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Properties of Solutions Solution: Homogenous mixture of 2 or more substances PowerPoint Presentation
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Properties of Solutions Solution: Homogenous mixture of 2 or more substances

Properties of Solutions Solution: Homogenous mixture of 2 or more substances

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Properties of Solutions Solution: Homogenous mixture of 2 or more substances

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  1. Properties of Solutions Solution: Homogenous mixture of 2 or more substances Solutions can be liquid, solid or gaseous Examples: Ocean, sugar water Gold alloy Air, humid oxygen

  2. Solvent: Substance present in a solution in the greatest amount Example: Water in the ocean; nitrogen in air Solute: Substance present in a solution in lesser amounts than the solvent Example: Salt in ocean; oxygen in air Solutes can be electrolytes or nonelectrolytes Electrolytes: solutes that dissociate in solution into ions that carry charge (ionic compounds) Nonelectrolytes: solutes that do not dissociate in solution, and do not carry any charge

  3. Colloid: Homogenous mixture of 2 or more substances in which the substances are larger than those in solutions Suspension: Heterogeneous mixture, with very large particles capable of settling out of solution Solubility Soluble substance: Substance that is able to dissolve in a solvent Insoluble substance: Substance that does not dissolve in a solvent

  4. Solubility: Maximum amount of solute that can be dissolved in a specific amount of solvent under specific conditions of temperature and pressure • Saturated Solution: Solution containing maximum amount of solute that will dissolve under current conditions • Supersaturated Solution: Unstable solution containing amount of solute greater than the solubility value

  5. Solubility of liquids and solids in water typically increases with increasing temperature • Example: More sugar will dissolve in warm water than in cold water • Solubility of gases in water decreases with temperature • Solubility of gases in water increases with increasing pressure (Henry’s Law)

  6. “Like dissolves like:” • polar solvents will dissolve polar solutes • nonpolar solvents will dissolve nonpolar solutes • Examples: wax in CCl4, sugar in water; oil in water? • Solutes fail to dissolve when: • 1) forces between solute particles out-weigh attractions between solute and solvent • 2) solvent particles are more attracted to each other than to solute

  7. Examples of Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2 (ionic) CH2Cl2 (nonpolar) I2 (nonpolar)

  8. Solutes dissolve faster when: Concentration: Relationship between amount of solute contained in a specific amount of solution • Solute particles are small • Solvent is heated • Solution is stirred

  9. Concentration as Percent • Percent: Solution concentration giving the amount of solute in 100 parts of solution • % = part/total x 100 • Weight/weight percent: Concentration giving the mass of solute in 100 mass units of solution • %(w/w) = solute mass/solution mass x 100 • Example: 12.0%(w/w) sugar solution • 12 g sugar per 100 g solution

  10. Weight/volume percent: Concentration giving the grams of solute contained in 100 mL of solution • %(w/v) = grams solute/mL solution x 100 • Example: 12.0%(w/v) sugar solution • 12 g sugar per 100 mL solution

  11. Molarity: Unit of concentration used with solutions; number of moles of solute per liter of solution • Molarity (M) = moles of solute/liters of solution • Examples: 2 moles of NaCl dissolved in 1 L of water • M = 2 moles/1 L = 2 M • 1.5 moles NaCl dissolved in 2 L of water: • M = 1.50 moles/2.00 L = .750 M

  12. Preparing Solutions • Measure proper amount of solute into container, then add solvent to proper volume. • Example: 1 L of 1.50 M CoCl2 Solution • M = moles/Liter  M x Liter = moles • 1.5 moles/L x 1L = 1.5 moles needed • 1.5 moles CoCl2 = 195g CoCl2 • 1.5 mol CoCl2 x 130 g CoCl2/mol = 195g CoCl2 • Place 195 g CoCl2 in a flask. Add water to fill to 1-L mark

  13. Dilute a more concentrated solution with solvent to give a solution of lower concentration. • M1 V1 =M2 V2 • Example: Prepare 250 mL of 0.100 M NaCl solution from a 2.00 M NaCl solution. • M1 = molarity of starting solution (in this case 2.00M NaCl) • V1 = volume of starting solution required (always unknown) • M2 = molarity of final solution after dilution (in this case 0.100M NaCl) • V2 = volume of final solution, after dilution (in this case 250ml)

  14. Use M1 V1 = M2 V2 to solve for the unknown variable (V1); this is the amount of the starting solution you need. Add solvent (usually water) to reach the desired total volume (V2). Prepare 250 mL of 0.100 M NaCl solution from a 2.00 M NaCl solution. M1 = molarity of starting solution (in this case 2.00M NaCl) V1 = volume of starting solution required (always unknown) M2 = molarity of final solution after dilution (in this case 0.100M NaCl) V2 = volume of final solution, after dilution (in this case 250ml) Answer: Prepare solution by adding 0.0125 L of 2.0 M NaCl to flask and adding water up to 250 mL mark.

  15. Osmotic Pressure Osmosis: Movement of water through a semipermeable membrane, from more dilute solution towards more concentrated solution Osmotic pressure: amount of pressure required to stop flow of water due to osmosis Isotonic solutions: solutions with identical osmotic pressure; no urge for water to flow

  16. Example: During osmosis, water flows across the semi-permeable membrane from the 4% starch solution into the 10% solution. 4% starch 10% starch H2O

  17. Eventually, the flow of water across the semi- permeable membrane becomes equal in both directions. 7% starch 7% starch H2O

  18. Hypotonic solution: the more dilute of 2 solutions separated by a semipermeable membrane; water leaves this solution and flows across membrane to the more concentrated solution Hypertonic solutions: the more concentrated of 2 solutions separated by a semipermeable membrane; water enters this solution, moving across the membrane from the more dilute solution Crenate Burst No Change (hypertonic) (hypotonic) (isotonic)

  19. Practice with Molarity • Calculate the molarity of 1.50 L of solution containing 0.294 moles of solute • Calculate the molarity of 500 mL of solution containing 0.304 moles of solute

  20. Calculate the molarity of a 100 mL solution containing 4 g of NaOH. How would you make 100 mL of 0.25 M Na2SO4 solution? How would you prepare 5 L of 6 M H2SO4 from 18 M H2SO4 solution?