Buffer Solutions and Henderson-Hasselbalch

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

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:

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

Or:

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

Where pK_a = -log K_a

For weak base buffer:

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

Then pH = 14 - pOH

Special Cases

When [A⁻] = [HA]:

pH=pKa+log⁡(1)=pKa\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:

  1. Concentrations: Higher = more capacity
  2. 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:

  1. Determine initial amounts (use moles, not M)
  2. Add strong acid/base (neutralization)
  3. Calculate new amounts after reaction
  4. Find new concentrations (divide by volume)
  5. 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:

pKa=−log⁡Ka\text{pK}_a = -\log K_a

pKa=−log⁡(1.8×10−5)\text{pK}_a = -\log(1.8 \times 10^{-5})

pKa=4.74\text{pK}_a = 4.74

Use Henderson-Hasselbalch:

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

pH=4.74+log⁡[CH3COO−][CH3COOH]\text{pH} = 4.74 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}

pH=4.74+log⁡0.300.50\text{pH} = 4.74 + \log\frac{0.30}{0.50}

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

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

pH=4.52\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⁺:

Ka=KwKb=1.0×10−141.8×10−5=5.6×10−10K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}

pKa=−log⁡(5.6×10−10)=9.25\text{pK}_a = -\log(5.6 \times 10^{-10}) = 9.25

Calculate total volume: 50.0 + 50.0 = 100.0 mL

Calculate concentrations:

[NH3]=(0.40)(50.0)100.0=0.20 M[NH_3] = \frac{(0.40)(50.0)}{100.0} = 0.20 \text{ M}

[NH4+]=(0.20)(50.0)100.0=0.10 M[NH_4^+] = \frac{(0.20)(50.0)}{100.0} = 0.10 \text{ M}

Henderson-Hasselbalch:

pH=pKa+log⁡[NH3][NH4+]\text{pH} = \text{pK}_a + \log\frac{[NH_3]}{[NH_4^+]}

pH=9.25+log⁡0.200.10\text{pH} = 9.25 + \log\frac{0.20}{0.10}

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

pH=9.25+0.30\text{pH} = 9.25 + 0.30

pH=9.55\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:

[NH3]=0.019500.105=0.186 M[NH_3] = \frac{0.01950}{0.105} = 0.186 \text{ M}

[NH4+]=0.010500.105=0.100 M[NH_4^+] = \frac{0.01050}{0.105} = 0.100 \text{ M}

New pH:

pH=9.25+log⁡0.1860.100\text{pH} = 9.25 + \log\frac{0.186}{0.100}

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

pH=9.25+0.27\text{pH} = 9.25 + 0.27

pH=9.52\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:

pKa=−log⁡(1.8×10−5)=4.74\text{pK}_a = -\log(1.8 \times 10^{-5}) = 4.74

Use Henderson-Hasselbalch to find ratio:

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

5.00=4.74+log⁡[CH3COO−][CH3COOH]5.00 = 4.74 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}

5.00−4.74=log⁡[CH3COO−][CH3COOH]5.00 - 4.74 = \log\frac{[CH_3COO^-]}{[CH_3COOH]}

0.26=log⁡[CH3COO−][CH3COOH]0.26 = \log\frac{[CH_3COO^-]}{[CH_3COOH]}

Take antilog:

[CH3COO−][CH3COOH]=100.26=1.82\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{0.26} = 1.82

Calculate [CH₃COO⁻]:

[CH3COO−]=1.82×[CH3COOH][CH_3COO^-] = 1.82 \times [CH_3COOH]

[CH3COO−]=1.82×0.20[CH_3COO^-] = 1.82 \times 0.20

[CH3COO−]=0.364 M[CH_3COO^-] = 0.364 \text{ M}

Calculate moles needed:

Moles CH₃COO⁻:

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

n=0.182 moln = 0.182 \text{ mol}

Calculate mass:

CH₃COONa → CH₃COO⁻ + Na⁺ (1:1 ratio)

Moles CH₃COONa needed = 0.182 mol

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

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

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

Answer: 15 g of sodium acetate

Verify:

After adding salt:

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

pH=4.74+log⁡0.3640.20\text{pH} = 4.74 + \log\frac{0.364}{0.20}

pH=4.74+0.26=5.00\text{pH} = 4.74 + 0.26 = 5.00 ✓

Note: Assumes volume change negligible when adding solid

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