1. Properties of Solutions AP Chemistry Chapter 13 JMS AP Chemistry Chapter 13 JMS

2. What is a solution  A solution is a homogeneous mixture of one or more pure substances (solutes) which disperse uniformly throughout another pure substance (solvent).  IMF’s determine the interactions between solute and solvent known as solvation.  When water is the solvent, solvation is usually referred to as hydration.  A solution is a homogeneous mixture of one or more pure substances (solutes) which disperse uniformly throughout another pure substance (solvent).  IMF’s determine the interactions between solute and solvent known as solvation.  When water is the solvent, solvation is usually referred to as hydration.

3. Hydration of NaCl  In the Ion-dipole attractions, we see the positive hydrogen ends of the water attracting the Cl - ions, while the negative Oxygen ends of water attract the Na + ion.

4. Crystallization  The reverse process of solvation is crystallization, where solids come out of solution.

5. Enthalpy and Solvation  Sometimes addition of a solute releases heat, and sometimes it absorbs heat.

6. Enthalpy and Nature  Why does oxygen form diatomic molecules?  Nature favors more stable compounds; this means that lower enthalpy states are more likely to occur.  So nature prefers exothermic reactions.  Why does oxygen form diatomic molecules?  Nature favors more stable compounds; this means that lower enthalpy states are more likely to occur.  So nature prefers exothermic reactions.

7. Entropy and Nature  So why would water evaporate? (evaporation is endothermic).  Nature also favors higher entropy.  Entropy is a measure of disorder or randomness.  By evaporating, the molecules of water (structured with hydrogen bonding) can become much more random as a gas.  So why would water evaporate? (evaporation is endothermic).  Nature also favors higher entropy.  Entropy is a measure of disorder or randomness.  By evaporating, the molecules of water (structured with hydrogen bonding) can become much more random as a gas.

8. Entropy or Chaos  In the solution process, solids become solute particles.  Is this an increase or decrease in entropy?  In the solution process, solids become solute particles.  Is this an increase or decrease in entropy?

9. Spontaneity  In chemistry, spontaneity means a process can occur without added energy.  It does not mean “instantaneous”.  If I put a glass of water out at room temperature, will it evaporate on its own?  Yes eventually. This is a spontaneous process at room temperature.  If I put a lump of salt into water, will it dissolve spontaneously?  In chemistry, spontaneity means a process can occur without added energy.  It does not mean “instantaneous”.  If I put a glass of water out at room temperature, will it evaporate on its own?  Yes eventually. This is a spontaneous process at room temperature.  If I put a lump of salt into water, will it dissolve spontaneously?

10. Spontaneity  Since nature prefers exothermic reactions and more random states:  A reaction that decreases in enthalpy and increases in entropy is always spontaneous.  A reaction that increases in enthalpy and decreases in entropy is never spontaneous.  Any other reaction may be spontaneous, depending on the room temperature.  Since nature prefers exothermic reactions and more random states:  A reaction that decreases in enthalpy and increases in entropy is always spontaneous.  A reaction that increases in enthalpy and decreases in entropy is never spontaneous.  Any other reaction may be spontaneous, depending on the room temperature.

11. Key Terms  Concentration - the amount of solute in a solvent.  Solubility - the maximum amount of solute that can be dissolved in a given amount of solvent.  Concentration - the amount of solute in a solvent.  Solubility - the maximum amount of solute that can be dissolved in a given amount of solvent.  Unsaturated - a solution under its solubility.  Saturated - a solution at its solubility.  Supersaturated - a solution above its solubility.  Unsaturated - a solution under its solubility.  Saturated - a solution at its solubility.  Supersaturated - a solution above its solubility.

12. How can a solution be above its maximum?  Solubility varies with temperature.  At 298 K, 91 g of salt dissolved.  At 323 K, over 100 g can dissolve.  If we dissolve 100 g at 298 K, then lower the temperature, the solution is now supersturated.  Solubility varies with temperature.  At 298 K, 91 g of salt dissolved.  At 323 K, over 100 g can dissolve.  If we dissolve 100 g at 298 K, then lower the temperature, the solution is now supersturated.

13. Miscible  Liquids that mix in all proportions  Dry Gas is added to gasoline to eliminate any moisture in the line. The alcohol based substance is miscible in water and gasoline.  Liquids that mix in all proportions  Dry Gas is added to gasoline to eliminate any moisture in the line. The alcohol based substance is miscible in water and gasoline.

14. Immiscible  Liquids that do not mix.  Oil and water are the common example of immiscible liquids.  Liquids that do not mix.  Oil and water are the common example of immiscible liquids.

15. Solubility vs Temperature  At different temperatures, the solubility of a given solute/solvent varies.  For example, most solids are more soluble in water at higher temperatures.  At different temperatures, the solubility of a given solute/solvent varies.  For example, most solids are more soluble in water at higher temperatures.

16. Solubility of Solids  In this case, the solubility is on the amount of solute in a saturated solution based on 100 g (or 100 ml) of water.

17. Solubility of Gases  Generally, the solubility of gases decreases as the temperature increases.  Why?  Generally, the solubility of gases decreases as the temperature increases.  Why?

18. Calculating Concentration  Molarity(M)=moles of solute/liters of solution. The unit is Molars.  molality(m)=moles of solute/kilograms of solvent. The unit is molals.  Mole Fraction()=moles of solute/total moles of solution. There is no unit.  Mass percent (%)=mass of solute/total mass of solution. This is expressed as a percent.  Parts per Million (ppm) = (mass of solute/total mass of solution)*1,000,000. The units are ppm.  Molarity(M)=moles of solute/liters of solution. The unit is Molars.  molality(m)=moles of solute/kilograms of solvent. The unit is molals.  Mole Fraction()=moles of solute/total moles of solution. There is no unit.  Mass percent (%)=mass of solute/total mass of solution. This is expressed as a percent.  Parts per Million (ppm) = (mass of solute/total mass of solution)*1,000,000. The units are ppm.

19. Sample Problem  A solution is prepared from 117 g of NaCl in 900 g of water. The volume of the solution was 1.00 liter. Calculate the:  Molarity  molality  Mole fraction  Mass percent  A solution is prepared from 117 g of NaCl in 900 g of water. The volume of the solution was 1.00 liter. Calculate the:  Molarity  molality  Mole fraction  Mass percent

20. Molarity  Molarity = moles of solute/liter of solution.  117 g of NaCl/(58.5 g/mol) = 2.00 mol NaCl.  Molarity = 2.00 mol NaCl/ 1.00 liter = 2.00 M  Molarity = moles of solute/liter of solution.  117 g of NaCl/(58.5 g/mol) = 2.00 mol NaCl.  Molarity = 2.00 mol NaCl/ 1.00 liter = 2.00 M

21. molality  molality = moles of solute/kilograms of solvent  We already established that there are 2.00 moles NaCl.  900 g H 2 O =.900 kg H 2 O  molality = 2.00 moles NaCl /.900 kg H 2 O = 2.22 m  molality = moles of solute/kilograms of solvent  We already established that there are 2.00 moles NaCl.  900 g H 2 O =.900 kg H 2 O  molality = 2.00 moles NaCl /.900 kg H 2 O = 2.22 m

22. Mole Fraction   = moles of solute/total number of moles  We already established that there are 2.00 moles NaCl.  900 g H 2 O / (18 g/mol) = 50 mol H 2 O  X = (2 moles NaCl)/(2 moles NaCl + 50 moles H 2 O) =.0385   = moles of solute/total number of moles  We already established that there are 2.00 moles NaCl.  900 g H 2 O / (18 g/mol) = 50 mol H 2 O  X = (2 moles NaCl)/(2 moles NaCl + 50 moles H 2 O) =.0385

23. Mass percent  Mass percent = mass of solute/total moles of solution  Mass percent = (117 g NaCl)/(117 g NaCl + 900 g H 2 O) =.115=11.5%  Mass percent = mass of solute/total moles of solution  Mass percent = (117 g NaCl)/(117 g NaCl + 900 g H 2 O) =.115=11.5%

24. Colligative Properties  Certain properties of solutions are independent on the type of solute molecules, but are instead dependent on the amount of solute particles. These are called colligative properties.

25. Analyzing Colligative Properties  Let’s say we put some sugar in water. Now we want to boil the water. For the water to boil, it has to leave the liquid phase and become random gaseous particles.  However, the sugar particles are attracted to multiple water molecules, holding the water molecules together with additional force.  The more sugar particles present, the more attractions to water, and the harder it will be to boil the water.  Let’s say we put some sugar in water. Now we want to boil the water. For the water to boil, it has to leave the liquid phase and become random gaseous particles.  However, the sugar particles are attracted to multiple water molecules, holding the water molecules together with additional force.  The more sugar particles present, the more attractions to water, and the harder it will be to boil the water.

26. So Why Doesn’t the Type of Solute Matter?  First of all we have to understand “like dissolves like”. This means that a polar solvent will best dissolve a polar solute and a nonpolar solvent will best dissolve a nonpolar solute.  Water is polar as is sugar (sucrose).  First of all we have to understand “like dissolves like”. This means that a polar solvent will best dissolve a polar solute and a nonpolar solvent will best dissolve a nonpolar solute.  Water is polar as is sugar (sucrose).

27. So?  Since both water and sugar are polar, they form a solution. This means there are already strong attractions between solute and solvent.  What is more important now is the number of sugar particles bonded to water molecules. The more sugar particles present, the harder it will be to boil.  Since both water and sugar are polar, they form a solution. This means there are already strong attractions between solute and solvent.  What is more important now is the number of sugar particles bonded to water molecules. The more sugar particles present, the harder it will be to boil.

28. Boiling Point Elevation  T b = k b im  T b is the change in boiling point.  k b is the boiling point constant for the solvent. Each solvent has its own constant. For water, k b =.51 C o /m.  i is the vant hoff factor.  m is the molality.  T b = k b im  T b is the change in boiling point.  k b is the boiling point constant for the solvent. Each solvent has its own constant. For water, k b =.51 C o /m.  i is the vant hoff factor.  m is the molality.

29. Vant Hoff Factor  We said that colligative properties depend on the number of solute particles.  Well, strong electrolytes tend to dissociate in water, thus giving more particles.  The vant hoff factor is a relative number indicating how many particles are present per molecule of solute.  We said that colligative properties depend on the number of solute particles.  Well, strong electrolytes tend to dissociate in water, thus giving more particles.  The vant hoff factor is a relative number indicating how many particles are present per molecule of solute.

30. For example  For sucrose (a nonelectrolyte) there is no dissociation so there is 1 particle in every molecule. i =1.  For NaCl (a strong electrolyte), there is dissociation, so there are 2 particles in every molecule. i = 2.  For PbCl 2 (a poor electrolyte), there is only slight dissociation, so the number of particles is harder to determine. 1<i<3  For sucrose (a nonelectrolyte) there is no dissociation so there is 1 particle in every molecule. i =1.  For NaCl (a strong electrolyte), there is dissociation, so there are 2 particles in every molecule. i = 2.  For PbCl 2 (a poor electrolyte), there is only slight dissociation, so the number of particles is harder to determine. 1<i<3

31. Try This  424 g of K 3 PO 4 are dissolved in 1.00 liter of water. At what temperature will the water boil?

32. The process  First find the molality.  424 g K 3 PO 4 / (212 g/mol) = 2.00 mol K 3 PO 4.  1.00 liter H 2 O = 1.00 kg H 2 O  m = 2.00 mol/1.00 kg = 2.00 m  Now solve: T b = k b im  T b = (.51 C o /m)(4)(2.00 m) = 4.08 C o  Since water normally boils at 100 o C, it would now boil at 104.08 o C.  First find the molality.  424 g K 3 PO 4 / (212 g/mol) = 2.00 mol K 3 PO 4.  1.00 liter H 2 O = 1.00 kg H 2 O  m = 2.00 mol/1.00 kg = 2.00 m  Now solve: T b = k b im  T b = (.51 C o /m)(4)(2.00 m) = 4.08 C o  Since water normally boils at 100 o C, it would now boil at 104.08 o C.

33. Freezing Point Depression To freeze water, we have to force the liquid molecules to line up in orderly bonds. This is harder to do when solute particles hold the water in irregular patters.  T f = k f im  k f is the freezing point constant for the solvent. Each solvent has its own constant. For water, k f = 1.86 C o /m. To freeze water, we have to force the liquid molecules to line up in orderly bonds. This is harder to do when solute particles hold the water in irregular patters.  T f = k f im  k f is the freezing point constant for the solvent. Each solvent has its own constant. For water, k f = 1.86 C o /m.

34. Now Try This  424 g of K 3 PO 4 are dissolved in 1.00 liter of water. At what temperature will the water freeze?

35. The Same Process  First find the molality.  424 g K 3 PO 4 / (212 g/mol) = 2.00 mol K 3 PO 4.  1.00 liter H 2 O = 1.00 kg H 2 O  m = 2.00 mol/1.00 kg = 2.00 m  Now solve: T f = k f im  T f = (1.86 C o /m)(4)(2.00 m) = 14.88 C o  Since water normally freezes at 0 o C, it would now freeze at -14.88 o C.  First find the molality.  424 g K 3 PO 4 / (212 g/mol) = 2.00 mol K 3 PO 4.  1.00 liter H 2 O = 1.00 kg H 2 O  m = 2.00 mol/1.00 kg = 2.00 m  Now solve: T f = k f im  T f = (1.86 C o /m)(4)(2.00 m) = 14.88 C o  Since water normally freezes at 0 o C, it would now freeze at -14.88 o C.

36. Vapor Pressure Depression  Like we discussed with boiling, molecules have a harder time leaving the liquid phase when a solute is present.  The Boiling Point increased because more heat is needed to vaporize the liquid.  At any given temperature, there will be less gas molecules above the liquid, so the vapor pressure will have decreased.  Like we discussed with boiling, molecules have a harder time leaving the liquid phase when a solute is present.  The Boiling Point increased because more heat is needed to vaporize the liquid.  At any given temperature, there will be less gas molecules above the liquid, so the vapor pressure will have decreased.

37. Calculating Vapor Pressure Depression  P A = X A *P A o  P A is the vapor pressure above the solution.  X A is the mole fraction of the solvent.  P A o is the vapor pressure of the pure solvent at the same temperature.  P A = X A *P A o  P A is the vapor pressure above the solution.  X A is the mole fraction of the solvent.  P A o is the vapor pressure of the pure solvent at the same temperature.

38. Osmotic Pressure Elevation  Osmosis is the net movement of solvent through a semipermeable membrane toward the solution with greater solute concentration.   = T*R*i*M   is the osmotic pressure.  M is the molarity of the solution.  R is the ideal gas law costant (.0821 latm/molK)  T is the absolute temperature.  Osmosis is the net movement of solvent through a semipermeable membrane toward the solution with greater solute concentration.   = T*R*i*M   is the osmotic pressure.  M is the molarity of the solution.  R is the ideal gas law costant (.0821 latm/molK)  T is the absolute temperature.

39. Last Problem  What is the osmotic pressure of a saline solution made from 117 g of NaCl in 1.00 liter of solution at 298 K?

40. The answer is:   = i*M*R*T  117 g NaCl /(58.5 g/mol) = 2.00 mol  M = 2.00 mol/ 1.00 liter = 2.00 M   = (2)*(2.00 M)*(.0821 latm/molK)*(298 K)   = 97.9 atm   = i*M*R*T  117 g NaCl /(58.5 g/mol) = 2.00 mol  M = 2.00 mol/ 1.00 liter = 2.00 M   = (2)*(2.00 M)*(.0821 latm/molK)*(298 K)   = 97.9 atm

41. Here’s Another Problem  What is the vapor pressure above a solution made from dissolving 360 g of glucose (C 6 H 12 O 6 ) in 144 g of water (H 2 O) at 20 o C?  The vapor pressure of pure water at 20 o C is 17.5 Torr.  What is the vapor pressure above a solution made from dissolving 360 g of glucose (C 6 H 12 O 6 ) in 144 g of water (H 2 O) at 20 o C?  The vapor pressure of pure water at 20 o C is 17.5 Torr.

42. This Process is Different  First find the mole fraction of water  360 g of glucose/(180 g/mol) = 2 mol  144 g of water/(18 g/mol) = 8 mol  X A = 8 mol H 2 O/(8 mol H 2 O + 2 mol C 6 H 12 O 6 ) =.800  P A = X A *P A o = (.800)(17.5 Torr)  P A = 14.0 Torr  First find the mole fraction of water  360 g of glucose/(180 g/mol) = 2 mol  144 g of water/(18 g/mol) = 8 mol  X A = 8 mol H 2 O/(8 mol H 2 O + 2 mol C 6 H 12 O 6 ) =.800  P A = X A *P A o = (.800)(17.5 Torr)  P A = 14.0 Torr

43. Colloids  Although they often appear to be solutions, there are heterogeneous mixtures called colloids in which larger particles appear to dissolve.

44. Colloidal Suspension  Yeah I know it’s a lava lamp, but its also a colloidal suspension.  A colloidal suspension is a colloid in which the particles are visible to the unaided eye.  Yeah I know it’s a lava lamp, but its also a colloidal suspension.  A colloidal suspension is a colloid in which the particles are visible to the unaided eye.

45. The Tyndall Effect  Colloids can appear similar to a solution until we shine a beam of light in them.  In the solution, the beam of light is coherent.  In the colloid, the beam of light bounces off the particles, scattering the light.  Colloids can appear similar to a solution until we shine a beam of light in them.  In the solution, the beam of light is coherent.  In the colloid, the beam of light bounces off the particles, scattering the light.

46. There are many types of Colloids

47. Hydrophobic Colloids  Hydrophobic colloids do not mix well with water; they form clusters within the water and hang there (thus the term suspension).  They are usually particles that are charged or become charged in water and use their own localized attractions to avoid the water.  Hydrophobic colloids do not mix well with water; they form clusters within the water and hang there (thus the term suspension).  They are usually particles that are charged or become charged in water and use their own localized attractions to avoid the water.

48. Hydrophilic Colloids  These “water loving” colloids disperse well with water, generally due to the similar IMF attractions.

49. Associative (Amphipathic) Colloids  These chemicals are usually “designed” to have a hydrophobic end and a hydrophillic end. One end allows it to attract to nonpolar substance, the other end attracts polar substances, like water.  Sometimes these colloids are called surfactants.  These chemicals are usually “designed” to have a hydrophobic end and a hydrophillic end. One end allows it to attract to nonpolar substance, the other end attracts polar substances, like water.  Sometimes these colloids are called surfactants.