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 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Weak Acids, Weak Bases, and K_a/K_b study guide free?▾
Yes — all study notes, flashcards, and practice problems for Weak Acids, Weak Bases, and K_a/K_b on Study Mondo are 100% free. No account is needed to access the content.
What course covers Weak Acids, Weak Bases, and K_a/K_b?▾
Weak Acids, Weak Bases, and K_a/K_b is part of the AP Chemistry course on Study Mondo, specifically in the Acids and Bases section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Weak Acids, Weak Bases, and K_a/K_b?
For pH: Usually only first ionization matters (K_a1 >> K_a2)
CH₃COO⁻
I
0.10
0
0
C
-x
+x
+x
E
0.10-x
x
x
Write K_a expression:
Ka=[CH3COOH][H+][CH3COO−]=0.10−xx2
Check for approximation:
Ka[HA]0=1.8×10−50.10=5.6×103
Since ratio > 100, try small x approximation:
1.8×10−5=0.10x2
x2=(1.8×10−5)(0.10)
x2=1.8×10−6
x=1.34×10−3
Check validity:
[HA]0x=0.101.34×10−3=0.0134=1.34%
1.34% < 5% ✓ Approximation valid!
Calculate pH:
[H⁺] = x = 1.34 × 10⁻³ M
pH=−log(1.34×10−3)
pH=2.87
Answer: pH = 2.87
Percent ionization:
%=0.101.34×10−3×100%=1.3%
Only 1.3% of acetic acid ionized (weak acid!)
2Problem 2medium
❓ 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:
Kb=[NH
Check approximation:
Kb[B] ✓
Use approximation:
1.8×10−5=0.25x2
x2=(1.8×10−5)(0.25)
x2=4.5×10−6
x=2.12×10−3
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)
pOH=2.67
Calculate pH:
pH=14.00−pOH
pH=14.00−2.67
pH=11.33
Answer: pH = 11.33
Interpretation:
pH > 7: basic solution ✓
Weak base partially ionizes
Only ~0.85% ionized
3Problem 3hard
❓ 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+]=2.09×10−3 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:
Ka=[HA][H
Ka=0.498(2.09×10
Ka=0.4984.37×10−
Ka=8.8×10−6
Answer (a): K_a = 8.8 × 10⁻⁶
(b) Calculate percent ionization
% ionization=[HA]0[H
% ionization=0.502.09×10−3×
% ionization=0.418%
Answer (b): 0.42% ionization
(c) Calculate K_b for conjugate base A⁻
Use relationship:
Ka×Kb=Kw
Kb=KaK
Kb=8.8×10−6
Kb=1.1×10−9
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)
▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.