brdy-6ed-ch12-solution
DESCRIPTION
Kimia 2ATRANSCRIPT
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Chapter 12Mixtures at the Molecular
Level: Properties of Solutions
Chemistry, 7th Edition International Student Version
Brady/Jespersen/Hyslop
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Chapter 12: SolutionsSolution
Homogeneous mixture Composed of solvent and solute(s)
Solvent More abundant component of mixture
Solute(s) Less abundant or other component(s) of mixture
e.g., Lactated Ringers solution NaCl, KCl, CaCl2, NaC3H5O3 in water
2
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Why do Solutions Form?
Two driving forces behind formation of solution
1. Disordering/Randomizing*2. Intermolecular ForcesWhether or not a solution forms depends on
both opposing forces
*In the thermodynamics chapter this is called entropy and is defined more precisely
3
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Spontaneous Mixing Two gases mix
spontaneously Due to random motions Mix without outside
work Never separate
spontaneously Tendency of system
left to itself, to become increasingly disordered
4
Gas A Gas B
separate mixed
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Intermolecular Forces Attractive forces between molecules in pure
substances (solvents and solutes) were considered in the last chapter
Now we consider the attractive forces between a solvent molecule and a solute molecule
Strength of intermolecular attractive forces in a mixture, compared to the pure substances, will indicate if a mixture can form or if the substances will remain separate
5
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Formation of Solutions Depends on Intermolecular Forces
When mixed, for a solution to form, Solvent-to-solute attractions must be similar to
attractions between solute alone and solvent alone
Initially solute and solvent separate to make room for each other Solute molecules held together by intermolecular
forces: energy added Solvent molecules held together by
intermolecular forces: energy added
6
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Formation of Solutions Depends on Intermolecular Forces
When the separated solute and solvent are mixed to make a solution Solute molecules are attracted to solvent
molecules: energy released Solvent molecules are attracted to solute
molecules: energy released If there is a net release of energy, the
solution will form.
7
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Miscible LiquidsMiscible liquids
Two liquids that dissolve in one another in all proportions
Form solution Strengths of intermolecular attractions are similar in
solute and solvent Similar polaritye.g., Ethanol and water
8
Ethanol; polar bondHOCC
H
H
H
H
H
+
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Immiscible Liquids Two insoluble liquids Do not mix Get two separate phases Strengths of intermolecular forces are
different in solute and solvent Different polaritye.g., Benzene and water
9
Benzene;no polar bond
C
CC
C
CC
H
H
H
H
H
H
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Rule of Thumb Like dissolves like Use polar solvent for polar solute Use nonpolar solvent for nonpolar solute
When strengths of intermolecular attractions are similar in solute and solvent, solutions form because net energy exchange is about the same
10
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckWhich of the following are miscible in water?
11
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!Which of the following molecules is most soluble in hexane, C6H14?A. NH3B. CH3NH2C. CH3OHD. CH3CH3E. H2O
12
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!Which of the following molecules is most soluble in methanol CH3OH?A. N(CH3)3B. CH3NH2C. CH3COCH3D. CH3CH2CH2CH2CH2CH3E. HOCH2CH2OH
13
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solutions of Solids in Liquids Basic principles remain the same when
solutes are solids Sodium chloride (NaCl)
Ionic bonding Strong intermolecular forces Ions dissolve in water because ion-dipole forces
of water with ions strong enough to overcome ion-ion attractions
14
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Hydration of Solid Solute Ions at edges, fewer
ion-ion attractions Multiple ion-dipole
attractions formed with water
These overcome stronger ion-ion electrostatic attractions
New ion at surface Process continues until
all ions in solution Hydration of ions
Completely surrounded by solvent
15
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Hydration vs. Solvation
Hydration Ions surrounded by water molecules
Solvation General term for surrounding solute particle by
solvent molecules
16
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Polar Molecules Dissolve in Water
17
H2O reorients so Positive H atoms are near negative ends of solute Negative O atoms are near positive ends of solute
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solvation in Nonpolar Solvents?
Wax is a mixture of long chain alkanes and is nonpolar
Benzene is a nonpolar solvent Benzene can dissolve waxes
Weak London dispersion forces in both solute and solvent. Weak London forces will form between solute and solvent.
Wax molecules Easily slip from solid Slide between benzene molecules Form new London forces between solvent and solute
18
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Heat of Solution Energy change associated with formation of
solution Difference in intermolecular forces between
isolated solute and solvent and mixture Cost of mixing Energy exchanged between system and
surroundingsMolar Enthalpy of Solution (Hsoln)
Amount of enthalpy exchanged when one mole of solute dissolves in a solvent at constant pressure to make a solution
19
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Heat of Solution
Hsoln > 0 (positive) Costs energy to make solution Endothermic Increase in potential energy of system
Hsoln < 0 (negative) Energy given off when solution is made Exothermic Decrease in potential energy of system
Which occurs depends on your system
20
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Modeling Formation of Solution
Formation of solution from solid and liquid can be modeled as two-step process
Step 1: Separate solute and solvent molecules Break intermolecular forces Endothermic, increase in potential energy of
system, cost Step 2: Mix solute and solvent
Come together Form new intermolecular forces Exothermic, decrease in potential energy of
system, profit21
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Two-Step Process Application of Hess Law
Way to take two things we can measure and use to calculate something we cant directly measure Hsoln is path independent
Method works because enthalpy is state function Hsoln = Hsoln + Hsolute + Hsolvent
Overall, steps take us from solid solute + liquid solvent final solution
These steps are not the way solution would actually be made in lab
22
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Enthalpy Diagram 1. Break up solid lattice
Hlattice = lattice enthalpy Increase in potential energy
2. Dissolve gas in solvent HSolvation = solvation enthalpy Decrease in potential energy
Hsolution = Hlattice + HSolvation
Whether Hsolution is positive or negative depends on two steps
In lab, solution formed directly
23
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Figure 12.5 Dissolving KI in H2OHlattice (KI) = 632 kJ mol1 Hhydration (KI) = 619 kJ mol1
Hsolution = Hlatt + HhydrHsoln = 632 kJ mol1 619 kJ mol1 Hsoln = +13 kJ mol1 Formation of KI(aq) is endothermic
24
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Dissolving NaBr in H2OHlattice (NaBr) = 728 kJ mol1 Hhydration (NaBr) = 741 kJ mol1
Hsolution = Hlatt + Hhydr
25
Hsoln = 728 kJ mol1 741 kJ mol1 Hsoln = 13 kJ mol1 Formation of NaBr(aq) is exothermic
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!Of the following ions which would have the most exothermic enthalpy of hydration?A. Ga3+
B. Ca2+
C. Li+
D. Na+
26
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!When 2.50 g of solid sodium hydroxide is added to 100.0 g of water in a calorimeter, the temperature of the mixture increases by 6.50 C. Determine the molar enthalpy of solution for sodium hydroxide. Assume the specific heat of the mixture is 4.184 J g1 K1.A. +44.6 kJ/molB. +43.5 kJ/molC. 44.6 kJ/molD. 43.5 kJ/mol
27
mol NaOH = 2.5 g 1 mol NaOH40.0 g NaOH
= 0.0625 mol NaOH
Hsoln
= 102.5 g( ) 4.184 J g1 K 1( ) 6.5 K( ) 1 kJ1000 J
0.0625 mol NaOH= 44.6 kJ/mol NaOH
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solutions Containing Liquid SoluteSimilar treatment but with three step path1. Solute expanded to gas H is positive Increase in potential
energy2. Solvent expanded to
gas H is positive Increase in potential
energy3. Solvation occurs H is negative Decrease in potential
energy28
Hsoln = Hsolute + Hsolvent + Hsolvation
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ideal Solution One in which inter-
molecular attractive forces are identical Hsoln = 0
e.g., Benzene in CCl4 All London forces Hsoln ~ 0
Step 1 + Step 2 = Step 3 or
Hsolute + Hsolvent = Hsolvation 29
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Gaseous Solutes in Liquid Solution Only very weak attractions exist between
gas molecules There are no intermolecular attractions in ideal
gases When making solution with gas solute
Energy required to expand solute is negligible
Heat absorbed or released when gas dissolves in liquid has two contributions:1.Expansion of solvent Hsolvent 2.Solvation of gas Hsolvation
30
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Gas dissolves in organic solvent
Generally endothermic Hsolvation > 0
Gas dissolves in H2O Generally exothermic Hsolvation < 0
31
Gaseous Solutes in Liquid Solution
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!
The solubility of a substance increases with increased temperature if:A. Hsolution > 0B. Hsolution < 0C. Hsolution = 0
32
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!
In which of the following situations would the hsolution be exothermic:A. Hhydration > HlatticeB. Hhydration< HlatticeC. Hhydration = Hlattice
33
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solubility Mass of solute that forms saturated solution with
given mass of solvent at specified temperature
If extra solute added to saturated solution, extra solute will remain as separate phase
34
solubility = g solute100 g solvent
soluteundissolved
solutedissolved
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Effect of Temperature on Solubility
Generally substances become more soluble at elevated temperatures Increased disorder in the solution is often the reason
for this
35
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solubility of Most Substances Increases with Temperature
Most substances become more soluble as T increases
Amount solubility increases Varies considerably Depends on
substance
36
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Effect of T on Gas Solubility in Liquids Solubility of gases usually decreases as T increases
Solubilities of Common Gases in Water
37
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Case Study: Dead Zones
During the industrial revolution, factories were built on rivers so that the river water could be used as a coolant for the machinery. The hot water was dumped back into the river and cool water recirculated. After some time, the rivers began to darken and many fish died. The water was not found to be contaminated by the machinery. What was the cause of the mysterious fish kills?
38
Increased temperature, lowered the amounts of dissolved oxygen
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Henrys Law Pressure-Solubility Law Concentration of gas in liquid at any given
temperature is directly proportional to partial pressure of gas over solution
Cgas = kHPgas (T is constant)
Cgas = concentration of gas Pgas = partial pressure of gas kH = Henrys Law constant
Unique to each gas Tabulated
39
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Henrys Law True only at low concentrations and
pressures where gases do NOT react with solvent
Alternate form
C1 and P1 refer to an initial set of conditions C2 and P2 refer to a final set of conditions
40
2
2
1
1PC
PC
=. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Effect of Pressure on Gas Solubility
Solubility increases as P increases
41
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Effect of Pressure on Gas SolubilityA. At some P, equilibrium exists between vapor phase and
solution ratein = rateout
B. Increase in P puts stress on equilibrium Increase in frequency of collisions so ratein > rateout
C. More gas dissolved rateout will increase until rateout = ratein again
42
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Using Henrys LawCalculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 5 atm over the liquid at 25 C. The Henrys Law constant for CO2 in water at this temperature is 3.12 102 mol L1 atm1.
43
= (3.12 102 mol/L atm) (5.0 atm)= 0.156 mol/L 0.16 mol/L
CCO2
= kH(CO
2)P
CO2
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Using Henrys LawCalculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25 C under a partial pressure of CO2 of 4.0 104 atm.
44
C2 = 1.2 104 mol CO2/LCompared to the concentration of CO2 at
5 atm: 0.16 mol CO2/L
C2 =
0.156 mol/L( ) 4.0 104 atm( )5.0atm
C1
P1
=C
2
P2
C2=
P2C
1
P1. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning Check
What is the concentration of dissolved nitrogen in a solution that is saturated in N2 at 2.0 atm? kH= 8.42 107 mol L1 atm1
45
Cg=kHPg Cg= (8.42 107 mol L1 atm1) 2.0 atm Cg=1.7 106 mol L1
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!How many grams of oxygen gas at 1.0 atm will dissolve in 10.0 L of water at 25 C if Henrys constant is 1.3 103 mol L1 atm1 at this temperature?A. 0.42 gB. 0.013 gC. 0.042 gD. 0.21 gE. 2.4 g
46
1.0 atm O2 .0013 mol O2
L atm 32.00 g
1.00 mol10 L
1 =0.416 g
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn! The solubility of O2 in water is approximately
0.00380 g L-1 when the temperature is 25.0 C and the partial pressure of gaseous oxygen is 760 torr. What will the solubility of oxygen be if the pressure is adjusted to 1000 torr?
A. 0.00289 g L-1
B. 0.00500 g L-1
C. 1.49 g L-1
D. 2.89 x 103 g L-1
E. 3.46 x 103 g L-1
47
0.00380 g/L760 torr
= X g/L1000 torr
X = 3.80760
= 0.00500 g/L
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solubility of Polar vs. Nonpolar Gases Gas molecules with polar bonds are much more
soluble in water than nonpolar molecules like oxygen and nitrogen CO2, SO2, NH3 >> O2, N2, Ar
Form H-bonds with H2O Some gases have increased solubility because they
react with H2O to some extente.g., CO2(aq) + H2O H2CO3(aq) H+(aq) + HCO3(aq)
SO2(aq) + H2O H2SO3(aq) H+(aq) + HSO3(aq)
NH3(aq) + H2O NH4+(aq) + OH(aq)
48
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Case Study
When you open a bottle of seltzer, it fizzes. How should you store it to increase the time before it goes flat?
49
Gases are more soluble at low temperature and high pressure.
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Concentration What units we use depends on situation Stoichiometric calculations Molarity
Units: mol/LProblem: M varies with temperature
Volume varies with temperature Solutions expand and contract when heated
and cooled If temperature independent concentration is
needed, must use other units50
solution of Lsolute of mol
=M
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Temperature Independent Concentration
1. Percent Concentrations Also called percent by mass or percent by
weight
This is sometimes indicated %(w/w) where w stands for weight or mass
The (w/w) is often omitted
51
%100solution of masssolute of mass massby percent =. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Percent by Mass What is the percent by mass of NaCl in a solution consisting of 12.5 g of NaCl and 75.0 g water?
52
percent by massNaCl
= 14.3% NaCl
percent by massNaCl
= 12.5 g12.5 g + 75.0 g
100%
%100solution of masssolute of mass
massby percent =. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckSeawater is typically 3.5% sea salt and has a density of 1.03 g/mL. How many grams of sea salt would be needed to prepare enough seawater solution to fill a 62.5 L aquarium?What do we need to find?
62.5 L ? g sea saltWhat do we know?
3.5 g sea salt 100 g solution 1.03 g solution 1.00 mL solution 1000 mL 1.00 L
53
= 2.2 103 g sea saltsoln g 100
salt sea g 5.3solnmL 1.00
soln g 1.03L 1mL 1000
L 5.62
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
More Temperature Independent Concentration Units
Molality (m) Number of moles of solute per kilogram solvent
Also molal concentration Independent of temperature m vs. M Similar when d = 1.00 g/mL Different when d >> 1.00 g/mL or
d
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Concentration CalculationIf you prepare a solution by dissolving 25.38 g of I2 in 500.0 g of water, what is the molality (m) of the solution? What do we need to find? 25.38 g ? m
What do we know? 253.8 g I2 1 mol I2 m = mol solute/kg solvent 500.0 g 0.5000 kg
Solve it
55
= 0.2000 mol I2/kg water = 0.2000 mwater kg .50000
1I g 253.8
I mol 1 I g 5.382
2
22
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Concentration Calcn. (cont)What is the molarity (M) of this solution? The density of this solution is 1.00 g/mL. What do we need to find? 25.38 g ? M
What do we know? 253.8 g I2 1 mol I2 M = mol solute/L soln 1.00 g soln 1 mL soln
Solve it
56
= 0.1903 mol I2/ L soln = 0.1903 M
25.38 g I2
1 mol I2
253.8 g I2
1525.38 g soln
1.00 g1.00 mL
1000 mL1 L
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. M and m in CCl4What is the molality (m) and molarity (M) of a solution prepared by dissolving 25.38 g of I2 in 500.0 g of CCl4? The density of this solution is 1.59 g/mL. What do we need to find? 25.38 g ? m
What do we know? 253.8 g I2 1 mol I2 m = mol solute/kg solvent 500.0 g soln 0.5000 kg soln
Solve it
57
= 0.2000 mol I2/kg CCl4= 0.2000 m
25.38 g I2
1 mol I2
253.8 g I2
1500.0 g CCl
4
1000 g1 kg
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. M and m in CCl4 (cont)What is the molarity (M) of this solution? The density of this solution is 1.59 g/mL.What do we need to find? 25.38 g ? M
What do we know? 253.8 g I2 1 mol I2 M = mol solute/L solution 1.59 g solution 1 mL solution g of solution = g I2 + g CCl4 = 500.0 g + 25.38 g
Solve it
58
= 0.3030 mol I2/L soln = 0.303 M
25.38 g I2
1 mol I2
253.8 g I2
1525.38 g soln
1.59 g1.00 mL
1000 mL1 L
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Converting between ConcentrationsCalculate the molarity and the molality of a 40.0% HBr solution. The density of this solution is 1.38 g/mL.
If we assume 100.0 g of solution, then 40.0 g of HBr.
If 100 g solution, then
mass H2O = 100.0 g solution 40.0 g HBr = 60.0 g H2O
59
mol HBr= 40.0 g HBr80.91 g HBr/mol
=0.494 mol HBr
40.0% HBr = wt%= 40.0 g HBr100 g solution
100%
m= mol HBrkg of H
2O
m = molHBrkgH
2O
= 0.494molHBr0.0600kgH
2O=8.24m
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Converting between Concentrations (cont.)
Now calculate molarity of 40% HBr
60
mol HBr = 0.494 mol
= 72.46 mL
= 6.82 M
Volume solution= massdensity
= 100 g1.38 g/mL
M = 0.494 mol HBr72.46 mL solution
1000 mL1 L
M = mol HBrL solution
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!What is the molality of 50.0% (w/w) sodium hydroxide solution? A. 0.500 mB. 1.25 mC. 0.025 mD. 25 mE. 50 m
61
Molar Masses H2O: 18.02 g/mol ; NaOH: 40.00 g/mol
100.0 g soln = 50.0 g NaOH + 50.0 g water
= 25 m
50.0 g NaOH50.0 g water
1 mol NaOH40.00 g NaOH
1000 g water1 kg water
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!What is the molarity of the 50%(w/w) solution if its density is 1.529 g/mL?A. 19 MB. 1.25 MC. 1.9 MD. 0.76 M
62
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn! - Solution
63
?M = 50 g NaOH 100 g solution
= 50 g NaOH 100 g solution
1 mol NaOH40.0 g NaOH
1.529 g solutionmL solution
1000 mLL
= 19 M
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!What mass of NH4Cl must be dissolved in 100 mL of H2O to make a 0.250 M solution?
A. 1340 gB. 21.40 gC. 13.4 gD. 1.34 g
64
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn! - Solution
65
=0.250 moles NH
4Cl
1 L solution0.100 L
1
53.49 g NH4Cl
1 Mole
= 1.34 g
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
More Temperature-Independent Concentration Units
Mole Fraction
Mole %
66
XA= # mol A
Total moles of all components
mol %A = XA 100%
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Colligative Properties Physical properties of solutions Depend mostly on relative populations of particles
in mixtures Dont depend on their chemical identities
Effects of solute on vapor pressure of solvents Solutes that cant evaporate from solution are
called nonvolatile solutes
Fact: All solutions of nonvolatile solutes have lower vapor pressures than their pure solvents
67
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Raoults Law Vapor pressure of solution, Psolution, equals
product of mole fraction of solvent, Xsolvent and its vapor pressure when pure, Psolvent
Applies for dilute solutions
68
solvent pure of pressurevapor
solvent the of fraction mole solution the of pressure vapor
=
=
=
solvent
solvent
solution
P
XP
solventsolventsolution PXP =
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Alternate form of Raoults Law Plot of Psoln vs. Xsolvent
should be linear
Slope =
Intercept = 0 Change in vapor
pressure can be expressed as
Usually more interested in how solutes mole fraction changes the vapor pressure of solvent
69
P = change in P = (Psolvent P
solution)
P = Xsolute
Psolvent
Xsolvent
Psolvent
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Glycerin (using Raoults Law)Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 C. Calculate vapor pressure at 25 C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 C is 23.8 torr.To solve use:
First we need Xsolvent, so we need to determine the moles of glycerin and moles of water.
70
Psolution
= Xsolvent
Psolvent
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Glycerin (cont.)
71
Mole glycerin
Mole water
Mole fraction glycerin
50.0 mL C3H
8O
3 1.26 g
mL
1 mol C3H
8O
3
92.1 g C3H
8O
3
= 0.684 mol C3H
8O
3
500.0 mL H2O 1.00 g
mL
1 mol H2O
18.02 g H2O
= 27.75 mol H2O
XC3H8O3
= 27.75 mol27.75 mol + 0.684 mol
= 0.976
Psolution
= csolvent
Psolvent
= 0.9759( ) 23.8 torr = 23.2 torr
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckThe vapor pressure of 2-methylhexane is 37.986 torr at 15C. What would be the pressure of the mixture of 78.0 g 2-methylhexane and 15 g naphthalene, which is nearly non-volatile at this temperature?
72
Psolution = solvent Posolvent
= 33.02 torr = 33 torrP = 0.869 37.986 torr( )
mole 2-methylhexane = 78.0 g100.2 g/mol
= 0.7784 mol
mole naphthalene = 15 g128.17 g/mol
= 0.117 mol
c2methylhexane
= 0.7784 mol 0.7784 mol + 0.117 mol
= 0.869
naphthaleneC10H8
MM 128.17
2-methylhexaneC7H16MM 100.2
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Why Nonvolatile Solute Lowers Vapor Pressure
A. Lots of solvent molecules in liquid phase Rate of evaporation and
condensation high and equal
B. Fewer solvent molecules in liquid Rate of evaporation is
lower so rate of condensation must be lower which is achieved by fewer gas particles
At equilibrium, fewer gas molecules give lower pressure
73
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solutions That Contain Two or More Volatile Components
Vapor contains two components Partial pressure of each component A and B is
given by Raoults Law
Total pressure of solution of components A and B given by Daltons Law of Partial Pressures
Since XA + XB = 1 then 1 XA = XB and we can write
74
PA= X
AP
A and P
B= X
BP
B
Ptotal
= PA+P
B= X
AP
A + X
BP
B
Ptotal
= XAP
A + (1 X
A)P
B
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
For Ideal, Two Component Solution of Volatile Components
75
PA= X
AP
A
PB= X
BP
Bo
Ptotal
= PA+P
B= X
AP
A + X
BP
B
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Benzene and Toluene
Consider a mixture of benzene, C6H6, and toluene, C7H8, containing 1.0 mol benzene and 2.0 mol toluene. At 20 C, the vapor pressures of the pure substances are:P benzene = 75 torr and P toluene = 22 torr
Assuming the mixture obeys Raoults law, what is the total pressure above this solution?
76
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Benzene and Toluene (cont.)1. Calculate mole fractions of A and B
2. Calculate partial pressures of A and B
3. Calculate total pressure
77
Pbenzene
= Xbenzene
Pbenzeneo = 0.33 75 torr = 25 torr
Ptoluene
= Xtoluene
Ptolueneo = 0.67 22 torr = 15 torr
Xbenzene
= 1.0 mol benzene1.0 mol + 2.0 mol
= 0.33 benzene
Xtoluene
= 2.0 mol1.0 + 2.0( ) mol
= 0.67 toluene
Ptotal
= Pbenzene
+Ptoluene
= (25 +15) torr = 40 torr
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckThe vapor pressure of 2-methylheptane is 233.95 torr at 55 C. At the same temperature, 3-ethylpentane has a vapor pressure of 207.68 torr. What would be the pressure of the mixture of 78.0 g 2-methylheptane and 15 g 3-ethylpentane?
78
Psolution = XAP oA + XBP oB
2-methylheptaneC8H18
MM 114.23 g/mol
3-ethylpentaneC7H16
MM 100.2 g/mol
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning Check
79
P = 230 torr
P = 0.827 233.95 torr( )+ 0.173207.68 torr( )
mole 2-methylheptane = 78.0 g114.23 g/mol
= 0.683 mol
mole 3-ethylpentane = 15.0 g100.2 g/mol
= 0.150 mol
X3-ethylpentane =0.150 mol
(0.683 mol + 0.150 mol) = 0.173
X2-methylpentane =0.683 mol
(0.683 mol + 0.150 mol) = 0.827. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!n-Hexane and n-heptane are miscible. If the vapor pressure of pure hexane is 151.28 mm Hg, and heptane is 45.67 at 25 C, which equation can be used to determine the mole fraction of hexane in the mixture if the mixtures vapor pressure is 145.5 mm Hg?
A. X (151.28 mmHg) = 145.5 mmHgB. X (151.28 mmHg) + (X )(45.67 mm Hg)
= 145.5 mmHgC. X (151.28 mmHg) + (1 X )(45.67 mm Hg) =
145.5 mm HgD. None of these
80
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Solutes also Affect Freezing and Boiling Points of Solutions
Facts: Freezing point of solution always lower than
pure solvent Boiling point of solution always higher than
pure solventWhy? Consider the phase diagram of H2O
Solid, liquid, gas phases in equilibrium Blue lines P vs. T
81
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
SolutionEffect of Solute Solute molecules stay
in solution only None in vapor None in solid
Crystal structure prevents from entering
Liquid/vapor Decrease number solvent
molecules entering vapor Need higher T to get all
liquid to gas Line at higher T along phase
border (red)
82
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
SolutionEffect of Solute Triple point moves
lower and to the left Solid/liquid
Solid/liquid line to left (red)
Lower T all along phase boundary
Solute keeps solvent in solution longer
Must go to lower T to form crystal
83
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Freezing Point Depression and Boiling Point Elevation
Solution Observe increase in boiling point and decrease in
freezing point compared to pure solvent Both Tf and Tb depend on relative amounts of
solvent and solute
Colligative properties Boiling Point Elevation (Tb)
Increase in boiling point of solution vs. pure solvent
Freezing Point Depression (Tf ) Decrease in freezing point of solution vs. pure solvent
84
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Freezing Point Depression (Tf)
Tf = Kf mwhere Tf = (Tf Tsoln)m = concentration in MolalityKf = molal freezing point depression constant
Units of C/molalDepend on solvent, see Table 12.4
85
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Boiling Point Elevation (Tb)
Tb = Kb mwhere Tb = (Tsoln Tb)m = concentration in MolalityKb = molal boiling point elevation constant
Units of C/mDepend on solvent, see Table 12.4
86
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Table 12.3 Kf and Kb
87
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Freezing Point DepressionEstimate the freezing point of a permanent type of antifreeze solution made up of 100.0 g ethylene glycol, C2H6O2, (MM = 62.07 g/mol) and 100.0 g H2O (MM = 18.02 g/mol).
88
Tf = Kfm = (1.86 C/m) 16.11 m Tf = (Tfp Tsoln)30.0 C = 0.0 C Tsoln Tsoln = 0.0 C 30.0 C
= 1.611 mol C2H6O2
= 16.11 m C2H6O2
= 30.0 C
= 30.0 C
100.0 g C2H
6O
2
1 mol C2H
6O
2
62.07 g C2H
6O
2
m = mol solutekg solvent
=1.611 mol C
2H
6O
2
0.100 kg water
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!When 0.25 g of an unknown organic compound is added to 25.0 g of cyclohexane, the freezing point of cyclohexane is lowered by 1.6 C. Kf for the solvent is 20.2 C m1. Determine the molar mass of the unknown.A. 505 g/molB. 32 g/molC. 315 g/molD. 126 g/mol
89
Tf= K
fm
1.6 oC = 20.2oCm
0.250 gmolar mass
0.025 kgMW = 126 g/mol
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Boiling Point ElevationA 2.00 g sample of a substance was dissolved in 15.0 g of CCl4. The boiling point of this solution was determined to be 77.85 C. Calculate the molar mass of the compound. [For CCl4, the Kb = 5.07 C/m and BP = 76.50 C]
90
Tb= K
bm m = Tb
Kb=
77.85 76.50( ) C5.07 C /m
= 0.2684 m
m = mol solutekg solvent
mol solute = 0.2684 m0.0150 kg CCl4= 4.026103 mol
molar mass = 2.00 g compound4.026103 mole
= 497 g/mol
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckAccording to the Sierra Antifreeze literature, the freezing point of a 40/60 solution of Sierra antifreeze and water is 4 F. What is the molality of the solution?
91
m = 11 m
tF = 1.8 tC + 32
4 F = 1.8 tC + 32
tC = 20. C
T = Tf Tsoln
Tsolution = m Kfp
0 (20.)( ) C = m 1.86 C m 1
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckIn the previous sample of a Sierra antifreeze mixture, 100 mL is known to contain 42 g of the antifreeze and 60. g of water. What is the molar mass of the compound found in this antifreeze if it has a freezing point of 4 F?
92
From before:m = 11 m
MM solute = 64 g/mol
= 0.66 mol solutemol solute = 11 m 0.060 kg solvent
solute mol 0.66solute g 42
MM =
solvent kgsolute mol
=m. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!If all of the following generic compounds have the same Kb, which will cause a greater change in the boiling point of a 0.25 m nonelectrolyte solution
(Assume all dissociate completely in solution)
A. MXB. MX2C. M2X3D. MX3E. Not enough information given
93
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Membranes and PermeabilityMembranes
Separators Example: cell walls Keep mixtures separated
Permeability Ability to pass substances
through membrane
Semipermeable Some substances pass,
others dont Membranes are
semipermeable Selective
94
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Membranes and Permeability Degree of permeability depends on type of
membrane Some pass water only Some pass water and small ions only
Membranes separating two solutions of different concentration Two similar phenomena occur Depends on membrane
Dialysis When semipermeable membrane lets both H2O and
small solute particles through Membrane called dialyzing membrane Keeps out large molecules such as proteins
95
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
OsmosisOsmotic Membrane
Semipermeable membrane that lets only solvent molecules through
Osmosis Net shift of solvent molecules (usually water)
through an osmotic membrane Direction of flow in osmosis,
Solvent flows from dilute to more concentrated side Flow of solvent molecules across osmotic membrane
Increase in concentration of solute on dilute side Decrease in concentration of solute on more
concentrated side
96
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Osmosis and Osmotic Pressure
A. Initially, solution B separated from pure water, A, by osmotic membrane. No osmosis occurred yet
B. After a while, volume of fluid in tube higher. Osmosis has occurred.
C. Need back pressure to prevent osmosis = osmotic pressure.
97
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Osmotic Pressure Exact back pressure needed to prevent osmotic
flow when one liquid is pure solvent.Why does osmosis eventually stop? Extra weight of solvent as rises in column
generates this opposing pressure When enough solvent transfers to solution so that
when osmotic pressure is reached, flow stops If osmotic pressure is exceeded, then reverse
process occurssolvent leaves solution Reverse osmosisused to purify sea water
98
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Equation for Osmotic Pressure Similar to ideal gas law
V = nRT or = MRT = osmotic pressure V = volume n = moles M = molarity of solution (n/V) T = Temperature in Kelvins R = Ideal gas constant
= 0.082057 L atm mol1 K1
99
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Osmometer Instrument to measure osmotic pressure Very important in solutions used for
biological samplesIsotonic solution
Same salt concentration as cells Same osmotic pressure as cells
Hypertonic solution Higher salt concentration than cells Higher osmotic pressure than cells Will cause cells to shrink and dehydrate
Hypotonic solution Lower salt concentration than cells Lower osmotic pressure than cells Will cause cells to swell and burst
100
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Osmotic PressureEye drops must be at the same osmotic pressure as the human eye to prevent water from moving into or out of the eye. A commercial eye drop solution is 0.327 M in electrolyte particles. What is the osmotic pressure in the human eye at 25 C?
101
= MRT T = 25 C + 273.15
= 0.327 M( ) 0.08206 L atmK mol
298 K( ) = 8.00 atm
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Ex. Using to determine MMThe osmotic pressure of an aqueous solution of certain protein was measured to determine its molar mass. The solution contained 3.50 mg of protein in sufficient H2O to form 5.00 mL of solution. The measured osmotic pressure of this solution was 1.54 torr at 25 C. Calculate the molar mass of the protein.
102
M = PRT
= 1.54 torr 1 atm
760 torr
0.08206 L atm mol1 K 1( )298 K = 8.28 105 mol
L
mol = 8.28 105 M( ) 5.00 103 L( ) = 4.14 107 mol
Molar mass = gmol
= 3.50 103 g
4.14 107 mol = 8.45 103 g/mol
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning Check: OsmosisA solution of 5% dextrose (C6H1206) in water is placed into the osmometer shown at right. It has a density of 1.0 g/mL. The surroundings are filled with distilled water. What is the expected osmotic pressure at 25 C?
103
= 7 atm
= 0.277 molL
0.082057 L atmmol K
298 K( )
5 g C6H
12O
6
100 g solution
mol C6H
12O
6
180.16 g
1.0 g solnmL soln
1000 mLL
= M
= MRT. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckFor a typical blood plasma, the osmotic pressure at body temperature (37 C) is 5409 mm Hg. If the dominant solute is serum protein, what is the concentration of serum protein?
104
Molarity = 0.280 M
7.117 atm = ? molL
0.082057 L atmmol K
310. K( )
5409 mm Hg( ) 1 atm760 mm Hg
=
= MRT. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!Which aqueous solution has the highest osmotic pressure. A)0.100 molar Al(NO3)3B) 0.150 molar Ba(NO3)2C) 0.100 molar CaCl2D) 0.150 molar NaClE) 0.200 molar NH3
105
Molar Mass H2S: 34.076 g/mol
All solutions have the same temperature so the only influence is number of ions.
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Colligative Properties of Electrolyte Solutions Differ
Kf (H2O) = 1.86 C/m Expect 1.00 m solution of NaCl to freeze at
1.86 C Actual freezing point = 3.37 C About twice expected T
Why? Colligative properties depend on
concentration (number) of particles One NaCl dissociates to form two particles
NaCl(s) Na+(aq) + Cl(aq)
106
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Colligative Properties of Electrolyte Solutions Depend on Number of Ions Actual concentration of ions = 2.00 m
(Started with 1.00 m NaCl) Now use this to calculate T Tf = 1.86 C/m 2.00 m = 3.72 COr Tfinal = Tinitial Tf = 0.00 3.72 C
= 3.72 C Not exactly = to actual Tf = 3.37 C This method for ions gives rough estimate if
you assume that all ions dissociate 100%.107
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Why isnt this Exact for Electrolytes? Assumes 100% dissociation of ions
Electrolytes dont dissociate 100%, especially in concentrated solutions
Some ions exist in ion pairs Closely associated pairs of oppositely charged ions that
behave as a single particle in solution So, fewer particles than predicted Result: freezing point depression and boiling point
elevation not as great as expected As you go to more dilute solutions, electrolytes
more fully dissociated and observe freezing point and boiling point closer to calculated value. Model works better at dilute concentrations
108
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
vant Hoff Factor = i Scales solute molality to correct number of
particles Measure of dissociation of electrolytes vant Hoff factor is equivalent to percent
ionization In general, it varies with concentration (see
Table 12.5, page 613)
109
i =(T
f)
measured
(Tf)
calcd as nonelectrolyte
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Table 12.5 vant Hoff Factors vs. Concentration
110
Note: 1. As concentration decreases, iobserved iexpected 2. MgSO4 much less dissociated than NaCl or KCl
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Your Turn!What is the vant Hoff factor for FeCl3 if a solution of 81.1 g dissolved in 500 g H2O has a boiling point of 101.74 C?A. 4B. 1C. 3.4D. 2.7E. Not enough information given
111
MM: FeCl3 = 162.2 g/mol; Kb=0.512 C/m
i = tKbm
= 1.74C0.512 C/m * 1.00 m
= 3.4. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Nonelectrolytes Some molecular solutes produce weaker
colligative effects than predicted by their molal concentrations
Evidence of solute molecule clustering or associating
Result: only half the number of particles expected based on molality of solution, so Tf only half of what expected
112
C6H5 C
O
O H
2 C6H5 C
O
O H
C6H5C
O
OH. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Nonelectrolytes
Result: calculated molar mass double what is expected Concentration of particles is about half of
what is expected and Tf only half of what expected
So size of solute particles is important Common with organic acids and alcohols
113
C6H5 C
O
O H
2 C6H5 C
O
O H
C6H5C
O
OH
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Learning CheckIn preparing pasta, 2 L of water at 25 C are combined with about 15 g salt (NaCl, MM= 58.44 g/mol) and the solution brought to a boil. What is the change in the boiling point of the water?
114
T = i m Kbpmass of water = volume density = 2000 mL 1.0 g/mL
= 2000 g water = 2 kg
mNaCl = 0.26 mol / 2 kg = 0.1 m
T = 0.1 CT = 2 ionmol
0.1 m1
0.51 Cm
mol NaCl = 15 g NaCl58.44 g mol1
= 0.257 mol
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Case StudySuppose you run out of salt. What mass of sugar (C12H22O11, MM = 342.30 g/mol) added to 2 L of water would raise the temperature of water by 0.10 C?
115
T=i m Kbp
mass of water = volume density = 2000 mL 1.0 g/mL
= 2000 g water = 2 kg 0.196 m = y mol / 2 kg
molality = 0.196 m
y = 0.392 mol
mass of sucrose =130 g
0.10 oC = 1 ionmol
x m( ) 0.51
oCm
0.392 mol = mass sucrose342.30 g mol1
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Colligative Properties SummaryColligative properties depend on number
of particles i = mapp/mmolecular Raoults Law Freezing point depression Boiling point elevation Osmotic pressure
Must look at solute and see if molecular or ionic If molecular, i = 1 If ionic, must include i > 1 in equations
116
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Colligative Properties
Raoults law
Freezing point depression Tf = iKf m
Boiling point elevation Tb = iKb m
Osmotic pressure = iMRT
117
Psolution
= csolvent
Psolvento
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Heterogeneous Mixtures
118
Dispersions: 2 or more phases that are not homogeneous Many consumer products are of this type
Two major types Suspensions
Large particles mixed in a solvent Two phases
Colloidal dispersions Very small particles Do not settle or have separate phases Look like true solutions but scatter visible light
. Personal Used Only - Use Wisely .
-
Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.
Colloidal Dispersions
119
Tyndall effect Scattering of light observed in colloidal
dispersions
. Personal Used Only - Use Wisely .
Slide 1Chapter 12: SolutionsWhy do Solutions Form?Spontaneous MixingIntermolecular ForcesFormation of Solutions Depends on Intermolecular ForcesFormation of Solutions Depends on Intermolecular ForcesMiscible LiquidsImmiscible LiquidsRule of ThumbLearning CheckYour Turn!Your Turn!Solutions of Solids in LiquidsHydration of Solid SoluteHydration vs. SolvationPolar Molecules Dissolve in WaterSolvation in Nonpolar Solvents?Heat of SolutionHeat of SolutionModeling Formation of SolutionTwo-Step ProcessEnthalpy DiagramFigure 12.5 Dissolving KI in H2ODissolving NaBr in H2OYour Turn!Your Turn!Solutions Containing Liquid SoluteIdeal SolutionGaseous Solutes in Liquid SolutionSlide 31Your Turn!Your Turn!SolubilityEffect of Temperature on SolubilitySolubility of Most Substances Increases with TemperatureEffect of T on Gas Solubility in LiquidsCase Study: Dead ZonesHenrys LawHenrys LawEffect of Pressure on Gas SolubilityEffect of Pressure on Gas SolubilityEx. Using Henrys LawEx. Using Henrys LawLearning CheckYour Turn!Your Turn!Solubility of Polar vs. Nonpolar GasesCase StudyConcentrationTemperature Independent ConcentrationEx. Percent by MassLearning CheckMore Temperature Independent Concentration UnitsEx. Concentration CalculationEx. Concentration Calcn. (cont)Ex. M and m in CCl4Ex. M and m in CCl4 (cont)Converting between ConcentrationsConverting between Concentrations (cont.)Your Turn!Your Turn!Your Turn! - SolutionYour Turn!Your Turn! - SolutionMore Temperature-Independent Concentration UnitsColligative PropertiesRaoults LawAlternate form of Raoults LawEx. Glycerin (using Raoults Law)Ex. Glycerin (cont.)Learning CheckWhy Nonvolatile Solute Lowers Vapor PressureSolutions That Contain Two or More Volatile ComponentsFor Ideal, Two Component Solution of Volatile ComponentsEx. Benzene and TolueneEx. Benzene and Toluene (cont.)Learning CheckLearning CheckYour Turn!Solutes also Affect Freezing and Boiling Points of SolutionsSolutionEffect of SoluteSolutionEffect of SoluteFreezing Point Depression and Boiling Point ElevationFreezing Point Depression (Tf)Boiling Point Elevation (Tb)Table 12.3 Kf and KbEx. Freezing Point DepressionYour Turn!Ex. Boiling Point ElevationLearning CheckLearning CheckYour Turn!Membranes and PermeabilityMembranes and PermeabilityOsmosisOsmosis and Osmotic PressureOsmotic PressureEquation for Osmotic PressureOsmometerEx. Osmotic PressureEx. Using to determine MMLearning Check: OsmosisLearning CheckYour Turn!Colligative Properties of Electrolyte Solutions DifferSlide 107Why isnt this Exact for Electrolytes?vant Hoff Factor = iTable 12.5 vant Hoff Factors vs. ConcentrationYour Turn!NonelectrolytesNonelectrolytesLearning CheckCase StudyColligative Properties SummaryColligative PropertiesHeterogeneous MixturesColloidal Dispersions