Dilution Calculations and Concepts

Concentrated vs. Dilute Solutions

  • Concentrated Solution: A solution with a large amount of solute.
  • Dilute Solution: A solution with a small amount of solute.

Visual Representation

  • Concentration can be visually assessed by the intensity of color in a solution.
  • More color intensity indicates a more concentrated solution with more solute particles.

Molarity Calculation

M<em>1V</em>1=M<em>2V</em>2M<em>1V</em>1 = M<em>2V</em>2

  • M1M_1 = initial molarity
  • V1V_1 = initial volume
  • M2M_2 = final molarity
  • V2V_2 = final volume
  • The number of moles of solute remains the same during dilution, but the volume changes.
  • The equation is based on the principle of conservation of mass.
Example Calculation 1
  • Problem: Diluting 100 mL of a 2.5 M potassium bromide solution to 0.55 M.
  • Goal: Find the final volume (V2V_2).
  • Given:
    • V1=100 mLV_1 = 100 \text{ mL}
    • M1=2.5 MM_1 = 2.5 \text{ M}
    • M2=0.55 MM_2 = 0.55 \text{ M}
  • Equation: M<em>1V</em>1=M<em>2V</em>2M<em>1V</em>1 = M<em>2V</em>2
  • Plug in values:
    • (2.5 M)(100 mL)=(0.55 M)V2(2.5 \text{ M}) (100 \text{ mL}) = (0.55 \text{ M}) V_2
  • Solve for V2V_2:
    • V2=(2.5 M)(100 mL)0.55 M=454.5 mLV_2 = \frac{(2.5 \text{ M}) (100 \text{ mL})}{0.55 \text{ M}} = 454.5 \text{ mL}
  • Calculate the amount of water to add:
    • 454.5 mL100 mL=354.5 mL454.5 \text{ mL} - 100 \text{ mL} = 354.5 \text{ mL}
    • Final volume: 454.5 mL
    • Water to add: 354.5 mL
Example Calculation 2
  • Diluting 18 M sulfuric acid to 1.0 M.
  • Goal: Determine the volume of 18 M sulfuric acid needed to create 500 mL of a 1.0 M solution and the amount of water needed.
  • Given:
    • M1=18 MM_1 = 18 \text{ M}
    • V2=500 mLV_2 = 500 \text{ mL}
    • M2=1.0 MM_2 = 1.0 \text{ M}
  • Find: V1V_1
  • Equation: M<em>1V</em>1=M<em>2V</em>2M<em>1V</em>1 = M<em>2V</em>2
  • Plug in values:
    • (18 M)V1=(1.0 M)(500 mL)(18 \text{ M}) V_1 = (1.0 \text{ M}) (500 \text{ mL})
  • Solve for V1V_1:
    • V1=(1.0 M)(500 mL)18 M=27.777 mL27.8 mLV_1 = \frac{(1.0 \text{ M}) (500 \text{ mL})}{18 \text{ M}} = 27.777 \text{ mL} \approx 27.8 \text{ mL}
  • Amount of 18 M sulfuric acid needed: 27.8 mL
  • Calculate amount of water needed:
    • Water=500 mL27.8 mL=472.2 mL\text{Water} = 500 \text{ mL} - 27.8 \text{ mL} = 472.2 \text{ mL}
  • Always add acid to water to dissipate heat of the reaction and prevent boiling or splashing.

Diluting Acids Safely

  • Use a volumetric flask.
  • Add most of the water first (e.g., 400 mL).
  • Carefully add the required amount of concentrated acid (e.g., 27.8 mL).
  • Add the remaining water to reach the final desired volume (e.g., add the last bit of water drop wise).