Acid Dissociation Constant (Ka) - Chemistry

1. Fundamental Concepts

Ka is the acid dissociation equilibrium constant for weak acids in aqueous solution, quantitatively describing the extent of dissociation.
Only applies to weak acids; strong acids dissociate completely and do not use Ka.
For a general weak acid dissociation:
$HA \rightleftharpoons H^+ + A^-$

$Ka = \frac{[H^+][A^-]}{[HA]}$
Ka depends on temperature, not concentration.

Strong vs. Weak Acids:

p :

This is a more convenient way to express the 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.

Easy

A weak acid has $K_a = 1.8 \times 10^{-5}$.

Compare its strength to an acid with $K_a = 1.0 \times 10^{-6}$.

Solution:

Larger $K_a$ means stronger acid.

$1.8 \times 10^{-5} > 1.0 \times 10^{-6}$.

Conclusion: Acid with $K_a = 1.8 \times 10^{-5}$ is stronger.

Medium

Calculate $[H^+]$ in a 0.10 M HA solution, $K_a = 1.0 \times 10^{-6}$.

Step 1: Write equilibrium

$HA \rightleftharpoons H^+ + A^-$

Step 2: Define variables

Let $[H^+] = [A^-] = x$

$[HA] = 0.10 - x \approx 0.10$ (valid because $c/K_a \gg 100$)

Step 3: Set up $K_a$

\[K_a = \frac{x^2}{0.10} = 1.0 \times 10^{-6}\]

\[x^2 = 1.0 \times 10^{-7}\]

\[x = \sqrt{1.0 \times 10^{-7}} \approx 3.16 \times 10^{-4}\ \text{M}\]

Answer:$[H^+] \approx 3.16 \times 10^{-4}\ \text{mol/L}$

Hard

A 0.15 M weak acid solution has pH = 2.50. Calculate $K_a$.

Step 1: Find $[H^+]$

\[[H^+] = 10^{-pH} = 10^{-2.50} \approx 3.16 \times 10^{-3}\ \text{M}\]

Step 2: Equilibrium concentrations

$[H^+] = [A^-] = 3.16 \times 10^{-3}\ \text{M}$

$[HA] = 0.15 - 3.16 \times 10^{-3} \approx 0.1468\ \text{M}$

Step 3: Solve for $K_a$

\[K_a = \frac{[H^+][A^-]}{[HA]}= \frac{(3.16 \times 10^{-3})^2}{0.1468}\approx 6.8 \times 10^{-5}\]

Answer:$K_a \approx 6.8 \times 10^{-5}$

Write the dissociation equation and Ka expression first.
Let $[H^+] = [A^-] = x$. Use approximation $[HA] \approx c_0$ if $c/Ka \ge 100$.
When solving for Ka or pH, make sure to explicitly mention that approximation can be used when $[H^+] \ll c_0$ , as it's crucial for simplifying the calculations in most weak acid problems.
Use core formulas:
$[H^+] = \sqrt{Ka \cdot c},\quad pH = -\log[H^+]$
Do not confuse Ka and pKa.
For common ion problems, include the common ion concentration in the Ka expression.