10 - Molarity and Titration

Molarity

  • Definition of Molarity:

    • Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters.

    • Formula: M = moles of solute / liters of solution.

  • Understanding Concentration:

    • Concentration refers to the amount of solute present in a certain volume of solution.

Molarity Triangle

  • Visual representation for calculating molarity:

    • Top: Moles of solute

    • Bottom Left: Molarity

    • Bottom Right: Volume (liters)

  • Using the Triangle:

    • M = moles/V

    • moles = M × V

    • V = moles/M

Problem Example: Sugar Cube Problem

  • Problem Details:

    • 4g of sucrose (C12H22O11) in 350 mL of water.

  • Calculating Molarity:

    • Convert grams to moles using molar mass (342 g/mol):

      • 4g × (1 mol / 342 g) = 0.0117 moles.

    • Convert 350 mL to liters: 350 mL = 0.350 L.

    • Molarity: M = 0.0117 mol / 0.350 L = 0.033 M.

Key Concepts about Molarity

  • Molarity is Independent of Quantity:

    • Dilute or concentrated solutions of the same substance have the same molarity regardless of volume.

  • Dilution Changes Molarity:

    • To change molarity, dilution (addition of solvent) is required.

Additional Problems

  1. Grams of H3PO4 in a Solution:

    • Given a 3.5 M solution and a volume of 175 mL, find grams of the solute:

    • Convert volume to liters: 0.175 L.

    • Calculate moles: Moles = M × V = 3.5 mol/L × 0.175 L = 0.6125 moles.

    • Convert moles to grams using molar mass (98 g/mol):

      • Grams = 0.6125 mol × 98 g/mol = 60.0 g.

  2. Calculating Concentration of Phosphate Ions:

    • Given 50 mL of 0.25 M calcium phosphate and needed to find phosphate molarity:

    • Moles of Ca3(PO4)2 = 0.25 M × 0.050 L = 0.0125 moles.

    • Moles of phosphate ions (P04) = 2 × 0.0125 = 0.025 moles.

    • Molarity of phosphate ions = 0.025 moles / 0.050 L = 0.50 M.

  3. Final Concentration After Mixing Solutions:

    • Mixing results of calcium phosphate with K3PO4.

    • Calculate individual moles from given volumes and molarities, then combine for final concentration.

Dilution Calculations

  • Dilution Formula: M1V1 = M2V2

    • Use this to find unknown concentration after dilution.

    • Example Problem: If 175 mL of 1.6 M is diluted to 1 L, calculate new concentration.

Titration Basics

  • Titration Process:

    • Neutralizes an acid with a base to find unknown concentration.

    • Use stoichiometry to relate moles of acid and base.

    • Common example: HCl and NaOH reaction.

Oxalic Acid Percent Composition Problem

  • Calculate grams from a neutralization reaction to find % composition.

    • Required NaOH volume and concentration gives moles of H2C2O4.

    • Convert moles to grams and calculate percent:

      • % Oxalic acid = (mass of oxalic acid / total mass) × 100.

Practice Problems

  • Review all aforementioned problems and exercises to solidify understanding of concepts.

  • Focus on calculations involving molarity, dilution, and stoichiometry.