Notes One Unit Seven– Chapter 13 Solutions

•Definitions•Types of Mixtures•Example Solutions•Factors Affecting Solubility•Like Dissolves Like•Solubility of Solids Changes with Temperature•Solubility of Gases Changes with Temperature•Pressure Factor•Molar Concentration•Finding Molarity From Mass and Volume•Finding Mass from Molarity and Volume•Finding Volume from Molarity and Mass

Pages 466-486

Definitions• Solutions are

homogeneous mixtures.

• Uniform throughout.• Solvent.• Determines the

state of solution• Largest component• Solute.• Dissolved in solvent

Common Mixtures

liquid liquid emulsion mayonnaisegas liquid liquid foam whipped cream

liquid gas aerosol hair spraysolid solid solid ruby glass

SOLUTE SOLVENT Type EXAMPLE

solid gas aerosol dust in air

liquid solid emulsion pearlgas solid solid foam Styrofoam

Solution Types

gas gas gas a i rgas liquid liquid soda popliquid liquid liquid antifreeze

solid liquid liquid seawatersolid solid solid brass

SOLUTE SOLVENT PHASE EXAMPLE

liquid solid solid filling

Factors Affecting Solubility• 1. Nature of Solute / Solvent1. Nature of Solute / Solvent.

• 2. Temperature Increase2. Temperature Increase

• i) Solid/Liquid

• ii) gas

• 3. Pressure Factor -3. Pressure Factor -

• i) Solids/Liquids - Very little

• ii) gas

• iii) squeezes gas into solution.

Like Dissolves Like

• Non-polar in Non-polar• Butter in Oil

• Non-polar in polar• Oil in H2O

• Polar in Polar• C2H5OH in H2O

• Ionic compounds in polar solvents• NaCl in H2O

Solubility of solids Changes with Temperature

19gx _____250g100g =48g

solids gases

101g

82g

• How does the solubility Δ with temperature Increase?

• How many grams of potassium chromate will dissolve in100g water at 70oC?

• 70g• How many grams of lead(II)

nitrate will precipitate from 250g water cooling from 70oC to 50oC?

Solubility of Gases Changes with Temperature

• a) Why are fish stressed, if the temperature of the water increases?

• How much does the solubility of oxygen change, for a 20oC to 60oC change?

• 0.90-0.60=0.30mg

0.60mg

0.90mg

Pressure FactorPressure FactorGreater pressure… more dissolved gas

Pressure FactorPressure Factor

Molar Concentration

• M=n/V

• n=MxV

• V=n/M

• Calculate molarity for 25.5 g of NH3 in 600. mL solution.

• 1) Calculate Formula Mass:

• 2) Calculate the moles of solute:

• 3) Calculate the Moles/Liters Ratio

Finding Molarity From Mass and Volume

25.5g ÷ 17.0g/m=1.50m

NH

1x3x

14.0 =1.0 =

14.0 3.017.0g/m

E # Mass

M =1.50m / 0.600LM = 2.52 mol/L

Finding Volume from Molarity and Mass• How many milliliters of 2.50M solution can be

made using 25.5grams of NH3?

• 1)Calculate formula mass:

• 2)Calculate the moles of solute:

• 3)Calculate Volume:

• V=

NH

1x3x

14.0 =1.0 =

14.0 3.017.0g/m

E # Mass

25.5g ÷ 17.0g/m= 1.50m

V=(1.50m) /(2.50M)

V=n/M

0.600L solution600.mL

Finding Mass from Molarity and Volume• How many grams of NH3 are in 600. mL solution at

2.50M?• 1) Calculate formula mass:

• 2) Calculate moles

• n=1.50m• 3) Calculate mass

• g=25.5g NH3

n= 2.50M x 0.600L

g=(17.0g/m)x(1.50m)

NH

1x3x

14.0 =1.0 =

14.0 3.017.0g/m

E # Mass

g÷ fm= molM x Ln =

fm x ng =

Notes Two Unit Seven– Chapter 13 Solutions

• Saturated versus Unsaturated• Colligative properties of water• Forming a Saturated Solution• How Does a Solution Form?• Colligative Properties• Vapor Pressure• Boiling and Freezing Point• BP Elevation and Freezing FP Depression• Calculating Freezing Point Depression Mass

Pages 487-501

Characteristics of Saturated Solutions

Soliddissolve

water

dissolve

precipitate

dissolve

precipitate

Unsaturated Unsaturated Saturated

DynamicEquilibrium

Cooling causes precipitation.Warming causes dissolving.

• As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

Solvation

Colligative Properties• Colligative properties depend on moles dissolved

particles. Vapor pressure lowering Boiling point elevationMelting point depressionOsmotic pressure

How do you get from this…

…to this?

Add an ionic compound!

Vapor Pressure

Vapor Pressure Lowering

• The particles of solute are surrounded by and attracted to particles of solvent.

• Now the solvent particles have less kinetic energy and tend less to escape into the space above the liquid.

• So the vapor pressure is less.

Ionic vs Molecular Solutes

• Ionic solutes produce two or more ion particles in solution.

• They affect the colligative properties proportionately more than molecular solutes (that do not ionize).

• The effect is proportional to the number of particles of the solute in the solution.

How many particles do each of the following give upon solvation?

• NaCl

• CaCl2• Glucose

Freezing Point Depression

Example

• Salt is added to melt ice by reducing the freezing point of water.

Boiling Point Elevation

Example• Addition of ethylene glycol C2H6O2

(antifreeze) to car radiators.

Freezing Point Depression and Boiling Point Elevation

Boiling Point Elevation

• ∆Tb =imkb (for water kb=0.51 oC/m)

• Freezing Point Depression

• ∆Tf=imkf (for water kf=1.86 oC/m)

• Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to. (i is # of particles called van hoff factor)

BP Elevation Constants (Kb)FP Depression Constants( Kf)

Which is more effective for lowering the freezing point of water?

• NaCl or CaCl2

Example 1:

• Find the new freezing point of 3m NaCl in water.

Example 2:

• Find the new boiling point of 3m NaCl in water.

Molarity versus Molality

Molality (m) = ________________moles of solutekilograms solvent

Molarity (M) = ________________moles of soluteliters of solution

Calculating Tf andTb• Calculate the freezing and boiling points of a

solution made using 1000.g antifreeze (C2H6O2) in 4450g water.

• 1) Calculate Moles

• 2) Calculate molality

• 3) Calculate Temperature Change • Δt=Kxm • ΔTf =• Tf = • ΔTb = • Tb =

1000.g ÷62.0g/mol = 16.1 moles

16.1 mole ÷ 4.45 Kg water =3.62m

(1.858oC/m) (3.62 m) = 6.73oC0.000oC- 6.73oC= -6.73oC(0.512oC/m) (3.62 m) =1.96oC

100.000oC + 1.96oC = 101.96oC

C

H

2x

6x

12.0 =

1.0 =

24.0

6.0

E # Mass

O 2x 16.0 = 32.0

62.0g/m

Calculating Boiling Point Elevation Mass• A solution containing 18.00 g of glucose in 150.0 g

of water boils at 100.34oC. Calculate the molecular weight of glucose.

• 1.)Calculate Temperature Change• ΔTb =

• 2.)Calculate moles per Kilograms• ΔTb = Kb x m m = ΔTb /Kb

• m =0.67m/kg

• 3.)Calculate grams / kilograms• g =• g =120 g/kg• MW=120 g/0.67m• 180g/m

100.34oC-100.00oC= 0.34oC

0.34÷ 0.512oC/ m = m

18.00 g ÷ 0.1500kg

One Molal Solution of WaterP

ress

ure

Temperature

solid

Liquid

gas

Kf Kb

1 atm

1.858oC 0.512oC

Notes Three Unit Seven• Ice-cream Lab A Calculating Freezing Point

• Depression Mass

• Colligative Properties of Electrolytes

• Distillation

• Osmotic Pressure

• Dialysis

Pages 487-501

Ice-cream

Calculating Freezing Point Depression Mass

• 1.)Calculate Temperature Change• ΔTf =

• 2.)Calculate moles per Kilograms• ΔTf = Kfx m m = ΔTf /Kf

• m =1.83mol/kg

• 3.)Calculate grams / kilograms• g = • g =36.4 g/kg• MW=36.4 g/1.83m• 19.9g/m

1.1oC- (-2.3oC)= 3.4oC

3.4oC÷1.858oC/ m = m

1.89 g ÷ 0.05196kg

Colligative Properties of Electrolytes• Colligative properties depend on the number of particles

dissolved.

• NaClNa+1+Cl-1 CH3OHAl2(SO4)32Al+3 + 3SO4

-2 C6H12O6

Distillation

Distillation

Osmotic Pressure• Hypertonic

• > 0.92% (9.g/L)

• Crenation

• Isotonic Saline

• = 0.92% (9.g/L)

• Hypotonic

• < 0.92% (9.g/L)

• Rupture

Dialysis

Kidney

Dialysis

Final Quiz Notes

Finding Molarity From Mass and Volume• Calculate molarity for 14.0 g of sodium peroxide(Na2O2)

in 615 mL solution.

• 1) Calculate Formula Mass:

• 2) Calculate the moles of solute:

• 3) Calculate the Moles/Liters Ratio

14.0g ÷ 78.0g/m= 0.179m

Na 2x23.0 =46.0E # Mass

O 2x 16.0 = 32.078.0g/m

M = 0. 179moles ÷0.615L = 0.289MM = n/ v

Finding Volume From Mass and Molarity • Find the volume for a solution having 14.0 g of sodium

peroxide(Na2O2) in a 0.289M solution.

• 1) Calculate Formula Mass:

• 2) Calculate the moles of solute:

• 3) Calculate the Liters.

14.0g ÷ 78.0g/m= 0.179m

Na 2x23.0 =46.0E # Mass

O 2x 16.0 = 32.078.0g/m

v = 0. 179moles ÷ 0.289M= 0.615Lv = n / M

Finding Mass from concentration and Volume• How many grams of sodium peroxide(Na2O2) would be

needed for a 0.289M solution of 615mL volume?

• 1) Calculate Formula Mass:

• 2) Calculate the moles of solute:• M=n/V MxV=n

• 3) Calculate the Grams.

0.289mol/LX0.615L = 0.179m

Na 2x23.0 =46.0E # Mass

O 2x 16.0 = 32.078.0g/m

g =0. 179moles x 78.0g/m= 14.0gg = n x fm

Calculating Freezing Point Depression Mass• A solution containing 7.67 g of ethanol in 333.0 g

of water freezes at -0.929oC. Calculate the molecular weight of ethanol.

• 1.)Calculate Temperature Change• ΔTf =

• 2.)Calculate moles per Kilograms• ΔTf = Kf x m m = ΔTf /Kf

• m =0.500m/kg

• 3.)Calculate grams / kilograms• g/kg =• g/Kg =23.0g/kg• fm=• 46.0g/m

0.000oC- (-0.929oC)= 0.929oC

0.929÷ 1.858oC/ m = m

7.67 g ÷ 0.3330kg

23.0 g/0.500m

Solubility of solids Changes with Temperature

• How many grams Cs2SO4 will precipitate from 267g water as it cools from 60oC to 25oC?

18gx _____267g100g =48g

200g

182g

Phase DiagramP

ress

ure

Temperature

solid

Liquid

gas

meltingfreezing

NFP NBP

1 atm

Triple point

Critical point

0.0oC 100.0oC

vaporizingcondensing

sublimationdepostion

end

0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 9.0 10.00

10

20

30

40

50

60

70

8.0

80

Mg

of

gas

per

100

gra

ms

of

wat

er

Gas pressure in atmospheres

Calculating Tf andTb• Calculate the freezing and boiling points of a

solution made using 36.0g glucose(C6H12O6) in 225.0g water.

• 1) Calculate Moles

• 2) Calculate molality of H2O

• 3) Calculate Temperature Change • Δt= • ΔTf =• Tf = • ΔTb = • Tb =

36.0g ÷180.0g/mol =0.200 moles

0.200 mol ÷0.2250 Kg=0.889m

(1.858oC/m) (0.889m) = 1.65oC0.000oC- 1.65oC= -1.65oC(0.512oC/m) (0.889 m) = 0.455oC

100.000oC + 0.455oC =100. 455oC

C

H

6x

12x

12.0 =

1.0 =

72.0

12.0

E # Mass

O 6x 16.0 = 96.0

180.0g/m

K xm

m = n/ Kg

Finding Molarity From Mass and Volume• Calculate molarity for 24.0 g of antifreeze(C2H6O2) in

445. mL solution.

• 1) Calculate Formula Mass:

• 2) Calculate the moles of solute:

• 3) Calculate the Moles/Liters Ratio

24.0g ÷ 62.0g/m= 0.387m

C

H

2x

6x

12.0 =

1.0 =

24.0

6.0

E # Mass

O 2x 16.0 = 32.0

62.0g/m

M = 0. 387moles ÷0.445L = 0.870MM = n/ v

Seven/Eight Rows

Calculating Freezing Point Depression Mass• A solution containing 1.89 g of methanol in 51.96 g

of water freezes at -3.4oC. Calculate the molecular weight of methanol .

• 1.)Calculate Temperature Change• ΔTb =

• 2.)Calculate moles per Kilograms• ΔTf = Kf x m m = ΔTf /Kf

• m =0.500m/kg

• 3.)Calculate grams / kilograms• g =• g =23.0g/kg• fm=• 46.0g/m

0.000oC- 0.929oC= 0.929oC

0.929÷ 1.858oC/ m = m

7.67 g ÷ 0.3330kg

23.0 g/0.500m

Calculating Freezing Point Depression Mass

• A solution containing 1.89 g of ethanol in 51.96 g of water freezes at -3.4oC. Calculate the molecular weight of ethanol .

• 1.)Calculate Temperature Change• ΔTb =

• 2.)Calculate moles per Kilograms• ΔTf = Kf x m m = ΔTf /Kf

• m =0.500m/kg

• 3.)Calculate grams / kilograms• g =• g =23.0g/kg• fm=• 46.0g/m

0.000oC- 0.929oC= 0.929oC

0.929÷ 1.858oC/ m = m

7.67 g ÷ 0.3330kg

23.0 g/0.500m

Calculating Freezing Point Depression Mass• A solution containing 1.89 g of methanol in 51.96 g

of water freezes at -3.4oC. Calculate the molecular weight of methanol .

• 1.)Calculate Temperature Change• ΔTb =

• 2.)Calculate moles per Kilograms• ΔTf = Kf x m m = ΔTf /Kf

• m =0.500m/kg

• 3.)Calculate grams / kilograms• g =• g =23.0g/kg• fm=• 46.0g/m

0.000oC- 0.929oC= 0.929oC

0.929÷ 1.858oC/ m = m

7.67 g ÷ 0.3330kg

23.0 g/0.500m