Buffers: Properties, Calculations, and Titration Applications

Buffers of Aqueous Acids and Bases

Introduction to Buffers

A buffer is a solution that resists significant changes in pH when small amounts of strong acid or base are added. Buffers are essential in many chemical and biological systems to maintain a stable pH environment.

• Composition: Buffers contain relatively large amounts of a weak acid (HA) and its conjugate base (A-).

• Action: When H3O+ (strong acid) is added, it reacts with the conjugate base. When OH- (strong base) is added, it reacts with the weak acid.

• pH Determination: The pH is determined by the equilibrium between the weak acid and its conjugate base.

How Buffers Work: Visual Representation

The following diagrams illustrate the effect of adding strong acid or base to a buffer composed of acetic acid (CH3COOH) and its conjugate base (CH3COO-):

• Initial Buffer: Equal concentrations of CH3COOH and CH3COO-.

• After Addition of H3O+: The added acid reacts with CH3COO-, decreasing its concentration and increasing CH3COOH.

• After Addition of OH-: The added base reacts with CH3COOH, decreasing its concentration and increasing CH3COO-.

Buffer Capacity

Buffer capacity is the measure of a buffer's ability to resist pH change. It increases as the concentrations of the buffer components increase (assuming equal volumes).

• High buffer capacity means the solution can neutralize more added acid or base without a significant pH change.

The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution:

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

• pKa is the negative logarithm of the acid dissociation constant (Ka).

• This equation is especially useful for buffer calculations and when preparing buffer solutions of a desired pH.

Buffer Calculations: Example Problems

Consider a buffer made from 0.50 M CH3COOH and 0.50 M CH3COONa (Ka = 1.8 × 10-5):

1. Initial pH: Use the Henderson-Hasselbalch equation with equal concentrations of acid and base.

   • $pH = pK_a + \log(1) = pK_a$

   • $pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$

   • pH = 4.74

2. After Adding 0.020 mol NaOH to 1.0 L Buffer:

   • NaOH reacts with CH3COOH, converting it to CH3COO-.

   • New concentrations: CH3COOH = 0.48 M, CH3COO- = 0.52 M

   • Apply Henderson-Hasselbalch equation:

   • $pH = 4.74 + \log \left( \frac{0.52}{0.48} \right)$

3. After Adding 0.020 mol HCl to 1.0 L Buffer:

   • HCl reacts with CH3COO-, converting it to CH3COOH.

   • New concentrations: CH3COOH = 0.52 M, CH3COO- = 0.48 M

   • Apply Henderson-Hasselbalch equation:

   • $pH = 4.74 + \log \left( \frac{0.48}{0.52} \right)$

Preparing Buffers of Desired pH

To prepare a buffer of a specific pH, use the Henderson-Hasselbalch equation to determine the required ratio of conjugate base to acid. Then, calculate the amounts or volumes needed based on the desired total volume and concentrations.

• Example: To prepare a carbonate buffer (CO32- and HCO3-) at pH 10.00, use the Ka of HCO3- and solve for the required moles of Na2CO3 to add to a given volume of NaHCO3.

Buffers in Titration

During the titration of a weak acid with a strong base (or vice versa), a buffer is formed in the region before the equivalence point. The solution contains both the weak acid and its conjugate base, which resists drastic pH changes.

• At half-equivalence (Ve/2): The concentrations of acid and conjugate base are equal, so pH = pKa.

• Equivalence point (Ve): All weak acid has been converted to its conjugate base.

Key Equations and Constants

• $K_w = [H_3O^+][OH^-] = 1 \times 10^{-14}$ at 25°C

• $K_w = K_a K_b$

• $pH = -\log[H_3O^+]$

• $14 = pH + pOH$ at 25°C

• Quadratic formula for equilibrium calculations: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Summary Table: Buffer Properties and Calculations

Concept	Key Points
Buffer Composition	Weak acid + conjugate base (or weak base + conjugate acid)
Buffer Capacity	Increases with higher concentrations of buffer components
Henderson-Hasselbalch Equation	$pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)$
pH at Half-Equivalence	pH = pKa (when [acid] = [base])
Buffer in Titration	Exists before equivalence point; resists pH change