Buffer Solutions and Henderson-Hasselbalch

Master buffer systems, calculate buffer pH using Henderson-Hasselbalch equation, and understand buffer capacity.

Written and reviewed by the Study Mondo Education TeamLast updated February 14, 2026

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Buffer Solutions and Henderson-Hasselbalch

What is a Buffer?

Buffer: Solution that resists pH change when small amounts of acid or base are added

Composition:

Weak acid (HA) + conjugate base (A⁻), OR
Weak base (B) + conjugate acid (BH⁺)

Examples:

CH₃COOH/CH₃COO⁻ (acetic acid/acetate)
NH₃/NH₄⁺ (ammonia/ammonium)
H₂PO₄⁻/HPO₄²⁻ (phosphate buffer)

How Buffers Work

Weak acid buffer: HA/A⁻

Add H⁺: A⁻ + H⁺ → HA (conjugate base neutralizes)

Add OH⁻: HA + OH⁻ → A⁻ + H₂O (weak acid neutralizes)

Result: pH stays relatively constant

Key: Need significant amounts of BOTH components

Henderson-Hasselbalch Equation

For weak acid buffer:

$\text{pH} = \text{pK}_a + \log\frac{[A^-]}{[HA]}$

Or:

$\text{pH} = \text{pK}_a + \log\frac{[\text{base}]}{[\text{acid}]}$

Where pK_a = -log K_a

For weak base buffer:

$\text{pOH} = \text{pK}_b + \log\frac{[BH^+]}{[B]}$

Then pH = 14 - pOH

Special Cases

When [A⁻] = [HA]:

$\text{pH} = \text{pK}_a + \log(1) = \text{pK}_a$

Equal amounts: pH = pK_a (best buffering!)

Ratio of 10:1 or 1:10:

pH = pK_a ± 1
Still effective buffer

Preparing Buffers

Two methods:

Method 1: Mix weak acid and its salt

Example: CH₃COOH + NaCH₃COO
Calculate pH from Henderson-Hasselbalch

Method 2: Partial neutralization

Start with weak acid, add strong base (or vice versa)
Creates conjugate base in situ
Common in lab

Example: 1.0 M HA + 0.5 M NaOH

Forms 0.5 M A⁻ and leaves 0.5 M HA
pH = pK_a (equal amounts)

Buffer Capacity

Capacity: Amount of acid/base buffer can neutralize

Factors:

Concentrations: Higher = more capacity
Ratio: 1:1 ratio = maximum capacity

Best buffer:

pH ≈ pK_a (ratio close to 1:1)
High concentrations of both

Effective range: pK_a ± 1

Calculating pH Changes in Buffers

General approach:

Determine initial amounts (use moles, not M)
Add strong acid/base (neutralization)
Calculate new amounts after reaction
Find new concentrations (divide by volume)
Use Henderson-Hasselbalch

Or use ICE for neutralization:

Add OH⁻ to HA/A⁻ buffer:

HA + OH⁻ → A⁻ + H₂O (goes to completion)
Decreases [HA], increases [A⁻]

Add H⁺ to HA/A⁻ buffer:

A⁻ + H⁺ → HA (goes to completion)
Increases [HA], decreases [A⁻]

Common Ion Effect Revisited

Buffer is example of common ion effect:

HA ⇌ H⁺ + A⁻

Adding A⁻ (from salt):

Shifts left
Suppresses ionization of HA
Lower [H⁺] than HA alone
Higher pH than HA alone

Choosing a Buffer

Select pK_a close to desired pH:

Want pH = 7.0?

Choose acid with pK_a ≈ 7.0
H₂PO₄⁻/HPO₄²⁻ (pK_a = 7.2) ✓

Want pH = 4.7?

Choose acid with pK_a ≈ 4.7
CH₃COOH/CH₃COO⁻ (pK_a = 4.76) ✓

Rule: Buffer effective within pK_a ± 1

Biological Buffers

Blood pH = 7.4 (tightly controlled):

Major buffer: H₂CO₃/HCO₃⁻

pK_a = 6.1
CO₂(g) ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻
Respiratory system controls CO₂
Kidneys control HCO₃⁻

Phosphate buffer: H₂PO₄⁻/HPO₄²⁻

pK_a = 7.2
Important in cells

Protein buffers: Hemoglobin, albumin

📚 Practice Problems

1Problem 1easy

❓ Question:

Calculate the pH of a buffer containing 0.50 M CH₃COOH and 0.30 M CH₃COO⁻. K_a = 1.8 × 10⁻⁵.

💡 Show Solution

Given:

[CH₃COOH] = 0.50 M (weak acid)
[CH₃COO⁻] = 0.30 M (conjugate base)
K_a = 1.8 × 10⁻⁵

This is a buffer! HA/A⁻ system

Calculate pK_a:

$\text{pK}_a = -\log K_a$

$\text{pK}_a = -\log(1.8 \times 10^{-5})$

$\text{pK}_a = 4.74$

Use Henderson-Hasselbalch:

$\text{pH} = \text{pK}_a + \log\frac{[A^-]}{[HA]}$

$\text{pH} = 4.74 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

$\text{pH} = 4.74 + \log\frac{0.30}{0.50}$

$\text{pH} = 4.74 + \log(0.60)$

$\text{pH} = 4.74 + (-0.22)$

$\text{pH} = 4.52$

Answer: pH = 4.52

Interpretation:

pH < pK_a because [acid] > [base]

Ratio: 0.30/0.50 = 0.60

More acid than base
pH slightly below pK_a
Still effective buffer (ratio between 0.1 and 10)

Buffer range: 4.74 ± 1 = 3.74 to 5.74

pH 4.52 is well within range ✓

2Problem 2medium

❓ Question:

A buffer is made by mixing 50.0 mL of 0.40 M NH₃ with 50.0 mL of 0.20 M NH₄Cl. K_b(NH₃) = 1.8 × 10⁻⁵. (a) Calculate the pH. (b) Calculate pH after adding 5.0 mL of 0.10 M HCl.

💡 Show Solution

Given:

50.0 mL of 0.40 M NH₃
50.0 mL of 0.20 M NH₄Cl
K_b = 1.8 × 10⁻⁵

(a) Initial pH

Find K_a for NH₄⁺:

$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

$\text{pK}_a = -\log(5.6 \times 10^{-10}) = 9.25$

Calculate total volume: 50.0 + 50.0 = 100.0 mL

Calculate concentrations:

$[NH_3] = \frac{(0.40)(50.0)}{100.0} = 0.20 \text{ M}$

$[NH_4^+] = \frac{(0.20)(50.0)}{100.0} = 0.10 \text{ M}$

Henderson-Hasselbalch:

$\text{pH} = \text{pK}_a + \log\frac{[NH_3]}{[NH_4^+]}$

$\text{pH} = 9.25 + \log\frac{0.20}{0.10}$

$\text{pH} = 9.25 + \log(2.0)$

$\text{pH} = 9.25 + 0.30$

$\text{pH} = 9.55$

Answer (a): pH = 9.55

(b) After adding HCl

Calculate moles before HCl:

NH₃: 0.20 M × 0.100 L = 0.020 mol
NH₄⁺: 0.10 M × 0.100 L = 0.010 mol

Moles HCl added:

0.10 M × 0.005 L = 0.00050 mol H⁺

Reaction: NH₃ + H⁺ → NH₄⁺

	NH₃	H⁺	NH₄⁺
Before	0.020	0.00050	0.010
Change	-0.00050	-0.00050	+0.00050
After	0.01950	0	0.01050

New volume: 100.0 + 5.0 = 105.0 mL = 0.105 L

New concentrations:

$[NH_3] = \frac{0.01950}{0.105} = 0.186 \text{ M}$

$[NH_4^+] = \frac{0.01050}{0.105} = 0.100 \text{ M}$

New pH:

$\text{pH} = 9.25 + \log\frac{0.186}{0.100}$

$\text{pH} = 9.25 + \log(1.86)$

$\text{pH} = 9.25 + 0.27$

$\text{pH} = 9.52$

Answer (b): pH = 9.52

Summary:

Before HCl: pH = 9.55 After HCl: pH = 9.52 Change: ΔpH = -0.03 (tiny!)

Without buffer: 5.0 mL of 0.10 M HCl in 100 mL water → pH ≈ 2

Buffer resists pH change!

3Problem 3hard

❓ Question:

What mass of sodium acetate (CH₃COONa, MW = 82.0 g/mol) must be added to 500.0 mL of 0.20 M acetic acid to create a buffer with pH = 5.00? K_a = 1.8 × 10⁻⁵.

💡 Show Solution

Given:

Volume acetic acid: 500.0 mL = 0.500 L
[CH₃COOH] = 0.20 M
Desired pH = 5.00
K_a = 1.8 × 10⁻⁵
MW(CH₃COONa) = 82.0 g/mol

Calculate pK_a:

$\text{pK}_a = -\log(1.8 \times 10^{-5}) = 4.74$

Use Henderson-Hasselbalch to find ratio:

$\text{pH} = \text{pK}_a + \log\frac{[A^-]}{[HA]}$

$5.00 = 4.74 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

$5.00 - 4.74 = \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

$0.26 = \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

Take antilog:

$\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{0.26} = 1.82$

Calculate [CH₃COO⁻]:

$[CH_3COO^-] = 1.82 \times [CH_3COOH]$

$[CH_3COO^-] = 1.82 \times 0.20$

$[CH_3COO^-] = 0.364 \text{ M}$

Calculate moles needed:

Moles CH₃COO⁻:

$n = M \times V = 0.364 \text{ mol/L} \times 0.500 \text{ L}$

$n = 0.182 \text{ mol}$

Calculate mass:

CH₃COONa → CH₃COO⁻ + Na⁺ (1:1 ratio)

Moles CH₃COONa needed = 0.182 mol

$\text{mass} = n \times \text{MW}$

$\text{mass} = 0.182 \text{ mol} \times 82.0 \text{ g/mol}$

$\text{mass} = 14.9 \text{ g}$

Answer: 15 g of sodium acetate

Verify:

After adding salt:

[CH₃COOH] = 0.20 M
[CH₃COO⁻] = 0.364 M

$\text{pH} = 4.74 + \log\frac{0.364}{0.20}$

$\text{pH} = 4.74 + 0.26 = 5.00$ ✓

Note: Assumes volume change negligible when adding solid

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections

Section	Format	Questions	Time	Weight	Calculator
Multiple Choice	MCQ	60	90 min	50%	✅
Free Response (Long)	FRQ	3	69 min	30%	✅
Free Response (Short)	FRQ	4	36 min	20%	✅

📊 Scoring: 1-5

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Well Qualified

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💡 Key Test-Day Tips

✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws

⚠️ Common Mistakes: Buffer Solutions and Henderson-Hasselbalch

Avoid these 3 frequent errors

🌍 Real-World Applications: Buffer Solutions and Henderson-Hasselbalch

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

2Determine the limiting reagent

3Calculate moles of product

4Convert moles to grams

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Acid-Base Theories and pH Scale Weak Acids, Weak Bases, and K_a/K_b Acid-Base Titrations and Indicators

❓ Frequently Asked Questions

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Master buffer systems, calculate buffer pH using Henderson-Hasselbalch equation, and understand buffer capacity.

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Buffer Solutions and Henderson-Hasselbalch

Try the Interactive Version!

Buffer Solutions and Henderson-Hasselbalch

What is a Buffer?

How Buffers Work

Henderson-Hasselbalch Equation

Special Cases

Preparing Buffers

Method 1: Mix weak acid and its salt

Method 2: Partial neutralization

Buffer Capacity

Calculating pH Changes in Buffers

Common Ion Effect Revisited

Choosing a Buffer

Biological Buffers

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3hard

❓ Question:

📋 AP Chemistry — Exam Format Guide

📊 Scoring: 1-5

💡 Key Test-Day Tips

⚠️ Common Mistakes: Buffer Solutions and Henderson-Hasselbalch

🌍 Real-World Applications: Buffer Solutions and Henderson-Hasselbalch

📝 Worked Example: Stoichiometry — Limiting Reagent

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Acids and Bases

❓ Frequently Asked Questions