Weak Acids and Weak Bases - Chemistry

1. Fundamental Concepts

Definition: Weak acids and weak bases are substances that partially dissociate in water, releasing a small fraction of their ions.
Dissociation Constant (Ka for acids, Kb for bases): A measure of the strength of an acid or base, where a smaller value indicates a weaker acid or base.
pH Scale: A logarithmic scale used to specify the acidity or basicity of an aqueous solution, with pH 7 being neutral, below 7 acidic, and above 7 basic.

Dissociation Equation for Weak Acids: $$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$$

Dissociation Equation for Weak Bases: $$\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$$

Acid Dissociation Constant (Ka): $$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$$

Base Dissociation Constant (Kb): $$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$$

Relationship between Ka and Kb: $$K_w = K_a \cdot K_b = 1.0 \times 10^{-14} \text{ at } 25^\circ\text{C}$$

Application: Used in buffer solutions to maintain pH stability in biological and chemical systems

Problem: Calculate the pH of a 0.1 M solution of acetic acid (CH₃COOH) with a Ka of 1.8 × 10⁻⁵.

Write the dissociation equation: $$\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$$
Set up the equilibrium expression: $$K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 1.8 \times 10^{-5}$$
Assume x is the concentration of H⁺ and CH₃COO⁻ at equilibrium: $$K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1}$$
Solve for x: $$x^2 = 1.8 \times 10^{-6} \Rightarrow x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$$
Calculate pH: $$\text{pH} = -\log[\text{H}^+] = -\log(1.34 \times 10^{-3}) = 2.87$$

Validation: The calculated pH of 2.87 is consistent with the expected range for a weak acid.

Problem: Calculate the pOH of a 0.05 M solution of ammonia (NH₃) with a Kb of 1.8 × 10⁻⁵.

Write the dissociation equation: $$\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$$
Set up the equilibrium expression: $$K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = 1.8 \times 10^{-5}$$
Assume x is the concentration of OH⁻ and NH₄⁺ at equilibrium: $$K_b = \frac{x^2}{0.05 - x} \approx \frac{x^2}{0.05}$$
Solve for x: $$x^2 = 9.0 \times 10^{-7} \Rightarrow x = \sqrt{9.0 \times 10^{-7}} = 9.49 \times 10^{-4} \text{ M}$$
Calculate pOH: $$\text{pOH} = -\log[\text{OH}^-] = -\log(9.49 \times 10^{-4}) = 3.02$$

Validation: The calculated pOH of 3.02 is consistent with the expected range for a weak base.

ICE Table Method: Use an Initial, Change, Equilibrium (ICE) table to organize the concentrations of reactants and products.
Approximation: For weak acids and bases, assume the change in concentration (x) is small compared to the initial concentration, simplifying the calculations.
Logarithmic Calculations: Use the properties of logarithms to simplify pH and pOH calculations, such as $$\text{pH} = -\log[\text{H}^+]$$ and $$\text{pOH} = -\log[\text{OH}^-]$$.
Buffer Solutions: Understand how to calculate the pH of a buffer solution using the Henderson-Hasselbalch equation: $$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$.