FAD1018 W2 — Ionic Equilibria (Part 2)

Week 2 lecture covering the pH scale, ionic product of water, degree of dissociation, and acid–base equilibrium calculations. Lecturer: Dr. Fauzani Md. Salleh (Chemistry Division, Centre for Foundation Studies in Science).

Objectives

Calculate the pH and read the pH scale
Explain the acid–base properties of water and the concept of auto-ionisation
Describe the degree of dissociation ($\alpha$)
Explain the relationship between $K_a$ and $K_b$

The pH Scale

Introduced by Danish biochemist Søren Sørensen (1909). The pH scale measures the concentration of hydrogen ions in aqueous solution.

Neutral: $[H^+] = 10^{-7}$ mol dm⁻³ and pH = 7
Acidic: $[H^+] > 10^{-7}$ mol dm⁻³ and pH < 7
Basic: $[H^+] < 10^{-7}$ mol dm⁻³ and pH > 7

Definitions

Acidic solution: $$pH = -\log[H^+] = \log\frac{1}{[H^+]}$$

Or, using the hydronium ion: $$pH = -\log[H_3O^+] = \log\frac{1}{[H_3O^+]}$$

Basic solution: $$pOH = -\log[OH^-] = \log\frac{1}{[OH^-]}$$

Relationship between pOH and pH

$$pH + pOH = 14 \quad \text{(at 25 °C)}$$

Or, more generally: $$pH + pOH = pK_w$$

[OH3+]

[OH-]

Ionic Product of Water ($K_w$)

The auto-ionisation of water: $$H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$$

From this equilibrium, the dissociation constant for water (the ionic product) is: $$K_w = [H_3O^+][OH^-] = [H^+][OH^-]$$

At 298 K (25 °C): $$K_w = 1.0 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$$

Taking negative logarithms: $$-\log K_w = -\log[H^+] - \log[OH^-]$$ $$14 = pH + pOH$$

[!note] Temperature Dependence The ionisation of water is an endothermic process. The ionic product increases rapidly with temperature.

Temperature (°C) $K_w$ (mol² dm⁻⁶)

0 $1.1 \times 10^{-15}$

20 $6.8 \times 10^{-15}$

50 $5.5 \times 10^{-14}$

100 $5.1 \times 10^{-13}$

Temperature (°C)	$K_w$ (mol² dm⁻⁶)
0	$1.1 \times 10^{-15}$
20	$6.8 \times 10^{-15}$
50	$5.5 \times 10^{-14}$
100	$5.1 \times 10^{-13}$

Degree of Dissociation ($\alpha$)

The degree of dissociation applies to weak acids and weak bases, which only partially dissociate in water.

Weak Acid

$$\alpha = \frac{[H_3O^+]{\text{equilibrium}}}{[\text{acid}]{\text{Initial}}} = \frac{[H^+]_\rightleftharpoons}{[\text{acid}]_0}$$

Or as percentage ionisation: $$%\alpha = \frac{[H_3O^+]_\rightleftharpoons}{[\text{acid}]_0} \times 100%$$

For weak acids: $\alpha < 1$ or $%\alpha < 100%$
$\alpha$ is affected by the concentration of the acid
$K_a$ is not affected by concentration

Weak Base

$$\alpha = \frac{[OH^-]{\text{equilibrium}}}{[\text{base}]{\text{Initial}}}$$

Or as percentage ionisation: $$%\alpha = \frac{[OH^-]_\rightleftharpoons}{[\text{base}]_0} \times 100%$$

For strong bases: $\alpha = 1$ or $%\alpha = 100%$
For weak bases: $\alpha < 1$ or $%\alpha < 100%$
$\alpha$ is affected by the concentration of the base

Acid Dissociation Constant ($K_a$) and Weak Acids

For a generic weak acid HA: $$HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$$

$$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$

pKa

$$pK_a = -\log K_a$$

Assumption Method

For weak acids where $K_a$ is small, the change in concentration $x$ is very small compared to the initial concentration $C_0$:

$$(C_0 - x) \approx C_0$$

[!warning] When is the assumption valid? The assumption is typically valid when the resulting degree of dissociation is small ($%\alpha < 10%$). If $K_a$ or $K_b$ is given and is very small, the lecture explicitly applies $(C - x) \approx C$.

Example: Nitrous Acid ($HNO_2$)

O=NO

Calculate the molarity of $H_3O^+$ in 0.1 M $HNO_2$ ($K_a = 5 \times 10^{-4}$):

	$HNO_2$	$H_3O^+$	$NO_2^-$
Initial	0.1	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.1 - x$	$x$	$x$

$$K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1}$$

$$x = 7.07 \times 10^{-3} \text{ mol dm}^{-3}$$

$$pH = -\log(7.07 \times 10^{-3}) = 2.15$$

Example: Ethanoic Acid ($CH_3COOH$)

CC(=O)O

CC(=O)[O-]

Given 1.0 M $CH_3COOH$ with $K_a = 1.8 \times 10^{-5}$:

$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$
Using ICE table and assumption: $x = 4.24 \times 10^{-3}$ M
$pH = -\log(4.24 \times 10^{-3}) = 2.37$
Degree of dissociation: $\alpha = \dfrac{4.24 \times 10^{-3}}{1.0} = 4.24 \times 10^{-3}$

Base Dissociation Constant ($K_b$) and Weak Bases

For a generic weak base B: $$B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)$$

$$K_b = \frac{[BH^+][OH^-]}{[B]}$$

pKb

$$pK_b = -\log K_b$$

Example: Ammonia ($NH_3$)

[NH4+]

What is the pH of 0.40 M ammonia? ($K_b = 1.8 \times 10^{-5}$)

	$NH_3$	$NH_4^+$	$OH^-$
Initial	0.4	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.4 - x$	$x$	$x$

$$K_b = \frac{x^2}{0.4 - x} \approx \frac{x^2}{0.4}$$

$$x = 2.68 \times 10^{-3} \text{ mol dm}^{-3}$$

$$pOH = -\log(2.68 \times 10^{-3}) = 2.57$$ $$pH = 14 - 2.57 = 11.43$$

Example: Aniline ($C_6H_5NH_2$)

Nc1ccccc1

Given 0.1 M aniline with $K_b = 3.8 \times 10^{-10}$:

$pK_b = -\log(3.8 \times 10^{-10}) = 9.42$
Using assumption: $x = 6.164 \times 10^{-6}$ M
$pOH = -\log(6.164 \times 10^{-6}) = 5.21$
$pH = 14 - 5.21 = 8.79$
$\alpha = \dfrac{6.164 \times 10^{-6}}{0.1} = 6.164 \times 10^{-5}$

Relationship between $\alpha$, $K_a$ and $K_b$

When the degree of dissociation is small ($%\alpha < 10%$), the following approximations are valid:

For weak acids: $$\alpha = \sqrt{\frac{K_a}{c}}$$

For weak bases: $$\alpha = \sqrt{\frac{K_b}{c}}$$

[!important] Validity Condition These formulae are only valid when $%\alpha < 10%$.

The degree of dissociation is inversely proportional to the square root of the molarity / concentration.

Worked Exercises from Lecture

Strong Acid pH Calculations

0.001 mol dm⁻³ nitric acid ($HNO_3$) — strong acid, fully dissociates: $$pH = -\log(0.001) = 3$$
$[H^+] = 3.5 \times 10^{-3}$ mol dm⁻³: $$pH = -\log(3.5 \times 10^{-3}) = 2.5$$
0.01 M hydrochloric acid ($HCl$): $$pH = -\log(0.01) = 2$$

O=[N+]([O-])O

Cl

Strong Base pH Calculations

0.05 M potassium hydroxide ($KOH$): $$pOH = -\log(0.05) = 1.3$$ $$pH = 14 - 1.3 = 12.7$$
NaOH solution with pH = 13.5: $$pOH = 0.5$$ $$[OH^-] = 10^{-0.5} = 0.32 \text{ M}$$

[K+].[OH-]

[Na+].[OH-]

Mixed Calculations

Coffee with $[OH^-] = 2.5 \times 10^{-9}$ mol dm⁻³:
- Method 1: $pOH = 8.6$, $pH = 5.4$, $[H^+] = 10^{-5.4} = 3.98 \times 10^{-6}$ M
- Method 2: $[H^+] = \dfrac{K_w}{[OH^-]} = \dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-9}} = 4.00 \times 10^{-6}$ M
$2.0 \times 10^{-3}$ M HCl at 25 °C:
- $[H^+] = 2.0 \times 10^{-3}$ M
- $pH = 2.70$
- $pOH = 11.30$
- $[OH^-] = 5.01 \times 10^{-12}$ M

Weak Acid Calculations

0.1 M $HNO_2$ ($K_a = 5 \times 10^{-4}$):
- $[H_3O^+] = 7.07 \times 10^{-3}$ M, $pH = 2.15$
1.0 M $CH_3COOH$ ($K_a = 1.8 \times 10^{-5}$):
- $pK_a = 4.74$, $pH = 2.37$, $\alpha = 4.24 \times 10^{-3}$
0.1 M $CH_3COOH$, % ionisation = 1.34 %:
- $[H^+] = 1.34 \times 10^{-3}$ M
0.10 M HCOOH, pH = 2.39:
- $[H^+] = 10^{-2.39} = 4.07 \times 10^{-3}$ M
- $K_a = \dfrac{(4.07 \times 10^{-3})^2}{0.10 - 4.07 \times 10^{-3}} = 1.73 \times 10^{-4}$
0.30 M HA, % dissociation = 2.0 %:
- $[H_3O^+] = \dfrac{2.0}{100} \times 0.30 = 6.00 \times 10^{-3}$ M
- $K_a = \dfrac{(6.00 \times 10^{-3})^2}{0.30 - 6.00 \times 10^{-3}} = 1.22 \times 10^{-4}$
- $pH = -\log(6.00 \times 10^{-3}) = 2.22$
0.025 M HF ($K_a = 7.1 \times 10^{-4}$) — solved without assumption (quadratic equation):
- $[H_3O^+] = 3.85 \times 10^{-3}$ M
- $pH = 2.41$

[F-]

O=CO

Oc1ccccc1

[O-]c1ccccc1

0.5 M phenol ($C_6H_5OH$, $K_a = 1.3 \times 10^{-10}$):
- $[H_3O^+] = 8.06 \times 10^{-6}$ M
- $pH = 5.09$

Weak Base Calculations

0.40 M $NH_3$ ($K_b = 1.8 \times 10^{-5}$):
- $[OH^-] = 2.68 \times 10^{-3}$ M
- $pOH = 2.57$, $pH = 11.43$
0.1 M aniline ($K_b = 3.8 \times 10^{-10}$):
- $pK_b = 9.42$, $pH = 8.79$, $\alpha = 6.164 \times 10^{-5}$
1.0 M dimethylamine ($(CH_3)_2NH$, $K_b = 5.4 \times 10^{-4}$):
- $[OH^-] = 2.30 \times 10^{-2}$ M
- $pOH = 1.64$, $pH = 12.36$

CNC

CN

c1ccncc1

0.26 M methylamine ($CH_3NH_2$, $K_b = 4.4 \times 10^{-4}$):
- $[OH^-] = 1.05 \times 10^{-2}$ M
- $pOH = 1.98$, $pH = 12.02$
0.062 M pyridine ($C_5H_5N$, $K_b = 1.7 \times 10^{-9}$):
- $[OH^-] = 1.02 \times 10^{-5}$ M
- $pOH = 4.99$, $pH = 9.01$

Degree of Dissociation Applications

0.4 M acetic acid, $K_a = 1.8 \times 10^{-5}$: $$\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.4}} = 6.71 \times 10^{-3}$$
0.50 M $NH_3$, % dissociation = 6.3 %:
- $[OH^-] = \dfrac{6.3}{100} \times 0.5 = 3.15 \times 10^{-2}$ M
- $pOH = 1.5$, $pH = 12.5$

Key Takeaways

Strong acids/bases: Completely dissociate; no need for ICE tables or $K_a$/$K_b$
Weak acids/bases: Partially dissociate; use ICE tables with $K_a$ or $K_b$
Assumption: For weak acids/bases with small $K$ values, $(C - x) \approx C$
Temperature matters: $K_w$ changes with temperature; $pH + pOH = 14$ only at 25 °C
Degree of dissociation: Inversely proportional to concentration; use $\alpha = \sqrt{K/c}$ when $%\alpha < 10%$