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Buffer Solutions and Henderson-Hasselbalch | Study Mondo

Buffer Solutions and Henderson-Hasselbalch

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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

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

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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:

📚 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:

${pK}_{a} = - \log_{}$

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

Extremely Qualified

~12%

Well Qualified

~16%

Qualified

~24%

Possibly Qualified

~24%

No Recommendation

~24%

💡 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

← Previous Topic

Weak Acids, Weak Bases, and K_a/K_b

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Acid-Base Titrations and Indicators

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📌 Related Topics in Acids and Bases

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

❓ Frequently Asked Questions

What is Buffer Solutions and Henderson-Hasselbalch?▾

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

How can I study Buffer Solutions and Henderson-Hasselbalch effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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Buffer Solutions and Henderson-Hasselbalch is part of the AP Chemistry course on Study Mondo, specifically in the Acids and Bases section. You can explore the full course for more related topics and practice resources.

K

a

\text{pK}_a = -\log K_a

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

Buffer Solutions and Henderson-Hasselbalch

Try the Interactive Version!

Buffer Solutions and Henderson-Hasselbalch

What is a Buffer?

How Buffers Work

Henderson-Hasselbalch Equation

📚 Practice Problems

1Problem 1easy

❓ 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

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

2Problem 2medium

❓ Question:

3Problem 3hard

❓ Question: