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Chapter 11

Chapter 11. Properties of Solutions. 11.1 Solution composition. A solution is a homogenous mixture of 2 or more substances. The solute is(are) the substance(s) present in the smaller amount(s). The solvent is the substance present in the larger amount. Various types of solutions.

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Chapter 11

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  1. Chapter 11 Properties of Solutions

  2. 11.1 Solution composition A solutionis a homogenous mixture of 2 or more substances The soluteis(are) the substance(s) present in the smaller amount(s) Thesolventis the substance present in the larger amount

  3. Various types of solutions

  4. moles of A XA = sum of moles of all components x 100% massof solute x 100% = mass ofsolution mass of solute mass of solute + mass of solvent Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = Mole Fraction(X)

  5. moles of solute m = mass of solvent (kg) Molarity(M) molesof solute M = volume of solution (liters) Molality(m)

  6. Example • 1.00 g C2H5OH is added to 100.0 g of water to make 101 mL of solution. Find the molarity, mass %, mole fraction and molality of ethanol.

  7. Molarity • Number of moles of solute per L or solution

  8. Mass Percent • also called weight percent • percent by mass of the solute in the solution

  9. Mole Fraction • ratio of number of moles of a part of solution to total number of moles of solution

  10. Molality • Number of moles of solute per kg of solvent

  11. Example An aqueous antifreeze solution is 40% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity and mole fraction of ethylene glycol

  12. Mass of water (solvent) = 100-40 = 60.0 g where EG = ethylene glycol (C2H6O2) # mol EG = 0.644 mol EG # mol water = 60.0/18.0 = 3.33 mol

  13. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? Assume 1 L of solution: Mass of solute = mass of 5.86 moles ethanol = 270 g ethanol Mass of solvent = mass of 1 L solution= (1000 mL x 0.927 g/mL) = 927 g of solution mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg 5.86 moles C2H5OH = 8.92 m = 0.657 kg solvent

  14. 11.2 Energies of Solution Formation Dissolution of a solute in liquids “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. • Non-polar molecules are soluble in non-polar solvents CCl4 in C6H6 • Polar molecules are soluble in polar solvents C2H5OH in H2O • Ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l) • Why this behavior occur?

  15. Solubility Process The formation of a liquid solution takes place in 3 steps 1. Expand solute molecules 2. Expand solvent molecules 3. Mixing solute and solvent Endothermic Endothermic Often Exothermic

  16. Energy of Solubility Process • Steps 1 and 2 require energy to overcome IMFs • endothermic • Step 3 usually releases energy • exothermic • enthalpy of solution • sum of ∆H values • can be – or + • ∆Hsoln = ∆H1 + ∆H2 + ∆H3

  17. Energy of Solubility Process

  18. Case 1: oil and water • Oil is nonpolar (LD forces) • Water is polar (H bonding) • ∆H1 will be small for typical size • ∆H2 will be large • ∆H3 will be small since there won’t be much interaction between the two • ∆Hsoln will be large and +ve because energy required by steps 1 and 2 is larger than the amount released by 3

  19. Case 2: NaCl and water • NaCl is ionic • water is polar (H bonding) • ∆H1 will be large • ∆H2 will be large • ∆H3 will be large and –ve because of the strong interaction between ions and water • ∆Hsoln will be close to zero- small but +ve

  20. Energy of solubility process • Enthalpy of hydration - ∆Hhyd enthalpy change associated with dispersal of gaseous solute in water NaCl(s)  Na+(g) + Cl-(g) ∆H1=786 kJ/mol H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) ∆Hhyd=∆H2 + ∆H3=-783 kJ/mol ∆Hsoln=3 kJ/mol

  21. Solute and Solvent DH3 DH2 Solvent DH1 DH3 Solution Solution • Size of DH3 determines whether a solution will form Energy Reactants

  22. Mixing solvent and solute • Two factors explains the solubility: • An increase in the probability of mixing favors the process • Processes that require large amounts of energy tend not to occur • If DHsoln is small and positive, a solution will still form because of entropy. • There are many more ways for them to become mixed than there is for them to stay separate.

  23. Energy of Solubility Process

  24. 1. Structure and Solubility Water soluble molecules must have dipole moments -polar bonds. To be soluble in non polar solvents the molecules must be non polar. 11.3 Factors Affecting Solubility • Pentane C5H12 is miscible with hexane C6H14 • and immiscible with water • Solubility of alcohols decreases with the molar • mass? • Cl3OH CH3(CH2)3OH CH3 (CH2)6OH • Soluble Insoluble • (Hydrophilic) (Hydrophobic) Polarity decreases OH is smaller portion Hydrocarbon is larger

  25. Structure effects: Vitamins and the body • hydrophobic: water- fearing: nonpolar • Insoluble in water • hydrophilic: water-loving: polar • Soluble in water Hydrophilic, excreted by the body and must be consumed regularly Hydrophobic, accumulates in the body. The body can tolerate a diet deficient in vitamin

  26. 2. Pressure effects Changing the pressure doesn’t effect the amount of solid or liquid that dissolves They are incompressible. It does effect gases. Pressure effects the amount of gas that can dissolve in a liquid. The dissolved gas is at equilibrium with the gas above the liquid.

  27. The gas is at equilibrium with the dissolved gas in this solution. The equilibrium is dynamic.

  28. If the pressure is increased the gas molecules dissolve faster. The equilibrium is disturbed.

  29. The system reaches a new equilibrium with more gas dissolved. Henry’s Law. P= kC Pressure = constant x Concentration of gas

  30. Henry’s Law C  P C = kP C is concentration and P is partial pressure of gaseous solute The law is obeyed best by dilute solutions of gases that don’t dissociate or react with solvent • Amount of gas dissolved is directly proportional to P of gas above solution

  31. Example A soft drink bottled at 25°C contains CO2 at pressure of 5.0 atm over liquid. Assume that PCO2 in atmosphere is 4.0 x 10-4 atm. Find the equilibrium concentration in soda before and after opening. k=32 L*atm/mol at 25°C

  32. Example before opening: after opening:

  33. Example Solubility of pure N2 in blood at body temp, 37oC and 1 atm is 6.2X10-4 M. If a diver breaths air ( = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of N2 in his body.

  34. 3. Temperature Effects (for aqueous solutions) Increased temperature increases the rate at which a solid dissolves. But it is not possible to predict whether it will increase the amount of solid that dissolves. Solubility can be predicted only from a graph of experimental data.

  35. 100 40 60 80 20

  36. Temperature effects dissolving a solid occurs faster at higher T but the amount able to be dissolved does not change

  37. Temperature effects Solubility of gas in water decreases with T

  38. solubility increases with increasing temperature solubility decreases with increasing temperature Temperature and Solubility Solid solubility and temperature 12.4

  39. Solubility and environment Thermal pollution Water used as a coolant when pumped again into the source (lakes and rivers) floats on the cold water causing a decrease in solubility of O2 and consequently affecting the aquatic life. CO2 dissolves in water that contains CO32- causing formation of HCO3- that is soluble in water. When temp increases CO2 will be driven off the water causing precipitation of CO32- again forming scales on the wools that blocks the pipes and reduce the heating efficiency CO32- (aq) + CO2(aq) 2HCO3-

  40. 11.4 The vapor pressures of solutions A nonvolatile solvent lowers the vapor pressure of the solution. For the molecules of the solventto escape, they must overcome the forces of both the other solvent molecules and the solute molecules. Nonvolatile solute decreases # of solvent molecules per unit volume. Thus, # of solute molecules escaping will be lowered.

  41. Water has a higher vapor pressure than a solution How a nonvolatile solute affects a solvent? P1 Po1 Aqueous Solution Pure water

  42. Water molecules evaporate faster from pure water than from the solution Po1 P1 < Aqueous Solution Pure water

  43. The water molecules condense faster in the solution so it should all end up there. empty Aqueous Solution Pure water

  44. Raoult’s Law The presence of a nonvolatile solute lowers the vapor pressure of the solvent. Psolution = Observed Vapor pressure of the solution (vapor pressure of solvent in solution) solvent = Mole fraction of the solvent P0solvent = Vapor pressure of the pure solvent Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure (solute and solvent are alike: solute-solute, solvent-solvent and solute-solvent interactions Are very similar ). Solute acts to dilute the solvent

  45. Only one solute Vapor pressure lowering

  46. Deviations If solvent has a strong affinity for solute (H bonding). Lowers solvents ability to escape. Lower vapor pressure than expected. Negative deviation from Raoult’s law. If DHsoln is large and negative (exothermic): strong interaction exists between the solute and the solvent; thus a negtive deviation from Rault’s law because both components will have lower escaping tendency in the solution than in pure liquids. If the two liquids mix endothermically (solute-solvent interactions are weaker than interactions among molecules in the pure liquids. More energy is required to expand liquids than is released when liquids are mixed). In this case molecules in the solution have higher tendency to escape than expected and positive deviations from Raul’s law are observed.

  47. Nonideal solutions Liquid-liquid solutions in which both components are volatile Modified Raoult's Law: P0 is the vapor pressure of the pure solvent PAand PB are the partial pressures

  48. Raoult’s Law – Ideal Solution A solution of two liquids that obeys Raoult’s Law is called an ideal solution

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