1. 1 Solutions Chapter 12

2. 2 Definitions Definitions Solution = homogeneous mixture of two or more substances  Major component = solvent  Minor component = solute Solubility = amount of solute that dissolves in a given volume of solvent

3. 3 Why solutions form Entropy: natural tendency to randomize

4. 4 Why solutions form Intermolecular forces

5. 5 Like dissolves like Polar solvents (water) dissolve polar/ionic solutes Nonpolar solvents (oil) dissolve nonpolar solutes

6. 6 Energy & dissolving: molecular Separate solute (endothermic) Separate solvent (endothermic) Mix solute + solvent (exothermic)  H solution =  H solute  H solvent +  H mix

7. 7 Energy & dissolving: salts in water Energy to separate ions (∆H solute ) = lattice energy ∆H water + ∆H mixing = ∆H hydration 

8. 8 Energy & dissolving: salts in water Salts with high lattice energy may not dissolve in water  High ion charge  Small ion size   H solution =  H lattice  H hydration

9. 9 The limit of dissolving: equilibrium

10. 10 Solubility & temperature Most solids dissolve to higher concentrations as temperature increases Gases are less soluble as temperature increases

11. 11 Gas solubility & pressure Gas solubility increases as the pressure of that gas increases Henry’s Law

12. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 12 Henry’s Law the solubility of a gas (S gas ) is directly proportional to its partial pressure, (P gas ) S gas = k H P gas k H is called Henry’s Law Constant

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14. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 14 Concentrations solutions have variable composition to describe a solution, need to describe components and relative amounts the terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution concentration = amount of solute in a given amount of solution  occasionally amount of solvent

15. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 15 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar molarity = moles of solute liters of solution

16. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 16 Solution Concentration Molality, m moles of solute per 1 kilogram of solvent  defined in terms of amount of solvent, not solution S like the others does not vary with temperature  because based on masses, not volumes

17. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 17 Percent parts of solute in every 100 parts solution mass percent = mass of solute in 100 parts solution by mass  if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution S or 0.9 kg solute in every 100 kg solution

18. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 18 Percent Concentration

19. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 19 Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent  12%(m/m) sugar(aq) means 12 g sugar  100 g solution S or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution  5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution  22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa

20. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 20 Preparing a Solution Preparing a Solution need to know amount of solution and concentration of solution calculate the mass of solute needed  start with amount of solution  use concentration as a conversion factor  5% by mass  5 g solute  100 g solution  “Dissolve the grams of solute in enough solvent to total the total amount of solution.”

21. 21 Solution Concentration PPM grams of solute per 1,000,000 g of solution mg of solute per 1 kg of solution 1 liter of water = 1 kg of water  for water solutions we often approximate the kg of the solution as the kg or L of water grams solute grams solution x 106 mg solute kg solution mg solute L solution

22. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 22 Solution Concentrations Mole Fraction, X A the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution total of all the mole fractions in a solution = 1 unitless the mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution  = mole fraction x 100% mole fraction of A = X A = moles of components A total moles in the solution

23. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 23 Raoult’s Law the vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P° P solvent in solution =  solvent ∙ P°  since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent

24. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 24 Ionic Solutes and Vapor Pressure according to Raoult’s Law, the effect of solute on the vapor pressure simply depends on the number of solute particles when ionic compounds dissolve in water, they dissociate – so the number of solute particles is a multiple of the number of moles of formula units the effect of ionic compounds on the vapor pressure of water is magnified by the dissociation  since NaCl dissociates into 2 ions, Na + and Cl , one mole of NaCl lowers the vapor pressure of water twice as much as 1 mole of C 12 H 22 O 11 molecules would

25. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 25 Effect of Dissociation

26. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 26 Ex 12.6 – What is the vapor pressure of H 2 O when 0.102 mol Ca(NO 3 ) 2 is mixed with 0.927 mol H 2 O @ 55°C? the unit is correct, the pressure lower than the normal vapor pressure makes sense Check: Solve: P =  ∙P° Concept Plan: Relationships: 0.102 mol Ca(NO 3 ) 2, 0.927 mol H 2 O, P° = 118.1 torr P of H 2 O, torr Given: Find:  H2O, P° H2O P H2O Ca(NO 3 ) 2  Ca2+ + 2 NO 3   3(0.102 mol solute)

27. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 27 Raoult’s Law for Volatile Solute when both the solvent and the solute can evaporate, both molecules will be found in the vapor phase the total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent  for an ideal solution P total = P solute + P solvent the solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s P solute =  solute ∙ P° solute and P solvent =  solvent ∙ P° solvent

28. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 28 Ideal vs. Nonideal Solution in ideal solutions, the made solute-solvent interactions are equal to the sum of the broken solute-solute and solvent-solvent interactions  ideal solutions follow Raoult’s Law effectively, the solute is diluting the solvent if the solute-solvent interactions are stronger or weaker than the broken interactions the solution is nonideal

29. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 29 Vapor Pressure of a Nonideal Solution when the solute-solvent interactions are stronger than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s Law  because the vapor pressures of the solute and solvent are lower than ideal when the solute-solvent interactions are weaker than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be larger than predicted by Raoult’s Law

30. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 30 Deviations from Raoult’s Law

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32. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 32 Freezing Point Depression the freezing point of a solution is lower than the freezing point of the pure solvent  for a nonvolatile solute  therefore the melting point of the solid solution is lower the difference between the freezing point of the solution and freezing point of the pure solvent is directly proportional to the molal concentration of solute particles  FP solvent – FP solution ) =  T f = m ∙ K f the proportionality constant is called the Freezing Point Depression Constant, K f  the value of K f depends on the solvent  the units of K f are °C/m

33. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 33 KfKfKfKf

34. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 34 Ex 12.8 – What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C 2 H 6 O 2 ? the unit is correct, the freezing point lower than the normal freezing point makes sense Check: Solve:  T f = m ∙K f, K f for H 2 O = 1.86 °C/m, FP H2O = 0.00°C Concept Plan: Relationships: 1.7 m C 2 H 6 O 2 (aq) T f, °C Given: Find: m TfTf

35. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 35 Boiling Point Elevation the boiling point of a solution is higher than the boiling point of the pure solvent  for a nonvolatile solute the difference between the boiling point of the solution and boiling point of the pure solvent is directly proportional to the molal concentration of solute particles  BP solution – BP solvent ) =  T b = m ∙ K b the proportionality constant is called the Boiling Point Elevation Constant, K b  the value of K b depends on the solvent  the units of K b are °C/m

36. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 36 Ex 12.9 – How many g of ethylene glycol, C 2 H 6 O 2, must be added to 1.0 kg H 2 O to give a solution that boils at 105°C? Solve:  T b = m ∙K b, K b H 2 O = 0.512 °C/m, BP H2O = 100.0°C MM C2H6O2 = 62.07 g/mol, 1 kg = 1000 g Concept Plan: Relationships: 1.0 kg H 2 O, T b = 105°C mass C 2 H 6 O 2, g Given: Find: TbTb mkg H 2 Omol C 2 H 6 O 2 g C 2 H 6 O 2

37. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 37 Osmosis osmosis is the flow of solvent through a semi- permeable membrane from solution of low concentration to solution of high concentration the amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure the osmotic pressure, , is directly proportional to the molarity of the solute particles  R = 0.08206 (atm ∙ L)/(mol ∙ K)  = MRT

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39. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 39 Ex 12.10 – What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25°C? Solve:  MRT, T(K)=T(°C)+273.15, R= 0.08206atm∙L/mol∙K M = mol/L, 1 mL = 0.001 L, MM = g/mol, 1 atm = 760 torr Concept Plan: Relationships: 5.87 mg/10 mL, P = 2.45 torr, T = 25°C molar mass, g/mol Given: Find:  T MmLLmol protein

40. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 40 Colligative Properties colligative properties are properties whose value depends only on the number of solute particles, and not on what they are  Vapor Pressure Depression, Freezing Point Depression, Boiling Point Elevation, Osmotic Pressure the van’t Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved measured van’t Hoff factors are often lower than you might expect due to ion pairing in solution

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42. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 42 An isosmotic solution has the same osmotic pressure as the solution inside the cell – as a result there is no net the flow of water into or out of the cell. A hyperosmotic solution has a higher osmotic pressure than the solution inside the cell – as a result there is a net flow of water out of the cell, causing it to shrivel A hyposmotic solution has a lower osmotic pressure than the solution inside the cell – as a result there is a net flow of water into the cell, causing it to swell

43. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 43 Colloids a colloidal suspension is a heterogeneous mixture in which one substance is dispersed through another  most colloids are made of finely divided particles suspended in a medium the difference between colloids and regular suspensions is generally particle size – colloidal particles are from 1 to 100 nm in size

44. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 44 Properties of Colloids the particles in a colloid exhibit Brownian motion colloids exhibit the Tyndall Effect  scattering of light as it passes through a suspension  colloids scatter short wavelength (blue) light more effectively than long wavelength (red) light

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47. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 47 Soap and Micelles soap molecules have both a hydrophilic (polar/ionic) “head” and a hydrophobic (nonpolar) “tail” part of the molecule is attracted to water, but the majority of it isn’t the nonpolar tail tends to coagulate together to form a spherical structure called a micelle

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49. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 49 Soap Action most dirt and grease is made of nonpolar molecules – making it hard for water to remove it from the surface soap molecules form micelles around the small oil particles with the polar/ionic heads pointing out this allows the micelle to be attracted to water and stay suspended

50. Tro, Ch em istr y: A Mo lec ula r Ap pro ac h 50 The polar heads of the micelles attract them to the water, and simultaneously repel other micelles so they will not coalesce and settle out.

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