GENERAL CHEMISTRY II WORK OUT EXAMPLES
Equilibrium
Example 1. Calculating K₍c₎ from Equilibrium Concentrations
Problem: For the reaction
A+B⇌C
if at equilibrium [A] = 0.20 M, [B] = 0.30 M, and [C] = 0.10 M, calculate K₍c₎.
Formula:
K₍c₎ = [C]⁄([A][B])
Solution:
K₍c₎ = 0.10 / (0.20 × 0.30) = 0..06 = 1.67
Answer: K₍c₎ = 1.67
Example 2. Using an ICE Table to Determine Equilibrium Concentrations
Problem: For the reaction
X₂(g) ⇌ 2X(g)
if the initial concentration of X₂ is 0.50 M and no X is present initially, and 40% of X₂
dissociates at equilibrium, calculate [X₂] and [X].
Solution Outline:
1. Initial: [X₂] = 0.50 M; [X] = 0
2. Change: Let dissociation = 0.40 × 0.50 = 0.20 M; then [X₂] decreases by 0.20 and [X]
increases by 2(0.20) = 0.40 M.
3. Equilibrium: [X₂] = 0.50 – 0.20 = 0.30 M; [X] = 0.40 M
Answer: [X₂] = 0.30 M and [X] = 0.40 M
Example 3. Predicting the Direction of Reaction Using Q vs. K
Problem: For A ⇌ B, if K = 2.0 and the reaction mixture has [A] = 0.5 M and [B] = 0.5 M,
determine which direction the reaction will proceed.
Formula:
Q = [B]⁄[A] = 0.5/0.5 = 1.0
Comparison: Since Q (1.0) < K (2.0), the reaction will proceed to produce more B.
Answer: The reaction shifts to the right (toward B).
Acids and Bases
Example 4. Calculating pH of a Strong Acid Solution
Problem: Calculate the pH of a 0.010 M HCl solution.
Formula:
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, pH = –log[H⁺]
Solution:
pH = –log(0.010) = 2.00
Answer: pH = 2.00
Example 5. pH of a Weak Acid Solution
Problem: A 0.10 M solution of acetic acid (CH₃COOH) has a Ka = 1.8 × 10⁻⁵. Estimate its pH.
Approach: Use the approximation for a weak acid:
[H⁺] ≈ √(Ka·C)
Calculation:
[H⁺] ≈ √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M
pH = –log(1.34×10⁻³) ≈ 2.87
Answer: pH ≈ 2.87
Example 6. Using the Henderson–Hasselbalch Equation for a Buffer
Problem: A buffer is made from 0.20 M acetic acid (pKa = 4.76) and 0.10 M sodium acetate.
Calculate the pH.
Formula:
pH = pKa + log([A⁻]/[HA])
Solution:
pH = 4.76 + log(0.10/0.20) = 4.76 + log(0.5) ≈ 4.76 – 0.30 = 4.46
Answer: pH ≈ 4.46
Example 7. Titration of a Weak Acid with a Strong Base
Problem: When 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH, find the pH at the
equivalence point.
Concept: At the equivalence point, the conjugate base (acetate) is formed. Its Kb = Kw/Ka =
1.0×10⁻¹⁴/1.8×10⁻⁵ ≈ 5.56×10⁻¹⁰.
Calculation (using an approximation):
Volume at equivalence = 25.0 mL + 25.0 mL = 50.0 mL; [CH₃COO⁻] = (0.10 M × 25.0
mL) / 50.0 mL = 0.05 M.
For base hydrolysis, [OH⁻] ≈ √(Kb × [CH₃COO⁻]) ≈ √(5.56×10⁻¹⁰ × 0.05) ≈ √(2.78×10⁻¹¹)
≈ 5.27×10⁻⁶ M.
pOH = –log(5.27×10⁻⁶) ≈ 5.28, so pH = 14 – 5.28 ≈ 8.72
Answer: pH ≈ 8.72
Solubility Equilibria
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Equilibrium
Example 1. Calculating K₍c₎ from Equilibrium Concentrations
Problem: For the reaction
A+B⇌C
if at equilibrium [A] = 0.20 M, [B] = 0.30 M, and [C] = 0.10 M, calculate K₍c₎.
Formula:
K₍c₎ = [C]⁄([A][B])
Solution:
K₍c₎ = 0.10 / (0.20 × 0.30) = 0..06 = 1.67
Answer: K₍c₎ = 1.67
Example 2. Using an ICE Table to Determine Equilibrium Concentrations
Problem: For the reaction
X₂(g) ⇌ 2X(g)
if the initial concentration of X₂ is 0.50 M and no X is present initially, and 40% of X₂
dissociates at equilibrium, calculate [X₂] and [X].
Solution Outline:
1. Initial: [X₂] = 0.50 M; [X] = 0
2. Change: Let dissociation = 0.40 × 0.50 = 0.20 M; then [X₂] decreases by 0.20 and [X]
increases by 2(0.20) = 0.40 M.
3. Equilibrium: [X₂] = 0.50 – 0.20 = 0.30 M; [X] = 0.40 M
Answer: [X₂] = 0.30 M and [X] = 0.40 M
Example 3. Predicting the Direction of Reaction Using Q vs. K
Problem: For A ⇌ B, if K = 2.0 and the reaction mixture has [A] = 0.5 M and [B] = 0.5 M,
determine which direction the reaction will proceed.
Formula:
Q = [B]⁄[A] = 0.5/0.5 = 1.0
Comparison: Since Q (1.0) < K (2.0), the reaction will proceed to produce more B.
Answer: The reaction shifts to the right (toward B).
Acids and Bases
Example 4. Calculating pH of a Strong Acid Solution
Problem: Calculate the pH of a 0.010 M HCl solution.
Formula:
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, pH = –log[H⁺]
Solution:
pH = –log(0.010) = 2.00
Answer: pH = 2.00
Example 5. pH of a Weak Acid Solution
Problem: A 0.10 M solution of acetic acid (CH₃COOH) has a Ka = 1.8 × 10⁻⁵. Estimate its pH.
Approach: Use the approximation for a weak acid:
[H⁺] ≈ √(Ka·C)
Calculation:
[H⁺] ≈ √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M
pH = –log(1.34×10⁻³) ≈ 2.87
Answer: pH ≈ 2.87
Example 6. Using the Henderson–Hasselbalch Equation for a Buffer
Problem: A buffer is made from 0.20 M acetic acid (pKa = 4.76) and 0.10 M sodium acetate.
Calculate the pH.
Formula:
pH = pKa + log([A⁻]/[HA])
Solution:
pH = 4.76 + log(0.10/0.20) = 4.76 + log(0.5) ≈ 4.76 – 0.30 = 4.46
Answer: pH ≈ 4.46
Example 7. Titration of a Weak Acid with a Strong Base
Problem: When 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH, find the pH at the
equivalence point.
Concept: At the equivalence point, the conjugate base (acetate) is formed. Its Kb = Kw/Ka =
1.0×10⁻¹⁴/1.8×10⁻⁵ ≈ 5.56×10⁻¹⁰.
Calculation (using an approximation):
Volume at equivalence = 25.0 mL + 25.0 mL = 50.0 mL; [CH₃COO⁻] = (0.10 M × 25.0
mL) / 50.0 mL = 0.05 M.
For base hydrolysis, [OH⁻] ≈ √(Kb × [CH₃COO⁻]) ≈ √(5.56×10⁻¹⁰ × 0.05) ≈ √(2.78×10⁻¹¹)
≈ 5.27×10⁻⁶ M.
pOH = –log(5.27×10⁻⁶) ≈ 5.28, so pH = 14 – 5.28 ≈ 8.72
Answer: pH ≈ 8.72
Solubility Equilibria
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