Lecture on Titrations by Robert Smith at New River Community College

Introduction

The lecture focuses on understanding and solving titration problems.
Objective: Work through two example problems to gain a better understanding of titration calculations.

Problem Statement: 32.84 mL of KOH is needed to neutralize 0.4586 g of HCl. Find the molarity of KOH.

Convert Volume to Liters:
- 32.84 mL to liters:
  - Calculation: ( 32.84 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.03284 \text{ L} )
Convert Grams to Moles:
- Find molar mass of HCl:
  - ( 35.45 \text{ g/mol (Cl)} + 1.008 \text{ g/mol (H)} = 36.46 \text{ g/mol} )
- Convert grams to moles:
  - ( 0.4586 \text{ g} \times \frac{1 \text{ mol}}{36.46 \text{ g}} = 1.258 \times 10^{-2} \text{ mol} )
Calculate Molarity:
- Formula: ( M = \frac{moles}{volume} )
- ( M = \frac{1.258 \times 10^{-2} \text{ mol}}{0.03284 \text{ L}} = 0.3831 \text{ M} )
- 1:1 ratio between HCl and KOH.

Problem Statement: Calculate volume of 0.160 M KOH needed to neutralize 30.0 mL of 0.425 M H₃PO₄.

Convert Volume to Liters:
- 30.0 mL to liters:
  - ( 30.0 \text{ mL} = 0.0300 \text{ L} )
Find Moles of H₃PO₄:
- ( 0.425 \text{ moles/L} \times 0.0300 \text{ L} = 0.0128 \text{ moles} )
Determine Moles of KOH Needed:
- Consider the equivalency: 1 mole of H₃PO₄ produces 3 moles H⁺.
- Calculate moles of KOH:
  - ( 3 \times 0.0128 \text{ moles} = 0.0384 \text{ moles} )
Calculate Volume of KOH:
- Use molarity to find volume:
  - Formula: ( V = \frac{moles}{molarity} )
  - ( V = \frac{0.0384 \text{ moles}}{0.160 \text{ M}} = 0.240 \text{ L} )
Convert Volume to Milliliters:
- ( 0.240 \text{ L} \times 1000 = 240 \text{ mL} )

The importance of unit conversions and correct expression of answers in desired units (liters to milliliters).
Always double-check the specific requirement for units before finalizing the answer.
The lecture demonstrates the application of basic chemistry principles and arithmetic in solving titration problems.