Strong vs weak acids - bases; degree of dissociation α

Overview

Not all acids and bases are created equal. Some completely ionize in water (strong), while others only partially dissociate (weak). The degree of dissociation (α) quantifies this behavior and determines solution pH, buffer capacity, and reaction dynamics.

Core Concepts

[!intuition] Why Some Acids Are "Stronger" Than Others

Think of an acid as a company holding onto hydrogen ions (H⁺) as employees. A strong acid is a terrible employer—the moment it dissolves in water, all the H⁺ "employees" quit and leave (100% dissociation). A weak acid is a better employer—most H⁺ ions stay bonded, only a few leave at any time.

Why does this happen?

• Strong acids: The conjugate base (A⁻) is extremely stable after losing H⁺. The ion-dipole interactions with water strongly favor the dissociated state. The reverse reaction (recombination) has a negligible rate.
• Weak acids: The conjugate base is less stable or the acid form is relatively stable. An equilibrium forms where bothHA and H⁺ + A⁻ coexist.

Physical basis: Bond polarity, charge delocalization in the conjugate base, and solvation energy determine strength.

[!definition] Strong vs Weak Acids and Bases

Strong Acid: An acid that completely dissociates in aqueous solution. $\text{HA} \xrightarrow{\text{H}_2\text{O}} \text{H}^+ + \text{A}^- \quad (\text{100\% dissociation})$

Common strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄

Weak Acid: An acid that partially dissociates in solution, establishing an equilibrium. $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

Common weak acids: CH₃COOH, HF, H₂CO₃, H₃PO₄

Strong Base: A base that completely dissociates in solution. $\text{BOH} \xrightarrow{\text{H}_2\text{O}} \text{B}^+ + \text{OH}^-$

Common strong bases: NaOH, KOH, Ba(OH)₂, Ca(OH)₂

Weak Base: A base that partially accepts protons, establishing equilibrium. $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$

Common weak bases: NH₃, CH₃NH₂, pyridine

[!definition] Degree of Dissociation (α)

The degree of dissociation (α) is the fraction of the original acid or base molecules that dissociate into ions at equilibrium.

$\alpha = \frac{\text{amount dissociated}}{\text{initial amount}} = \frac{c - c_{\text{eq}}}{c}$

Where:

• $c$ = initial concentration
• $c_{\text{eq}}$ = equilibrium concentration of undissociated species

Alternative form (often more useful): $\alpha = \frac{[\text{H}^+]}{c_0} \quad \text{(for acids)}$ $\alpha = \frac{[\text{OH}^-]}{c_0} \quad \text{(for bases)}$

Key insights:

• For strong acids/bases: $\alpha \approx 1$ (essentially complete dissociation)
• For weak acids/bases: $\alpha \ll 1$ (typically 0.01 to 0.1 for moderate concentrations)
• $\alpha$ depends on concentration—weaker solutions have higher α (Le Chatelier's principle)

[!formula] Relationship Between α and Ka (Derivation)

Starting point: Consider a weak acid HA with initial concentration $c$.

Step 1: Write the dissociation equilibrium $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

Step 2: Set up ICE table (Initial, Change, Equilibrium)

Why this works: If α is the fraction dissociated, then $c\alpha$ moles dissociate, leaving $c(1-\alpha)$ undissociated.

Step 3: Write the equilibrium constant expression $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}$

Step 4: For weak acids where $\alpha \ll 1$, we can approximate $1 - \alpha \approx 1$: $K_a \approx c\alpha^2$

Solving for α: $\boxed{\alpha \approx \sqrt{\frac{K_a}{c}}}$

Why does α decrease with concentration? According to Le Chatelier's principle, adding moreHA shifts equilibrium left (toward undissociated form). The system "resists" complete dissociation at higher concentrations.

Exact formula (no approximation): $\boxed{\alpha = \frac{-K_a + \sqrt{K_a^2 + 4cK_a}}{2c}}$

Derivation: From $K_a = \frac{c\alpha^2}{1-\alpha}$, multiply through: $K_a(1-\alpha) = c\alpha^2$ $K_a - K_a\alpha = c\alpha^2$ $c\alpha^2 + K_a\alpha - K_a = 0$

Using quadratic formula with $\alpha$ as the variable: $\alpha = \frac{-K_a \pm \sqrt{K_a^2 + 4cK_a}}{2c}$

We take the positive root since α must be positive.

[!example] Worked Examples

Example 1: Finding α for Acetic Acid

Problem: Calculate the degree of dissociation of 0.1 M acetic acid (CH₃COOH). Given: $K_a = 1.8 \times 10^{-5}$.

Solution:

Step 1: Check if approximation is valid $\frac{c}{K_a} = \frac{0.1}{1.8 \times 10^{-5}} = 5555 \gg 100$

The "rule of 100" is satisfied, so we can use the approximation.

Why this step? The approximation $1 - \alpha \approx 1$ is only valid when α is small (typically < 5%). The ratio $c/K_a$ tells us if dissociation will be minimal.

Step 2: Apply the approximate formula $\alpha \approx \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = \sqrt{1.8 \times 10^{-4}} = 0.0134$

Step 3: Convert to percentage $\alpha = 1.34\%$

Interpretation: Only 1.34% of acetic acid molecules dissociate. At equilibrium:

• $[\text{H}^+] = c\alpha = 0.1 \times 0.0134 = 1.34 \times 10^{-3}$ M
• pH = $-\log(1.34 \times 10^{-3}) = 2.87$

Why this matters: This explains why vinegar (5% acetic acid) doesn't burn your skin like HCl would—most molecules remain undissociated.

Example 2: Concentration Dependence of α

Problem: Calculate α for acetic acid at 0.01 M and 1.0 M. Compare with 0.1 M (from Example 1).

Solution:

For 0.01 M: $\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.01}} = \sqrt{1.8 \times 10^{-3}} = 0.0424 = 4.24\%$

For 1.0 M: $\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{1.0}} = \sqrt{1.8 \times 10^{-5}} = 0.00424 = 0.424\%$

Summary table:

Why does this happen?

• Lower concentration → More dilution → Water molecules can stabilize ions better → Equilibrium shifts right → Higher α
• Higher concentration → Ions collide more frequently → More recombination → Lower α

Important: Even though α decreases, [H⁺] still increases with concentration (just not linearly).

Example 3: Strong vs Weak Acid Comparison

Problem: Compare 0.1 M HCl (strong) and 0.1 M acetic acid (weak).

HCl (strong acid):

• $\alpha = 1$ (complete dissociation)
• $[\text{H}^+] = 0.1$ M
• pH = 1.0

CH₃COOH (weak acid):

• $\alpha = 0.0134$
• $[\text{H}^+] = 1.34 \times 10^{-3}$ M
• pH = 2.87

Key insight: At the same concentration, strong acid produces 75 times more H⁺ ions than acetic acid. This is why HCl is far more corrosive and dangerous.

[!mistake] Common Errors

Mistake 1: Confusing α with Ka

Wrong thinking: "A larger Ka means larger α, so just compare Ka values to compare α."

Why this feels right: Ka does relate to acid strength, and stronger acids do tend to have higher α.

The fix: α depends on both Ka and concentration: $\alpha = \sqrt{\frac{K_a}{c}}$

Two acids can have the same Ka but different α if concentrations differ. Or the same α but different Ka.

Example:

• 0.1 M acetic acid ($K_a = 1.8 \times 10^{-5}$): α = 1.34%
• 0.01 M formic acid ($K_a = 1.8 \times 10^{-4}$): α = $\sqrt{\frac{1.8 \times 10^{-4}}{0.01}}$ = 13.4%

Formic is a stronger acid (higher Ka), but at the right concentrations, both can have arbitrary α values.

Mistake 2: Treating Weak Acids as Strong

Wrong thinking: "CH₃COOH has formula CH₃COOH, so it releases 1 H⁺ per molecule. For0.1 M solution, [H⁺] = 0.1 M."

Why this feels right: The stoichiometry says one H⁺ per molecule, and we learned to use stoichiometry in early chemistry.

The fix: Equilibrium overides stoichiometry. Only α fraction dissociates: $[\text{H}^+] = c \cdot \alpha, \text{ not } c$

For 0.1 M acetic acid, [H⁺] = 0.00134 M, not 0.1 M.

Consequence of this mistake: You'd calculate pH = 1instead of pH = 2.87—a massive error that would make you think vinegar is as acidic as stomach acid.

Mistake 3: Ignoring Water Autoionization

Wrong thinking: "For a very weak acid or very dilute solution, I can calculate [H⁺] from Ka alone."

The fix: When calculated [H⁺] from acid dissociation approaches $10^{-7}$ M, you must account for water's autoionization: $[\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{acid}} + 10^{-7}$

When does this matter?

• Very weak acids ($K_a < 10^{-7}$)
• Very dilute solutions (< $10^{-6}$ M)

Example: 0.001 M acetic acid gives calculated [H⁺] ≈ $4.2 \times 10^{-7}$ M from acid alone, but water contributes $10^{-7}$ M, so actual [H⁺] ≈ $5.2 \times 10^{-7}$ M. Ignoring water gives pH = 6.37 instead of pH = 6.28.

[!recall]- Explain to a 12-Year-Old

Imagine you have a box of Legos stuck together. Some Lego sets are held together with super glue (weak acids)—when you put them in water, only a few pieces come apart. Other sets are just loosely snapped together (strong acids)—the moment water touches them, ALL the pieces fall apart instantly.

The "degree of dissociation" (α) is like asking: "What percentage of my Lego pieces came apart?" For super-glued sets, maybe only 1-5% come apart. For loose sets, 100% come apart.

Here's the cool part: if you have MORE Lego sets in the water, the pieces that DO come apart have more chance of bumping into their partners and snapping back together. So bigger piles of Legos → lower percentage falling apart (even though the total number of loose pieces is still higher).

Strong acids are like Legos with NO glue. Weak acids are like Legos with varying amounts of glue. The Ka number tells you how much glue there is—smaller Ka means stronger glue, so fewer pieces come apart.

[!mnemonic] Memory Aids

Strong Acids: "Hary Clarke Brought Ice Nightly, So Please Clap Often"

• HCl, Bromide (HBr), Iodide (HI), Nitric (HNO₃), Sulfuric (H₂SO₄), Percloric (*HClO₄)

α Formula: "Square root because of squared term in equilibrium expression"

• $K_a = c\alpha^2/(1-\alpha)$ → when solved, α has a square root

Concentration Effect: "Dilute → Dissociates more" (Higher α at lower c)

Connections

• Acid-base equilibria and Ka, Kb—α is directly calculated from Ka
• pH and pOH calculations—α determines [H⁺], which determines pH
• Buffer solutions—bufers work because weak acids have α < 1
• Common ion effect—adding A⁻ suppresses α of HA
• Ostwald's dilution law—formalizes the $\alpha = \sqrt{K_a/c}$ relationship
• Polyprotic acids—multiple α values for each dissociation step
• Hydrolysis of salts—related concept for ions acting as acids/bases

Key Takeaways

1. Strong acids/bases completely dissociate (α ≈ 1); weak acids/bases partially dissociate (α ≪ 1)
2. ==Degree of dissociation α = (amount dissociated)/(initial amount)==
3. For weak acids: $\alpha \approx \sqrt{K_a/c}$ (when $c/K_a > 100$)
4. α decreases with increasing concentration due to Le Chatelier's principle
5. Never assume complete dissociation unless the acid/base is strong
6. The relationship $K_a = c\alpha^2/(1-\alpha)$ connects thermodynamics (Ka) with kinetics/extent (α)

#flashcards/chemistry

What is the degree of dissociation (α)? :: The fraction of acid or base molecules that dissociate into ions at equilibrium. Formula: α = (amount dissociated)/(initial amount).

If two acids have the same Ka but different concentrations, what can you say about their α values? :: The more dilute solution will have higher α, since α = √(Ka/c) and α is inversely related to √c.

Concept Map

Hinglish (regional understanding)