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Topic 8: Part II - Properties of Solutions

Topic 8: Part II - Properties of Solutions. Solutions. Are homogeneous mixtures that can occur in all phases – solid, liquid, or gas. Terms: solvent - does the dissolving. solute - is substance being dissolved. There are examples of all types of solvents dissolving all types of solutes.

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Topic 8: Part II - Properties of Solutions

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  1. Topic 8: Part II - Properties of Solutions

  2. Solutions • Are homogeneous mixtures that can occur in all phases – solid, liquid, or gas. • Terms: • solvent - does the dissolving. • solute - is substance being dissolved. • There are examples of all types of solvents dissolving all types of solutes. • We will focus on aqueous solutions.

  3. Methods of describing solution composition • Molarity= moles of soluteLiters of solution • % mass = Mass of solute x 100 Mass of solution • Mole fraction of component A • cA= NA NA + NB

  4. Methods of describing solution composition • Molality = moles of solute Kilograms of solvent • Molality is abbreviated m • Normality - read but don’t focus on it.

  5. Example 1 • A solution is prepared by mixing 1.00g ethanol (C2H5OH) with 100.0g of water to give a final volume of 101mL. Calculate: • Molarity • Mass percent • Mole fraction • molality

  6. Example 2 • The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. • Calculate the mass percent and molality.

  7. Structure and Solubility • Water soluble molecules must have dipole moments -polar bonds. • To be soluble in non polar solvents the molecules must be non polar.

  8. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap

  9. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap • Hydrophobic non-polar end

  10. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap • Hydrophilic polar end

  11. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 _

  12. A drop of grease in water • Grease is non-polar • Water is polar • Soap lets you dissolve the non-polar in the polar.

  13. Hydrophobic ends dissolve in grease

  14. Hydrophilic ends dissolve in water

  15. Water molecules can surround and dissolve grease. • Helps get grease out of your way.

  16. Factors Affecting SolubilityPressure Affects • Changing the pressure doesn’t effect the amount of solid or liquid that dissolves • They are incompressible. • It does effect gases.

  17. Dissolving Gases • Pressure effects the amount of gas that can dissolve in a liquid. • The dissolved gas is at equilibrium with the gas above the liquid.

  18. The gas is at equilibrium with the dissolved gas in this solution. • The equilibrium is dynamic.

  19. If you increase the pressure the gas molecules dissolve faster. • The equilibrium is disturbed.

  20. The system reaches a new equilibrium with more gas dissolved. • Henry’s Law. C = kP C = concentration of gas k = constant P = pressure of gas

  21. Temperature Effects • Increased temperature increases the rate at which a solid dissolves. • We can’t predict whether it will increase the amount of solid that dissolves. • We must read it from a graph of experimental data.

  22. g solute / 100g H2O 100 40 60 80 20 Temp. oC

  23. Gases are predictable • As temperature increases, solubility decreases. • Gas molecules can move fast enough to escape.

  24. Vapor Pressure of Solutions • A nonvolatile solute lowers the vapor pressure of the solution. • Lowers the # of solvent molecules at the surface and therefore lowers escaping tendency.

  25. Raoult’s Law: Psoln = csolventx Psolvent • Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent • Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

  26. Water has a higher vapor pressure than a solution Aqueous Solution Pure water

  27. Pure water has a higher vapor pressure and emits vapor to attempt to reach equilibrium. Aqueous Solution Pure water

  28. The solution tries to lower the vapor pressure toward its equilibrium. This continues until the pure water is gone. Aqueous Solution Pure water

  29. Example Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0g of sucrose (molar mass = 342.3 g/mol) in 643.5 cm3 of water. At 25oC, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.

  30. Ideal Solutions • Any solution that obeys Raoult’s Law • If a solvent has an affinity for the solute, the tendency of the solvent molecules to escape will be lowered • A negative deviation from Raoult’s Law

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