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Weak Acids, Weak Bases, and K_a/K_b

Understand weak acid/base equilibria, acid and base dissociation constants (K_a and K_b), and percent ionization.

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Weak Acids, Weak Bases, and K_a/K_b

Weak Acids

Definition: Partially ionizes in water (< 100%)

General equilibrium:

HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)HA\text{(aq)} + H2O\text{(l)} \rightleftharpoons H3O^+\text{(aq)} + A^-\text{(aq)}

Simplified: HA ⇌ H⁺ + A⁻

Examples: CH₃COOH, HF, HNO₂, H₃PO₄

Acid Dissociation Constant (K_a)

Equilibrium expression:

Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}

K_a magnitude:

  • Larger K_a = stronger acid
  • Typical range: 10⁻² to 10⁻¹⁴
  • Strong acids: K_a >> 1

Common weak acids:

AcidK_a
HF6.8 × 10⁻⁴
HNO₂4.5 × 10⁻⁴
CH₃COOH1.8 × 10⁻⁵
H₂CO₃4.3 × 10⁻⁷

Weak Acid pH Calculation

Use ICE table:

Example: HA ⇌ H⁺ + A⁻

HAH⁺A⁻
I[HA]₀00
C-x+x+x
E[HA]₀-xxx

K_a expression:

Ka=x2[HA]0−xK_a = \frac{x^2}{[HA]_0 - x}

Solve for x = [H⁺], then pH = -log x

Small x Approximation (5% Rule)

When [HA]₀/K_a > 100:

Assume: [HA]₀ - x ≈ [HA]₀

Ka≈x2[HA]0K_a \approx \frac{x^2}{[HA]_0}

x=Ka⋅[HA]0x = \sqrt{K_a \cdot [HA]_0}

Check validity:

  • Calculate x
  • If x/[HA]₀ < 5%, approximation valid
  • If x/[HA]₀ > 5%, use quadratic

Weak Bases

Definition: Partially ionize by accepting H⁺

General equilibrium:

B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)B\text{(aq)} + H2O\text{(l)} \rightleftharpoons BH^+\text{(aq)} + OH^-\text{(aq)}

Examples: NH₃, CH₃NH₂, pyridine

Base Dissociation Constant (K_b)

Equilibrium expression:

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

Common weak bases:

BaseK_b
NH₃1.8 × 10⁻⁵
CH₃NH₂4.4 × 10⁻⁴
C₅H₅N (pyridine)1.7 × 10⁻⁹

Weak Base pH Calculation

ICE table: B + H₂O ⇌ BH⁺ + OH⁻

BBH⁺OH⁻
I[B]₀00
C-x+x+x
E[B]₀-xxx

K_b expression:

Kb=x2[B]0−xK_b = \frac{x^2}{[B]_0 - x}

Solve for x = [OH⁻]:

  1. Calculate x from K_b
  2. pOH = -log[OH⁻]
  3. pH = 14.00 - pOH

Relationship between K_a and K_b

For conjugate acid-base pair:

Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0 \times 10^{-14}

Example: NH₃/NH₄⁺

  • K_b(NH₃) = 1.8 × 10⁻⁵
  • K_a(NH₄⁺) = K_w/K_b = 5.6 × 10⁻¹⁰

Stronger acid → weaker conjugate base Stronger base → weaker conjugate acid

Percent Ionization

Measure of acid/base strength:

\text{% ionization} = \frac{[H^+]_{\text{eq}}}{[HA]_0} \times 100\%

For weak acids:

  • Typically < 5%
  • Increases with dilution
  • Larger K_a → larger % ionization

Example: 0.10 M acetic acid

  • K_a = 1.8 × 10⁻⁵
  • [H⁺] = 1.3 × 10⁻³ M
  • % ionization = (1.3 × 10⁻³/0.10) × 100% = 1.3%

pK_a and pK_b

Analogous to pH:

pKa=−log⁡Ka\text{pK}_a = -\log K_a

pKb=−log⁡Kb\text{pK}_b = -\log K_b

Relationship:

pKa+pKb=14.00\text{pK}_a + \text{pK}_b = 14.00

Interpretation:

  • Smaller pK_a = stronger acid
  • Smaller pK_b = stronger base

Polyprotic Acids

Multiple ionizable protons:

Example: H₂SO₃

First ionization: H₂SO₃ ⇌ H⁺ + HSO₃⁻

  • K_a1 (larger)

Second ionization: HSO₃⁻ ⇌ H⁺ + SO₃²⁻

  • K_a2 (smaller)

Pattern: K_a1 >> K_a2 >> K_a3

For pH: Usually only first ionization matters (K_a1 >> K_a2)

📚 Practice Problems

1Problem 1easy

❓ Question:

Calculate the pH of a 0.10 M solution of acetic acid (CH₃COOH). K_a = 1.8 × 10⁻⁵.

💡 Show Solution

Given:

  • [CH₃COOH]₀ = 0.10 M
  • K_a = 1.8 × 10⁻⁵

Equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻

Set up ICE table:

CH₃COOHH⁺CH₃COO⁻
I0.1000
C-x+x+x
E0.10-xxx

Write K_a expression:

Ka=[H+][CH3COO−][CH3COOH]=x20.10−xK_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x^2}{0.10-x}

Check for approximation:

[HA]0Ka=0.101.8×10−5=5.6×103\frac{[HA]_0}{K_a} = \frac{0.10}{1.8 \times 10^{-5}} = 5.6 \times 10^3

Since ratio > 100, try small x approximation:

1.8×10−5=x20.101.8 \times 10^{-5} = \frac{x^2}{0.10}

x2=(1.8×10−5)(0.10)x^2 = (1.8 \times 10^{-5})(0.10)

x2=1.8×10−6x^2 = 1.8 \times 10^{-6}

x=1.34×10−3x = 1.34 \times 10^{-3}

Check validity:

x[HA]0=1.34×10−30.10=0.0134=1.34%\frac{x}{[HA]_0} = \frac{1.34 \times 10^{-3}}{0.10} = 0.0134 = 1.34\%

1.34% < 5% ✓ Approximation valid!

Calculate pH:

[H⁺] = x = 1.34 × 10⁻³ M

pH=−log⁡(1.34×10−3)\text{pH} = -\log(1.34 \times 10^{-3})

pH=2.87\text{pH} = 2.87

Answer: pH = 2.87

Percent ionization:

%=1.34×10−30.10×100%=1.3%\% = \frac{1.34 \times 10^{-3}}{0.10} \times 100\% = 1.3\%

Only 1.3% of acetic acid ionized (weak acid!)

2Problem 2hard

❓ Question:

Calculate the pH of a 0.10 M solution of acetic acid (CH₃COOH) given K_a = 1.8 × 10⁻⁵.

💡 Show Solution

Solution:

Equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻

ICE table:

CH₃COOHH⁺CH₃COO⁻
Initial0.1000
Change-x+x+x
Equil.0.10-xxx

K_a expression: K_a = [H⁺][CH₃COO⁻] / [CH₃COOH] 1.8 × 10⁻⁵ = x² / (0.10 - x)

Simplification: Assume x << 0.10, so 0.10 - x ≈ 0.10 1.8 × 10⁻⁵ = x² / 0.10 x² = 1.8 × 10⁻⁶ x = 1.34 × 10⁻³ M = [H⁺]

Check assumption: 1.34 × 10⁻³ / 0.10 = 0.0134 = 1.34% < 5% ✓

Calculate pH: pH = -log(1.34 × 10⁻³) = 2.87

3Problem 3medium

❓ Question:

Calculate the pH of a 0.25 M ammonia (NH₃) solution. K_b = 1.8 × 10⁻⁵.

💡 Show Solution

Given:

  • [NH₃]₀ = 0.25 M
  • K_b = 1.8 × 10⁻⁵

Equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

Set up ICE table:

NH₃NH₄⁺OH⁻
I0.2500
C-x+x+x
E0.25-xxx

Write K_b expression:

Kb=[NH4+][OH−][NH3]=x20.25−xK_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{x^2}{0.25-x}

Check approximation:

[B]0Kb=0.251.8×10−5=1.4×104>100\frac{[B]_0}{K_b} = \frac{0.25}{1.8 \times 10^{-5}} = 1.4 \times 10^4 > 100 ✓

Use approximation:

1.8×10−5=x20.251.8 \times 10^{-5} = \frac{x^2}{0.25}

x2=(1.8×10−5)(0.25)x^2 = (1.8 \times 10^{-5})(0.25)

x2=4.5×10−6x^2 = 4.5 \times 10^{-6}

x=2.12×10−3x = 2.12 \times 10^{-3}

Check: x/[B]₀ = 2.12×10⁻³/0.25 = 0.85% < 5% ✓

Calculate pOH:

[OH⁻] = x = 2.12 × 10⁻³ M

pOH=−log⁡(2.12×10−3)\text{pOH} = -\log(2.12 \times 10^{-3})

pOH=2.67\text{pOH} = 2.67

Calculate pH:

pH=14.00−pOH\text{pH} = 14.00 - \text{pOH}

pH=14.00−2.67\text{pH} = 14.00 - 2.67

pH=11.33\text{pH} = 11.33

Answer: pH = 11.33

Interpretation:

  • pH > 7: basic solution ✓
  • Weak base partially ionizes
  • Only ~0.85% ionized

4Problem 4hard

❓ Question:

Calculate the pH of a 0.50 M solution of ammonia (NH₃) given K_b = 1.8 × 10⁻⁵.

💡 Show Solution

Solution:

Equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

ICE table:

NH₃NH₄⁺OH⁻
Initial0.5000
Change-x+x+x
Equil.0.50-xxx

K_b expression: K_b = [NH₄⁺][OH⁻] / [NH₃] 1.8 × 10⁻⁵ = x² / (0.50 - x)

Assume x << 0.50: 1.8 × 10⁻⁵ = x² / 0.50 x² = 9.0 × 10⁻⁶ x = 3.0 × 10⁻³ M = [OH⁻]

Check: 3.0 × 10⁻³ / 0.50 = 0.006 = 0.6% < 5% ✓

Calculate pOH: pOH = -log(3.0 × 10⁻³) = 2.52

Calculate pH: pH = 14.00 - 2.52 = 11.48

5Problem 5hard

❓ Question:

The pH of a 0.50 M solution of a weak acid (HA) is 2.68. Calculate: (a) K_a, (b) percent ionization, (c) K_b for the conjugate base A⁻.

💡 Show Solution

Given:

  • [HA]₀ = 0.50 M
  • pH = 2.68

Equilibrium: HA ⇌ H⁺ + A⁻

(a) Calculate K_a

Find [H⁺] from pH:

[H+]=10−pH=10−2.68[H^+] = 10^{-\text{pH}} = 10^{-2.68}

[H+]=2.09×10−3 M[H^+] = 2.09 \times 10^{-3} \text{ M}

ICE table:

HAH⁺A⁻
I0.5000
C-2.09×10⁻³+2.09×10⁻³+2.09×10⁻³
E0.4982.09×10⁻³2.09×10⁻³

At equilibrium:

  • [HA] = 0.50 - 0.00209 = 0.498 M
  • [H⁺] = [A⁻] = 2.09 × 10⁻³ M

Calculate K_a:

Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}

Ka=(2.09×10−3)(2.09×10−3)0.498K_a = \frac{(2.09 \times 10^{-3})(2.09 \times 10^{-3})}{0.498}

Ka=4.37×10−60.498K_a = \frac{4.37 \times 10^{-6}}{0.498}

Ka=8.8×10−6K_a = 8.8 \times 10^{-6}

Answer (a): K_a = 8.8 × 10⁻⁶

(b) Calculate percent ionization

% ionization=[H+]eq[HA]0×100%\% \text{ ionization} = \frac{[H^+]_{\text{eq}}}{[HA]_0} \times 100\%

% ionization=2.09×10−30.50×100%\% \text{ ionization} = \frac{2.09 \times 10^{-3}}{0.50} \times 100\%

% ionization=0.418%\% \text{ ionization} = 0.418\%

Answer (b): 0.42% ionization

(c) Calculate K_b for conjugate base A⁻

Use relationship:

Ka×Kb=KwK_a \times K_b = K_w

Kb=KwKaK_b = \frac{K_w}{K_a}

Kb=1.0×10−148.8×10−6K_b = \frac{1.0 \times 10^{-14}}{8.8 \times 10^{-6}}

Kb=1.1×10−9K_b = 1.1 \times 10^{-9}

Answer (c): K_b = 1.1 × 10⁻⁹

Summary:

PropertyValue
K_a (HA)8.8 × 10⁻⁶
% ionization0.42%
K_b (A⁻)1.1 × 10⁻⁹

Note: HA is moderately weak acid; A⁻ is very weak base (small K_b)

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple ChoiceMCQ6090 min50%✅
Free Response (Long)FRQ369 min30%✅
Free Response (Short)FRQ436 min20%✅

📊 Scoring: 1-5

5
Extremely Qualified
~12%
4
Well Qualified
~16%
3
Qualified
~24%
2
Possibly Qualified
~24%
1
No Recommendation
~24%

💡 Key Test-Day Tips

  • ✓Memorize common polyatomic ions
  • ✓Practice dimensional analysis
  • ✓Know your gas laws

⚠️ Common Mistakes: Weak Acids, Weak Bases, and K_a/K_b

Avoid these 3 frequent errors

🌍 Real-World Applications: Weak Acids, Weak Bases, and K_a/K_b

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

22 mol of H2H_2 reacts with 11 mol of O2O_2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O2H_2 + O_2 \to 2H_2O)

2Determine the limiting reagent
3Calculate moles of product
4Convert moles to grams
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Buffer Solutions and Henderson-Hasselbalch
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📌 Related Topics in Acids and Bases

Acid-Base Theories and pH ScaleBuffer Solutions and Henderson-HasselbalchAcid-Base Titrations and Indicators

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Understand weak acid/base equilibria, acid and base dissociation constants (K_a and K_b), and percent ionization.
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