Lecture Notes: Acid-Base Titration Problems

Introduction

• Focus: Solving acid-base titration problems.
• Example Problems: Finding concentration, determining volumes, and calculating mass.

Problem 1: Concentration of HCl Solution

• Given: 24.7 mL of HCl is neutralized by 35.8 mL of 0.25 M NaOH.
• Methods:
  1. Using the Formula:
     • Identify substances: HCl is monoprotic; NaOH has one OH- ion.
     • Use formula: (M_1V_1 = M_2V_2), where M = molarity, V = volume.
     • Calculate: (M_1 = \frac{M_2V_2}{V_1} = \frac{0.25 \times 35.8}{24.7})
     • Result: (M_1 = 0.362 \text{ M}).
  2. Dimensional Analysis:
     • Write balanced equation: HCl + NaOH → H2O + NaCl.
     • Calculate moles of NaOH, then moles of HCl (1:1 ratio).
     • Convert volumes to liters.
     • Result: Same concentration, 0.362 M.

Problem 2: Volume of Barium Hydroxide Solution

• Given: 0.15 M Ba(OH)2 required to neutralize 45 mL of 0.29 M HNO3.
• Methods:
  1. Modified Equation:
     • Consider Ba(OH)2 has 2 hydroxide ions.
     • Use (M_1V_1 = 2M_2V_2) for base.
     • Calculate: (V_2 = \frac{M_1V_1}{2M_2} = \frac{0.29 \times 45}{0.3})
     • Result: 43.5 mL.
  2. Stoichiometry:
     • Write balanced equation: Ba(OH)2 + 2HNO3 → 2H2O + Ba(NO3)2.
     • Convert moles using volume in liters and concentration.
     • Use molar ratio (1:2) for conversion.
     • Result: Same volume, 43.5 mL.

Problem 3: Mass of KHP

• Given: 32.57 mL of 0.175 M NaOH to neutralize KHP.
• Approach:
  • Balanced equation: KHP + NaOH → H2O + KNaP.
  • Calculate moles of NaOH → moles of KHP (1:1 ratio).
  • Use molar mass of KHP (204.22 g/mol) to find mass.
  • Result: 1.164 grams of KHP.

Problem 4: Concentration of KOH Solution

• Given: 42.6 mL KOH required to titrate 0.137 g KHP.
• Approach:
  • Balanced equation: KHP + KOH → H2O + K2P.
  • Convert grams of KHP to moles.
  • Use molar ratio (1:1) for moles of KOH.
  • Find concentration (M = \frac{\text{moles}}{\text{liters}}).
  • Result: 0.01 M KOH.

Conclusion

• Multiple methods (formula, stoichiometry) applicable depending on the problem.
• Importance of balanced chemical equations and molar ratios in titration calculations.