# Solutions - Winona State University

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```					                           Properties of Solutions

Terminology

Solvation                          Compositions                      Solubility Rules

Heats of Solution

Structure/Intermolecular Forces               Temperature                          Pressure

Henry's Law              Vapor Pressures             Raoult's Law

Colligative Properties

B. pt. Elevation    Fr. pt. Depression      Osomotic Pressure
Terminology

• Solution = A homogeneous mixture.
• Solute = a substance dissolved in a solvent
to form a solution; usually the smaller
portion.
• Solvent = The dissolving medium of a
solution; usually the greater portion.
• Solubility = Amount of substance dissolved.
• Dilute , Concentrated , Saturated .
The Solution Process
Figure 13.2: The Solution Process
Hydration or Solvation
Figure 13.3:
Three Steps of
Solution Formation
The Solution Process
Solution Formation, Spontaneity, and
Disorder
• If the process leads to a greater state of disorder, then
the process is spontaneous.
• Example: a mixture of CCl4 and C6H14 is less ordered
than the two separate liquids. Therefore, they
spontaneously mix even though Hsoln is very close to
zero.
• There are solutions that form by physical processes and
those by chemical processes.
Figure 13.6: The Solution Process
Solution Formation, Spontaneity, and
Disorder
Factors Affecting Solubility

Solute-Solvent Interaction
• Miscible liquids: mix in any proportions.
• Immiscible liquids: do not mix. [ Oil and Vinegar ]
• Generalization: “like dissolves (in) like”.

• Polar liquids tend to dissolve in polar solvents.
Factors Affecting Solubility

Solute-Solvent Interaction
Ways of Expressing Concentration

Mass Percentage
• All methods involve quantifying amount of solute per
amount of solvent (or solution).
• Generally amounts or measures are masses, moles or
liters.

mass of component in solution
mass % of component                                100
total mass of solution
Ways of Expressing Concentration
• Mole Fraction, Molarity, and Molality

moles of component in solution
Mole fraction of component 
total moles of solution

moles solute
Molarity 
liters of solution

moles solute
Molality, m 
kg of solvent
Solution Compositions
• s = solute ; A = solvent; V = Tot. Vol. of solution.
• Weight %:                          ws
ws %              x 100
ws  wA
• Mole Fraction:
ns
s 
ns  nA
• Molarity:
ns
Ms 
V

• Molality:
ns
ms 
kg A
Example of Solution Compositions
• A solution is prepared by mixing 78.9 g of ethanol
(C2H5OH) with 100.0 g of water to give 190.5 mL of
solution. Calculate the solution compositions.

• The electrolyte in automobile lead storage batteries is
a 3.75 M H2SO4 solution that has a density of 1.230
g/mL. Calculate mass %, molality, and mole fraction
in terms of H2SO4 .
• [Hint: Assume exactly one liter of solution.]
• [Answers: 29.9% , 4.35 molal, 0.0727 ]
s = ethanol (solute);      A = water (solvent);

78 .9 g
ws %             x 100  44 .1%
178 .9 g

 78 .9 
        
s 
ns
       46 .07 

1.71
 0.236
ns  n A  78 .9 100 .0  1.71  5.55
 46 .07  18 .02 
                 

ns 1.713 mol
Ms              8.99 mol L1
V 190 .5 mL

ns     1.713 mol
ms                         17 .1 mol kg 1
kg A 0.1000 kg water
Exam ple 2 (Solution Com positions)

The e lectrolyte in autom obile le ad s torage batterie s is a 3.75 M HSO 4 solution that has a
2
-1
de nsity of 1.230 g m L . Calculate w eight%, m olality, and m ole fraction in te rm s of H SO 4 .
2

Hint: As sum e exactly one liter of solution in your calculations.
De fine:     s = H2 SO 4     A=H                              2O

In one liter of solution, from 3.75 mol/L solution, there are 3.75 moles of H       2 SO 4   .
In one liter of solution, from d=1.230 g/mL, there are 1230 grams of solution.

Weight % of H2 SO 4 Calculation:

  98.07gm 

  
3.75mol            368gm H2SO 4
 1 mol 

368 gm
Wt %                  100           Wt %  29.9

1230gm

Molality of H2 SO 4 Calculation :

In one liter of solution => 1230 gm soln - 368 gm H          2 SO 4   = 862 gm H 2 O


n s  3.75mol                            
kgA  0.862kg

ns                                  mol
ms                             ms  4.35
kgA                                  kg

Mole Fraction of H2 SO 4 Calculation :

862
n A          mol
18.015
ns
 s                              s  0.0727
ns  nA
Concentrations of Solutions

In the Dilution process of a more concentrated
solution:
• The number of moles are the same in diluted and
concentrated solutions.
• So:
MdiluteVdilute = moles = MconcentratedVconcentrated
General Properties of Aqueous Solutions

Electrolytic Properties
• Three types:
• Strong electrolytes,
• Weak electrolytes, and
• Nonelectrolytes.
General Properties of Aqueous Solutions

Strong and Weak Electrolytes
• Strong electrolytes: completely dissociated in solution.
e.g.
HCl(aq)           H+(aq) + Cl-(aq)

• Weak electrolytes: produce small concentration of ions
when dissolved.
• e.g.

HC2H3O2(aq)             H+(aq) + C2H3O2-(aq)
Precipitation Reactions
Precipitation Reactions

Exchange (Metathesis) Reactions
• Metathesis reactions involve swapping ions in solution:
AX + BY  AY + BX.
Precipitation Reactions
Ionic Equations
• Ionic equation: used to highlight reaction between ions.

• Molecular equation: all species listed as molecules:
AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq)

• Complete ionic equation(CIE): lists all ions:

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq)  AgI(s) + Na+(aq) + NO3-(aq)

• Net ionic equation: cancel spectator ions from CIE
Ag+(aq) + I-(aq)  AgI(s)
Figure 13.14: Factors Affecting Solubility

Intermolecular Forces

Pressure

Temperature
Figure 13.14: Factors Affecting Solubility

Pressure Effects
Factors Affecting Solubility

Pressure Effects
• If Sg is the solubility of a gas, k is a constant, and Pg is
the partial pressure of a gas, then Henry’s Law gives:
S g  k  Pg
• Carbonated beverages are bottled with a partial pressure
of CO2 >1 atm, ( pressure inside can ~4 atm above liq).
• As bottle is opened, partial pressure of CO2 decreases and
solubility of CO2 decreases.
• Therefore, bubbles of CO2 escape from solution.
CyberChem Diving Gases
Factors Affecting Solubility

Temperature Effects: Solids in Liquids
• Generally, as temperature increases, solubility of solids
generally increases, BUT
• Sometimes, solubility decreases as temperature increases
(e.g. Ce2(SO4)3).
Figure 13.17
Factors Affecting Solubility

Temperature Effects: Gases in Liquids
• Gases get less soluble as temperature increases.

• Thermal pollution: if lakes get too warm, CO2 and O2
become less soluble and are not available for plants or
animals.
Figure 13.18
Colligative Properties

• Colligative properties depend on quantity of solute
molecules. (e.g. freezing point depression and boiling
point elevation.)
Lowering Vapor Pressure
• Non-volatile solutes reduce the ability of the surface
solvent molecules to escape the liquid.
• Therefore, vapor pressure is lowered.
• The amount of vapor pressure lowering depends on the
amount of solute.
Figure 11.22: Vapor Pressure
Vapor Pressure on the Molecular Level
Figure 11.24
Figure 11.26: Phase Diagrams
Figure 11.27: Phase Diagrams

The Phase Diagrams of H2O and CO2
Figure 13.20: Colligative Properties

Lowering Vapor Pressure
Colligative Properties

Lowering Vapor Pressure
• Raoult’s Law:
PA   A  P A
• Where: PA = vapor pressure with solute,
• PA = vapor pressure without solute (pure
solvent), and
• A = mole fraction of A.
Vapor Pressure Examples
• Calculate the expected vapor pressure at 25oC for a
solution prepared by dissolving 158.0 g of common
table sugar (sucrose – MW=342.3) in 643.5 mL of
water. At 25oC, the density of water is 0.9971 g/mL
and the vapor pressure is 23.76 torr.
PA   A  P A
• A solution was prepared by adding 20.0 g of urea to
125 g of water at 25oC, a temperature at which pure
water has a vapor pressure of 23.76 torr. The
observed vapor pressure of the solution was found to
be 22.67 torr. Calculate the molecular weight of
urea.
Example 1:            Sugar in Water
1
s = sugar             A = w ater           ws  158.0
gm                            
MW s  342.3gm mol
3                               3                                                            1

VA  643.5cm                            
d A  0.9971gm cm                     
PoA  23.76t orr                      
MW A  18.015gm mol

VA  d A
643.5cm  
0.9971gm 
          
3                          1 mol
n A 
MW A
n A  35.62mol         nA         
        3

  18.015gm 
   
      cm        
ws
158.0gm               
n s                            n s  0.4616mol                                   1 mol
MW s                                             ns                            
 342.3gm 


nA
 A                             A  0.9872                          35.62                35.62
ns  nA                                         A
0.4616 35.62            36.08

PA   A  PoA                PA  23.46t orr             PA                
0.9872( 23.76t orr)
Example 2:               Urea Molecular Weight

•                                                                                              o
A solution was prepared by adding 20.0 g of urea to 125 g of water at 25 C, a temp erature
at which pure water has a vapor pressure of 23.76 torr. T he observed vapor p ressure of the
solution was found to be 22.67 torr. Calculate the molecular weight of urea.

s = urea                 A = w ater


ws  20.0gm                 wA  125 gm                         
PoA  23.76t orr                       
PA  22.67t orr

PA                                                        1
PA      A  PoA                   A                  A  0.9541                         
MW A  18.015gm mol
PoA

nA                               wA
A                                n A                        n A  6.939mol
ns  nA                             MW A

nA                 solving for n has solution of:
s                                      A  1
A                                                                       n A 
ns  nA                                                                   A

n s 

nA  1   A                     n s  0.334mol
A

ws                               ws                                    1
ns                               MW s                       MW s  60gm mol
MW s                              ns
Figure 13.22
Colligative Properties

Boiling-Point Elevation
• Molal boiling-point-elevation constant, Kb, expresses
how much Tb changes with molality, mS :

Tb  K b  mS
• Decrease in freezing point (Tf) is directly proportional
to molality (Kf is the molal freezing-point-depression
constant):
Tf  K f  mS
Colligative Properties
Applications of Colligative Properties
• A solution was made by dissolving 18.00 g of glucose
in 150.0 g of water. The resulting solution was found to
have a boiling point of 100.34oC. Calculate the
molecular weight of glucose.
Tb  K b  mS
• How many kg of ethylene glycol (C2H6O2, MW=62.1),
antifreeze, must be added to 10.0 L of water to produce
a solution for use in a car’s radiator that freezes at
–23.3oC? Assume that the density of water is 1.00
g/mL.
[Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)]
• A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution
was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.

Tb  K b  mS
How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L
of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume
that the density of water is 1.00 g/mL.
Tf  K f  mS
Free zing Point De pre ssion Example

•      How many kg of ethylene glycol (C2H6O2, M W=62.1), antifreeze, must be added to 10.0 L
of water to produce a solution for use in a car’s radiator that freezes at           –23.3 oC? Assume
that the density of water is 1.00 g/mL.
[Answer:         7.78 kg (with d = 1.18 g/mL => 6.59 L)]

Let:     s = ethylene glycol            A=water

 1 gm    1 mL    10  kg 
3
1
Kf  1.86K kg mol
                          Tf  23.3       kg_A  10.0                                 
 1 mL  10 3 L   1 gm 
K                  L
       
1
kg_A  10.0kg                                
MW s  62.1gm mol

 ns                ws       
Tf    Kf ms   Kf            Kf              
kg_A           MW s  kg_A 

Tf MW s  kg_A
ws 
Kf

1
      
23.3K 62.1 gm mol            
 10.0 kg
ws
1                                  ws  7.78kg

1.86K kg mol

1                     ws

d s  1.18gm mL                    Vs               Vs  6.59L
ds
Colligative Properties
Osmosis
• movement of a solvent from low solute concentration to
high solute concentration across a semipermeable
membrane.

Figure 13.23
Colligative Properties

Osmosis
• Osmotic pressure, , is the pressure required to stop
osmosis:
 V  n  R  T
n
     R T
V 

  M S  R T
Application of Osmotic Pressure

• To determine the molecular weight of a
certain protein, 1.00 mg of it was dissolved
in enough water to make 1.00 mL of
solution. The osmotic pressure of this
solution was found to be 1.12 torr at 25.0oC.
Calculate the molecular weight of the
protein.

  M S  R T
To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to
make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 oC.
Calculate the molecular weight of the protein.
  M S  R T
Colligativ e Prope rties: Osmotic Pressure Example

•      To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in
enough water to make 1.00 mL of solution. T he osmotic pressure of this solution was found
to be 1.12 torr at 25.0 oC. Calculate the molecular weight of the protein.

Def ine:              s = protein         A = w ater

3                        3                                                                          1     1

ws  1.0010        gm               
V  1.0010    L                 1.12 orr
t                                
R  0.08206L at m mol       K

  1.12t orr            
1at m                                 3
T  ( 25.0  273.15 K
)                                                                      1.474 10        at m
 760 t orr 

 ns       ws 
   M s  R T      R T             R T
V         MW s  V


The only unknow n is the molecular w eight of the protein ( MW s ).

MW s 
ws  R T
MW s 

1.0010
3            
 gm 0.08206L at m mol
1
K
1     298.15K

 V

1.47410
3            
3
at m 1.0010    L  

4          1
MW s  1.66 10 gm mol
Colligative Properties

Osmosis
• Isotonic solutions: two solutions with the same 
separated by a semipermeable membrane.

• Osmosis is spontaneous.
• Red blood cells are surrounded by semipermeable
membranes.
Figure 13.25: Colligative Properties
Osmosis

Crenation: hypertonic solution     Hemolysis: hypotonic solution

IV(intravenous) fluids must be Isotonic.
CyberChem: Desalination
Colloids

• Colloids are suspensions in which the suspended particles
are larger than molecules but too small to drop out of the
suspension due to gravity.
• Particle size: 10 to 2000 Å.
• Tyndall effect: ability of a Colloid to scatter light. The
beam of light can be seen through the colloid.
Colloids
Figure 13.31: Colloids
Properties of Solutions
Terminology

Solvation                          Compositions                      Solubility Rules

Heats of Solution

Structure/Intermolecular Forces               Temperature                          Pressure

Henry's Law              Vapor Pressures            Raoult's Law
PA   A  PA
o

Colligative Properties                                      nA
A 
nS  nA
B. pt. Elevation    Fr. pt. Depression      Osomotic Pressure
nS
mS 
kg A                  Tb, f  Kb, f  mS                        M S  R T           MS 
nS
V

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