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Chemistry Chapter 16

Chemistry Chapter 16. Dissociation of ions in solution. solubility increases with increasing temperature. solubility decreases with increasing temperature. Temperature and Solubility. Solid solubility and temperature. 12.4. Solubility of Gases in Solution.

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Chemistry Chapter 16

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  1. Chemistry Chapter 16

  2. Dissociation of ions in solution

  3. solubility increases with increasing temperature solubility decreases with increasing temperature Temperature and Solubility Solid solubility and temperature 12.4

  4. Solubility of Gases in Solution

  5. The following fishkill occurred September 2010, in Louisiana. What could have killed the fish?

  6. Solubility of Gases in Solution

  7. Molarity (M) = moles of solvent Liters of solution 4.5

  8. Moles of solute before dilution (1) Moles of solute after dilution (2) = Dilution Add Solvent = M2V2 M1V1 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. 4.5

  9. Concentrations of Solutions

  10. Molecular Solids in solution The sugar is a molecular solid, in which the individual molecules are held together by relatively weak intermolecular forces. When sugar dissolves in water, the weak bonds between the individual glucose molecules are broken, and these C6H12O6 molecules are released into solution.

  11. Colligative Properties • Colligative properties depend on quantity of solute molecules. Lowering the Vapor Pressure • Non-volatile solvents reduce the ability of the surface solvent molecules to escape the liquid. • Therefore, vapor pressure is lowered. • The amount of vapor pressure lowering depends on the amount of solute.

  12. Vapor Pressures of Pure Water and an Aqueous Solution The vapor pressure of water over pure water is greater than the vapor pressure of water over an aqueous solution containing a nonvolatile solute.

  13. Boiling Point Elevation When you add a nonvolatile solute to a solvent the vp of the solution is lowered. Therefore you will have to heat the solution to a higher temperature to cause it to have a high enough vp to boil Thus a solution has a higher bp than the pure solvent Boiling Point Elevation: the difference in temperature between the boiling points of a pure solvent and of a solution. The chance in bp is proportional to the number of particles added to the solvent, regardless of the type of particle.

  14. Freezing Point Depression Adding solute particles attracts water molecules around those particles. This disrupts the process of solid formation, so more energy must be removed from the solution to cause the substance to freeze. Thus the freezing point of a solution is always lower than the freezing point of pure solvent. Changing the freezing point is proportional to the number of particles added, regardless of the type of particle.

  15. Molality moles of solute Molality (m) = kilograms of solvent Molal conc. Molality: What is the molality of a solution with 25 grams of NaOH and 2 kg of water? 25 grams = .625 moles .625 moles NaOH= .3125 mNaOH 2 kg of water

  16. MOLE FRACTION!! Xa

  17. Boiling-Point Elevation ΔTb= Kb·m· i Freezing-Point Depression ΔTf= Kf·m· i ColligativeProperties of Molecular Solids 12.6

  18. 0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b DTb = Kbm Boiling-Point Elevation T b is the boiling point of the solution DTb > 0 m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 12.6

  19. 0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf DTf = Kfm Freezing-Point Depression T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 12.6

  20. 12.6

  21. 0 DTf = T f – Tf moles of solute m= mass of solvent (kg) = 3.202 kg solvent 1 mol 62 g 478 g x 0 Tf = T f – DTf What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62 g. DTf = Kfm Kf water = 1.86 0C/m = 2.41 m DTf = Kfm = 1.86 0C/m x 2.41 m = 4.48 0C = 0.00 0C – 4.48 0C = -4.48 0C 12.6

  22. Boiling-Point Elevation DTb = iKbm Freezing-Point Depression DTf = iKfm Colligative Properties of Electrolyte Solutions 12.7

  23. actual # of particles in soln after dissociation van’t Hoff factor (i) = # of formula units initially dissolved in soln Colligative Properties of Electrolyte Solutions NaClsolution Na+ ions + Cl- ions i should be nonelectrolytes 1 NaCl 2 CaCl2 3 12.7

  24. Freezing Point Depression At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1oC FP = 0 – 20.1 = -20.1oC

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