ICE Tables

Approximations in Equilibrium Calculations

This note covers approximations used to simplify equilibrium calculations, particularly when dealing with weak acids/bases and solubility. The goal is to avoid using the quadratic formula.

Key Assumption: The approximation relies on the assumption that the change in concentration ( $ x $ ) is negligible compared to the initial concentration. This is valid when $ K $ is small and the initial concentration is relatively large.

1. Percent Ionization/Dissociation:

The percent ionization (or dissociation) is a measure of how much of a weak acid or base dissociates in solution. It’s calculated as:

$ \text{Percent ionization} = \frac{[H^+]]_{eq}}{[HA]]_i} \times 100% $ for a weak acid $ HA $

$ \text{Percent dissociation} = \frac{[A^-]]_{eq}}{[HA]]_i} \times 100% $ for a weak acid $ HA $

A similar equation applies to weak bases.

2. The 5% Rule:

The 5% rule is a common criterion for determining if the approximation is valid. If the percent ionization (or dissociation) is less than 5%, then the approximation is considered acceptable.

3. Weak Acid Equilibrium:

Consider the dissociation of a weak acid $ HA $ :

$$ HA(aq) \rightleftharpoons H^+(aq) + A^-(aq) $$
$ K_a = \frac{[H^+]][A^-]]}{[HA]]} $

Approximation: If we assume that $ x $ (the change in concentration) is negligible compared to $ [HA]]i $ , then $ [HA]]{eq} \approx [HA]]_i $ . This simplifies the $ K_a $ expression to:

$ K_a \approx \frac{x^2}{[HA]]_i} $

where $ x = [H^+]]=[A^-]] $ . Solving for $ x $ gives:

$ x \approx \sqrt{K_a[HA]]_i} $

After solving for $ x $ , check the 5% rule to confirm the validity of the approximation.

**4. Weak Base Equilibrium:** (Weak Base Equilibrium

5. Solubility Equilibria: (Solubility Equilibria)

6. When the Approximation Fails:

If the 5% rule is not met, the quadratic formula must be used to solve for $ x $ exactly.

7. Example:

Calculate the pH of a 0.10 M solution of acetic acid ( $ K_a = 1.8 \times 10^{-5} $ ).

Using the approximation:

$ x = [H^+]] \approx \sqrt{(1.8 \times 10^{-5})(0.10)} \approx 1.34 \times 10^{-3} M $

Percent ionization = $ \frac{1.34 \times 10^{-3}}{0.10} \times 100% \approx 1.34% $

Since this is less than 5%, the approximation is valid.

$ pH = -\log(1.34 \times 10^{-3}) \approx 2.87 $

8. Further Reading: (Quadratic Formula and Equilibrium)