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:

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

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

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⁺

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!

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