properties of solutions chapter 11. a solution is a homogenous mixture of 2 or more substances the...
TRANSCRIPT
Properties of Solutions
Chapter 11
A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount
11.1 Solution composition
Various types of solutions
Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass
% by mass = x 100%mass of solutemass of solute + mass of solvent
= x 100%mass of solutemass of solution
Mole Fraction (X)
XA = moles of A
sum of moles of all components
M = moles of solute
volume of solution (liters)
Molarity (M)
Molality (m)
m =moles of solute
mass of solvent (kg)
BA
AA nn
nAoffractionMole
Example
1.00 g C2H5OH is added to 100.0 g of water to make 101 mL of solution. Find the molarity, mass %, mole fraction and molality of ethanol.
Molarity Number of moles of solute per L or
solution
solution of liters
solute of molesM
M 0.215L 0.101
mol 102.17
mL 1000L 1
mL 101
g 46.07mol 1
OHHC g 1.00M
252
Mass Percent also called weight percent percent by mass of the solute in the
solution
100solution of mass
solute of mass% Mass
% 0.990100solution g) 100.0g (1.00
OHHC g 1.00% Mass 52
Mole Fraction ratio of number of moles of a part of
solution to total number of moles of solution
BA
AA nn
n
00389.0
1017.202.18
10.100
1017.2
22
522
52
molg
molOgH
OHHmolCOHHC
Molality
Number of moles of solute per kg of solvent
solvent of kg
solute of molesm
0.217m
solution1000g
1kg100.0g
OHHmolC102.17m 52
2
Example
An aqueous antifreeze solution is 40% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity and mole fraction of ethylene glycol
mol/kg 07.11000
0.6007.62
mol10.40
2
kg
g
OHggEG
EGEGg
Molality
Lmol
mL
L
solutiong
mLsolutiong
EGg
EGmolEGg
Molarity /77.6
1000
1
05.1
10.100
07.62
10.40
162.0644.033.3
644.0EG
where EG = ethylene glycol (C2H6O2)
# mol water = 60.0/18.0 = 3.33 mol
# mol EG = 0.644 mol EG
Mass of water (solvent) = 100-40 = 60.0 g
What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?
m =moles of solute
mass of solvent (kg)M =
moles of solute
liters of solution
Assume 1 L of solution:Mass of solute = mass of 5.86 moles ethanol = 270 g ethanolMass of solvent = mass of 1 L solution= (1000 mL x 0.927 g/mL) = 927 g of solution
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
m =moles of solute
mass of solvent (kg)=
5.86 moles C2H5OH
0.657 kg solvent= 8.92 m
“like dissolves like”
Two substances with similar intermolecular forces are likely to be soluble in each other.
• Non-polar molecules are soluble in non-polar solvents CCl4 in C6H6
• Polar molecules are soluble in polar solvents C2H5OH in H2O
• Ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l)
11.2 Energies of Solution FormationDissolution of a solute in liquids
• Why this behavior occur?
Solubility Process
The formation of a liquidsolution takes place in 3 steps
1. Expand solutemolecules
2. Expand solventmolecules
3. Mixing solute andsolvent
Endothermic
Endothermic Often Exothermic
Energy of Solubility Process Steps 1 and 2 require energy to overcome
IMFs
• endothermic Step 3 usually releases energy
• exothermic enthalpy of solution
• sum of ∆H values• can be – or +• ∆Hsoln = ∆H1 + ∆H2 + ∆H3
Energy of Solubility Process
Case 1: oil and water Oil is nonpolar (LD forces) Water is polar (H bonding) ∆H1 will be small for typical size ∆H2 will be large ∆H3 will be small since there won’t be
much interaction between the two ∆Hsoln will be large and +ve because
energy required by steps 1 and 2 is larger than the amount released by 3
Case 2: NaCl and water NaCl is ionic water is polar (H bonding) ∆H1 will be large
∆H2 will be large
∆H3 will be large and –ve because of the strong interaction between ions and water
∆Hsoln will be close to zero- small but +ve
Energy of solubility process Enthalpy of hydration - ∆Hhyd
enthalpy change associated with dispersal
of gaseous solute in water
NaCl(s) Na+(g) + Cl-(g)
∆H1=786 kJ/mol
H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq)
∆Hhyd=∆H2 + ∆H3=-783 kJ/mol
∆Hsoln=3 kJ/mol
Energy
Reactants
Solution
H1
H2
H3
Solvent
Solute and Solvent
Size of H3 determines whether a solution will form
H3
Solution
Mixing solvent and solute
wo factors explains the solubility:
1. n increase in the probability of mixing favors the process
2. Processes that require large amounts of energy tend not to occur
If Hsoln is small and positive, a solution will still form because of entropy.
There are many more ways for them to become mixed than there is for them to stay separate.
Energy of Solubility Process
1 .Structure and Solubility
Water soluble molecules must have dipole moments -polar bonds.
To be soluble in non polar solvents the molecules must be non polar.
11.311.3 Factors Affecting SolubilityFactors Affecting Solubility
• Pentane C5H12 is miscible with hexane C6H14
and immiscible with water• Solubility of alcohols decreases with the molar mass? Cl3OH CH3(CH2)3OH CH3 (CH2)6OH Soluble Insoluble (Hydrophilic) (Hydrophobic)Polarity decreases
OH is smaller portionHydrocarbon is larger
Structure effects: Vitamins and the body
hydrophobic: water- fearing: nonpolar• Insoluble in water
hydrophilic: water-loving: polar• Soluble in water
Hydrophobic, accumulatesin the body. The body cantolerate a diet deficientin vitamin
Hydrophilic, excreted bythe body and must be consumed regularly
2 .Pressure effects Changing the pressure doesn’t effect the amount
of solid or liquid that dissolves They are incompressible. It does effect gases.
Pressure effects the amount of gas that can dissolve in a liquid.
The dissolved gas is at equilibrium with the gas above the liquid.
The gas is at equilibrium with the dissolved gas in this solution.
The equilibrium is dynamic.
If the pressure is increased the gas molecules dissolve faster.
The equilibrium is disturbed.
The system reaches a new equilibrium with more gas dissolved.
Henry’s Law.
P= kC
Pressure = constant x Concentration
of gas
Henry’s LawHenry’s Law
C C P C = kP P C = kP• C is concentration and P is partial pressure C is concentration and P is partial pressure
of gaseous solute of gaseous solute • The law is obeyed best by dilute solutions of The law is obeyed best by dilute solutions of
gases that don’t dissociate or react with gases that don’t dissociate or react with solventsolvent
•Amount of gas dissolved is directly Amount of gas dissolved is directly proportional to P of gas above solutionproportional to P of gas above solution
2
1
2
1
P
P
C
C
ExampleExample
A soft drink bottled at 25°C contains A soft drink bottled at 25°C contains COCO22 at pressure of 5.0 atm over at pressure of 5.0 atm over
liquid. Assume that Pliquid. Assume that PCO2CO2 in in
atmosphere is 4.0 x 10atmosphere is 4.0 x 10-4-4 atm. Find the atm. Find the equilibrium concentration in soda equilibrium concentration in soda before and after opening. k=32 before and after opening. k=32 L*atm/mol at 25°CL*atm/mol at 25°C
ExampleExample before opening:before opening:
after opening:after opening:
LmolmolatmL
atm
k
PCCkP CO
COCOCO /16.032
0.52
222
LmolmolatmL
atm
k
PC CO
CO /102.132
104 54
2
2
Example Solubility of pure N2 in blood at body
temp, 37oC and 1 atm is 6.2X10-4 M. If a diver breaths air ( = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of N2 in his body.
2N
.......22
ONtotalPPP
22 NtotNPP
atmatmPN
0.278.05.22
2
1
2
1
P
P
C
C
atm
atm
C
M
0.2
0.1102.6
2
4
M10 02.1blood) sdiver' in the N concof( -3
22C
3 .Temperature Effects (for aqueous solutions)
Increased temperature increases the rate at which a solid dissolves.
But it is not possible to predict whether it will increase the amount of solid that dissolves.
Solubility can be predicted only from a graph of experimental data.
20 40 60 80 100
Temperature effectsTemperature effects
dissolving a solid dissolving a solid occurs faster at occurs faster at higher Thigher T
but the amount able but the amount able to be dissolved to be dissolved does not changedoes not change
Temperature effectsTemperature effects
Solubility of Solubility of gas in water gas in water decreases decreases
with Twith T
Temperature and Solubility
Solid solubility and temperature
solubility increases with increasing temperature
solubility decreases with increasing temperature
12.4
Solubility and environment
Thermal pollution• Water used as a coolant when pumped again
into the source (lakes and rivers) floats on the cold water causing a decrease in solubility of O2 and consequently affecting the aquatic life.
CO2 dissolves in water that contains CO32-
causing formation of HCO3- that is soluble in
water. When temp increases CO2 will be driven off the water causing precipitation of CO3
2- again forming scales on the wools that blocks the pipes and reduce the heating efficiency
CO32- (aq) + CO2(aq) 2HCO3
-
11.411.4 The vapor pressures of solutionsThe vapor pressures of solutions A nonvolatile solvent lowers the vapor pressure
of the solution. For the molecules of the solvent
to escape, they must overcome the forces of both the other solvent
molecules and the solute molecules.
Nonvolatile solute decreases # of solvent molecules per unit volume. Thus, # of solute molecules escaping will be lowered.
Aqueous Solution
Pure water
Water has a higher vapor pressure than a solution
How a nonvolatile solute affects a solvent?
P1 Po1
Aqueous Solution
Pure water
Water molecules evaporate faster from pure water than from the solution
P1Po
1<
The water molecules condense faster in the solution so it should all end up there.
Aqueous Solution
Pure water
empty
Raoult’s LawRaoult’s LawThe presence of a nonvolatile solute lowers the vapor pressure of the solvent.
Psolution = Observed Vapor pressure of the solution (vapor pressure of solvent in solution)
P0solvent = Vapor pressure of the pure solvent
solvent = Mole fraction of the solvent
0
solventsolventsolutionPP
Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure (solute and solvent are alike: solute-solute, solvent-solvent and solute-solvent interactionsAre very similar ). Solute acts to dilute the solvent
0)1(solventsolutesolution
PP Only one solute
solvento
solutesosolvent
o PPP ln
solvento
solutePVPL
Vapor pressure lowering
Deviations If solvent has a strong affinity for solute (H bonding). Lowers solvents ability to escape. Lower vapor pressure than expected. Negative deviation from Raoult’s law. If Hsoln is large and negative (exothermic): strong
interaction exists between the solute and the solvent; thus a negtive deviation from Rault’s law because both components will have lower escaping tendency in the solution than in pure liquids.
If the two liquids mix endothermically (solute-solvent interactions are weaker than interactions among molecules in the pure liquids. More energy is required to expand liquids than is released when liquids are mixed). In this case molecules in the solution have higher tendency to escape than expected and positive deviations from Raul’s law are observed.
Liquid-liquid solutions in which both Liquid-liquid solutions in which both components are volatilecomponents are volatile
Modified Raoult's Law:Modified Raoult's Law:
00BBAABATOTAL PPPPP
P0 is the vapor pressure of the pure solvent PA and PB are the partial pressures
Nonideal solutions
Raoult’s Law – Ideal SolutionRaoult’s Law – Ideal Solution
A solution of two liquids that obeys Raoult’s Law is called an ideal solution
Negative Deviations from Raoult’s LawNegative Deviations from Raoult’s LawStrong solute-solvent interaction results in a vapor pressure lower than predicted
Exothermic mixing = Negative deviation
Positive Deviations from Raoult’s LawPositive Deviations from Raoult’s Law
Weak solute-solvent interaction results in a vapor pressure higher than predicted
Endothermic mixing = Positive deviation
PA = XA P A0
PB = XB P B0
PT = PA + PB
PT = XA P A0 + XB P B
0
Ideal Solution
An ______________
is one that obeys Raoult’s Law
PT is ______________ predicted by Raoults’s law
PT is ______________predicted by Raoults’s law
ForceA-B
ForceA-A
ForceB-B< &
ForceA-B
ForceA-A
ForceB-B> &
Vapor pressure lowering when ionic compounds are dissolved
VPL, when NaCl is dissolved in water, is twise as much as expected.
1 mol NaCl dissociates into 1 mol Na+ and 1 mol Cl- # mols of solute = 2 X # mols NaCl Thus, vapor pressure measurement can give information
about the nature of the solute When Na2SO4 is dissolved VPL is 3Xexpected
ExampleExample
Find the vapor pressure at 25°C for Find the vapor pressure at 25°C for solution of 158.0 g of sucrose solution of 158.0 g of sucrose (C(C1212HH2222OO1111) in 643.5 mL of water. ) in 643.5 mL of water.
The density of water at 25°C is 0.9971 The density of water at 25°C is 0.9971 g/mL and the partial pressure of g/mL and the partial pressure of water vapor at 25°C is 23.76 torr.water vapor at 25°C is 23.76 torr.
ExampleExample
solventsolventso PP ln
9872.0
3.3421
0.15802.18
119971.0
5.643
02.181
19971.0
5.643
gmol
gg
molmL
gmL
gmol
mLg
mL
water
torrtorrPso 46.23)76.23)(9872.0(ln
Colligative Properties of solutions
Colligative properties depend only on the number of solute particles (molecules or ions) in a solution
They do not depend on the identity of particles
11.5 Boiling point elevation and 11.5 Boiling point elevation and
freezing Point depressionfreezing Point depression
• Colligative properties include: 1. Vapor pressure lowering2. Boiling point elevation and freezing point depression3. Osmotic pressure
• Each of these properties is a consequence of a decrease in the escaping tendency of solvent molecules brought by the presence of solute particles.
Boiling Point Elevation
A nonvolatile solute lowers the vapor pressure
A higher T is required to reach the 1 atm of pressure which defines boiling point
A nonvolatile solute elevates the boiling point of the solvent
The amount of the elevation depends on concentration of the solute
solutebmKT
Freezing Point Depression Vapor pressure of solid and liquid are equal
at freezing point nonvolatile solute lowers the vapor pressure
so a lower T is needed to decrease the vapor pressure to that of the solid
a nonvolatile solute depresses the freezing point of the solvent
the amount of the depression depends on concentration of the solute
solutef mKT
1 atm
Vapor Pressure of solution
Vapor Pressure of pure water
1 atm
Freezing and boiling points of water
1 atm
Freezing and boiling points of solution
1 atm
TfTb
Boiling-Point Elevation
Tb = Tb – T b0
Tb > T b0 Tb > 0
T b is the boiling point of the pure solvent
0
T b is the boiling point of the solution
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point elevation constant (0C/m)
Freezing-Point Depression
Tf = T f – Tf0
T f > Tf0 Tf > 0
T f is the freezing point of the pure solvent
0
T f is the freezing point of the solution
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point depression constant (0C/m)
Molal Boiling-Point Elevation and Freezing-Point Depression Constants of Several Common Liquids
Example 1 18.00 g of glucose are added to 150.0 g of
water. The boiling point of the solution is 100.34 C. The boiling point constant is 0.51 C*kg/mol.
Find the molar mass of glucose.
Example 1
soluteb mKT
soluteif mmol
kgCTTT
)51.0(00.10034.100
kg
mol
molkgC
Cmsolute 67.0
/51.0
34.0
Example 1
waterkg 0.1500670 glucosen
kg
mol.
molkgkg
moln eglu 10.0)1500.0)(67.0(cos
molgmol
g
moles
mass/180
10.0
00.18massmolar
Example 2
What mass of C2H6O2 (M=62.1 g/mol) needs to be added to 10.0 L H2O to make a solution that freezes at -23.3°C? density is 1.00 g/mL; boiling point constant is 1.86°C*kg/mol
solutef mKT
soluteif mmol
kgCTTT
)86.1(0.03.23
kg
mol
molkgC
Cmsolute 5.12
/86.1
3.23
2623
262
262
2
2622 OHC g 107.76
OHC mol 1
OHC g 62.1
OH kg 1
OHC mol 12.5OH kg 10.0
kg 10.0g 1000
kg 1
mL 1.00
g 1
L 1
mL 1000OH L 10.0 2
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
m =moles of solute
mass of solvent (kg)= 2.41 m=
3.202 kg solvent
478 g x 1 mol62.01 g
Kf water = 1.86 0C/m
Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
Tf = T f – Tf0
Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C
Understanding osmoses: Experimental approach
Semipermeable membrane is a partition with pores that allow small solvent particles to pass through but not bigger solute particles. Thus, it separates a solution and a pure solvent
11.6 Osmoses and Osmotic Pressure11.6 Osmoses and Osmotic Pressure
Initial Final
Osmosis The flow of solvent molecules from
a pure solvent through a semi-permeable membrane into a solution
when the system has reached equilibrium, the water levels are different
Because the liquid levels are different, there is a greater hydrostatic pressurehydrostatic pressure on the solution than on pure solvent
BEFORE
AFTER
Osmotic pressure,
Osmotic pressure of a solutionOsmotic pressure of a solution: The minimum pressure that stops the osmosis
Osmotic PressureOsmotic Pressure
The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution
Osmotic pressure, , and concentration of nonelectrolytes
molarity of soultion PV = nRT (Ideal gas equation) Relation between and M is the
same: V = nRT
where • M is the molarity of the
solution• R is the gas law constant• T is the temperature in Kelvin
MRT RTV
n
Example When 1.00x10-3 g of a protein is mixed with
water to make 1.00 mL of solution, the osmotic pressure is 1.12 torr at 25.0°C.
Find the molar mass of this protein.
TRM
)298(8206.0760
112.1 K
Kmol
atmLM
torr
atmtorr
solution L 1
protein mol 106.01 5M
protein mol 106.01L 1
mol106.01
mL 1000
L 1soln mL 1.00 8
5
molgmol
g
moles
mass/1066.1
1001.6
1000.1massmolar 4
8
3
Osmosis and blood cells
hypotonicsolution
hypertonicsolution
isotonicsolution
Cell remains stable
0.95% NaCl & 0.31M glucose
0.95% NaCl & 0.31 M glucose
0.95% NaCl & 0.31M glucose
<0.95% NaCl <& 0.31 M glucose
<0.95% NaCl <& 0.31 M glucose
Cell swells and burst
Cell shrinks
HemolysesCrenation
Dialysis Separation of small species (molecules of solute and
solvent and ions) from big species in a solution by means of a semi permeable membrane
In the artificial kidney, the blood is circulated from the patient through cellophane tubes immersed in solution containing all essential ions and small molecules in blood at the appropriate concentrations.
Only waste products dialyze from blood through the membrane.
Functioning of an artificial kidney
Reverse osmosis
.
Reverse osmosis is the process of reversing the normal net flow of solvent molecules through a semipermeable membrane by applying to the solution a pressure exceeding the osmotic pressure
Reverse Osmosis
A net flow of Solvent molecules(blue) from solution to the solvent
Desalination (Water purification) is an application of reverse osmosis
Solutions of Electrolytes
The van’t Hoff factor (i) is used to modify the equations for colligative properties
FPD: Tf = –i Kfm BPE: Tb = i Kbm OP: = i M RT
EOS
i is dependent on solution molality
11.7 Collegative properties of electrolytes solutions
Since colligative properties only depend on the number of solute particles
Ionic compounds (salts) should have a bigger effect.
When they dissolve they dissociate. Individual Na+ and Cl- ions fall apart. 1 mole of NaCl makes 2 moles of ions. 1mole Al(NO3)3 makes 4 moles ions.
Electrolytes have a bigger impact on melting and freezing points per mole because they make more species.
Relationship is expressed using the
van’t Hoff factor i
The expected value can be determined from the formula of the salt.
dissolved particles solute of moles
solutionin particles of molesi
van’t Hoff Factor Observed i value is smaller than expected Ion pairing most affects i value for highly
charged ions affects colligative properties
The actual value (effect on colligative properties) is usually less because• At any given instant some of the ions in
solution will be paired.• Ion pairing tends to be higher for highly
charged ions• Ion pairing increases with concentration.• i decreases with concentration.
Thus, van’t Hoff factor should be added to the equations of collegative properties:
H = iKm
Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution 0.2 m ions in solution
van’t Hoff factor (i) = actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
nonelectrolytesNaCl
CaCl2
i should be
12
3
Dissociation Equations and Dissociation Equations and the Determination of the Determination of ii
NaCl(s)
AgNO3(s) MgCl2(s)
Na2SO4(s)
AlCl3(s)
Na+(aq) + Cl-(aq)
Ag+(aq) + NO3-(aq)
Mg2+(aq) + 2 Cl-(aq)
2 Na+(aq) + SO42-
(aq)Al3+(aq) + 3 Cl-(aq)
i = 2
i = 2
i = 3
i = 3
i = 4
Boiling-Point Elevation Tb = i Kb m
Freezing-Point Depression Tf = i Kf m
Osmotic Pressure () = iMRT
Colligative Properties of Electrolyte Solutions
Vapor pressure lowering, VPL ) ( solventpureo
solutePiVPL
Ideal vs. Real van’t Hoff FactorIdeal vs. Real van’t Hoff Factor
The ideal van’t Hoff Factor is only achieved in The ideal van’t Hoff Factor is only achieved in VERY DILUTEVERY DILUTE solution. solution.
Example 1 Osmotic pressure for 0.10 M solution of
Fe(NH2)2(SO4)2 at 25°C was 10.8 atm. Compare the van’t Hoff Factor observed and expected.
iexp= 1+2+2=5
iobs < iexp because of high ion pairing
4.4)298(08206.0)/10.0(
8.10
K
KmolatmL
Lmol
atm
MRTiobs
iMRT
Ch. 11 SolutionsCh. 11 Solutions
11.8 Colloids11.8 Colloids
ColloidsColloids
Tyndall EffectTyndall Effect scattering of light particles used to scattering of light particles used to
determine whether something is a determine whether something is a solution or suspensionsolution or suspension
ColloidColloid also called colloidal dispersionalso called colloidal dispersion suspension of tiny but visible particles in suspension of tiny but visible particles in
a mediuma medium
Types of Colloids Types of Colloids
ColloidsColloids
particles in a colloid are 1-1000nm in particles in a colloid are 1-1000nm in diameter and overall neutraldiameter and overall neutral
they have layer of negative ions on they have layer of negative ions on outsideoutside
so repel each otherso repel each other CoagulationCoagulation
when a colloid is destroyed through when a colloid is destroyed through heating or addition of electrolyteheating or addition of electrolyte