Chem 106, Prof. T.L. Heise

CHE 106: General Chemistry

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CHAPTER THIRTEEN

Copyright © Tyna L. Heise 2001 - 2002 All Rights Reserved

The Solution Process

Chem 106, Prof. T.L. Heise 2

• Solution - homogeneous mixture of two

or more substances

• Solvent - substance of greater amount in

the homogeneous mixture (solution)

• Solute - compounds “dissolved” in the

solvent

• Concentration - the amount of solute

dissolved in a solvent. Expressed in molarity (M) Chap 13.1

The Solution Process

Chem 106, Prof. T.L. Heise 3

• Solution -may be gases, liquids or solids –

Examples: air ocean water 10K gold (alloy) Chap 13.1

The Solution Process

Chem 106, Prof. T.L. Heise 4 A solution is formed when one substance disperses uniformly through another.

All solutions, except gas mixtures, involve substances in a condensed state Intermolecular forces are also going to operate between solute and solvent - interactions are known as solvation - when solvent is water, the interactions are known as hydration Chap 13.1

The Solution Process

Chem 106, Prof. T.L. Heise 5 Energy changes during solution formation is the sum of three energy changes:

D

H 1

=

separation of solute molecules

D

H 2 = separation of solvent molecules

D

H 3 = formation of solute-solvent interactions

D

H 1 and

D

H 2 are endothermic because you are breaking or overcoming interactive forces

D

H 3

The Solution Process

Chem 106, Prof. T.L. Heise 6 Chap 13.1

The Solution Process

Chem 106, Prof. T.L. Heise 7 Solution Formation depends on two factors 1) energy or enthalpy changes

D

H 2) chaos or entropy changes

D

S Natural phenomenon's occur to satisfy two basic laws - energy content decreases - disorder content increases ** Formation of solutions is favored by the increase in disorder that accompanies mixing Chap 13.1

The Solution Process

Chem 106, Prof. T.L. Heise 8 Solution Formation can occur during two basic processes 1) physical changes 2) chemical changes Our focus is on physical changes, key to recognizing difference is examining whether you can get the salt BACK unchanged when reaction is done.

Chap 13.1

Chem 106, Prof. T.L. Heise

Saturated Solutions and Solubility

9 As a solid solute is dissolved, the concentration of dissolved particles increases, as does the chance of a collision between to dissolved particles. If two dissolved particles collide, the attractive forces could cause recrystallization In any solution Solute + solvent dissolve solution crystallize Chap 13.2

Chem 106, Prof. T.L. Heise Saturated: in equilibrium, rate of dissolving equals rate of crystallization, concentration remains constant Solubility: the amount of solute needed to form a saturated solution in a given temperature Unsaturated: dissolved less the amount needed to form a saturated solution Supersaturated: dissolving more than the amount needed Chap 13.2

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

11

• Solute-Solvent Interactions » the stronger the attractions between solute

and solvent, the greater the solubility

» miscible and immiscible » like dissolves like

Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

12

• Pressure Effects » the solubility of a gas in any solvent is

increased as the pressure of the gas over the solvent is increased Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

13

• Pressure Effects » relationship between pressure and the

solubility of a gas is expressed in terms of a simple equations known as Henry’s Law Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

14 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm.

Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

15 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm.

1) C g = kP g Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

16 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm.

1) C g C g = x = kP g k = 3.1 x 10 P g -2 = 3.0 x 10 -4 mol/L-atm (given page 480) atm Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

17 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm.

1) C g = kP g x = 3.1 x 10 -2 mol/L-atm(3.0 x 10 -4 atm) Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

18 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm.

1) C g = kP g x = 3.1 x 10 -2 mol/L-atm(3.0 x 10 -4 x = 9.3 x 10 -6 mol/L atm) Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

19

• Temperature Effects » solubility of most solid solutes increases as

the temperature of the solution does

» solubility of most gases decreases as

temperature of the solution does

» look up solubility on a solubility table to

find exact trend Chap 13.3

Chem 106, Prof. T.L. Heise

Factors Affecting Solubility

20

• Temperature Effects

Chap 13.3

Chem 106, Prof. T.L. Heise

Expressing Concentration

21

• Dilute - relatively small concentration of

solute in solution

• Concentrated - relatively large concentration

of solute in solution

• Ways to express numerically- •mass percentage •mole fraction •molarity •molality

Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

22

• mass percentage

mass of component *100 total mass of soln

• ppm

mass of component *1,000,000 total mass of soln

• ppb

mass of component *1,000,000,000 total mass of soln Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

23 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

24 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100 total mass of soln Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

25 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100 total mass of soln = 1.50 g NaCl *100 51.50 g soln Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

26 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100 total mass of soln = 1.50 g NaCl *100 51.50 g soln Chap 13.4

=2.91 %

Chem 106, Prof. T.L. Heise

Expressing Concentration

27 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

28 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

mass percent = mass of component * 100 total mass of soln Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

29 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

mass percent = mass of component * 100 total mass of soln 3.62 % = 3.62 % * 2500g = x 100% x * 100 2500 g soln Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

30 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

mass percent = mass of component * 100 total mass of soln 3.62 % = 3.62 % * 2500g = x 100% x * 100 2500 g soln Chap 13.4

90.5 g NaOCl

Chem 106, Prof. T.L. Heise

Expressing Concentration

31

• mole fraction

mole fraction = moles of component total moles of all components

• symbol X is often used to represent mole

fraction

• molarity = moles of solute/L of solution

molality = moles of solute/kg of solvent Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

32 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8 , in 428 g of toluene, C 7 H 8 ?

Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

33 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8 , in 428 g of toluene, C 7 H 8 ?

1) molality = moles of solute kg of solvent Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

34 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8 , in 428 g of toluene, C 7 H 8 ?

1) molality = moles of solute kg of solvent moles = 36.5 g C 10 H 8 1 mol = 0.285 mol 128 g C 10 H 8 Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

35 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8 , in 428 g of toluene, C 7 H 8 ?

1) molality = moles of solute kg of solvent moles = 0.285 mol kg = 428g 1 kg = 0.428 kg 1000 g Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

36 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8 , in 428 g of toluene, C 7 H 8 ?

1) molality = moles of solute kg of solvent moles = 0.285 mol = 0.285 mol kg = 0.428 kg 0.428 kg Chap 13.4

= 0.670 mol/kg

Chem 106, Prof. T.L. Heise

Expressing Concentration

37 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality (b) the mole fraction of NaOCl Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

38 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality mass % = mass of solute *100 mass of solution Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

39 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality mass % = mass of solute *100 mass of solution 3.62% = mass of solute * 100 Chap 13.4

100 g solution * when not given an amount f solution, assume 100g.

Chem 106, Prof. T.L. Heise

Expressing Concentration

40 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality mass of solute = 3.62 g molality = moles kg 3.62 g NaOCl 1 mol NaOCl 74.44 g NaOCl = 0.0486 mol Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

41 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality molality = moles = 0.0486 mol = 0.505 mol kg 0.09638 kg kg Chap 13.4

*don’t forget we assumed 100 g of solution, so 3.62 g of NaOCl and 96.38 g water

Chem 106, Prof. T.L. Heise

Expressing Concentration

42 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl Mole fraction = moles of solute total moles of components Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

43 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl Mole fraction = moles of solute total moles of components 3.62 g NaOCl 1 mol NaOCl = 0.0486 mol 74.44 g NaOCl NaOCl 96.38 g H 2 O 1 mol H 2 O = 5.35 mol H 2 O 18 g H 2 O Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

44 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl Mole fraction = moles of solute total moles of components 0.0486 mol NaOCl 0.0486 mol NaOCl + 5.35 mol H 2 O = 0.00900 Chap 13.4

Colligative Properties

Chem 106, Prof. T.L. Heise 45 Certain physical properties of solutions differ from the pure solvent - lowering freezing point - raising boiling point - reduction of vapor pressure - alteration of osmotic pressure Chap 13.5

Colligative Properties

Chem 106, Prof. T.L. Heise 46 Vapor Pressure Vapor pressure is the pressure exerted by the vapor on the surface when closed flask achieves equilibrium Chap 13.5

Colligative Properties

Chem 106, Prof. T.L. Heise 47 Boiling Point

K b

is the molal boiling point constant and is equal to 0.52°C/m - this constant is related to the number of dissolved particles Chap 13.5

Colligative Properties

Chem 106, Prof. T.L. Heise 48 Freezing Point

K f

is the molal boiling point constant and is equal to 1.87°C/m - this constant is related to the number of dissolved particles Chap 13.5

Chem 106, Prof. T.L. Heise

Expressing Concentration

49 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C eucalyptus tree.

10 H 18 O, a fragrant substance found in the leaves of the Chap 13.5

Chem 106, Prof. T.L. Heise

Expressing Concentration

50 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C eucalyptus tree.

10 H 18 O, a fragrant substance found in the leaves of the molality = moles of solute kg of solvent Chap 13.5

Chem 106, Prof. T.L. Heise

Expressing Concentration

51 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C eucalyptus tree.

10 H 18 O, a fragrant substance found in the leaves of the molality = moles of solute kg of solvent 42.0 g C 10 H 18 O 1 mol C 10 H 18 O = 0.273 mol 154 g C 10 H 18 O Chap 13.5

Chem 106, Prof. T.L. Heise

Expressing Concentration

52 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C eucalyptus tree.

10 H 18 O, a fragrant substance found in the leaves of the molality = moles of solute kg of solvent = 0.273 mol 0.600 kg Chap 13.5

= 0.455 m

Chem 106, Prof. T.L. Heise

Expressing Concentration

53 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C eucalyptus tree.

10 H 18 O, a fragrant substance found in the leaves of the

D

T f = K f m

= 4.68(0.455) = 2.13°C Normal T f = -63.5°C -2.13°C Chap 13.5

New T f = -65.6°C

Chem 106, Prof. T.L. Heise

Expressing Concentration

54 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2 2 mol KCl 3 mol C 2 H 6 O 2 Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

55 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2

D

T f = K f m

= 1.87(1) = 1.87°C Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

56 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 2 mol KCl

D

T f = K f m

= 1.87(4) = 7.48°C Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

57 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 3 mol C 2 H 6 O 2

D

T f = K f m

= 1.87(3) = 5.61°C Chap 13.4

Chem 106, Prof. T.L. Heise

Expressing Concentration

58 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2 = -1.87 °C 2 mol KCl 3 mol C 2 H 6 O 2 = -7.48 °C = -5.61 °C Chap 13.4

Colligative Properties

Chem 106, Prof. T.L. Heise 59 Osmotic Pressure The net movement of solvent is always toward the solution with the higher solute concentration.

The pressure needed to stop such movement is called osmotic pressure Chap 13.5

Colligative Properties

Chem 106, Prof. T.L. Heise 60 Osmotic Pressure Chap 13.5

Colloids

Chem 106, Prof. T.L. Heise 61 Chap 13.6