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SOLUTIONS

SOLUTIONS. Chapter 13 and 14. VOCABULARY. Solution Solute Solvent Soluble Solubility Solution equilibrium Saturated solution Supersaturated solution Solvation (hydration) Electrolyte. The Solution Process. Factors affecting rate of dissolution 1. Increasing solute surface area

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SOLUTIONS

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  1. SOLUTIONS Chapter 13 and 14

  2. VOCABULARY • Solution • Solute • Solvent • Soluble • Solubility • Solution equilibrium • Saturated solution • Supersaturated solution • Solvation (hydration) • Electrolyte

  3. The Solution Process • Factors affecting rate of dissolution 1. Increasing solute surface area • Increase by crushing the solute • Exposes more solute to solvent and dissolution occurs more rapidly 2. Agitation • Helps disperse the solute particles and bring fresh solvent to the surface of the solute 3. Heating a solvent • Increases the kinetic energy of the solvent particles causing more collisions with the solute

  4. SOLUBILITY • Defined as: the amount of a substance required to form a saturated solution with a specific amount of solvent at a specified temperature. • Usually reported as g/ 100 g solvent • Ex. Sugar 204 grams per 100 g of water at 20ºC.

  5. What affects solubility?POLARITY • General rule: “likes dissolve likes” • Non-polar solvents dissolve non-polar solutes • Miscible Benzene CCl4 C6H5CH3 (tolulene) • Polar solvents dissolve polar and ionic solutes H2O and NaCl • C2H5OH – ethanol (slightly polar) • will NOT dissolve ionic compounds

  6. TEMPERATURE affects solubility of solids

  7. Gas solubility • Henry’s law – gas solubility increases as the partial PRESSURE of the gas above the solvent increases. • Gases become less soluble as temperature increases.

  8. CONCENTRATION of SOLUTIONS • A measure of the amount of solute in a given amount of solvent or solution. Can be expressed using: • Molarity • molality • Mass % • Parts per million (ppm) • Parts per billion (ppb)

  9. MOLARITY • Number of moles of solute/ Liters of solution. • Symbolized: M • M = moles solute/ L soln. • Ex. 6.0 M HCl

  10. MOLARITY 1. What is the molarity of a solution prepared by dissolving 150.0 grams of sodium carbonate in enough water to make 1.0 L of solution. 2. Given 35.0 ml of a 0.50 M solution of HCl, how many moles of HCl are present?

  11. examples 3. How many grams of NaCl are present in a 3.0 M solution? 4. What volume of 3.00 M H2SO4 is needed for a reaction that requires 146.3 grams of sulfuric acid?

  12. molality • The concentrations of a solution expressed in moles of solute per kilogram of solvent. • Used for freezing point and boiling point problems.

  13. examples • Ex. What is the molality of a solution prepared by mixing 15 grams of glucose into 500 grams of water? • What is the molality of a salt solution made by adding 5.0 kg of calcium chloride to 200 kg of ice?

  14. Colligative properties • What are the effects of adding a solute to a solvent? • Colligative properties – depend on the number of particles in solution, not the type of particles. • How many particles are present when the following dissolve? • NaCl • C6H12O6 • CaCl2

  15. Boiling point elevation • Boiling point – • The temperature at which the vapor pressure of the liquid = the atmospheric pressure • The boiling point of a solvent will increase when a non-volatile solute is added. • Why? Particles block the surface of the solution, making it harder for the solvent (g) to escape. More ke is required, so the temperature must increase. • How much? ∆Tb = i Kb m Where i = number of particles Kb = boiling point elevation constant (+.51°C/molal for water) m = molality

  16. examples • What is the boiling point elevation when 150.0 grams of sucrose is added to 850. grams of water? • What is the boiling point elevation when 0.50 grams of sodium chloride is added to 2.0 kg of water?

  17. Freezing point depression • Freezing point – the temperature at which a solution forms a solid from the (aq) phase • The freezing point of a solvent will decrease when a non-volatile solute is added. • Why? • Solute particles interfere with the organization of the solid. In order to form the solid around the solute particles, the kinetic energy must decrease, so the freezing point is lowered. • How much? ∆Tf = iKfm Where i = number of particles Kf = freezing point depression constant (-1.86°C/molal for water) m = molality

  18. examples What is the freezing point depression when 500. grams of sodium chloride is added to 2000. g of water? What is the freezing point depression when 50.0 grams of calcium chloride is added to 500. grams of water?

  19. STOICHIOMETRY • How many ml of 3.0 M NaOH are needed to neutralize 5.0 grams of acetic acid found in a vinegar solution? What is the M of the vinegar? • How many grams of Mg can be dissolved by 15.0 ml of 6.0 M HCl? How much gas will be produced by this reaction at STP?

  20. STOICHIOMETRY • 5.0 ml of 6.0 M HCl reacts with 25.0 ml .10 M NaOH. How much water will be produced? • How many ml 0.10 M NaOH should be used to neutralize the 5.0 ml of HCl?

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