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Solutions

Read chapter 12. Solutions. What determines solubility?. Temperature Pressure (when a gas is involved) Nature of the particles – “Like dissolves like”. d +. H. d -. O. d +. H. Aqueous Solutions. Water is a “polar” molecule. The O in the molecule has a partial negative charge

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Solutions

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  1. Read chapter 12 Solutions

  2. What determines solubility? • Temperature • Pressure (when a gas is involved) • Nature of the particles – • “Like dissolves like”

  3. d+ H d- O d+ H Aqueous Solutions • Water is a “polar” molecule. • The O in the molecule has a partial negative charge • The H’s have a partial positive charge.

  4. Aqueous Solutions • Water solvates (surrounds and stabilizes) the ions. • H’s (d+) interact with the negatively charged anions • O (d-) interacts with the positively charged cations

  5. + - Aqueous Solutions Solvation of anion Solvation of cation

  6. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/molvie1.swfhttp://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/molvie1.swf

  7. Henry’s Law • The solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. (at constant temp.) S1 S2 P1= P2

  8. Solubility is defined as grams per liter. (g/L)

  9. Example- • If 0.85 g of gas at 4.0 atm of pressure dissolves in 1.0L of water at 25°C, how much will dissolve in 1.0L of water at 1.2 atm of pressure and the same temperature?

  10. Units of Concentration: • Percent by Mass % = masssolute x 100 masssolution

  11. Example: • What is the percent by mass of NaHCO3 in a solution containing 20g NaHCO3 in 600g H2O?

  12. You have 1500. g of a bleach solution. The percent by mass of the solute sodium hypochlorite, NaOCl, is 3.62%. How many grams of NaOCl are in the solution? • How many grams of solvent were present?

  13. 1a. Percent by volume % = volumesolute X 100 volumesolution

  14. Example- • What is the percent by volume of ethanol in a solution that contains 35mL of ethanol dissolved in 115 mL of water?

  15. 2. Molarity M = mols solute literssolution

  16. Example - • What is the molarity of an aqueous solution containing 40.0g of glucose (C6H12O6) in 1.5 L of solution? • How many grams of NaOH are in 250mL of a 3.0M NaOH solution?

  17. 3.Dilution Formula M1V1 = M2V2 Example – What volume of a 3.00M KI stock solution would you use to make 0.300L of a 1.25M KI solution?

  18. 4. molality m = molssolute kilogramssolvent

  19. Example - • What is the molality of a solution containing 30.0g of naphthalene (C10H8) dissolved in 500.0g of toluene?

  20. How many grams of water would be necessary to make of 1.35 m salt solution from 25.8 g of NaCl?

  21. 5. Mole Fraction XA= molsA total mols

  22. Examples- • What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass? • An aqueous solution of NaCl has a mole fraction of 0.21. What is the mass of NaCl dissolved in 100.0mL of water?

  23. Colligative Properties • Properties of solutions that are affected by the number of particles, but not by the identity of the particles. (it doesn’t matter what is dissolved, only how much!)

  24. 1. Vapor Pressure Lowering Vapor pressure is caused by molecules that have evaporated from the surface of a liquid

  25. For a solution, the amount of vapor pressure (vapor above the solution) will be less.

  26. Why? • When solute is added, particles of solute replace some of the solvent at the surface. This reduces access for evaporation, thus reducing vapor pressure.

  27. Also, • In a solution the solvent-solute attraction is usually stronger than the original solvent-solvent attraction, further reducing evaporation.

  28. Raoult’s law PA = XAPA° PA= new vapor pressure of the solution XA = mol fraction of the solvent PA° = original vapor pressure of the solvent

  29. example • Calculate the vapor pressure of a solution prepared by dissolving 25.0 grams of NaCl in 100 grams of water. The original vapor pressure of the water is 19.8 torr.

  30. 2. Boiling point elevation • The difference between the normal boiling point of a pure solvent and the boiling point of a solution. ∆Tb = Kb m Kb H2O = 0.51 °C·Kg/mol

  31. example • What is the boiling point of a solution of 103.2 g of C6H12O6 in 0.75 Kg of H2O? • What is the molality of a water solution that boils at 106.5°C?

  32. 3. Freezing point depression • The difference between the normal freezing point of a pure solvent and the freezing point of a solution. ∆Tf = Kf m Kf H2O = -1.86 °C·Kg/mol

  33. example • What is the freezing point of water in a solution of 117.1g of sucrose (C12H22O11) and 200 g of water?

  34. 4. Osmotic Pressure Osmosis – the diffusion of solvent particles across a semipermeable membrane from an area of high solvent concentration to an area of lower solvent concentration. Why is this a colligative property?

  35. Hypertonic Hypotonic solution solution

  36. Π = MRT • The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose, C6H12O6, will be isotonic with blood?

  37. Types of Mixtures • Homogeneous mixture Solution - parts too small to see. (all looks the same) Solvated particles are less than 1 nm.

  38. 2. Heterogeneous mixtures • Suspension – a mixture containing particles large enough to settle out if left undisturbed. • Particles > 1000 nm

  39. Examples - • Muddy water • Medicines (shake well) • Paint

  40. Colloid – mixture with intermediate sized particles. • the particles are too small to filter out or to settle out on their own. • (between 1 nm and 1000 nm)

  41. Examples - Fog Blood Butter Foam Smoke

  42. The Tyndall Effect • Colloids can be cloudy (opaque) or as clear as solutions. • Particles in a colloid are always big enough to scatter light – the Tyndall effect • The Tyndall effect is used to distinguish between solution and colloids.

  43. Brownian motion This erratic movement of colloid particles resulting from collisions of particles of the dispersion medium with the dispersed particles. These collisions prevent the colloid particles from settling out of the mixture.

  44. Enthalpies of Solution • Three steps in the dissolving process • Breaking the solute-solute attraction *(energy required) • Breaking the solvent-solvent attraction *(energy required) • Formation of the solute-solvent attraction *(energy released)

  45. Enthalpy of Solution • If the sum of steps 1 & 2 is greater than step 3, the process is endothermic (feels cold) • If the sum of steps 1 & 2 is less than step 3, the process is exothermic (feels hot)

  46. Concepts to know… • Electrolyte vs. nonelectrolyte • Factors affecting rate of dissolution • Unsaturated, saturated, supersaturated • Miscible vs. immiscible

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