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: C<em>1/P</em>1=C<em>2/P</em>2C<em>1/P</em>1 = C<em>2/P</em>2
    • C<em>2=C</em>1imes(P<em>2/P</em>1)=(0.032extM)imes(5.0extatm/3.0extatm)=0.0533extMC<em>2 = C</em>1 imes (P<em>2/P</em>1) = (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: extPercentbymassofsolute=extmassofsoluteextmassofsolutionimes100ext{Percent by mass of solute} = \frac{ 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):
    • extVolumepercent=extvolumeofsoluteextvolumeofsolutionimes100ext{Volume percent} = \frac{ 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: extppm=extmassofsoluteextmassofsolutionimes106ext{ppm} = \frac{ ext{mass of solute}}{ ext{mass of solution}} imes 10^6
  • ppb: extppb=extmassofsoluteextmassofsolutionimes109ext{ppb} = \frac{ 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:
    • extppm=5.4imes10−6extg2.5extgimes106=2.16extppmext{ppm} = \frac{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:
    • extMolarity(M)=extNumberofmolesofsoluteextVolumeofsolutioninlitersext{Molarity (M)} = \frac{ 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 = 16extg32extg/mol=0.5extmol\frac{16 ext{ g}}{32 ext{ g/mol}} = 0.5 ext{ mol}
    • Volume = 0.2 L, so Molarity = 0.50.2=2.50extM\frac{0.5}{0.2} = 2.50 ext{ M}

Molality and Normality

  1. Molality (m): Moles of solute per kg of solvent:
    • extMolality=extmolesofsoluteextkgofsolventext{Molality} = \frac{ ext{moles of solute}}{ ext{kg of solvent}}
  2. Normality (N): Equivalent weight of solute per liter of solution:
    • extN=extnumberofequivalentsofsoluteextvolumeofsolution(L)ext{N} = \frac{ 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.