Acid Dissociation Constant (Ka) - Chemistry

1. Fundamental Concepts

Definition: The acid dissociation constant ($$ K_a $$) is a measure of the strength of an acid in solution. It quantifies the extent to which an acid dissociates into its ions.
Equilibrium Expression: For a general acid $$ HA $$ dissociating in water, the equilibrium can be written as: $$ HA + H_2O \rightleftharpoons H_3O^+ + A^- $$ The $$ K_a $$ expression is: $$ K_a = \frac{[H_3O^+][A^-]}{[HA]} $$
Units: $$ K_a $$ is dimensionless because it is a ratio of concentrations.

2. Key Concepts

Strong vs. Weak Acids: $$K_a \text{ for strong acids} \gg K_a \text{ for weak acids}$$

p$$ K_a $$: $$pK_a = -\log(K_a)$$

This is a more convenient way to express the $$ K_a $$ value, especially for very small or large values.

Application: Used to predict the pH of solutions and the behavior of acids in various chemical reactions.

3. Examples

Example 1 (Basic)

Problem: Calculate the $$ K_a $$ for acetic acid ($$ CH_3COOH $$) if the concentrations are: $$ [CH_3COOH] = 0.10 M $$, $$ [H_3O^+] = 1.34 \times 10^{-3} M $$, and $$ [CH_3COO^-] = 1.34 \times 10^{-3} M $$.

Step-by-Step Solution:

Write the $$ K_a $$ expression: $$ K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]} $$
Substitute the given values: $$ K_a = \frac{(1.34 \times 10^{-3})(1.34 \times 10^{-3})}{0.10} $$
Calculate the value: $$ K_a = \frac{1.7956 \times 10^{-6}}{0.10} = 1.7956 \times 10^{-5} $$

Validation: The calculated $$ K_a $$ value is consistent with the known $$ K_a $$ for acetic acid, which is approximately $$ 1.8 \times 10^{-5} $$ ✓

Example 2 (Intermediate)

Problem: Given the $$ K_a $$ of formic acid ($$ HCOOH $$) is $$ 1.8 \times 10^{-4} $$, calculate the concentration of $$ [H_3O^+] $$ in a 0.10 M solution of formic acid.

Step-by-Step Solution:

Set up the equilibrium expression: $$ HCOOH + H_2O \rightleftharpoons H_3O^+ + HCOO^- $$ $$ K_a = \frac{[H_3O^+][HCOO^-]}{[HCOOH]} $$
Let $$ x $$ be the concentration of $$ [H_3O^+] $$ at equilibrium. Then: $$ [H_3O^+] = x $$ $$ [HCOO^-] = x $$ $$ [HCOOH] = 0.10 - x $$
Substitute into the $$ K_a $$ expression: $$ 1.8 \times 10^{-4} = \frac{x \cdot x}{0.10 - x} $$
Solve for $$ x $$: $$ 1.8 \times 10^{-4} = \frac{x^2}{0.10 - x} $$ $$ 1.8 \times 10^{-4} (0.10 - x) = x^2 $$ $$ 1.8 \times 10^{-5} - 1.8 \times 10^{-4}x = x^2 $$ $$ x^2 + 1.8 \times 10^{-4}x - 1.8 \times 10^{-5} = 0 $$
Use the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$ a = 1 $$, $$ b = 1.8 \times 10^{-4} $$, and $$ c = -1.8 \times 10^{-5} $$: $$ x = \frac{-1.8 \times 10^{-4} \pm \sqrt{(1.8 \times 10^{-4})^2 - 4 \cdot 1 \cdot (-1.8 \times 10^{-5})}}{2 \cdot 1} $$ $$ x = \frac{-1.8 \times 10^{-4} \pm \sqrt{3.24 \times 10^{-8} + 7.2 \times 10^{-5}}}{2} $$ $$ x = \frac{-1.8 \times 10^{-4} \pm \sqrt{7.20324 \times 10^{-5}}}{2} $$ $$ x = \frac{-1.8 \times 10^{-4} \pm 8.487 \times 10^{-3}}{2} $$ $$ x \approx 4.15 \times 10^{-3} $$

Validation: The calculated $$ [H_3O^+] $$ is approximately $$ 4.15 \times 10^{-3} M $$, which is reasonable for a weak acid like formic acid ✓

4. Problem-Solving Techniques

ICE Tables: Use Initial, Change, Equilibrium (ICE) tables to organize and solve equilibrium problems.
Assumptions: For weak acids, assume that the change in concentration due to dissociation is small compared to the initial concentration.
Quadratic Formula: When the assumption of small $$ x $$ is not valid, use the quadratic formula to solve for $$ x $$.
Logarithms: Use logarithms to convert between $$ K_a $$ and $$ pK_a $$.