Lecture Notes: Acid-Base Titration Problems

Overview

In this lecture, we focus on solving acid-base titration problems using different methods: an equation-based approach and dimensional analysis or stoichiometry.

Problem 1: Concentration of HCl Solution

• Problem Statement: 24.7 mL of an HCl solution is completely neutralized by 35.8 mL of a 0.25 M NaOH solution. Calculate the concentration of the HCl solution.
• Methods:
  • Equation-Based Method:
    • Recognize HCl as monoprotic and NaOH as having one hydroxide ion.
    • Use the formula M1V1 = M2V2 where M and V represent molarity and volume for the acid and base.
    • Calculation: M1 = (M2 * V2) / V1
      • M1 = (0.25 mol/L * 35.8 mL) / 24.7 mL
      • Result: M1 = 0.362 M
  • Dimensional Analysis:
    • Write a balanced chemical equation: HCl + NaOH → H2O + NaCl.
    • Convert volume of NaOH to liters: 35.8 mL = 0.0358 L.
    • Use molarity to find moles: 0.25 mol/L * 0.0358 L = moles of NaOH.
    • Convert to moles of HCl using a 1:1 ratio.
    • Convert volume of HCl to liters: 24.7 mL = 0.0247 L.
    • Divide moles of HCl by its volume to find molarity: Result is 0.362 M.*

Problem 2: Volume of Barium Hydroxide Solution

• Problem Statement: Determine the volume of 0.15 M Ba(OH)_2 solution required to neutralize 45 mL of 0.29 M HNO3.
• Methods:
  • Equation-Based Method:
    • Recognize HNO3 as monoprotic and Ba(OH)_2 as having two hydroxide ions.
    • Adjust the formula: (M1V1)acid = 2(M2V2)base.
    • Calculation: V2 = (M1V1)/(2M2)
      • Result: V2 = 43.5 mL
  • Stoichiometry:
    • Write a balanced chemical equation: Ba(OH)_2 + 2 HNO3 → Ba(NO3)_2 + 2 H2O.
    • Calculate moles of HNO3 and use the stoichiometric ratio to find moles of Ba(OH)_2.
    • Calculate the volume of Ba(OH)_2 in milliliters.
    • Result: V2 = 43.5 mL

Problem 3: Mass of KHP

• Problem Statement: What mass of KHP (potassium hydrogen phthalate) is neutralized by 32.57 mL of 0.175 M NaOH?
• Solution:
  • Recognize KHP as monoprotic.
  • Use stoichiometry since M1V1 = M2V2 does not apply to mass.
  • Balanced equation: KHP + NaOH → H2O + NaKP.
  • Calculate moles of NaOH and convert to moles of KHP.
  • Use the molar mass of KHP to find mass.
  • Result: Mass = 1.164 g

Problem 4: Concentration of KOH Solution

• Problem Statement: 42.6 mL of KOH is required to titrate 0.137 g of KHP. Find the concentration of KOH.
• Solution:
  • Recognize the 1:1 molar ratio between KHP and KOH.
  • Convert mass of KHP to moles using molar mass.
  • Find moles of KOH and divide by its volume in liters to find molarity.
  • Result: Molarity = 0.01 M

Key Concepts

• Monoprotic Acids and Bases: Acids and bases that donate or accept one proton per molecule.
• Balanced Chemical Equations: Essential for determining stoichiometric relationships and ratios used in calculations.
• Dimensional Analysis and Stoichiometry: Useful for finding unknown concentrations or volumes when dealing with masses or solutions not directly related to the molarity equation.

These problems illustrate the application of titration concepts and methods in solving different acid-base reaction scenarios.