Introduction to Weak Base Solutions

Determining the pH of a weak base solution is an essential skill in chemistry, particularly in applications related to environmental science, pharmaceuticals, and chemical engineering. Weak bases do not completely dissociate in water, which means their pH can be calculated using the base dissociation constant (Kb) and the concentration of the base. In this article, we will explore three practical examples that demonstrate how to determine the pH of a weak base solution.

Example 1: Ammonium Hydroxide in Household Cleaning

Context

Ammonium hydroxide, commonly found in household cleaning products, is a weak base. Understanding its pH is crucial for ensuring safe usage and optimal cleaning efficacy.

To determine the pH of a 0.10 M ammonium hydroxide (NH₄OH) solution, we need to know its Kb value, which is approximately 1.8 × 10⁻⁵.

First, we set up the equilibrium expression for the dissociation of NH₄OH:

NH₄OH (aq) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

Using the Kb expression:

Kb = [NH₄⁺][OH⁻] / [NH₄OH] = 1.8 × 10⁻⁵

Assuming x is the concentration of OH⁻ ions produced, we can write:

1.8 × 10⁻⁵ = (x)(x) / (0.10 - x)

Since Kb is small compared to the initial concentration, we can approximate:

1.8 × 10⁻⁵ = (x²) / 0.10

This simplifies to:

x² = (1.8 × 10⁻⁵)(0.10)

x² = 1.8 × 10⁻⁶

Taking the square root gives:

x ≈ 0.00134 M (which is the concentration of OH⁻)

Next, we calculate the pOH:

pOH = -log[OH⁻] = -log(0.00134) ≈ 2.87

Finally, we find the pH:

pH = 14 - pOH = 14 - 2.87 ≈ 11.13

Notes

• Ammonium hydroxide is often used at various concentrations; adjusting the initial concentration will affect pH.
• Safety precautions should be taken when handling strong cleaning agents, regardless of pH.

Example 2: Sodium Bicarbonate in Cooking

Context

Sodium bicarbonate (NaHCO₃), commonly known as baking soda, is a weak base used extensively in cooking and baking. Knowing its pH can help in understanding its effectiveness in recipes and its interaction with other ingredients.

For a 0.50 M sodium bicarbonate solution, we need to determine its pH. The Kb value for bicarbonate (HCO₃⁻) is approximately 4.8 × 10⁻⁴.

Setting up the equilibrium reaction:

HCO₃⁻ (aq) + H₂O (l) ⇌ H₂CO₃ (aq) + OH⁻ (aq)

Using the Kb expression:

Kb = [H₂CO₃][OH⁻] / [HCO₃⁻] = 4.8 × 10⁻⁴

Letting x be the concentration of OH⁻ formed, we have:

4.8 × 10⁻⁴ = (x)(x) / (0.50 - x)

Approximating:

4.8 × 10⁻⁴ = (x²) / 0.50

This simplifies to:

x² = (4.8 × 10⁻⁴)(0.50)

x² = 2.4 × 10⁻⁴

Taking the square root gives:

x ≈ 0.0155 M (concentration of OH⁻)

Calculating pOH:

pOH = -log(0.0155) ≈ 1.81

Finding pH:

pH = 14 - pOH = 14 - 1.81 ≈ 12.19

Notes

• Sodium bicarbonate is often used in recipes to neutralize acidity; its pH contributes to the overall flavor and texture of baked goods.
• The effectiveness can vary based on other ingredients present in the mixture.

Example 3: Pyridine in Organic Chemistry

Context

Pyridine is a common weak base used in organic synthesis. Its pH is crucial for reactions that involve nucleophilic attack or protonation.

To determine the pH of a 0.10 M pyridine (C₅H₅N) solution, we refer to its Kb, which is approximately 1.7 × 10⁻⁹.

The dissociation reaction for pyridine is:

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH⁻

Using the Kb expression:

Kb = [C₅H₅NH⁺][OH⁻] / [C₅H₅N] = 1.7 × 10⁻⁹

Letting x be the concentration of OH⁻, we have:

1.7 × 10⁻⁹ = (x)(x) / (0.10 - x)

Approximating:

1.7 × 10⁻⁹ = (x²) / 0.10

This simplifies to:

x² = (1.7 × 10⁻⁹)(0.10)

x² = 1.7 × 10⁻¹⁰

Taking the square root gives:

x ≈ 1.30 × 10⁻⁵ M (concentration of OH⁻)

Calculating pOH:

pOH = -log(1.30 × 10⁻⁵) ≈ 4.87

Finding pH:

pH = 14 - pOH = 14 - 4.87 ≈ 9.13

Notes

• Pyridine’s pH can have significant effects on reaction pathways; understanding its pH helps in predicting reaction outcomes.
• Care should be taken when handling pyridine due to its toxicity and potential health hazards.