Buffer solutions resist changes in pH by neutralizing small amounts of added acid (H⁺) or base (OH⁻).

• In an acidic buffer, the salt's anion (e.g., CH₃COO⁻) neutralizes added H⁺, while the weak acid (e.g., CH₃COOH) neutralizes added OH⁻.
• In a basic buffer, the salt's cation (e.g., NH₄⁺) neutralizes added OH⁻, while the weak base (e.g., NH₃) neutralizes added H⁺.

1. Acidic buffers are used in fermentation processes, pharmaceuticals, and biochemical assays.
2. Basic buffers are used in detergents, photographic solutions, and maintaining pH in aquariums.

Consider the buffer system of acetic acid and sodium acetate:

CH₃COOH ⇌ CH₃COO⁻ + H⁺

Sodium acetate (the salt) dissociates completely into CH₃COO⁻, increasing its concentration and shifting the equilibrium to the left. This suppresses the dissociation of acetic acid due to the common ion effect. The solution will contain:

• A large amount of unionized acetic acid
• A large amount of acetate ions (CH₃COO⁻) from the salt
• Enough H⁺ ions to make the solution acidic

When a small amount of H⁺ is added:

CH₃COO⁻ + H⁺ → CH₃COOH

The acetate ions react with the added H⁺, preventing a significant change in pH.

When a small amount of OH⁻ is added:

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

The OH⁻ neutralizes some of the H⁺ ions, causing the equilibrium to shift to the right to replace them. This prevents a significant change in pH.

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

$pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$

Example

A buffer solution contains 1M of acid and has a pKa of 4.742:

1. Determine the salt concentration in the buffer given the pH is 5.

   Solution:

   From the Henderson-Hasselbalch equation:

   $pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$

   $5 = 4.742 + \log\left(\frac{[\text{salt}]}{1}\right)$

   $\log([\text{salt}]) = 5 - 4.742 = 0.258$

   $[\text{salt}] = 10^{0.258} \approx 1.81 \text{ M}$

2. Calculate the pH of the buffer when 0.01 moles of H⁺ are added.

   Solution:

   Updated [salt] = 1.81 - 0.01 = 1.80 M

   Updated [acid] = 1 + 0.01 = 1.01 M

   $pH = 4.742 + \log(1.80/1.01) \approx 4.99$

3. Calculate the pH when 0.01 moles of OH⁻ are added.

   Solution:

   Updated [salt] = 1.81 + 0.01 = 1.82 M

   Updated [acid] = 1 - 0.01 = 0.99 M

   $pH = 4.742 + \log(1.82/0.99) \approx 5.02$

Important notes

• Adding H⁺ increases the acid concentration but decreases the salt concentration by the same amount.
• Adding OH⁻ decreases the acid concentration but increases the salt concentration by the same amount.
• Water is not a buffer system.