Chapter 9 - Stoichiometry (Dilution)


Dilution Practice Problems (Dimensional Analysis Method)

Problem 1: Diluting Concentrated HCl

  • Objective: Determine the volume of 3.5 M HCl stock solution needed to prepare 400 mL of a 1.5 M dilute solution.

  • Note: Solved via Dimensional Analysis instead of M₁V₁ = M₂V₂.

  • Setup & Calculation:

    ? mL = 400 mL (dilute) × (1 L / 1000 mL) × (1.5 mol / 1 L) × (1 L / 3.5 mol) × (1000 mL / 1 L) = 170 mL

  • Steps Breakdown:

    1. Convert target dilute volume (400 mL) to Liters (L).

    2. Use the target dilute concentration (1.5 M) to find required moles of HCl.

    3. Use the inverted stock concentration (3.5 M) to convert moles back to Liters of stock.

    4. Convert stock Liters back to milliliters (mL).

  • Final Answer: 170 mL

Problem 2: Diluting NaCl Solution

  • Objective: Determine the final volume needed when diluting 5.00 mL of a 6.0 M NaCl stock solution into a 1.33 M dilute solution.

  • Setup & Calculation:

    ? mL = 5.00 mL (stock) × (1 L / 1000 mL) × (6.0 mol / 1 L) × (1 L / 1.33 mol) × (1000 mL / 1 L) = 23 mL

  • Steps Breakdown:

    1. Convert initial stock volume (5.00 mL) to Liters (L).

    2. Multiply by stock molarity (6.0 M) to get total moles of solute.

    3. Divide by target dilute molarity (1.33 M) to find required dilute volume in Liters.

    4. Convert Liters to milliliters (mL).

  • Final Answer: 23 mL

Problem 3: Calculating Final Molarity After Dilution

  • Objective: Determine the molarity (mol/L) of a solution when 65 mL of 8.5 M NaOH stock is diluted to a total volume of 2.0 L.

  • Setup & Calculation:

    ? mol/L = 65 mL × (1 L / 1000 mL) × (8.5 mol / 1 L) × (1 / 2.0 L) = 0.28 M NaOH

  • Steps Breakdown:

    1. Convert initial stock volume (65 mL) to Liters (L).

    2. Multiply by stock molarity (8.5 M) to calculate total moles of solute.

    3. Divide by the new total diluted volume (2.0 L).

  • Final Answer: 0.28 M NaOH

Supplemental Structural Notation

  • Hydrate Coefficient Example: 3 XY · 2H₂O

    • Demonstrates how a scalar coefficient (3) distributes to all components of a hydrate formula unit.