Acid-Base Equilibrium and Titration Practice Problems

Multiple Choice Answer Key

• 1) Answer: B. H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
• 2) Answer: B. NH₄⁺(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NH₃(aq)
• 3) Answer: B. Basic since K_a < K_b
• 4) Answer: C. 3.33 M
• 5) Answer: D. K_b = [H₂CO₃][OH⁻] / [HCO₃⁻]

Additional Answers: 6. B, 7. A, 8. B, 9. B, 10. B, 11. A, 12. C, 13. D, 14. D, 15. B, 16. B, 17. A

Detailed Solutions for Chemistry Problems

Question 1: Titration of Oxalic Acid

Part A: Determining NaOH Molarity

Trial 1 Volume: 24.00 mL − 5.85 mL = 18.15 mL (Outlier)
Trial 2 Volume: 40.05 mL − 24.00 mL = 16.05 mL
Trial 3 Volume: 21.45 mL − 5.50 mL = 15.95 mL

Average Volume: (16.05 mL + 15.95 mL) / 2 = 16.00 mL = 0.01600 L

Chemical Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

• Moles H₂C₂O₄: (0.175 M)(0.01600 L) = 0.00280 mol
• Moles NaOH: 0.00280 mol × (2 mol NaOH / 1 mol H₂C₂O₄) = 0.00560 mol
• [NaOH] Concentration: 0.00560 mol / 0.02500 L = 0.224 M

Answer: [NaOH] = 0.224 M

Part B: pH Prediction

Reaction: C₂O₄²⁻(aq) + H₂O(l) &rightleftharpoons; HC₂O₄⁻(aq) + OH⁻(aq)

The production of OH⁻ makes the solution basic; therefore, the pH > 7.

Question 2: Calculating the Acid Dissociation Constant

[H₃O⁺]: 10^−3.518 = 3.034 × 10⁻⁴ M

ICE Table for HX

HX + H₂O &rightleftharpoons; H₃O⁺ + X⁻

• Initial (I): 0.090 M (HX), 0 (H₃O⁺), 0 (X⁻)
• Change (C): −3.034 × 10⁻⁴, +3.034 × 10⁻⁴, +3.034 × 10⁻⁴
• Equilibrium (E): 0.0897 M, 3.034 × 10⁻⁴ M, 3.034 × 10⁻⁴ M

K_a Calculation:
K_a = ([H₃O⁺][X⁻]) / [HX]
K_a = (3.034 × 10⁻⁴)² / 0.0897 = 1.03 × 10⁻⁶

Answer: K_a = 1.03 × 10⁻⁶

Question 3: Calculating the Base Dissociation Constant

pOH: 14.00 − 11.68 = 2.32
[OH⁻]: 10^−2.32 = 4.786 × 10⁻³ M

ICE Table for CH₃NH₂

CH₃NH₂ + H₂O &rightleftharpoons; CH₃NH₃⁺ + OH⁻

• Initial (I): 0.0700 M, 0, 0
• Change (C): −4.786 × 10⁻³, +4.786 × 10⁻³, +4.786 × 10⁻³
• Equilibrium (E): 0.0652 M, 4.786 × 10⁻³ M, 4.786 × 10⁻³ M

K_b Calculation:
K_b = ([CH₃NH₃⁺][OH⁻]) / [CH₃NH₂]
K_b = (4.786 × 10⁻³)² / 0.0652 = 3.51 × 10⁻⁴

Answer: K_b = 3.51 × 10⁻⁴

Question 4: Calculating pH of an Ammonia Solution

K_b Calculation:
K_b = K_w / K_a = (1.0 × 10⁻¹⁴) / (5.6 × 10⁻¹⁰) = 1.79 × 10⁻⁵

ICE Table for NH₃

NH₃ + H₂O &rightleftharpoons; NH₄⁺ + OH⁻

• Initial (I): 2.00 M, 0, 0
• Change (C): −x, +x, +x
• Equilibrium (E): 2.00 − x, x, x

Solving for x:
1.79 × 10⁻⁵ = x² / (2.00 − x)
Assume 2.00 − x &approx; 2.00:
x² = 3.58 × 10⁻⁵
x = 5.98 × 10⁻³ M = [OH⁻]

Final pH Calculation:
pOH = −log(5.98 × 10⁻³) = 2.22
pH = 14.00 − 2.22 = 11.78

Answer: pH = 11.78

Final Answers Summary

• 1a) 0.224 M
• 1b) pH > 7 because C₂O₄²⁻ produces OH⁻
• 2) K_a = 1.03 × 10⁻⁶
• 3) K_b = 3.51 × 10⁻⁴
• 4) pH = 11.78