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1 Properties of Solutions Chapter 11. 2 Overview Introduce student to solution composition and energy of solution formation. Factor affecting solubilities.

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Presentation on theme: "1 Properties of Solutions Chapter 11. 2 Overview Introduce student to solution composition and energy of solution formation. Factor affecting solubilities."— Presentation transcript:

1 1 Properties of Solutions Chapter 11

2 2 Overview Introduce student to solution composition and energy of solution formation. Factor affecting solubilities will be discussed: structure, pressure and temperature effects. Vapor pressure of solutions, boiling point, freezing point affected by solute addition. Colligative properties of electrolyte solutions and colloids.

3 3 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

4 4 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.

5 5 Solution Composition 1.Molarity (M) = 2.Mass (weight) percent = 3.Mole fraction (  A ) = 4.Molality (m) =

6 6 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 = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution Mole Fraction (X) X A = moles of A sum of moles of all components

7 7 Concentration Units Continued M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg)

8 8 N = Number of equivalent liters of solution Normality (N) Acid-Base reaction: is the amount needed to accept one mole of H + Number of equivalent = mass Equivalent mass x 1 V Equivalent mass of H 2 SO 4 = MM 2 Equivalent mass of Ca(OH) 2 = MM 2

9 9 Redox reaction : is the amount needed to accept exactly one mole e - MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O Equivalent mass of KMnO 4 = MM 5

10 10

11 11 What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is 0.927 g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH 0.657 kg solvent = 8.92 m

12 12 Steps in Solution Formation Step 1 -Expanding the solute (endothermic) Step 2 -Expanding the solvent (endothermic) Step 3 -Allowing the solute and solvent to interact to form a solution (exothermic)  H soln =  H step 1 +  H step 2 +  H step 3

13 13 The formation of a liquid solution can be divided into three steps: (1) expanding the solute, (2) expanding the solvent, and (3) combining the expanded solute and solvent to form the solution.

14 14 HH < 0Exothermic Solution will occur

15 15 The driving factor that favor a process of solution formation is an increase in disorder

16 16

17 17 Factor Affecting Solubility 1.Structure 2.Pressure 3.Temperature

18 18 “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 CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l)

19 19 The molecular structures of (a) vitamin A (nonpolar, fat-soluble) and (b) vitamin C (polar, water-soluble). The circles in the structural formulas indicate polar bonds. Note that vitamin C contains far more polar bonds than vitamin A. Fat Soluble Hydrophobic Water Soluble Hydrophilic

20 20 Pressure Effects Pressure has little effect on solids and liquids. It increases the solubility of gases.

21 21 (a) A gaseous solute in equilibrium with a solution. (b) The piston is pushed in, increasing the pressure of the gas and number of gas molecules per unit volume. This causes an increase in the rate at which the gas enters the solution, so the concentration of dissolved gas increases. (c) The greater gas concentration in the solution causes an increase in the rate of escape. A new equilibrium is reached.

22 22

23 23 Henry’s Law P = kC P = partial pressure of gaseous solute above the solution C = concentration of dissolved gas k = a constant The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

24 24 Henry’s Law Applied for Dilute solutions Gases that do not dissociate or react with solvent: –O 2 /waterApplied –HCl/waterIs not applied

25 25 Temperature Effect on Gases The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.

26 26 The solubilities of several solids as a function of temperature. Temperature Effect on Gases

27 27 The Vapor Pressure of Solutions Solutions have different physical properties from pure solvent. Solutions of nonvolatile solutes differ from solutions of volatile solvents.

28 28 An aqueous solution and pure water in a closed environment. (a) Initial stage. (b) After a period of time, the water is transferred to the solution. Vapor pressure of pure water is higher VP of solution is lower

29 29 Raoult’s Law P soln =  solvent P  solvent P soln = vapor pressure of the solution  solvent = mole fraction of the solvent P  solvent = vapor pressure of the pure solvent The presence of a nonvolatile solute lowers the vapor pressure of a solvent.  solvent = Moles of solvents Moles of solvent + Moles of solute

30 30 Note Solutions that obey Raoult’s Law are called Ideal Solutions. Can be used to determine the molar mass of unknown. For ionic compounds you should multiply by the total number of ions per molecule e.g. Na 2 SO 4 n = 3 x Na 2 SO 4

31 31 For a solution that obeys Raoult's law, a plot of P soln versus x solvent gives a straight line.

32 32 Non-Ideal Solutions When a solution contains two volatile components, both contribute to the total vapor pressure.

33 33 P A = X A P A 0 P B = X B P B 0 P T = P A + P B P T = X A P A 0 + X B P B 0

34 34 (a) ideal liquid-liquid solution by Raoult's law. (b) This solution shows a positive deviation from Raoult's law. (c) This solution shows a negative deviation from Raoult's law.

35 35

36 36 P T is greater than predicted by Raoults’s law P T is less than predicted by Raoults’s law Force A-B Force A-A Force B-B <& Force A-B Force A-A Force B-B >&

37 37 Colligative Properties of Non-Electrolyte Solutions Depend only on the number, not on the identity, of the solute particles in an ideal solution. 4 Boiling point elevation 4 Freezing point depression 4 Osmotic pressure

38 38 Boiling Point Elevation A nonvolatile solute elevates the boiling point of the solvent.  T = K b m solute K b = molal boiling point elevation constant m = molality of the solute

39 39 Boiling-Point Elevation  T b = T b – T b 0 T b > T b 0  T b > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution  T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m)

40 40 Freezing Point Depression A nonvolatile solute depresses the freezing point of the solvent.  T = K f m solute K f = molal freezing point depression constant m = molality of the solute

41 41 Freezing-Point Depression  T f = T f – T f 0 T f > T f 0  T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution  T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m)

42 42 Osmotic Pressure Osmosis: The flow of solvent into the solution through the semipermeable membrane. Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent.

43 43 Osmotic Pressure (  ) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (  ) is the pressure required to stop osmosis.  = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K)

44 44 A tube with a bulb on the end that is covered by a semipermeable membrane.

45 45 The normal flow of solvent into the solution (osmosis) can be prevented by applying an external pressure to the solution. The minimum pressure required to stop the osmosis is equal to the osmotic pressure of the solution.

46 46 (a) A pure solvent and its solution (containing a nonvolatile solute) are separated by a semipermeable membrane through which solvent molecules (blue) can pass but solute molecules (green) cannot. The rate of solvent transfer is greater from solvent to solution than from solution to solvent. (b) The system at equilibrium, where the rate of solvent transfer is the same in both directions.

47 47 If the external pressure is larger than the osmotic pressure, reverse osmosis occurs. One application is desalination of seawater.

48 48 Reverse osmosis.

49 49 A cell in an: isotonic Solution (identical  ) hypotonic Solution (water in) hypertonic Solution (water out)

50 50 Colligative Properties of Electrolyte Solutions  T = imK  = iMRT van’t Hoff factor, “i”, relates to the number of ions per formula unit. NaCl = 2, K 2 SO 4 = 3 (Theoretically)

51 51 Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = iMRT Colligative Properties of Electrolyte Solutions

52 52 In an aqueous solution a few ions aggregate, forming ion pairs that behave as a unit: Ion Pairing

53 53 Colloids Colloidal Dispersion (colloid): A suspension of tiny particles in some medium. aerosols, foams, emulsions, sols Coagulation: The addition of an electrolyte, causing destruction of a colloid.

54 54 A representation of two colloidal particles. The opposing charge repel and prevent precipitation The destruction of colloids is called coagulation, achieved by 1.Increasing temperature where colliding at higher veloicities to penetrate the ion barriers and aggregate. 2.Adding electrolytes to neutralize ion layers, this is why clay deposits where rivers reach the oceans.

55 55 A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution colloidal particles are much larger than solute molecules colloidal suspension is not as homogeneous as a solution

56 56 The Cleansing Action of Soap

57 57 The Cottrell precipitator installed in a smokestack. The charged plates attract the colloidal particles because of their ion layers and thus remove them from the smoke.

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100 100 Chapter 11 Properties of Solutions

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