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>2
- M1 = initial molarity
- V1 = initial volume
- M2 = final molarity
- V2 = 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 (V2).
- Given:
- V1=100 mL
- M1=2.5 M
- M2=0.55 M
- Equation: M<em>1V</em>1=M<em>2V</em>2
- Plug in values:
- (2.5 M)(100 mL)=(0.55 M)V2
- Solve for V2:
- V2=0.55 M(2.5 M)(100 mL)=454.5 mL
- Calculate the amount of water to add:
- 454.5 mL−100 mL=354.5 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 M
- V2=500 mL
- M2=1.0 M
- Find: V1
- Equation: M<em>1V</em>1=M<em>2V</em>2
- Plug in values:
- (18 M)V1=(1.0 M)(500 mL)
- Solve for V1:
- V1=18 M(1.0 M)(500 mL)=27.777 mL≈27.8 mL
- Amount of 18 M sulfuric acid needed: 27.8 mL
- Calculate amount of water needed:
- Water=500 mL−27.8 mL=472.2 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).