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 Ka < Kb
- 4) Answer: C. 3.33 M
- 5) Answer: D. Kb = [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