Strong vs Weak Acids and Bases

Overview

Acids and bases differ in their degree of ionization in water. Strong acids/bases ionize completely, while weak acids/bases only partially ionize.

Strong Acids

Definition

Ionize completely (100%) in water.

Common Strong Acids

Formula	Name
HCl	Hydrochloric acid
HBr	Hydrobromic acid
HI	Hydroiodic acid
HNO₃	Nitric acid
H₂SO₄	Sulfuric acid (1st H)
HClO₄	Perchloric acid
HClO₃	Chloric acid

Calculations

$$[\text{H}^+] = [\text{Strong Acid}]$$ $$\text{pH} = -\log[\text{Strong Acid}]$$

Example: 0.025 M HCl

$$[\text{H}^+] = 0.025 \text{ M}$$ $$\text{pH} = -\log(0.025) = 1.60$$

Strong Bases

Definition

Ionize completely in water.

Common Strong Bases

Formula	Name	OH⁻ per formula
LiOH	Lithium hydroxide	1
NaOH	Sodium hydroxide	1
KOH	Potassium hydroxide	1
Ca(OH)₂	Calcium hydroxide	2
Sr(OH)₂	Strontium hydroxide	2
Ba(OH)₂	Barium hydroxide	2

Calculations

$$[\text{OH}^-] = n \times [\text{Strong Base}] \quad (n = \text{number of OH}^-)$$ $$\text{pOH} = -\log[\text{OH}^-]$$ $$\text{pH} = 14 - \text{pOH}$$

Example: 0.010 M Ba(OH)₂

$$[\text{OH}^-] = 2 \times 0.010 = 0.020 \text{ M}$$ $$\text{pOH} = -\log(0.020) = 1.70$$ $$\text{pH} = 14 - 1.70 = 12.30$$

Weak Acids

Definition

Only partially ionize in water; establish equilibrium.

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

Common Weak Acids

Formula	Name	$K_a$	p$K_a$
HF	Hydrofluoric	$6.8 \times 10^{-4}$	3.17
CH₃COOH	Acetic	$1.8 \times 10^{-5}$	4.74
HNO₂	Nitrous	$4.0 \times 10^{-4}$	3.40
H₂CO₃	Carbonic	$4.3 \times 10^{-7}$	6.37
H₂S	Hydrosulfuric	$1.0 \times 10^{-7}$	7.00
HCN	Hydrocyanic	$4.9 \times 10^{-10}$	9.31

Calculating pH of Weak Acids

ICE Table Method:

	HA	H⁺	A⁻
I	C	0	0
C	$-x$	$+x$	$+x$
E	$C-x$	$x$	$x$

$$K_a = \frac{x^2}{C-x}$$

Approximation (if $C/K_a > 100$):

$$x = [\text{H}^+] = \sqrt{K_a \times C}$$ $$\text{pH} = -\log[\text{H}^+]$$

Example: 0.10 M acetic acid ($K_a = 1.8 \times 10^{-5}$)

$$[\text{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$$ $$\text{pH} = 2.87$$

Check: $\frac{1.34 \times 10^{-3}}{0.10} = 1.3\% < 5\%$ ✓

Weak Bases

Definition

Only partially ionize in water.

$$\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq)$$ $$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$$

Common Weak Bases

Formula	Name	$K_b$
NH₃	Ammonia	$1.8 \times 10^{-5}$
CH₃NH₂	Methylamine	$4.4 \times 10^{-4}$
C₅H₅N	Pyridine	$1.7 \times 10^{-9}$

Calculating pH of Weak Bases

$$[\text{OH}^-] = \sqrt{K_b \times C}$$ $$\text{pOH} = -\log[\text{OH}^-]$$ $$\text{pH} = 14 - \text{pOH}$$

Example: 0.15 M ammonia ($K_b = 1.8 \times 10^{-5}$)

$$[\text{OH}^-] = \sqrt{1.8 \times 10^{-5} \times 0.15} = 1.64 \times 10^{-3} \text{ M}$$ $$\text{pOH} = 2.79$$ $$\text{pH} = 14 - 2.79 = 11.21$$

Percent Ionization

$$\% \text{ Ionization} = \frac{[\text{H}^+]}{C_0} \times 100\%$$

For weak acids:

$$\% \text{ Ionization} = \sqrt{\frac{K_a}{C_0}} \times 100\% \quad \text{(if } C/K_a > 100\text{)}$$

Key point: Percent ionization increases as concentration decreases.

Conjugate Acid-Base Pairs

Relationship

$$K_a \times K_b = K_w = 1.0 \times 10^{-14}$$ $$\text{p}K_a + \text{p}K_b = 14$$

Finding $K_b$ from $K_a$

$$K_b = \frac{K_w}{K_a}$$

Example: For acetic acid ($K_a = 1.8 \times 10^{-5}$)

$$K_b(\text{acetate}) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$

Polyprotic Acids

Acids with more than one ionizable hydrogen.

H₂SO₄ (diprotic)

$$\text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \quad \text{(strong, complete)}$$ $$\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \quad (K_{a2} = 1.2 \times 10^{-2})$$

H₃PO₄ (triprotic)

$$K_{a1} = 7.5 \times 10^{-3}$$ $$K_{a2} = 6.2 \times 10^{-8}$$ $$K_{a3} = 3.6 \times 10^{-13}$$

Note: $K_{a1} \gg K_{a2} \gg K_{a3}$ (each successive H⁺ is harder to remove)

Comparison Summary

Property	Strong	Weak
Ionization	100%	Partial
$K_a$ or $K_b$	Very large	Small (< 1)
[H⁺] or [OH⁻]	= Concentration	< Concentration
pH calculation	Direct	Requires $K_a$/$K_b$
Equilibrium arrow	→	⇌
Conductivity	High	Low