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Weak Acids, Weak Bases, and K_a/K_b | Study Mondo
Topics / Acids and Bases / Weak Acids, Weak Bases, and K_a/K_b 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
Definition: Partially ionizes in water (< 100%)
General equilibrium:
H A (aq) + H 2 O (l) ⇌ H 3 O + (aq) + A − (aq) HA\text{(aq)} + H2O\text{(l)} \rightleftharpoons H3O^+\text{(aq)} + A^-\text{(aq)} H A (aq) + H 2 O (l) ⇌ H 3 O
📚 Practice Problems
1 Problem 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:
Explain using: 📝 Simple words 🔗 Analogy 🎨 Visual desc. 📐 Example 💡 Explain
📋 AP Chemistry — Exam Format Guide⏱ 3 hours 15 minutes 📝 67 questions 📊 3 sections
Section Format Questions Time Weight Calculator Multiple Choice MCQ 60 90 min 50% ✅ Free Response (Long) FRQ 3 69 min 30% ✅ Free Response (Short) FRQ 4 36 min 20% ✅
💡 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_bAvoid these 3 frequent errors
1 Not balancing equations before doing stoichiometry
▾ 2 Confusing molarity (M) with molality (m)
▾ 3 Forgetting to convert temperature to Kelvin for gas law problems
▾ 🌍 Real-World Applications: Weak Acids, Weak Bases, and K_a/K_bSee how this math is used in the real world
🌍 Water Purification
Environment
▾ 💻 Battery Technology
Technology
▾
📝 Worked Example: Stoichiometry — Limiting ReagentProblem: 2 2 2 H 2 H_2 H 2 1 1 1 O 2 O_2 O 2 2 H 2 + O 2 → 2 H 2 O 2H_2 + O_2 \to 2H_2O 2 H 2 + O 2 → 2 H 2 O
1 Write the balanced equation Click to reveal →
2 Determine the limiting reagent
3 Calculate moles of product
🧪 Practice Lab Interactive practice problems for Weak Acids, Weak Bases, and K_a/K_b
▾ 📌 Related Topics in Acids and Bases❓ Frequently Asked QuestionsWhat is 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.
How can I study Weak Acids, Weak Bases, and K_a/K_b effectively?▾ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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💡 Study Tips✓ Work through examples step-by-step ✓ Practice with flashcards daily ✓ Review common mistakes +
(aq)
+
A − (aq)
Examples: CH₃COOH, HF, HNO₂, H₃PO₄
Acid Dissociation Constant (K_a) K a = [ H + ] [ A − ] [ H A ] K_a = \frac{[H^+][A^-]}{[HA]} K a = [ H A ] [ H + ] [ A − ]
Larger K_a = stronger acid
Typical range: 10⁻² to 10⁻¹⁴
Strong acids: K_a >> 1
Acid K_a HF 6.8 × 10⁻⁴ HNO₂ 4.5 × 10⁻⁴ CH₃COOH 1.8 × 10⁻⁵ H₂CO₃ 4.3 × 10⁻⁷
Weak Acid pH Calculation HA H⁺ A⁻ I [HA]₀ 0 0 C -x +x +x E [HA]₀-x x x
K a = x 2 [ H A ] 0 − x K_a = \frac{x^2}{[HA]_0 - x} K a = [ H A ] 0 − x x 2
Solve for x = [H⁺], then pH = -log x
Small x Approximation (5% Rule) Assume: [HA]₀ - x ≈ [HA]₀
K a ≈ x 2 [ H A ] 0 K_a \approx \frac{x^2}{[HA]_0} K a ≈ [ H A ] 0 x 2
x = K a ⋅ [ H A ] 0 x = \sqrt{K_a \cdot [HA]_0} x = K a ⋅ [ H A ] 0
Calculate x
If x/[HA]₀ < 5%, approximation valid
If x/[HA]₀ > 5%, use quadratic
Weak Bases Definition: Partially ionize by accepting H⁺
B (aq) + H 2 O (l) ⇌ B H + (aq) + O H − (aq) B\text{(aq)} + H2O\text{(l)} \rightleftharpoons BH^+\text{(aq)} + OH^-\text{(aq)} B (aq) + H 2 O (l) ⇌ B H + (aq) + O H − (aq)
Examples: NH₃, CH₃NH₂, pyridine
Base Dissociation Constant (K_b) K b = [ B H + ] [ O H − ] [ B ] K_b = \frac{[BH^+][OH^-]}{[B]} K b = [ B ] [ B H + ] [ O H − ]
Base K_b NH₃ 1.8 × 10⁻⁵ CH₃NH₂ 4.4 × 10⁻⁴ C₅H₅N (pyridine) 1.7 × 10⁻⁹
Weak Base pH Calculation ICE table: B + H₂O ⇌ BH⁺ + OH⁻
B BH⁺ OH⁻ I [B]₀ 0 0 C -x +x +x E [B]₀-x x x
K b = x 2 [ B ] 0 − x K_b = \frac{x^2}{[B]_0 - x} K b = [ B ] 0 − x x 2
Calculate x from K_b
pOH = -log[OH⁻]
pH = 14.00 - pOH
Relationship between K_a and K_b For conjugate acid-base pair:
K a × K b = K w = 1.0 × 10 − 14 K_a \times K_b = K_w = 1.0 \times 10^{-14} K a × K b = K w = 1.0 × 1 0 − 14
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\%
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 pK a = − log K a \text{pK}_a = -\log K_a pK a = − log K a
pK b = − log K b \text{pK}_b = -\log K_b pK b = − log K b
pK a + pK b = 14.00 \text{pK}_a + \text{pK}_b = 14.00 pK a + pK b = 14.00
Smaller pK_a = stronger acid
Smaller pK_b = stronger base
Polyprotic Acids Multiple ionizable protons:
First ionization: H₂SO₃ ⇌ H⁺ + HSO₃⁻
Second ionization: HSO₃⁻ ⇌ H⁺ + SO₃²⁻
Pattern: K_a1 >> K_a2 >> K_a3
For pH: Usually only first ionization matters (K_a1 >> K_a2)
K a = [ H + ] [ C H 3 C O O − ] [ C H 3 C O O H ] = x 2 0.10 − x K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x^2}{0.10-x} K a = [ C H 3 COO H ] [ H + ] [ C H 3 CO O − ] = 0.10 − x x 2
[ H A ] 0 K a = 0.10 1.8 × 10 − 5 = 5.6 × 10 3 \frac{[HA]_0}{K_a} = \frac{0.10}{1.8 \times 10^{-5}} = 5.6 \times 10^3 K a [ H A ] 0 = 1.8 × 1 0 − 5 0.10 = 5.6 × 1 0 3
Since ratio > 100, try small x approximation:
1.8 × 10 − 5 = x 2 0.10 1.8 \times 10^{-5} = \frac{x^2}{0.10} 1.8 × 1 0 − 5 = 0.10 x 2
x 2 = ( 1.8 × 10 − 5 ) ( 0.10 ) x^2 = (1.8 \times 10^{-5})(0.10) x 2 = ( 1.8 × 1 0 − 5 ) ( 0.10 )
x 2 = 1.8 × 10 − 6 x^2 = 1.8 \times 10^{-6} x 2 = 1.8 × 1 0 − 6
x = 1.34 × 10 − 3 x = 1.34 \times 10^{-3} x = 1.34 × 1 0 − 3
x [ H A ] 0 = 1.34 × 10 − 3 0.10 = 0.0134 = 1.34 % \frac{x}{[HA]_0} = \frac{1.34 \times 10^{-3}}{0.10} = 0.0134 = 1.34\% [ H A ] 0 x = 0.10 1.34 × 1 0 − 3 = 0.0134 = 1.34%
1.34% < 5% ✓ Approximation valid!
pH = − log ( 1.34 × 10 − 3 ) \text{pH} = -\log(1.34 \times 10^{-3}) pH = − log ( 1.34 × 1 0 − 3 )
pH = 2.87 \text{pH} = 2.87 pH = 2.87
% = 1.34 × 10 − 3 0.10 × 100 % = 1.3 % \% = \frac{1.34 \times 10^{-3}}{0.10} \times 100\% = 1.3\% % = 0.10 1.34 × 1 0 − 3 × 100% = 1.3%
Only 1.3% of acetic acid ionized (weak acid!)
2 Problem 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:
CH₃COOH H⁺ CH₃COO⁻ Initial 0.10 0 0 Change -x +x +x Equil. 0.10-x x x
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
3 Problem 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:
NH₃ NH₄⁺ OH⁻ I 0.25 0 0 C -x +x +x E 0.25-x x x
Write K_b expression:
K b = [ N H 4 + ] [ O H − ] [ N H 3 ] = x 2 0.25 − x K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{x^2}{0.25-x} K b = [ N H
Check approximation:
[ B ] 0 K b = 0.25 1.8 × 10 − 5 = 1.4 × 10 4 > 100 \frac{[B]_0}{K_b} = \frac{0.25}{1.8 \times 10^{-5}} = 1.4 \times 10^4 > 100 K b [ B ]
Use approximation:
1.8 × 10 − 5 = x 2 0.25 1.8 \times 10^{-5} = \frac{x^2}{0.25} 1.8 × 1 0 − 5 = 0.25 x 2
x 2 = ( 1.8 × 10 − 5 ) ( 0.25 ) x^2 = (1.8 \times 10^{-5})(0.25) x 2 = ( 1.8 × 1 0 − 5 ) ( 0.25 )
x 2 = 4.5 × 10 − 6 x^2 = 4.5 \times 10^{-6} x 2 = 4.5 × 1 0 − 6
x = 2.12 × 10 − 3 x = 2.12 \times 10^{-3} x = 2.12 × 1 0 − 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 = − log ( 2.12 × 1 0 − 3 )
pOH = 2.67 \text{pOH} = 2.67 pOH = 2.67
Calculate pH:
pH = 14.00 − pOH \text{pH} = 14.00 - \text{pOH} pH = 14.00 − pOH
pH = 14.00 − 2.67 \text{pH} = 14.00 - 2.67 pH = 14.00 − 2.67
pH = 11.33 \text{pH} = 11.33 pH = 11.33
Answer: pH = 11.33
Interpretation:
pH > 7: basic solution ✓
Weak base partially ionizes
Only ~0.85% ionized
4 Problem 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:
NH₃ NH₄⁺ OH⁻ Initial 0.50 0 0 Change -x +x +x Equil. 0.50-x x x
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
5 Problem 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:
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 + ] = 1 0 − pH = 1 0 − 2.68
[ H + ] = 2.09 × 10 − 3 M [H^+] = 2.09 \times 10^{-3} \text{ M} [ H + ] = 2.09 × 1 0 − 3 M
ICE table:
HA H⁺ A⁻ I 0.50 0 0 C -2.09×10⁻³ +2.09×10⁻³ +2.09×10⁻³ E 0.498 2.09×10⁻³ 2.09×10⁻³
At equilibrium:
[HA] = 0.50 - 0.00209 = 0.498 M
[H⁺] = [A⁻] = 2.09 × 10⁻³ M
Calculate K_a:
K a = [ H + ] [ A − ] [ H A ] K_a = \frac{[H^+][A^-]}{[HA]} K a = [ H A ] [ H
K a = ( 2.09 × 10 − 3 ) ( 2.09 × 10 − 3 ) 0.498 K_a = \frac{(2.09 \times 10^{-3})(2.09 \times 10^{-3})}{0.498} K a = 0.498 ( 2.09 × 1 0
K a = 4.37 × 10 − 6 0.498 K_a = \frac{4.37 \times 10^{-6}}{0.498} K a = 0.498 4.37 × 1 0 −
K a = 8.8 × 10 − 6 K_a = 8.8 \times 10^{-6} K a = 8.8 × 1 0 − 6
Answer (a): K_a = 8.8 × 10⁻⁶
(b) Calculate percent ionization
% ionization = [ H + ] eq [ H A ] 0 × 100 % \% \text{ ionization} = \frac{[H^+]_{\text{eq}}}{[HA]_0} \times 100\% % ionization = [ H A ] 0 [ H
% ionization = 2.09 × 10 − 3 0.50 × 100 % \% \text{ ionization} = \frac{2.09 \times 10^{-3}}{0.50} \times 100\% % ionization = 0.50 2.09 × 1 0 − 3 ×
% ionization = 0.418 % \% \text{ ionization} = 0.418\% % ionization = 0.418%
Answer (b): 0.42% ionization
(c) Calculate K_b for conjugate base A⁻
Use relationship:
K a × K b = K w K_a \times K_b = K_w K a × K b = K w
K b = K w K a K_b = \frac{K_w}{K_a} K b = K a K
K b = 1.0 × 10 − 14 8.8 × 10 − 6 K_b = \frac{1.0 \times 10^{-14}}{8.8 \times 10^{-6}} K b = 8.8 × 1 0 − 6
K b = 1.1 × 10 − 9 K_b = 1.1 \times 10^{-9} K b = 1.1 × 1 0 − 9
Answer (c): K_b = 1.1 × 10⁻⁹
Summary:
Property Value K_a (HA) 8.8 × 10⁻⁶ % ionization 0.42% K_b (A⁻) 1.1 × 10⁻⁹
Note: HA is moderately weak acid; A⁻ is very weak base (small K_b)
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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
3
]
[ N H 4 + ] [ O H − ]
=
0.25 − x x 2
0
=
1.8 × 1 0 − 5 0.25 =
1.4 ×
1 0 4 >
100
+
]
[
A −
]
− 3
)
(
2.09
×
1
0 − 3
)
6
+
] eq
×
100%
100%
w
1.0 × 1 0 − 14