Slide  1 

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Physical Properties of Solutions ___________________________________ 

Chapter 13

___________________________________  1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Slide  2 

A solution is a homogenous mixture of 2 or more substances

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The solute is(are) the substance(s) present in the smaller amount(s)

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The solvent is the substance present in the larger amount

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A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.

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An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.

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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.

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3

 

Slide  4 

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Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction

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Molecular view of the formation of solution

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∆Hsoln = ∆H1 + ∆H2 + ∆H3

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Slide 

“like dissolves like”



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Two substances with similar intermolecular forces are likely to be soluble in each other.

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non-polar molecules are soluble in non-polar solvents CCl4 in C6H6



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polar molecules are soluble in polar solvents C2H5OH in H2O



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ionic compounds are more soluble in polar solvents

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NaCl in H2O or NH3 (l) 5

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Concentration Units



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The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

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Percent by Mass % by mass = =

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mass of solute x 100% mass of solute + mass of solvent

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mass of solute x 100% mass of solution

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Mole Fraction (X) moles of A XA = sum of moles of all components

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Slide 

Concentration Units Continued



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Molarity (M)

M =

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moles of solute liters of solution

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Molality (m) m =

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moles of solute mass of solvent (kg)

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Slide  8 

What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

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moles of solute

moles of solute m =

M =

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liters of solution

mass of solvent (kg)

Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL)

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mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg moles of solute

=

m =

5.86 moles C2H5OH

___________________________________  = 8.92 m

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0.657 kg solvent

mass of solvent (kg)

8

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Slide  9 

Solution Stoichiometry ___________________________________ 

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity =

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moles of solute ___________________________________ 

liters of solution

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What mass of KI is required to make 500. mL of a 2.80 M KI solution? M KI

volume of KI solution

moles KI

M KI

___________________________________  grams KI

___________________________________  500. mL x

1L 1000 mL

x

2.80 mol KI 1 L soln

x

166 g KI 1 mol KI

= 232 g KI 9

 

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Slide  Preparing a Solution of Known Concentration

10 

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Slide  11 

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

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Dilution

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Add Solvent

___________________________________  Moles of solute before dilution (i)

=

Moles of solute after dilution (f)

MiVi

=

MfVf

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Slide  12 

How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3?

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MiVi = MfVf Mi = 4.00 M Mf = 0.200 M Vf = 0.0600 L Vi =

MfVf Mi

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Vi = ? L

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= 0.200 M x 0.0600 L = 0.00300 L = 3.00 mL 4.00 M

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Dilute 3.00 mL of acid with water to a total volume of 60.0 mL. 12

 

Slide  13 

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Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

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Equivalence point – the point at which the reaction is complete

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Indicator – substance that changes color at (or near) the equivalence point

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Slowly add base to unknown acid UNTIL

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the indicator changes color 13

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Titrations can be used in the analysis of

14 

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Acid-base reactions ___________________________________ 

H2SO4 + 2NaOH

2H2O + Na2SO4

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Redox reactions

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5Fe2+

-

+ MnO4 +

8H+

Mn2+

+

5Fe3+

+ 4H2O 14

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Slide  15 

What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution?

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WRITE THE CHEMICAL EQUATION! H2SO4 + 2NaOH

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2H2O + Na2SO4

___________________________________  M

volume acid

25.00 mL x

acid

rxn

moles red

4.50 mol H2SO4 1000 mL soln

x

coef.

M

moles base

2 mol NaOH 1 mol H2SO4

x

base

volume base

1000 ml soln 1.420 mol NaOH

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= 158 mL 15

 

Slide  16 

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16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution?

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WRITE THE CHEMICAL EQUATION! 5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

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M

volume red

red

rxn

moles red

coef.

V

moles oxid

oxid

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M oxid

___________________________________  16.42 mL = 0.01642 L

0.01642 L x

0.1327 mol KMnO4 1L

25.00 mL = 0.02500 L

x

5 mol Fe2+ 1 mol KMnO4

x

1 0.02500 L Fe2+

___________________________________  = 0.4358 M 16

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Temperature and Solubility

17 

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Solid solubility and temperature

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solubility increases with increasing temperature

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solubility decreases with increasing temperature

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17

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Temperature and Solubility

18  ___________________________________ 

O2 gas solubility and temperature

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solubility usually decreases with increasing temperature

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Slide 

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Pressure and Solubility of Gases

19 

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The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

___________________________________  c is the concentration (M) of the dissolved gas

c = kP

P is the pressure of the gas over the solution

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k is a constant for each gas (mol/L•atm) that depends only on temperature

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low P

high P

low c

high c

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Colligative Properties

20 

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Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

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Vapor-Pressure Lowering P1 = X1 P 10

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P 10 = vapor pressure of pure solvent

Raoult’s law

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X1 = mole fraction of the solvent

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If the solution contains only one solute: ___________________________________ 

X1 = 1 – X2 P 10 - P1 = ∆P = X2 P 10

X2 = mole fraction of the solute 20

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Slide  21 

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Ideal Solution

PA = XA P A0

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PB = XB P 0B PT = PA + PB

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PT = XA P A0 + XB P 0B

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Slide  22 

Boiling-Point Elevation ___________________________________ 

∆Tb = Tb – T b0 T b0 is the boiling point of the pure solvent T b is the boiling point of the solution

Tb > T b0

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∆Tb > 0

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∆Tb = Kb m

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m is the molality of the solution

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Kb is the molal boiling-point elevation constant (0C/m) for a given solvent 22

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Slide  23 

Freezing-Point Depression ___________________________________ 

∆Tf = T 0f – Tf T

0

Tf

f

is the freezing point of the pure solvent is the freezing point of the solution

T 0f > Tf

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∆Tf > 0

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∆Tf = Kf m

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m is the molality of the solution

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Kf is the molal freezing-point depression constant (0C/m) for a given solvent 23

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Slide  24 

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Slide  25 

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

∆Tf = Kf m

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Kf water = 1.86 oC/m 478 g x

moles of solute

1 mol 62.01 g

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=

m =

= 2.41 m 3.202 kg solvent

mass of solvent (kg)

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∆Tf = Kf m = 1.86 oC/m x 2.41 m = 4.48 oC

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∆Tf = T 0f – Tf Tf = T f – ∆Tf = 0.00 0

___________________________________  oC

– 4.48

oC

= -4.48

oC 25

  Slide  26 

Osmotic Pressure (π)

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Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

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A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

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Osmotic pressure (π) is the pressure required to stop osmosis.

___________________________________  ___________________________________  ___________________________________  more concentrated

dilute

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Osmotic Pressure (π)

27  ___________________________________  ___________________________________  time solvent

High P

solution

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Low P

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π = MRT

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M is the molarity of the solution R is the gas constant T is the temperature (in K)

27

 

Slide 

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A cell in an:

28 

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isotonic solution

hypotonic solution

hypertonic solution 28  

Slide 

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Colligative Properties

29 

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Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

___________________________________ 

Vapor-Pressure Lowering

P1 = X1 P 10

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Boiling-Point Elevation

∆Tb = Kb m

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Freezing-Point Depression

∆Tf = Kf m

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π = MRT

Osmotic Pressure (π)

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Colligative Properties of Electrolyte Solutions

30 

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0.1 m Na+ ions & 0.1 m Cl- ions

0.1 m NaCl solution

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

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0.2 m ions in solution

0.1 m NaCl solution van’t Hoff factor (i) =

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actual number of particles in soln after dissociation number of formula units initially dissolved in soln

___________________________________  i should be nonelectrolytes NaCl CaCl2

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1 2 3

30

 

Slide 

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Colligative Properties of Electrolyte Solutions

31 

Boiling-Point Elevation

∆Tb = i Kb m

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Freezing-Point Depression

∆Tf = i Kf m

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π = iMRT

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Osmotic Pressure (π)

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Slide  32 

CH-13 HW

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Questions and Problems Pages 131 - 132

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4.54, 4.56, 4.58, 4.60, 4.64, 4.66, 4.68, 4.78, 4.80.

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Pages 459 - 460

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13.8, 13.10, 13.14, 13.16, 13.18, 13.20 (molarity only), 13.22, 13.26.

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