Concentration of Solutions in Chemistry

Carbonated Beverages and Gas Solubility

• Carbonated drinks like Coca Cola are sealed under pressure to keep the dissolved CO₂ in solution, preventing it from escaping as gas.

Henry's Law and Concentration of Gases

• Henry’s Law: The concentration of a gas in a liquid is proportional to the pressure of that gas above the liquid.
• Example Calculation: Given CO₂ concentration of 0.032 M at 3.0 atm, find the concentration at 5.0 atm:
  • Using the formula: C1/P1 = C2/P2
  • C2 = C1 imes (P2/P1) = (0.032 ext{ M}) imes (5.0 ext{ atm}/3.0 ext{ atm}) = 0.0533 ext{ M}

Concentration Definitions

1. Concentration: The amount of solute in a given amount of solution or solvent.
2. Common Ways to Express Concentration:
   • Mass percentage (% mass)
   • Parts per million (ppm)
   • Parts per billion (ppb)
   • Molarity
   • Molality

Percent Concentration

• Percent by Mass:
  • Formula: ext{Percent by mass of solute} = rac{ ext{mass of solute}}{ ext{mass of solution}} imes 100 (% w/w)

Example 1: Mass Percentage Calculation

• Dissolving 13.5 g of glucose in 0.100 kg (100 g) of water gives:
  • Total mass = 13.5 g (solute) + 100 g (water) = 113.5 g
  • ext{Percent} = rac{13.5}{113.5} imes 100 = 11.9 ext{ %}

Parts by Volume (v/v)

• Commonly used % (v/v):
  • ext{Volume percent} = rac{ ext{volume of solute}}{ ext{volume of solution}} imes 100

Example 2: Volume Percentage Calculation

• For 70 mL of isopropanol in 100 mL solution, % (v/v) = rac{70}{100} imes 100 = 70 ext{ %}

Parts per Million (ppm) and Parts per Billion (ppb)

• ppm: ext{ppm} = rac{ ext{mass of solute}}{ ext{mass of solution}} imes 10^6
• ppb: ext{ppb} = rac{ ext{mass of solute}}{ ext{mass of solution}} imes 10^9

Example 3: ppm Calculation

• 5.4 µg of Pb²⁺ in 2.5 g of water yields:
  • ext{ppm} = rac{5.4 imes 10^{-6} ext{ g}}{2.5 ext{ g}} imes 10^6 = 2.16 ext{ ppm}

Calculating Molar Concentration

• Molarity: Number of moles of solute per liter of solution:
  • ext{Molarity (M)} = rac{ ext{Number of moles of solute}}{ ext{Volume of solution in liters}}

Example 4: Molarity Calculation

• For 16.0 g of CH₃OH in 200 mL:
  • Molar mass of CH₃OH = 32 g/mol, thus,
  • Moles = rac{16 ext{ g}}{32 ext{ g/mol}} = 0.5 ext{ mol}
  • Volume = 0.2 L, so Molarity = rac{0.5}{0.2} = 2.50 ext{ M}

Molality and Normality

1. Molality (m): Moles of solute per kg of solvent:
   • ext{Molality} = rac{ ext{moles of solute}}{ ext{kg of solvent}}
2. Normality (N): Equivalent weight of solute per liter of solution:
   • ext{N} = rac{ ext{number of equivalents of solute}}{ ext{volume of solution (L)}}

Exercises

1. How many grams of KCl in 54.6 g water for a 0.1% (w/w) solution?
2. Volume of 0.15 M KNO₃ prepared from stock solution.
3. Calculate mole fraction of isopropyl and water in a 150 g solution.
4. Determine mass of Na₂CO₃ for preparing 100 mL of 0.1 N solution.
5. Molarity and normality of Fe₂(SO₄)₃ in 200 mL at 16.2 g sample.