Properties of Solutions Chapter 11. Topics Concentration of solutions using different terms ...
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Transcript of Properties of Solutions Chapter 11. Topics Concentration of solutions using different terms ...
Properties of SolutionsProperties of Solutions
Chapter 11Chapter 11
TopicsTopics
Concentration of solutions using different Concentration of solutions using different termsterms
Solution process and solubilitySolution process and solubility Factors affecting the solubilityFactors affecting the solubility Vapor pressure of solutionVapor pressure of solution Collegative properties of solutionCollegative properties of solution
Boiling point elevationBoiling point elevation Freezing point depressionFreezing point depression Osmotic pressureOsmotic pressure
Section 11-8: Colloids-Self studySection 11-8: Colloids-Self study
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 solvent or solution.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
BA
AA nn
nAoffractionMole
M =
Molarity (M)
Molality (m)
m =moles of solute
mass of solvent (kg)
moles of solute
volume of solution (liters)
ExampleExample
1.00 g C1.00 g C22HH55OH is added to 100.0 g of water to OH is added to 100.0 g of water to
make 101 mL of solution. Find the molarity, make 101 mL of solution. Find the molarity, mass %, mole fraction and molality of mass %, mole fraction and molality of ethanol.ethanol.
MolarityMolarity
Number of moles of solute per L orNumber of moles of solute per L or
solutionsolution
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 PercentMass Percent
Also called weight percentAlso called weight percent Percent by mass of the solute in the solutionPercent 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 FractionMole Fraction
ratio of number of moles of a part ofratio of number of moles of a part of
solution to total number of moles of solutionsolution 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
MolalityMolality
Number of moles of solute per kg of solventNumber of moles of solute per kg of solvent
solvent of kg
solute of molesm
0.217m
solution1000g
1kg100.0g
OHHmolC102.17m 52
2
ExampleExample
An aqueous antifreeze solution is 40% An aqueous antifreeze solution is 40% ethylene glycol (Cethylene glycol (C22HH66OO22) by mass. The ) by mass. The
density of the solution is 1.05 g/cmdensity of the solution is 1.05 g/cm33. . Calculate the molality, molarity and Calculate the molality, molarity and mole fraction of ethylene glycolmole 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 solution= 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
• Any explanation for this behavior?
Solubility ProcessSolubility Process
The formation of aThe formation of aliquid solution takesliquid solution takesplace in 3 stepsplace in 3 steps
1. Expand solute1. Expand solutemoleculesmolecules
2. Expand solvent2. Expand solventmoleculesmolecules
3. Mixing solute and3. Mixing solute andsolventsolvent
Endothermic
Endothermic
Often Exothermic
Energy of Solubility ProcessEnergy of Solubility Process
Steps 1 and 2 require energy to overcome Steps 1 and 2 require energy to overcome intermolecular forces:intermolecular forces: endothermicendothermic
Step 3 usually releases energyStep 3 usually releases energy often exothermicoften exothermic
enthalpy of solutionenthalpy of solution sum of ∆H valuessum of ∆H values can be – or +can be – or + ∆∆HHsoln soln = ∆H= ∆H11 + ∆H + ∆H22 + ∆H + ∆H33
Energy of Solubility ProcessEnergy of Solubility Process
soln is negative
Case 1: oil and waterCase 1: oil and water
Oil is nonpolar (London forces)Oil is nonpolar (London forces) Water is polar (H-bonding)Water is polar (H-bonding) ∆∆HH11 will be small for typical molecular size will be small for typical molecular size ∆∆HH22 will be large will be large ∆∆HH33 will be small since there won’t be much will be small since there won’t be much
interaction between the twointeraction between the two ∆∆HHsolnsoln will be will be large and +velarge and +ve because because
energy required by steps 1 and 2 is larger energy required by steps 1 and 2 is larger than the amount released by 3than the amount released by 3
Case 2: NaCl and waterCase 2: NaCl and water
NaCl is ionic NaCl is ionic water is polar (H-bonding)water is polar (H-bonding) ∆∆HH11 will be large will be large
∆∆HH22 will be large will be large
∆∆HH33 will be large and –ve because of the will be large and –ve because of the
strong interaction between ions and waterstrong interaction between ions and water ∆∆HHsolnsoln will be close to zero- small but +ve will be close to zero- small but +ve
Enthalpy of hydrationEnthalpy of hydration - ∆H- ∆Hhydhyd
enthalpy change associated with dispersal enthalpy change associated with dispersal of gaseous solute species in waterof gaseous solute species in water
It combinesIt combines ∆H∆H22 (for expanding solvent) and ∆H (for expanding solvent) and ∆H3 3 (for (for
solvent-solute interaction)solvent-solute interaction)
NaCl(s) NaCl(s) Na Na++(g) + Cl(g) + Cl--(g)(g)
∆∆HH1 1 =786 kJ/mol=786 kJ/mol
HH22O(l) + NaO(l) + Na++(g) + Cl(g) + Cl--(g) (g) Na Na++(aq) + Cl(aq) + Cl--(aq)(aq)
∆∆HHhydhyd=∆H=∆H22 + ∆H + ∆H33=-783 kJ/mol=-783 kJ/mol
∆∆HHsolnsoln= ∆H= ∆H1 1 + ∆H+ ∆Hhydhyd = = 3 kJ/mol3 kJ/mol
Energy
Reactants
Solution
H1
H2
H3
Solvent
Solute and Solvent
Size of Size of HH33 determines whether a solution will form determines whether a solution will form
H3
Solution
∆∆HHsolnsoln is small but NaCl is highly soluble, Whyis small but NaCl is highly soluble, Why??
wo factors explains the solubility: wo factors explains the solubility:
1.1. n increase in the probability of mixing n increase in the probability of mixing favors the processfavors the process
2.2. Processes that require large amounts of Processes that require large amounts of energy tend not to occurenergy tend not to occur
If If HHsolnsoln is small and positive, a solution will still is small and positive, a solution will still
form because of high probability (entropy) for form because of high probability (entropy) for mixingmixing
There are many more ways for them to become There are many more ways for them to become mixed than there is for them to stay separate.mixed than there is for them to stay separate.
Energy of Solubility ProcessEnergy of Solubility Process
11 . .Structure EffectsStructure Effects
Water soluble molecules must have dipole moments Water soluble molecules must have dipole moments -polar bonds.-polar bonds.
To be soluble in non polar solvents the molecules To be soluble in non polar solvents the molecules must be non polar.must be non polar.
11.311.3 Factors Affecting SolubilityFactors Affecting Solubility
• Pentane C5H12 is misciblemiscible with hexane C6H14 and immiscibleimmiscible 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 bodyStructure effects: Vitamins and the body
Hydrophobic soluteHydrophobic solute: : water- fearingwater- fearing: : nonpolarnonpolar Insoluble in waterInsoluble in water
Hydrophilic soluteHydrophilic solute: : water-lovingwater-loving:: polarpolar Soluble in waterSoluble in water
Hydrophobic, accumulatesin the body. The body cantolerate a diet deficientin vitamin A
Hydrophilic, excreted bythe body and must be consumed regularly. The body cannot tolerate a diet deficientin vitamin C
22 . .Pressure effectsPressure effects
Changing the pressure Changing the pressure doesn’t effectdoesn’t effect the amount the amount of solid or liquid that dissolves; of solid or liquid that dissolves; they are incompressible.
Changing the pressure Changing the pressure effectseffects gases. gases. Pressure effects the amount of gas that can Pressure effects the amount of gas that can
dissolve in a liquid.dissolve in a liquid. The dissolved gas is at equilibrium with the gas The dissolved gas is at equilibrium with the gas
above the liquid.above the liquid.
The gas molecules above the The gas molecules above the liquid are at equilibrium with liquid are at equilibrium with the gas molecules dissolved the gas molecules dissolved in this solution.in this solution.
The equilibrium is dynamicThe equilibrium is dynamic..
If the pressure is increased If the pressure is increased the gas molecules dissolve the gas molecules dissolve faster.faster.
The equilibrium is disturbed.The equilibrium is disturbed.
The system reaches a new The system reaches a new equilibrium with more gas equilibrium with more gas dissolved.dissolved.
Henry’s Law.Henry’s Law. C= kPC= kPConcentration = Pressure X constantConcentration = Pressure X constant
Henry’s LawHenry’s Law
C C P P C = kPC = kP C is concentration and P is partial pressure of C is concentration and P is partial pressure of
gaseous solute gaseous solute The law is obeyed best by The law is obeyed best by dilute solutionsdilute solutions of of
gases that don’t dissociate or react with solventgases that don’t dissociate or react with solvent
•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
AA soft drink bottled at 25°C contains soft drink bottled at 25°C contains COCO22 at pressure of 5.0 atm over liquid. at pressure of 5.0 atm over liquid.
Assume that PAssume that PCO2CO2 in atmosphere is 4.0 x in atmosphere is 4.0 x
1010-4-4 atm. Find the equilibrium atm. Find the equilibrium concentration in soda before and after concentration in soda before and after opening. k=3.1 X 10opening. k=3.1 X 10-2-2 mol/L.atm at 25°C mol/L.atm at 25°C
ExampleExample
before opening:before opening:
after opening:after opening:
LmolXCPkCCOCOCO
/16.0)atm 5.0mol/L.atm(101.3 2
222
LmolatmxPkCCOCOCO
/102.1)104(mol/L.atm) (3.1x10 54-2
222
ExampleExample
Solubility of pure NSolubility of pure N22 in blood at body temp, in blood at body temp,
3737ooC and 1 atm is 6.2X10C and 1 atm is 6.2X10-4-4 M. If a diver M. If a diver breaths air ( = 0.78) at a depth where the breaths air ( = 0.78) at a depth where the total pressure is 2.5 atm, calculate the total pressure is 2.5 atm, calculate the concentration of Nconcentration of N22 in his body. 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
33 . .Temperature Effects (for aqueous solutions)Temperature Effects (for aqueous solutions)
Solids dissolve more rapidly at higher Solids dissolve more rapidly at higher temperaturestemperatures
But it is not possible to predict whether the But it is not possible to predict whether the solubility (amount dissolved) increases or solubility (amount dissolved) increases or decreasesdecreases
Solubility of most solids increases with Solubility of most solids increases with temperature but some solids showed a decrease temperature but some solids showed a decrease in solubility with temperaturein solubility with temperature
Temperature effect cannot be predicted; only by Temperature effect cannot be predicted; only by experiment experiment
Temperature effectsTemperature effects
dissolving a solid dissolving a solid occurs faster at occurs faster at higher Thigher T
but the amount to but the amount to be dissolved does be dissolved does not changenot change
Temperature effectsTemperature effects
Solubility of gas in Solubility of gas in water decreases water decreases
with Twith T
Solubility and EnvironmentSolubility and Environment
Thermal pollutionThermal pollution Water used as a coolant when pumped again into Water used as a coolant when pumped again into
the source (lakes and rivers) floats on the cold the source (lakes and rivers) floats on the cold water causing a decrease in solubility of Owater causing a decrease in solubility of O22 and and consequently affecting the aquatic life.consequently affecting the aquatic life.
COCO22 dissolves in water that contains CO dissolves in water that contains CO332-2- causing causing
formation of HCOformation of HCO33-- that is soluble in water. that is soluble in water.
COCO332-2- (aq) + CO (aq) + CO22(aq) 2HCO(aq) 2HCO33
--
When temp increases COWhen temp increases CO22 will be driven off the water will be driven off the water causing precipitation of COcausing precipitation of CO33
2-2- again forming scales again forming scales on the wools that blocks the pipes and reduce the on the wools that blocks the pipes and reduce the heating efficiency heating efficiency
11.411.4 The vapor pressures of The vapor pressures of solutionssolutions
A A nonvolatilenonvolatile solvent lowers the vapor pressure of solvent lowers the vapor pressure of the solution.the solution.
Nonvolatile solute Nonvolatile solute decreasesdecreases # of solvent molecules # of solvent molecules per unit volume. per unit volume.
Thus, # of solute molecules escaping will be loweredThus, # of solute molecules escaping will be lowered. .
Aqueous Solution
Pure water
Water has a higher vapor pressure than a solutionWater has a higher vapor pressure than a solution
Aqueous solution and pure ware in closed environment
P1 Po1
Aqueous Solution
Pure water
Water molecules evaporate faster from pure water than Water molecules evaporate faster from pure water than from the solutionfrom the solution
P1Po
1<
Water molecules condense faster in the solution so it Water molecules condense faster in the solution so it should all end up there.should all end up there.
Aqueous Solution
Pure water
empty
Vapor pressures of solutions containing nonvolatile solvents were studied by Raoult
RaoultRaoult’’s Laws Law
The 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
• This law applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure, i.e., • solute and solvent are alike: solute-solute, solvent-solvent and solute-solvent interactions are very similar ).
0)1(solventsolutesolution
PP Only one solute
solvento
solutesosolvent
o PPP ln
solvento
solutePVPL
Vapor pressure lowering
Vapor pressure lowering when ionic Vapor pressure lowering when ionic compounds are dissolvedcompounds are dissolved
When NaCl is dissolved in water, the VPL is twice as When NaCl is dissolved in water, the VPL is twice as much as expected.much as expected.
1 mol NaCl dissociates into 1 mol Na1 mol NaCl dissociates into 1 mol Na++ and 1 mol Cl- and 1 mol Cl- # mols of solute = 2 X # mols NaCl# mols of solute = 2 X # mols NaCl Thus, vapor pressure measurement can give Thus, vapor pressure measurement can give
information about the nature of the soluteinformation about the nature of the solute When NaWhen Na22SOSO44 is dissolved VPL is is dissolved VPL is 3 x expected
ExampleExample
Find the vapor pressure at 25°C for Find the vapor pressure at 25°C for solution of 158.0 g of sucrose (Csolution of 158.0 g of sucrose (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 water g/mL and the partial pressure of water vapor at 25°C is 23.76 torr.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
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 resulting from A
& B in the vapor above the solution
Nonideal solutions
RaoultRaoult’’s Law s Law –– Ideal Solution Ideal Solution
A solution of two liquids that obeys Raoult’s Law is called an ideal solution
Negative Deviations from RaoultNegative Deviations from Raoult’’s Laws Law
Strong solute-solvent interaction results in a vapor pressure lower than predicted
Exothermic mixing = Negative deviation
Positive Deviations from RaoultPositive Deviations from Raoult’’s Laws Law
Weak solute-solvent interaction results in a vapor pressure higher than predicted
Endothermic mixing = Positive deviation
Colligative Properties of solutionsColligative Properties of solutions
Colligative properties are properties that depend only on the number of solute particles (molecules or ions)(molecules or ions) in solution and not on the nature (identity) of the solute particles.
11.5 Boiling point elevation and freezing Point depression11.5 Boiling point elevation and freezing 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 ElevationBoiling Point Elevation
A nonvolatile solute A nonvolatile solute lowers the vaporlowers the vapor pressurepressure
A higher TA higher T is required to reach the 1 atm of is required to reach the 1 atm of pressure which defines boiling pointpressure which defines boiling point
A nonvolatile solute A nonvolatile solute elevates the boilingelevates the boiling point of the solventpoint of the solvent
The amount of the elevation depends on The amount of the elevation depends on concentration of the soluteconcentration of the solute
solutebmKT
Freezing Point DepressionFreezing Point Depression
Vapor pressure of solid and liquid are equal at Vapor pressure of solid and liquid are equal at freezing pointfreezing point
nonvolatile solute lowers the vapor pressure so a nonvolatile solute lowers the vapor pressure so a lower T is needed to decrease the vapor pressure to lower T is needed to decrease the vapor pressure to that of the solidthat of the solid
a nonvolatile solute depresses the freezing point of a nonvolatile solute depresses the freezing point of the solventthe solvent
the amount of the depression depends on the amount of the depression depends on concentration of the soluteconcentration 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
Example 1Example 1
18.00 g of glucose are added to 150.0 g of 18.00 g of glucose are added to 150.0 g of water. The boiling point of the solution is water. The boiling point of the solution is 100.34 C. The boiling point constant is 0.51 100.34 C. The boiling point constant is 0.51 C*kg/mol. C*kg/mol.
Find the molar mass of glucose.Find the molar mass of glucose.
Example 1Example 1
soluteb mKT
soluteif mmol
kgCTTT
)51.0(00.10034.100
kg
mol
molkgC
Cmsolute 67.0
/51.0
34.0
Example 1Example 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 2Example 2
What mass of CWhat mass of C22HH66OO22 (M=62.1 g/mol) needs (M=62.1 g/mol) needs
to be added to 10.0 L Hto be added to 10.0 L H22O to make a solution O to make a solution
that freezes at -23.3°C? density is 1.00 that freezes at -23.3°C? density is 1.00 g/mL; boiling point constant is 1.86°C*kg/molg/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
Experimental approachExperimental approach
Semipermeable membraneSemipermeable membrane It is a partition with pores that allow small solvent particles to pass It is a partition with pores that allow small solvent particles to pass
through but not bigger solute particles. Thus, it separates a solution through but not bigger solute particles. Thus, it separates a solution and a pure solventand a pure solvent
11.6 Osmoses and Osmotic Pressure11.6 Osmoses and Osmotic Pressure
Initial Final
OsmosisOsmosis
The flow of solvent molecules from The flow of solvent molecules from a pure solvent through a semi-a pure solvent through a semi-permeable membrane into a permeable membrane into a solutionsolution
when the system has reached when the system has reached equilibrium, the water levels are equilibrium, the water levels are differentdifferent
Because the liquid levels are Because the liquid levels are different, there is a greater different, there is a greater hydrostatic pressurehydrostatic pressure on the on the solutionsolution than on than on pure solventpure solvent
After
Before 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, Osmotic pressure, ,, and concentration of and concentration of nonelectrolytesnonelectrolytes
molarity of soultionmolarity of soultion PV = nRT (Ideal gas equation)PV = nRT (Ideal gas equation) Relation between Relation between and M is the and M is the
same: same: V = nRTV = nRT
where where M is the molarity of the solutionM is the molarity of the solution R is the gas law constantR is the gas law constant T is the temperature in KelvinT is the temperature in Kelvin
MRT RTV
n
ExampleExample When 1.00x10When 1.00x10-3-3 g of a protein is mixed with g of a protein is mixed with
water to make 1.00 mL of solution, the water to make 1.00 mL of solution, the osmotic pressure is 1.12 torr at 25.0°C.osmotic pressure is 1.12 torr at 25.0°C.
Find the molar mass of this protein. 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
DialysisDialysis
Separation of small species (molecules of solute and Separation of small species (molecules of solute and solvent and ions) from big species in a solution by solvent and ions) from big species in a solution by means of a semi permeable membranemeans of a semi permeable membrane
In the artificial kidney, the blood is circulated from In the artificial kidney, the blood is circulated from the patient through cellophane tubes immersed in the patient through cellophane tubes immersed in solution containing all essential ions and small solution containing all essential ions and small molecules in blood at the appropriate molecules in blood at the appropriate concentrations.concentrations.
Only waste products dialyze from blood through the Only waste products dialyze from blood through the membranemembrane. .
Functioning of an artificial kidneyFunctioning of an artificial kidney
Reverse osmosisReverse 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 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 OsmosisReverse OsmosisA net flow of Solvent molecules(blue) from solution to the solvent
Desalination (Water purification) is an application of reverse osmosis
11.7 Collegative properties of electrolytes solutions11.7 Collegative properties of electrolytes solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Since colligative properties only depend on the Since colligative properties only depend on the number of solute particlesnumber of solute particles
Ionic compounds (salts) should have a bigger Ionic compounds (salts) should have a bigger effect.effect.
When they dissolve they dissociate.When they dissolve they dissociate. NaCl dissociate into NaNaCl dissociate into Na++ and Cl and Cl-- ions ions 1 mole of NaCl makes 2 moles of ions.1 mole of NaCl makes 2 moles of ions. 1mole Al(NO1mole Al(NO33))33 makes 4 moles ions. makes 4 moles ions.
Electrolytes have a bigger impact on melting and Electrolytes have a bigger impact on melting and freezing points per mole because they make more freezing points per mole because they make more species.species.
Relationship is expressedRelationship is expressed using theusing the
van’t Hoff factorvan’t Hoff factor i i
The expected value can be determined from the formula The expected value can be determined from the formula of the salt.of the salt.
dissolved particles solute of moles
solutionin particles of molesi
van’t Hoff Factorvan’t Hoff Factor
Observed Observed ii value is smaller than expected value is smaller than expected Ion pairing most affects Ion pairing most affects ii value for highly charged value for highly charged
ionsions affects colligative propertiesaffects colligative properties
The actual value (effect on colligative The actual value (effect on colligative properties) is usually less becauseproperties) is usually less because
At any given instant some of the ions in solution At any given instant some of the ions in solution will be paired.will be paired.
Ion pairing tends to be higher for highly charged Ion pairing tends to be higher for highly charged ionsions
Ion pairing increases with concentration.Ion pairing increases with concentration. ii decreases with concentration. decreases with concentration.
Thus, van’t Hoff factor should be added to Thus, van’t Hoff factor should be added to the equations of collegativethe equations of collegative properties: properties:
H = H = iiKKmm
Dissociation Equations and theDissociation Equations and the Determination of 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
ExampleExample Osmotic pressure for 0.10 M solution of Fe(NHOsmotic pressure for 0.10 M solution of Fe(NH22))22(SO(SO44))22
at 25°C was 10.8 atm. Compare the van’t Hoff Factor at 25°C was 10.8 atm. Compare the van’t Hoff Factor observed and expected.observed and expected.
iiexpexp= 1+2+2=5= 1+2+2=5
iiobsobs < < iiexpexp because of high ion pairing because of high ion pairing
4.4)298(08206.0)/10.0(
8.10
K
KmolatmL
Lmol
atm
MRTiobs
iMRT