffer at pH 7.20 containing 0.0250 mol H₂PO₄⁻ and 0.0250 mol HPO₄²⁻ (pKa 7.20). How much 1.00 mol/dm³ NaOH should be added to raise the pH to 7.40?

• Target ratio: 10^(7.40 − 7.20) = 10⁰·²⁰ = 1.585
• Let x = moles of OH⁻ added. Then (0.0250 + x) ÷ (0.0250 − x) = 1.585
• 0.0250 + x = 1.585 × (0.0250 − x) = 0.03963 − 1.585x
• 2.585x = 0.01463 → x = 5.66 × 10⁻³ mol
• Volume of 1.00 mol/dm³ NaOH = 5.66 cm³

Practice questions

1. A buffer contains 0.050 mol HA and 0.080 mol A⁻ (pKa 4.20). Find its pH, then the pH after adding 0.010 mol NaOH.
2. The same buffer instead has 0.010 mol HCl added. Find the new pH.
3. How many moles of HCl would destroy this buffer completely?

Answers:

1. Start: 4.20 + log(0.080 ÷ 0.050) = 4.40. After NaOH: HA = 0.040, A⁻ = 0.090 → 4.20 + log(2.25) = 4.55.
2. HA = 0.060, A⁻ = 0.070 → 4.20 + log(1.167) = 4.27.
3. More than 0.080 mol, the amount of A⁻ available to react with added acid.

Summary table

Situation	What happens	pH change
Add a little strong acid	A⁻ → HA	small decrease
Add a little strong base	HA → A⁻	small increase
Add water	ratio unchanged	none (capacity falls)
Add more acid than there is A⁻	buffer destroyed	large decrease
Add more base than there is HA	buffer destroyed	large increase

Common mistakes

• Using concentrations before accounting for the added volume. Stick to moles and the problem disappears.
• Letting the added H⁺ react with the acid. Added acid reacts with the base component; added base reacts with the acid component.
• Applying Henderson–Hasselbalch after the buffer is exhausted.
• Rounding moles too early. Log ratios near 1 are sensitive to small errors.

Key takeaways

• Added H⁺ converts A⁻ to HA; added OH⁻ converts HA to A⁻. Do this neutralisation in moles first.
• Then apply pH = pKa + log(moles A⁻ ÷ moles HA).
• Small additions give small pH changes; the same amount in pure water gives huge changes.
• Dilution doesn’t change a buffer’s pH, but it lowers its capacity.
• Once one component runs out, the buffer fails and the excess strong acid or base sets the pH.