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  1. Characteristics of solutions • Solution – homogeneous mixture a) parts of a solution i) solute – substance being dissolved ii) solvent – substance doing dissolving both can be either solid, liquid, or gas

  2. Solubility • Soluble – substance can dissolve in a solvent ex: salt in water • Insoluble – substance cannot dissolve in a solvent ex: sand in water

  3. Solvation In Aqueous Solutions • Solvation – process of surrounding solute particles with solvent particles Why are some substances soluble in a solvent and some others are not? must be compatibility between solute and solvent

  4. Dissolution of sodium Chloride

  5. “like dissolves like” • Defn – rule used to determine if substance will dissolve in another - based on attractive forces between solute and solvent

  6. “like dissolves like” • polar solvents – dissolve polar molecular compounds and ionic compounds ex: salt and water, alcohol and vinegar • nonpolar solvents – dissolve nonpolar compounds only ex: oil and gasoline

  7. Factors Affecting Rate of Solvation • How can you dissolve something faster??? • increase temp of solvent this accelerates particles creating more particle collisions

  8. Factors Affecting Rate of Solvation • agitate the solution more particle collisions between solute and solvent • Increase surface area of solute breaking into smaller pieces allows more solute to be in contact w/ solvent

  9. Solubility • Defn – max amt of solute that can dissolve in a solvent at a specific temp how much solute can be put into solvent?

  10. Unsaturated Solution • Defn – less than max amt of solute dissolved if I put sugar into water and all sugar is dissolved, solution is unsaturated

  11. Saturated Solution • Defn – contains max amt of solute dissolved if I put sugar into water and not dissolves (you can see the sugar), the solution is saturated

  12. Supersaturated Solution • Defn – contains more solute than saturated solution at the same conditions a saturated solution made at high temp cools slowly. Slow cooling allows excess solute to remain dissolved in solution at lower temperature very unstable

  13. Saturated-Line represents max amount solute that will dissolve at a given temperature Solubility Curve (generic) Supersaturated (above line) Solubility (g solute/ 100 g H2O) Unsaturated (below line) Temperature

  14. How does temp affect solubility? • The higher the temp, higher the solubility (for most cases)

  15. Solution Concentration • Concentration – how much solute dissolved in amount of solvent what is difference between concentrated and diluted?

  16. Concentrated vs. Dilute

  17. Concentration • 3 different units of concentration a) percent by mass b) molarity (M) c) molality (m)

  18. Percent by mass • Formula OR

  19. Percent by mass • Ex prob: If 3.6 g NaCl is dissolved in 100 g H2O, what is the percent by mass? What is the solute? NaCl What is the solvent? H2O

  20. Percent by mass • Mass of solute (NaCl) = 3.6 g • Mass solvent (H2O) = 100 g • Mass solution = 3.6 + 100 = 103.6 g

  21. Percent by mass Percent by mass = 3.6 g x 100 103. 6 g = 3.5 % NaCl

  22. Molarity • Defn - # of moles per liter of solution • Formula mol solute L solution • unit mol = M (capital M) L

  23. Molarity Ex prob #1 • A solution has a volume of 250 mL and has 0.70 mol NaCl. What is the molarity? 2.8 mol/L or 2.8 M 0.70 mol = 0.250 L

  24. Molarity ex prob #2 • What is the molarity of a solution made of 47.3 g NaOH in 500 mL water? step 1: convert grams to moles 47.3 g NaOH 1 mol NaOH 1.1825 mol NaOH = 40 g NaOH

  25. Molarity ex prob #2 Step 2: divide moles by volume (L) 1.1825 mol 2.37 mol/L NaOH or 2.37 M NaOH = 0.500 L

  26. Molarity ex prob #3 • How many moles of solute are present in 1.5 L of 2.4 M NaCl? How many grams? # moles = volume x molarity 2.4 mol NaCl 1.5 L x = 3.6 mol NaCl L

  27. Molarity ex prob #3 • moles to grams 58.5 g NaCl 3.6 mol NaCl = 210.6 g NaCl 1 mol NaCl

  28. Diluting Solutions • Defn – add more solvent to original solution • Formula M1V1 = M2V2 M1 is more concentrated than M2

  29. Diluting Solutions • What volume of a 2.0 M stock solution is needed to make 0.50 L of a 0.300 M solution? M1= 2.0 M M2= 0.300 M V1= ? V2= 0.50 L (2.0 M) V1 = (0.300 M)(0.50 L) V1 = 0.075 L

  30. Diluting Solutions • If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution? M1= 3.5 M M2= ? V1= 20.0 mL V2= 100 mL (3.5 M)(20.0 mL) = M2 (100.0 mL) M2 = 0.7 M

  31. Molality • Defn - # moles of solute in one kg solvent • Formula mol solute kg solvent • Units mol = m (lower case m) kg

  32. Molality ex problem • What is the molality of a solution with 8.4 g NaCl in 255 g of water? Step 1: convert grams to moles 8.4 g NaCl 1 mol NaCl 0.14 mol NaCl = 58.5 g NaCl

  33. Molality ex problem Step 2: divide by mass (kg) 0.55 mol/kg NaCl or 0.55 m NaCl 0.14 mol NaCl = 0.255 kg

  34. Colligative Properties of Solutions • Solutes affect the physical properties of their solvents • Colligative properties (defn) – properties that depend only on the number of solute particles present, not their identity • Ex: boiling point, freezing point

  35. Electrolytes • Defn – substances that break up (ionize) in water to produce ions; can conduct electricity - consist of acids, bases, ionic compounds Ex: NaCl  Na1+ + Cl1- H2SO4  2 H+ + SO42-

  36. Nonelectrolytes • Defn – do not break up (ionize) in water, they stay the same; doesn’t conduct electricity - usually molecular/covalent compounds Ex: sugar C6H12O6 C6H12O6 ethanol C2H5OH  C2H5OH

  37. Determining # of solute particles • For ionic cmpds/acids  sum # moles of ions ex: NaCl  1 Na+ + 1 Cl1- = 2 particles CaBr2  1 Ca2+ + 2 Br1- = 3 particles

  38. Determining # of solute particles • For covalent compounds  # is always 1 ex: sugar C6H12O6  1 particle ethanol C2H5OH  1 particle

  39. Freezing Point Depression • Defn (ΔTf) – difference in temp between solution’s fp and pure solvent’s fp • Formula ΔTf = Kf x m x i # particles molal fp constant molality

  40. Ex problem • What is the freezing pt of water if 12.3 g NaCl is added to 200 g water? (Kf = 1.86 °C/m, fp = 0°C) 12.3 g NaCl 1 mol NaCl = 0.21 mol NaCl 58.5 g NaCl 0.21 mol NaCl = 1.05 m m = 0.200 kg water

  41. Ex problem • ΔTf = Kf x m x i (1.86 °C/m) (1.05 m) (2) ΔTf = = 3.91 °C New f.p. = 0° - 3.91° = -3.91°C

  42. Boiling Point Elevation • Defn (ΔTb) – difference in temp between solution’s bp and pure solvent’s bp • Formula ΔTb = Kb x m x i # particles molal bp constant molality

  43. Ex problem • What is the new boiling point of acetone if 13.7 g C6H12O6 is dissolved in 200 g acetone? (Kb = 1.71°C/m, bp = 56°C) 13.7 g C6H12O6 1 mol C6H12O6 = 0.076 mol 180 g C6H12O6 0.076 mol C6H12O6 = 0.381 m m = 0.200 kg acetone

  44. Ex problem • ΔTb = Kb x m x i (1.71 °C/m) (0.381 m) (1) ΔTb = = 0.65 °C New bp = 56° + 0.65° = 56.65°C