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Exploring Colligative Properties in Chemistry: Surfactants and Their Effects

Join us in our first chemistry seminar as we delve into colligative properties and surfactants. Learn about the impact of solute-solvent mixtures, with practical applications like cooking and antifreeze. We will cover freezing point depression and boiling point elevation, demonstrating calculations, and real-life examples including ethanol in wine. Gain insights into osmotic pressure and Henry's Law, essential concepts in understanding gas solubility in liquids. Prepare for upcoming ChemBoard assignments and get ahead by reading Chapter 12!

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Exploring Colligative Properties in Chemistry: Surfactants and Their Effects

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  1. Outline 1/12/07 • First Chem seminar today… • Pick up CAPA #2 - in front • ChemBoard Assignment = Monday • Fill in keypad response unit #’s… Today: Surfactants Colligative Properties Calculating Colligative Effects

  2. Surfactants: • Anything which alters behavior of solute-solvent mixture….(e.g. soap) Applications? NaCl in cooking Road salt for ice Antifreeze Saline drips Gatorade e.g. Demo of S on water

  3. Effect of solutes on solvent properties... Vapor pressures Boiling points Freezing points Osmotic pressure “Colligative properties”

  4. Calculations of Colligative Property effects • Freezing Point Depression • DT = Kfm • Boiling Point Elevation • DT = Kbm Kf = -1.858°C/m Kb = 0.512°C/m where Kfand Kbare constants and m is concentration of solute in terms of molality (mols/kg solvent)

  5. liquid solid

  6. Calculations of Colligative Property effects • Freezing Point Depression • DT = Kfm • Boiling Point Elevation • DT = Kbm Kf = -1.858°C/m Kb = 0.512°C/m where Kfand Kbare constants and m is concentration of solute in terms of molality (mols/kg solvent)

  7. Example #1: Freezing Point Depression: DT = Kfm • Estimate the freezing point of wine that is 12% by mass ethanol... Kf= -1.858 C kg/mol (for water) Assume 100g of wine, then you have 12g ethanol and 88g H2O... 12g C2H5OH  46 g/mol= 0.261 mol 0.261 molEtOH/88 g H2O = x mol/1000g x = 2.97 m DT = -5.5°C

  8. Example #2: • Compute the molar mass of A, if a solution containing 35.0 g of A dissolved in 135 mL of H2O freezes at -1.75C. • Assume A does not ionize in water. Freezing Point Depression: DT = Kfm Kf= -1.858 C kg/mol (for water) -1.75 = -1.858 m or m = 0.942 mol/kg

  9. Example #2 (cont’d): • m = 0.942mols solute/kg solvent • How many kg of solvent? 135 mL  1 g/mL = 135 g = 0.135 kg 0.942 mol/kg  0.135 kg = 0.127 mol Molar mass: 35.0 g / 0.127 mol = 275 g/mol

  10. Another useful application: • Osmotic Pressure (p) Passage of solute molecules through a semi-permeable membrane Start with something we know: PV = nRT (ideal gases) or P = (n/V)RT Adapt to solutions: (n/V)  (mol/L) = Molarity!

  11. p.107 p = M RT • Osmotic Pressure (p)

  12. p.107

  13. Example: • Compute the molar mass of hemoglobin, if a 25 mL solution containing 0.420 g of hemoglobin has an osmotic pressure of 4.6 torr at 27C. Osmotic pressure: p = M RT M= [m (g) / MM (g/mol)]/V p = m (RT)/ (MM) V = 68300 g/mol

  14. Practice…Worksheet #1 • Remember to keep solute & solvent separate when answering questions!

  15. Henry’s Law • Gases dissolved in a liquid: Concentration = KHp Examples: soda CO2, NOx, SOx and acid rain

  16. For Monday… • Do CAPA set #1 • Post a trial message on Chemboard • Read through Chapter 12 …get ahead! Have a great week-end!

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