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CHAPTER 11

CHAPTER 11. Properties of solutions. COMPOSITION of SOLUTIONS. Solute – the substance that is being dissolved Solvent – the substance doing the dissolving If both are liquids, the one that is present in higher proportions is the solvent. MOLARITY. M = moles of solute Liters of solution

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CHAPTER 11

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  1. CHAPTER 11 Properties of solutions

  2. COMPOSITION of SOLUTIONS • Solute – the substance that is being dissolved • Solvent – the substance doing the dissolving • If both are liquids, the one that is present in higher proportions is the solvent.

  3. MOLARITY • M = moles of solute Liters of solution Ex. Calculate the molarity of a solution of glucose, if 250.0 g are dissolved in 350.0 ml of solution.

  4. MASS % • % by mass = mass of solute x 100 mass of solution Ex. A solution of vinegar is 5.0% by mass acetic acid. Calculate the mass of acetic acid dissolved in 5.0 L water if the solution has a 1.08 g/ml.

  5. MOLE FRACTION • Recall from the gas laws chapter. Compares the number of moles of one part of the solution to the total number of moles in the solution. • X = na /nt

  6. Practice problem: • A solution was made by adding 5.84 g of H2CO to 100.0 g of water. The final volume of solution was 104.0 ml. • Calculate Molarity • Molality • % by mass • Mole fraction

  7. Energy of Solution Formation • 3 steps to solution formation: • Break the solid into individual components • Expand the solute • Endothermic • Overcome IMF in solvent to make room for the solute • Expand the solvent • endothermic • Solute/Solvent interaction • Endo or exo-thermic

  8. Heat of solution ΔHsoln • Overall endothermic = + ΔHsoln • Feels cold to the touch • Overall exothermic = - ΔHsoln • Feels warm to the touch

  9. FACTORS affecting SOLUBILITY • Structural effects • “Likes dissolve likes” • Polar substances are more soluble in polar solvents. • Hydrophillic – water loving • Ionic and polar covalent compounds • Hydrophobic – water fearing • Non-polar substances Which of the following would be miscible (mutually soluble)? C6H6 H20 MgCl2 CH3OH I2

  10. 2. Pressure • Has little effect on solid or liquid solubility. • Higher pressure increases the solubility of a gas. • Henry’s law • P = kC • P = partial pressure of the gas above the solution • K = constant for a particular solution • C = concentration of dissolved gas

  11. Henry’s law (cont.) • The amount of dissolved gas is directly proportional to the pressure of the gas above the solution. • This only applies to solutions where the gas does not react or dissociate in the solvent. Ex. The solubility of oxygen is 2.2x10-4 M at 0°C and 0.10 atm. Calculate the solubility at 0°C and 0.35 atm.

  12. 3. Temperature effects • FOR SOLIDS, dissolving always occurs more rapidly at higher temperatures BUT • The amount of solute able to be dissolved may increase or decrease with the increased temperature. • Solubility (the total amount of solute that may be dissolved at a certain temperature) must be determined experimentally.

  13. FOR GASES, solubility decreases with increasing temperature. • Higher kinetic energy of the gas causes higher Pvap of the dissolved gas, more gas molecules escape the surface to the solvent and the gas becomes less soluble. • Everyday example:

  14. Pvap of solutions • A non-volatile solute LOWERS the vapor pressure of the solvent • Molecules of the solute block the surface of the solvent, making it harder for the solvent molecules to escape into the gas phase. • The number of particles is directly proportional to the decrease in the amount of vapor pressure, so strong electrolytes (which completely dissociate into ions) have a greater effect on the Pvap. • A volatile solute INCREASES the vapor pressure of the solvent Raoult’s law

  15. RAOULT’S LAW • Psoln = Xsolvent (P°solvent) Psoln = observed vapor pressure Xsolvent = mole fraction of the solvent P°solvent = the pressure of the solvent alone

  16. Raoult’s law practice Glycerine, C3H8O3 – a non volatile solute What is the Psoln made by adding 164 g of glycerine to 338 ml of water at 39.8°C? Vapor pressure of water is 54.74 torr at this temperature. Density of water is 0.992 g/ml

  17. Raoult’s law with electrolytes • 52.9 g CuCl2 (a strong electrolyte) is added to 800.0 ml of water at 52.0°C. Vapor pressure of water at this temperature is 102.1 torr and the density is .987 g/ml. Hint: Write the dissociation reaction to determine the total number of moles of ions in solution.

  18. Finding MM using Raoult’s law • 29.6 °C P°H20 = 31.1 torr • 86.7 grams of an unknown non-volatile, non-electrolyte is added to 350.0 g of water and the P soln = 28.6 torr. What is the molar mass of the substance?

  19. VOLATILE SOLUTES • Contribute to vapor pressure • Ptotal = Psolute + Psolvent • Ptotal = (Xsolute )(P°solute) + (Xsolvent)(P°solvent)

  20. Practice problem • What is the vapor pressure when 58.9 g of hexame is mixed with 44.0 grams of benzene at 60.0°C? • P° Hexane C6H14 60.0°C is 573 torr • P° benzene C6H6 60.0°C is 391 torr

  21. COLLIGATIVE PROPERTIES • Boiling point elevation • Freezing point depression • Osmotic pressure • A colligative property only depends on the number, not the identity, of the solute particles.

  22. BP ELEVATION • A non-volatile solute increases the bp of the liquid (since the overall Psoln is lowered) • Recall that boiling occurs when Patm= Psoln • bp elevation depends on the molal concentration of the solute • If the bp elevation is known, the molar mass of a solute may be determined.

  23. BP ELEVATION ΔT = i Kbmsolute • ΔT = bp elevation • i = van t Hoff factor = number of ions that result from complete dissociation • Kb is the bp elevation constant for a solvent • Kb H20 = 0.51 °C kg/mol • msolute = moles solute/ kg solvent

  24. Practice problem • What is the boiling point elevation if 31.65 g of NaCl is added to 220.0 ml of water at 34.0 °C if the density of water is 0.994 g/ml. Assume complete dissociation of NaCl.

  25. FREEZING POINT DEPRESSION • The solvent must be cooled to a lower temperature to form crystals since solute particles are blocking the solid formation ΔT = iKfmsolute • KfH20 = -1.86 °C kg/mol

  26. Practice problem • How many grams of glycerin (C3H8O3) must be added to 350.0 grams of water to lower the freezing point to -3.84 °C?

  27. Vant Hoff factor i • i expected = the number of moles of ions that results from complete dissociation of a solute. • Ex. NaCl  Na+ + Cl- • i = 2 • But, i actually = 1.9 • Why is it lower? • Some ions in solution pair up momentarily – “ion pairing” • i = moles of particles in solution/ moles of solute dissolved

  28. i • Ion pairing is greater in solutions with higher charged particles since the ions will have a greater attraction for each other. • Ex. FeCl3 • i expected = 4 • i observed = 3.4

  29. OSMOSIS • Osmosis – flow of a solvent into a solution through a semi-permeable membrane due to differences in solute concentration. Flows into a region with higher solute concentration. • hypertonic – higher pressure than the surrounding solution due to higher solute concentration. • hypotonic – lower solute concentration and pressure than the surrounding solution • Isotonic – solutions with identical osmotic pressures.

  30. Low pressure High pressure Lid with pressure sensors Osmotic pressure

  31. OSMOTIC PRESSURE • Osmotic pressure – the pressure that just stops osmosis. • Measured in atm • Can be used to find molar mass • More accurate than bp or fpdetemination  = iM R T  = osmotic pressure (in atm) i = vant Hoff factor M = molarity of the solute R = 0.0821 Latm/mol K T = Kelvin temperature

  32. Practice problem • The osmotic pressure of a solution of 26.5 mg of aspartame per L is 1.70 torr @ 30.0 °C. What is the molar mass of aspartame?

  33. Everyday example • Desalination of water by reverse osmosis. • Apply a pressure greater than the osmotic pressure to a solution of salt water and the pure water will pass through the semi-permeable membrane towards the pure solvent side.

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