1 Solutions Properties of Water Solutions 2 Predict the % water in the following foods

1

Solutions

Properties of Water

Solutions

2

Predict the % water in the following foods

3

Predict the % water in the following foods

88%

water

94% water

85%

water

86% water

4

Water in the Body

water gain water loss

liquids1000 mL urine 1500 mL

food 1200 mL perspiring 300 mL

cells 300 mL exhaling 600 mL

feces 100 mL

Calculate the total water gain and water loss

Total ______ mL _____ mL

5

Water

Most common solventA polar molecule

O -

a hydrogen bond

H +

H +

6

Hydrogen Bonds Attract Polar Water Molecules

7

Explore: Surface Tension HW

Fill a glass to the brim with water

How many pennies can you add to the glass without causing any water to run over?

Predict _________________

Actual _________________Explain your results

8

Explore

1. Place some water on a waxy surface. Why do drops form?

2. Carefully place a needle on the surface of water. Why does it float? What happens if you push it through the water surface?

3. Sprinkle pepper on water. What does it do? Add a drop of soap. What happens?

9

Surface Tension

Water molecules within water hydrogen bond in all directions

Water molecules at surface cannot hydrogen bond above the surface, pulled inward

Water surface behaves like a thin, elastic membrane or “skin”

Surfactants (detergents) undo hydrogen bonding

10

Solute and Solvent

Solutions are homogeneous mixtures of two or more substances

Solute

The substance in the lesser amountSolvent

The substance in the greater amount

11

Nature of Solutes in Solutions

Spread evenly throughout the

solution

Cannot be separated by filtration

Can be separated by evaporation

Not visible, solution appears

transparent

May give a color to the solution

12

Types of Solutions

air O2 gas and N2 gas gas/gas

soda CO2 gas in water gas/liquid

seawater NaCl in water solid/liquid

brass copper and zinc solid/solid

13

Discussion

Give examples of some solutions and explain why they are solutions.

14

Learning Check SF1

(1) element (2) compound (3) solution

A. water 1 2 3

B. sugar 1 2 3

C. salt water 1 2 3

D. air 1 2 3

E. tea 1 2 3

15

Solution SF1

(1) element (2) compound (3) solution

A. water 2

B. sugar 2

C. salt water 3

D. air 3

E. tea 3

16

Learning Check SF2

Identify the solute and the solvent.

A. brass: 20 g zinc + 50 g copper

solute = 1) zinc 2) copper

solvent = 1) zinc 2) copper

B. 100 g H2O + 5 g KCl

solute = 1) KCl 2) H2O

solvent = 1) KCl 2) H2O

17

Solution SF2

A. brass: 20 g zinc + 50 g copper

solute = 1) zinc solvent = 2) copper

B. 100 g H2O + 5 g KCl

solute = 1) KCl

solvent = 2) H2O

18

Learning Check SF3

Identify the solute in each of the following solutions:

A. 2 g sugar (1) + 100 mL water (2)

B. 60.0 mL ethyl alcohol(1) and 30.0 mL

of methyl alcohol (2)

C. 55.0 mL water (1) and 1.50 g NaCl (2)

D. Air: 200 mL O2 (1) + 800 mL N2 (2)

19

Solution SF3

Identify the solute in each of the following solutions:

A. 2 g sugar (1)

B. 30.0 mL of methyl alcohol (2)

C. 1.5 g NaCl (2)

D. 200 mL O2 (1)

20

Like dissolves like

A ____________ solvent such as water is

needed to dissolve polar solutes such as

sugar and ionic solutes such as NaCl.

A ___________solvent such as hexane

(C6H14) is needed to dissolve nonpolar

solutes such as oil or grease.

21

Learning Check SF4

Which of the following solutes will dissolve in water? Why?

1) Na2SO4

2) gasoline

3) I2

4) HCl

22

Solution SF4

Which of the following solutes will dissolve in water? Why?

1) Na2SO4 Yes, polar (ionic)

2) gasoline No, nonnpolar

3) I2 No, nonpolar

4) HCl Yes, Polar

23

Formation of a Solution

Cl-

Na+ Cl-Na+

H2O

H2O

Na+

Cl-

solute

Dissolvedsolute

Hydration

24

Electrolyte and Non-electrolyte

• Electrolyte: a substance that conducts electricity when dissolved in water. – Acids, bases and soluble ionic

solutions are electrolytes.

• Non-electrolyte: a substance that does not conduct electricity when dissolved in water. – Molecular compounds and insoluble

ionic compounds are non-electrolytes.

25

Electrolytes

• Some solutes can dissociate into ions.

• Electric charge can be carried.

26

Types of solutes

Na+

Cl-

Strong Electrolyte -100% dissociation,all ions in solution

high conductivity

27

Types of solutes

CH3COOH

CH3COO-

H+

Weak Electrolyte -partial dissociation,molecules and ions in solution

slight conductivity

28

Types of solutes

SugarC6H12O6

Non-electrolyte -No dissociation,all molecules in solution

no conductivity

29

Types of Electrolytes

• Weak electrolyte partially dissociates.– Fair conductor of electricity.

• Non-electrolyte does not dissociate. – Poor conductor of electricity.

• Strong electrolyte dissociates completely.– Good electrical conduction.

30

Representation of Electrolytes using Chemical Equations

MgCl2(s) → Mg2+(aq) + 2 Cl- (aq)

A strong electrolyte:

A weak electrolyte:

CH3COOH(aq) ← CH3COO -(aq) +H+(aq)→

CH3OH(aq)

A non-electrolyte:

31

Electrolytes in Action

32

Strong Electrolytes

Strong acids: HNO3, H2SO4, HCl, HClO4

Strong bases: MOH (M = Na, K, Cs, Rb etc)

Salts: All salts dissolving in water are completely ionized.

Stoichiometry & concentration relationship

NaCl (s) Na+ (aq) + Cl– (aq)

Ca(OH)2 (s) Ca2+(aq) + 2 OH– (aq)

AlCl3 (s) Al3+ (aq) + 3 Cl– (aq)

(NH4)2SO4 (s) 2 NH4 + (aq) + SO42– (aq)

33

Writing An Equation for a Solution

When NaCl(s) dissolves in water, the reaction can be written as

H2O

NaCl(s) Na+ (aq) + Cl- (aq)

solid separation of ions in water

34

Learning Check SF5

Solid LiCl is added to some water. It dissolves because

A. The Li+ ions are attracted to the

1) oxygen atom(-) of water

2) hydrogen atom(+) of water

B. The Cl- ions are attracted to the

1) oxygen atom(-) of water

2) hydrogen atom(+) of water

35

Solution SF5

Solid LiCl is added to some water. It dissolves because

A. The Li+ ions are attracted to the

1) oxygen atom(-) of water

B. The Cl- ions are attracted to the

2) hydrogen atom(+) of water

36

Rate of Solution

You are making a chicken broth using a bouillon cube. What are some things you can do to make it dissolve faster?

Crush it

Use hot water (increase temperature)

Stir it

37

How do I get sugar to dissolve faster in my iced tea?

Stir, and stir, and stir

Add sugar to warm tea then add ice

Grind the sugar to a powder

Fresh solvent contact and interaction with solute

Greater surface area, more solute-solvent interaction

Faster rate of dissolution at higher temperature

38

Learning Check SF6

You need to dissolve some gelatin in water. Indicate the effect of each of the following on the rate at which the gelatin dissolves as (1) increase, (2) decrease,

(3) no change

A. ___Heating the water

B. ___Using large pieces of gelatin

C. ___Stirring the solution

39

Learning Check SF6

You need to dissolve some gelatin in water. Indicate the effect of each of the following on the rate at which the gelatin dissolves as (1) increase, (2) decrease,

(3) no change

A. 1 Heating the water

B. 2 Using large pieces of gelatin

C. 2 Stirring the solution

40

SolubilityPercent Concentration

Colloids and Suspensions

41

Solubility

The maximum amount of solute that can dissolve in a specific amount of solvent usually 100 g.

g of solute

100 g water

42

Saturated and Unsaturated

A saturated solution contains the maximum

amount of solute that can dissolve.

Undissolved solute remains.

An unsaturated solution does not contain all

the solute that could dissolve

SolubilitySATURATED SOLUTION

no more solute dissolves

UNSATURATED SOLUTIONmore solute

dissolves

SUPERSATURATED SOLUTION

becomes unstable, crystals form

increasing concentration increasing concentration

Polarity

Factors Affecting Solid Solubility

Temperature

Surface Area

Stirring

Polarity

Factors Affecting Solubility

Temperature

Pressure

Intramolecular Bonding• Intramolecular bonding refers to the chemical

bonding that holds atoms together within a molecule of a compound

Covalent bonding and ionic bonding are the two main types of intramolecular bonding

Covalent bonding involves the sharing of valence electrons involves the sharing of valence electrons between two atoms. POLAR- unequal sharing of electrons

NON POLAR – equal sharing of electrons

Ionic bonding involves the transference of valence electrons

47

SOLUTE POLAR SOLVENT

NONPOLAR SOLVENT

Ionic Soluble Insoluble

Polar Soluble Insoluble

Nonpolar Insoluble soluble

48

Learning Check S1

At 40C, the solubility of KBr is 80 g/100 g

H2O. Indicate if the following solutions are

(1) saturated or (2) unsaturated

A. ___60 g KBr in 100 g of water at 40C

B. ___200 g KBr in 200 g of water at 40C

C. ___25 KBr in 50 g of water at 40C

49

Solution S1

At 40C, the solubility of KBr is 80 g/100 g H2O.

Indicate if the following solutions are

(1) saturated or (2) unsaturated

A. 2 Less than 80 g/100 g H2O

B. 1 Same as 100 g KBr in 100 g of water

at 40C, which is greater than its solubility

C. 2 Same as 60 g KBr in 100 g of water,

which is less than its solubility

50

Temperature and Solubility of Solids

Temperature Solubility (g/100 g H2O)

KCl(s) NaNO3(s)

0° 27.6 74

20°C 34.0 88

50°C 42.6 114

100°C 57.6 182

The solubility of most solids (decreases or increases ) with an increase in the temperature.

51

Temperature and Solubility of Solids

Temperature Solubility (g/100 g H2O)

KCl(s) NaNO3(s)

0° 27.6 74

20°C 34.0 88

50°C 42.6 114

100°C 57.6 182

The solubility of most solids increases with an increase in the temperature.

52

Temperature and Solubility of Gases

Temperature Solubility (g/100 g H2O)

CO2(g) O2(g)

0°C 0.34 0.0070

20°C 0.17 0.0043

50°C 0.076 0.0026

The solubility of gases (decreases or increases) with an increase in temperature.

53

Temperature and Solubility of Gases

Temperature Solubility (g/100 g H2O)

CO2(g) O2(g)

0°C 0.34 0.0070

20°C 0.17 0.0043

50°C 0.076 0.0026

The solubility of gases decreases with an increase in temperature.

54

Learning Check S2

A. Why would a bottle of carbonated drink

possibly burst (explode) when it is left out

in the hot sun ?

B. Why would fish die in water that gets too warm?

55

Solution S2

A. Gas in the bottle builds up as the gas becomes less soluble in water at high temperatures, which may cause the bottle to explode.

B. Because O2 gas is less soluble in warm

water, the fish may not obtain the needed

amount of O2 for their survival.

Gas Solubility

56

CH4

O2

CO

He

Sol

ubili

ty (

mM

)

2.0

1.0

0 10 20 30 40 50

Higher Temperature

…Gas is LESS Soluble

57

Solubility Curves

Show the conditions that affect states of the solution: unsaturated, saturated, supersaturated.

Solubility Table

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

0 10 20 30 40 50 60 70 80 90 100

Solubility vs. Temperature for Solids

Sol

ubili

ty (

gram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

NaCl KClO3

SO2

shows the dependence

of solubility on temperature

gases

solids

How to determine the solubility of a given substance?

• Find out the mass of solute needed to make a saturated solution in 100 cm3 of water for a specific temperature(referred to as the solubility).

• This is repeated for each of the temperatures from 0ºC to 100ºC. The data is then plotted on a temperature/solubility graph, and the points are connected. These connected points are called a solubility curve.

How to use a solubility graph?A. IDENTIFYING A SUBSTANCE

( given the solubility in g/100 cm3 of water and the temperature)

• Look for the intersection of the solubility and temperature.

Learning Check SG1:What substance has a solubility of 90 g/100 cm3 in water at a temperature of 25ºC ?

Learning Check SG2:What substance has a solubility of 200 g/100 cm3 of water at a temperature of 90ºC ?

Look for the temperature or solubility

•Locate the solubility curve needed and see for a given temperature, which solubility it lines up with and visa versa.

Learning Check SG3:

What is the solubility of potassium nitrate at 80ºC ?

• What is the solubility of potassium nitrate at 80ºC ?

Learning Check SG4:

At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?

Learning Check SG4:

At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?

Learning Check SG5:

At what temperature will potassium iodide have a solubility of 230 g/100 cm3 ?

Learning Check SG5:

At what temperature will potassium iodide have a solubility of 130 g/100 cm3 ?

Using Solubility Curves:

What is the solubility of sodium chloride at 25ºC in 150 cm3 of water ?

From the solubility graph we see that sodium chlorides solubility is 36 g.

Solubility in grams = unknown solubility in grams

100 cm3 of water other volume of water

___36 grams____ = unknown solubility in grams

100 cm3 of water 150 cm3 water

Place this in the proportion below and solve for the unknown solubility. Solve for the unknown quantity by cross multiplying.

The unknown solubility is 54 grams. You can use this proportion to solve for the other volume of water

if you're given the other solubility.

C. Determine if a solution is saturated,unsaturated,or supersaturated.

• If the solubility for a given substance places it anywhere on it's solubility curve line it is saturated.• If it lies above the solubility curve line, then it's supersaturated,• If it lies below the solubility curve line it's an unsaturated solution. Remember though, if the volume of water isn't 100 cm3 to use a proportion first as shown above.

Temp. (oC)

Solubility(g/100 g

H2O)

KNO3 (s)

KCl (s)

HCl (g)

SOLUBILITYCURVE

Solubility how much solute dissolves in a given amt.of solvent at a given temp.

unsaturated: solution could hold more solute; belowbelow linesaturated: solution has “just right” amt. of solute; onon linesupersaturated:solution has “too much” solute dissolved in

it;aboveabove the line

To

Sol.

To

Sol.

Solids dissolved in liquids Gases dissolved in liquids

As To , solubility As To , solubility

Sometimes you'll need to determine how much additional solute needs to be added to a unsaturated solution in order to make it saturated.

For example,30 grams of potassium nitrate has been added to 100 cm3 of water at a temperature of 50ºC. How many additional grams of solute must be added in order to make it saturated?

How many additional grams of solute must be added in order to make it saturated?

From the graph you can see that the solubility for potassium nitrate at 50ºC is 84 grams

If there are already 30 grams of solute in the solution, all you need to get to 84 grams is 54 more grams ( 84g-30g )

Solubility Table

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

shows the dependence

of solubility on temperature

0 10 20 30 40 50 60 70 80 90 100

Solubility vs. Temperature for Solids

Sol

ubili

ty (

gram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

NaCl KClO3

SO2

gases

solids

Classify as unsaturated, saturated, or supersaturated.Classify as unsaturated, saturated, or supersaturated.

per100

gH2O

80 g NaNO3 @ 30oC

45 g KCl @ 60oC

50 g NH3 @ 10oC

70 g NH4Cl @ 70oC

=unsaturated

=saturated

=unsaturated

=supersaturated

0 10 20 30 40 50 60 70 80 90 100

Solubility vs. Temperature for Solids

Sol

ubili

ty (

gram

s of

sol

ute/

100

g H

2O)

KI

KCl

20

10

30

40

50

60

70

80

90

110

120

130

140

100

NaNO3

KNO3

HCl NH4Cl

NH3

NaCl KClO3

SO2

gases

solids

(A) Per 100 g H2O, 100 g Unsaturated; all

soluteNaNO3 @ 50oC. dissolves; clear

solution.

(B) Cool solution (A) very Supersaturated; extraslowly to 10oC. solute remains in solution;

still clear.

Describe each situation below.

(C) Quench solution (A) in Saturated; extra solute an ice bath to 10oC. (20 g) can’t remain in

solution, becomes visible.

84

Soluble and Insoluble Salts

A soluble salt is an ionic compound that dissolves in water.

An insoluble salt is an ionic compound that does not dissolve in water

85

How do we know ions are How do we know ions are present in aqueous present in aqueous solutions?solutions?

The solutions:_________The solutions:_________They are called They are called

ELECTROLYTESELECTROLYTESHCl, MgClHCl, MgCl22, and NaCl are , and NaCl are

strong electrolytesstrong electrolytes. . They dissociate They dissociate completely (or nearly so) completely (or nearly so) into ions.into ions.

Aqueous SolutionsAqueous Solutions

86

Solubility Rules

1. A salt is soluble in water if it contains any one of the following ions:

NH4+ Li+ Na+ K+ or NO3

-

Examples:

soluble salts

LiCl Na2SO4 KBr Ca(NO3)2

87

Cl- Salts

2. Salts with Cl- are soluble, but not if the

positive ion is Ag+, Pb2+, or Hg22+.

Examples:

soluble not soluble(will not dissolve)

MgCl2 AgCl

PbCl2

88

SO42- Salts

3. Salts with SO42- are soluble, but not if

the positive ion is Ba2+, Pb2+, Hg2+ or Ca2+.

Examples:

soluble not soluble

MgSO4 BaSO4

PbSO4

89

Other Salts

4. Most salts containing CO32-, PO4

3-, S2-

and OH- are not soluble.

Examples:

soluble not soluble

Na2CO3 CaCO3

K2S CuS

90

Learning Check S3

Indicate if each salt is (1)soluble or (2)not soluble:

A. ______ Na2SO4

B. ______ MgCO3

C. ______ PbCl2

D. ______ MgCl2

91

Solution S3

Indicate if each salt is (1) soluble or

(2) not soluble:

A. _1_ Na2SO4

B. _2_ MgCO3

C. _2_ PbCl2

D. _1_ MgCl2

92

Solutions

Molarity

93

Molarity (M)

A concentration that expresses the

moles of solute in 1 L of solution

Molarity (M) = moles of solute

1 liter solution

94

Units of Molarity

2.0 M HCl = 2.0 moles HCl

1 L HCl solution

6.0 M HCl = 6.0 moles HCl

1 L HCl solution

95

Molarity Calculation

NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to

remove potato peels commercially.

If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

96

Calculating Molarity

1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH

40.0 g NaOH

2) 500. mL x 1 L _ = 0.500 L

1000 mL

3. 0.10 mole NaOH = 0.20 mole NaOH

0.500 L 1 L

= 0.20 M NaOH

97

Learning Check M1

A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution?

1) 8 M

2) 5 M

3) 2 M Dra

no

98

Solution M1

A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution?

2) 5 M

M = 2 mole KOH = 5 M

0.4 L Dra

no

99

Learning Check M2

A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose

has a molar mass of 180. g/mole, what is the molarity of the glucose solution?

1) 0.20 M

2) 5.0 M

3) 36 M

100

Solution M2

A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose

has a molar mass of 180. g/mole, what is the molarity of the glucose solution?

1) 72 g x 1 mole x 1 = 0.20 M

180. g 2.0 L

101

Molarity Conversion Factors

A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors.

3.0 moles NaOH and 1 L NaOH soln

1 L NaOH soln 3.0 moles NaOH

102

Learning Check M3

Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution?

1) 15 moles HCl

2) 1.5 moles HCl

3) 0.15 moles HCl

103

Solution M3

3) 1500 mL x 1 L = 1.5 L

1000 mL

1.5 L x 0.10 mole HCl = 0.15 mole HCl

1 L

(Molarity factor)

104

Learning Check M4

How many grams of KCl are present in 2.5 L

of 0.50 M KCl?

1) 1.3 g

2) 5.0 g

3) 93 g

105

Solution M4

3)

2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl

1 L 1 mole KCl

106

Learning Check M5

How many milliliters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl?

1) 150 mL

2) 1500 mL

3) 5000 mL

107

Solution M5

2) 0.15 mole HCl x 1 L soln x 1000 mL

0.10 mole HCl 1 L

(Molarity inverted)

= 1500 mL HCl

108

Learning Check M6

How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

1) 12 g

2) 48 g

3) 300 g

109

Solution M6

2) 400. mL x 1 L = 0.400 L 1000 mL

0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1 L 1 mole NaOH

(molar mass)

= 48 g NaOH

110

Solution

Percent Concentration

111

Percent Concentration

Describes the amount of solute dissolved in 100 parts of solution

amount of solute

100 parts solution

112

Mass-Mass % Concentration

mass/mass % = g solute x 100% 100 g solution

113

Mixing Solute and Solvent

Solute + Solvent

4.0 g KCl 46.0 g H2O

50.0 g KCl solution

114

Calculating Mass-Mass %

g of KCl = 4.0 g

g of solvent = 46.0 g

g of solution = 50.0 g

%(m/m) = 4.0 g KCl (solute) x 100 = 8.0% KCl

50.0 g KCl solution

115

Learning Check PC1

A solution contains 15 g Na2CO3 and 235 g of

H2O? What is the mass % of the solution?

1) 15% (m/m) Na2CO3

2) 6.4% (m/m) Na2CO3

3) 6.0% (m/m) Na2CO3

116

Solution PC1

mass solute = 15 g Na2CO3

mass solution = 15 g + 235 g = 250 g

%(m/m) = 15 g Na2CO3 x 100

250 g solution

= 6.0% Na2CO3 solution

117

Mass-Volume %

mass/volume % = g solute x 100% 100 mL solution

118

Learning Check PC2

An IV solution is prepared by

dissolving 25 g glucose

(C6H12O6) in water to make 500.

mL solution. What is the

percent (m/v) of the glucose in

the IV solution?

1) 5.0% 2) 20.% 3) 50.%

119

Solution PC2

1) 5.0%

%(m/v) = 25 g glucose x 100

500. mL solution

= 5.0 %(m/v) glucose solution

120

Writing Factors from %

A physiological saline solution is a 0.85% (m/v) NaCl solution.

Two conversion factors can be written for the % value.

0.85 g NaCl and 100 mL NaCl soln

100 mL NaCl soln 0.85 g NaCl

121

% (m/m) Factors

Write the conversion factors for a 10 %(m/m) NaOH solution

NaOH and NaOH soln

NaOH soln NaOH

122

% (m/m) Factors

Write the conversion factors for a 10 %(m/m) NaOH solution

NaOH and NaOH soln

NaOH soln NaOH

10 g

10 g100 g

100 g

123

Learning Check PC 3

Write two conversion factors for each of the following solutions:

A. 8 %(m/v) NaOH

B. 12 %(v/v) ethyl alcohol

124

Solution PC 3

Write conversion factors for the following:

A. 8 %(m/v) NaOH

8 g NaOH and 100 mL 100 mL 8 g NaOH

B. 12 %(v/v) ethyl alcohol

12 mL alcohol and 100 mL 100 mL 12 mL alcohol

125

Using % Factors

How many grams of NaCl are needed to prepare 250 g of a 10.0% (m/m) NaCl solution?

Complete data:____________ g solution

____________% or (______/_100 g_) solution

____________ g solute

126

Clculation Using % Factors

250 g solution

10.0% or (10.0 g/100 g) solution

? g solute

250 g NaCl soln x 10.0 g NaCl = 25 g NaCl100 g NaCl soln

127

Learning Check PC4

How many grams of NaOH do you need to measure out to prepare 2.0 L of a 12%(m/v) NaOH solution?

1) 24 g NaOH

2) 240 g NaOH

3) 2400 g NaOH

128

Solution PC4

2.0 L soln x 1000 mL = 2000 mL soln

1 L

12 % (m/v) NaOH = 12 g NaOH

100 mL NaOH soln

2000 mL x 12 g NaOH = 240 g NaOH

100 mL NaOH soln

129

Learning Check PC5

How many milliliters of 5 % (m/v) glucose solution are given if a patient receives 150 g of glucose?

1) 30 mL

2) 3000 mL

3) 7500 mL

130

Solution PC5

5% m/v factor

150 g glucose x 100 mL = 3000 mL

5 g glucose

131

Preparing a Solution by Dilution

132

Units of Concentrations

amount of solute per amount of solvent or solution

Percent (by mass) =g solute

g solutionx 100

g solute

g solute + g solvent

x 100=

Molarity (M) =

moles of solute

volume in liters of solution

moles = M x VL

133

Solutions

Colloids and Suspensions

Osmosis and Dialysis

134

Solutions

Have small particles (ions or molecules)

Are transparent

Do not separate

Cannot be filtered

Do not scatter light.

135

Colloids

Have medium size particles

Cannot be filtered

Separated with semipermeable membranes

Scatter light (Tyndall effect)

136

Examples of Colloids

Fog

Whipped cream

Milk

Cheese

Blood plasma

Pearls

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Suspensions

Have very large particles

Settle out

Can be filtered

Must stir to stay suspended

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Examples of Suspensions

Blood platelets

Muddy water

Calamine lotion

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Osmosis

In osmosis, the solvent water moves

through a semipermeable membrane

Water flows from the side with the lower solute concentration into the side with the higher solute concentration

Eventually, the concentrations of the two

solutions become equal.

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Osmosis

semipermeable membrane

4% starch 10% starch

H2O

141

Equilibrium is reached.

water flow becomes equal

7% starch

7% starch

H2OO

142

Osmotic Pressure

Produced by the number of solute particles

dissolved in a solution

Equal to the pressure that would prevent

the flow of additional water into the more

concentrated solution

Increases as the number of dissolved

particles increase

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Osmotic Pressure of the Blood

Cell walls are semipermeable membranes

The osmotic pressure of blood cells

cannot change or damage occurs.

The flow of water between a red blood

cell and its surrounding environment

must be equal

144

Isotonic solutions

• Exert the same osmotic pressure as red blood cells.

• Medically 5% glucose and 0.9% NaCl are used their solute concentrations provide an osmotic pressure equal to that of red blood cells

H2O

145

Hypotonic Solutions

Lower osmotic pressure than red blood cells

Lower concentration of particles than RBCs

In a hypotonic solution, water flows into the

RBC

The RBC undergoes hemolysis; it swells and

may burst.

H2O

146

Hypertonic Solutions

Has higher osmotic pressure than RBCHas a higher particle concentration In hypertonic solutions, water flows out

of the RBCThe RBC shrinks in size (crenation)

H2O

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Dialysis

Occurs when solvent and small solute

particles pass through a semipermeable

membrane

Large particles retained inside

Hemodialysis is used medically (artificial

kidney) to remove waste particles such as

urea from blood

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Colligative Colligative PropertiesPropertiesOn adding a solute to a solvent, the On adding a solute to a solvent, the

properties of the solvent are modified.properties of the solvent are modified.• Vapor pressure Vapor pressure decreasesdecreases• Melting point Melting point decreasesdecreases• Boiling point Boiling point increasesincreases• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)

These changes are called These changes are called COLLIGATIVE COLLIGATIVE PROPERTIESPROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of solute of solute particles relative to solvent particles, not on particles relative to solvent particles, not on the the KINDKIND of solute particles. of solute particles.

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Change in Freezing Change in Freezing Point Point

The freezing point of a solution is The freezing point of a solution is

LOWERLOWER than that of the pure than that of the pure solventsolvent

Pure waterPure waterEthylene glycol/water Ethylene glycol/water

solutionsolution

150

Change in Freezing Point Change in Freezing Point Common Applications of Common Applications of

Freezing Point Freezing Point DepressionDepression

Propylene glycol

Ethylene glycol – deadly to small animals

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Common Applications of Common Applications of Freezing Point DepressionFreezing Point DepressionWhich would you use for the streets of

Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?

a) sand, SiO2

b) Rock salt, NaCl

c) Ice Melt, CaCl2

Change in Freezing Change in Freezing Point Point

152

Change in Boiling Change in Boiling Point Point

Common Applications of Common Applications of Boiling Point ElevationBoiling Point Elevation