AP Chemistry - Colligative Properties of Solutions
Electrolytes and Non-Electrolytes
Electrolytes are dissolved ions. These allow the flow of electricity between a positive anode and a negative cathode in a wet cell battery.  Aqueous solutions of ionic compounds such as NaCl or CaCl2 are able to conduct electricity. 
Not all chemical compounds behave this way, however. If sugar is dissolved in water the solution that results is not electrically conductive.  It is not really difficult to imagine why this happens.  In order for a solution to conduct electricity there must be electrical charges in the water that are able to move.  When an ionic compound is dissolved in water, the ions that are packed together in the solid become
separated. This is called dissociation.   As the ions enter the solution they become surrounded by water molecules that are attached to them.  This process is called hydration and we say that the ions become hydrated.  These hydrated ions are then able to migrate about in the solution and behave more or less independently, and it is this ability to move that accounts for the electrical conductivity of the solution.   When a molecular substance like sugar which is made from  covalent bonds, the molecules disperse through the water but the molecules stay intact.  There are no charged particles in the solution, so the solution is a nonconductor.
The ability to produce a conducting solution is a special property of a solute -- a property that is given a special name.  Any solute that dissolves in water to give a solution that contains ions, and which therefore conducts electricity, is called an electrolyte.  On the other hand, a solute that remains undissociated in solution is called a nonelectrolyte.

Dissociation Reactions
When a solute dissolves, the ionic bonds holding it together break down and the solid becomes individual ions.  For example:

                      NaCl(s) -------->  Na+(aq)  +  Cl-(aq)
 

The aq, which stands for aqueous, means that the ions are dissolved in water and are hydrated, and by writing the formulas of the ions separately we mean that in the solution the ions are essentially independent of each other.  A similar equation that represents what occurs when calcium chloride, CaCl2(s), dissolves in water is:

                CaCl2(s) ------->  Ca2+(aq)  +  2 Cl-(aq)
 

which shows that two Cl- ions are present in the solution for each Ca2+. Once again we have used s and aq in parentheses to point out the changes that take place.
Salts that contain polyatomic ions also dissociate in aqueous solutions, and it is important to remember that in the solution the polyatomic ions remain intact. For instance, when the salt sodium sulphate, Na2SO4, dissolves in water, the solution contains sodium ions and sulphate ions.

                   Na2SO4(s)  -------->  2 Na+(aq)  +  SO42-(aq)
 

You can see that it is especially important that you know both the formulas and charges of the polyatomic ions to write equations such as these properly.
Often, in writing chemical equations for dissociation reactions or for other reactions involving ions in solution, the symbols (s) and (aq) following the formulas are omitted.  When we're discussing reactions in an aqueous solution, they are "understood."  Therefore, you should not be fooled when you see an equation such as

          CaCl2    ---------->   Ca2+   +  2 Cl-1
 

Unless something is said to the contrary, it means

         CaCl2(s)   ---------->   Ca2+(aq)  +  2 Cl-1(aq)

 

Strong and Weak Electrolytes
Some compounds when dissolved in water dissolve completely.  These compounds are considered to be strong electrolytes.   They are strong because they dissolve 100% of the time.  Some examples are HCl, or NaOH.
Other compounds do not dissolve very well in water. Vinegar or acetic acid dissolves in water. But it does not dissociate completely.  Actually vinegar only dissociates roughly 1.8%.  Therefore only 1.8% of the total amount of vinegar put into the water actually breaks down into ionic form.

                  CH3COOH  --------->  CH3COO-  +    H+
before             100%                            0                 0
after                98.2%                         1.8%           1.8%

Go to the Dissociation Equations of Electrolytes Worksheet


Ionic Equations
In writing a chemical equation to describe an ionic reaction, we have several options. Consider this reaction.  When 100 mL of a 1.0 M cadmium nitrate, Cd(NO3)2, is added to 100 mL of a 1.0 M solution of sodium sulphide, Na2S, a bright orange-yellow solid precipitates as the two solutions mix.  The solid can be shown to be cadmium sulphide, CdS and if we filter the mixture and evaporate the filtrate, crystals of sodium nitrate are left behind.  From this equation we can write the chemical reaction:
          Cd(NO3)2(aq)  +  Na2S(aq)  ------>   CdS(s)   +    2 NaNO3(aq)
This is a molecular reaction because we have written it as complete molecules.
A more accurate description of a reaction as it actually takes place is provided by an ionic equation.  In an ionic equation the formula of any strong electrolytes are written in dissociated form to show that the solute exists in the solution in the form of separated ions.   On the other hand, the formulas of solids and weak electrolytes are written in molecular form. For a solid, this shows that the ions are not free to move away from each other, but instead are forced to stay together within the crystals of the solid.   For a weak electrolyte, writing the molecular formula shows that this is the predominant form of the solute in solution.  The ionic equation for the reaction that we've been discussing is
 Cd2+(aq)  + NO3-(aq)  +  2 Na+(aq)  +  S2-(aq)  ---->   CdS(s)   +   2 Na+(aq)  +  2 NO3-(aq)
To obtain the ionic equation we have divided each of the soluble, ionic compounds into the ions that are released when the salts dissolve, but we have left intact the formula of the solid.
An ionic equation, like any other equation, should be balanced, and this one is.  It satisfies, two criteria for a balanced ionic equation.
1.  Material balance:  All atoms on one side of the arrow must appear somewhere on the other side.
2.  Electrical balance: The net charge on the left equals the net charge on the right.

Net Ionic Equations
If you pay careful attention to the above reaction you'll see that two of the ions appear in exactly the same way on both sides of the equation.  They are in this case the nitrate, and sodium ions. This tells us that during the reaction, these ions are left unaffected. They are present before the reaction starts and they are still there after the reaction is over.  They do not actually participate in the formation of the CdS.  They are called spectator ions.   They are there only to carry the cadmium and sulphide ions.  A net ionic reaction is a reaction written without the spectator ions being shown. A typical net reaction is shown below.

Cd2+(aq)  + NO3-(aq)  + 2 Na+(aq)  +  S2-(aq
                                             ---->   CdS(s)   + 2 Na+(aq)  + 2 NO3-(aq)


resulting in:             Cd2+(aq)   +  S2-(aq)  ---->   CdS(s)
Go to the Molecular, Ionic and Net Ionic Worksheet

Colligative Properties of Solutions
Colligative properties are the physical properties of solutions.  These are properties that depend upon the relative amounts of solute and solvent in the solution mixture and not their chemical identities. 

Effect of Pressure on the Solubilities of Gases - Henry's Law
When the temperature is constant, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas on the solution.
A carbonated soft drink is a solution of carbon dioxide in water.  It is bottled under pressure so that more CO2 will be dissolved, because the solubility of any gas in any liquid is increased by pressure.  When the bottle of soda is opened, the pressure on the liquid drops back to atmospheric pressure, so the carbon dioxide fizzes out.  In a short time, the soda is "flat", because very little CO2 will remain in solution.
Any other carbonated drink, beer or even champagne will go flat for the same reason when opened.  If the beverages are warm they will go flat even faster because gases are even less soluble at higher temperatures.

Air is not very soluble in water under ordinary pressures, but at elevated pressures both the oxygen and nitrogen in the air become increasingly soluble.  This results in some dangerous implications for those people who must work or play under higher than normal pressures. 

When people work in a space where the air pressure is much above normal, they have to be careful to return slowly to the atmosphere.  Otherwise, they face the danger of the "bends."  The bends are severe pains in the joints and muscles, fainting, possible deafness, paralysis, and death.  Workers building deep tunnels, where higher  than normal air pressure is maintained to keep water out, are at risk, so are deep-sea divers and scuba divers who stay down for to prolonged a period. 

The bends develop because the solubilities of both nitrogen and oxygen are higher under higher pressure, as Henry's Law states.  Once the blood is enriched in these gases it  must not be allowed to lose them suddenly.  If the person returns to normal atmospheric pressure to rapidly the blood will act like a soda pop bottle and fizz.   If microbubbles of nitrogen and oxygen appear at blood capillaries, they will block the flow of blood.  Such a loss is particularly painful at joints, and any reduction in blood flow to the brain can be extremely serious.

For approximately each 10 m of depth that a diver descends, another atmosphere of pressure is exerted on their body.  For each atmosphere of pressure, about 20 minutes of careful decompression is usually recommended.  This allows time for the respiratory system to gather and expel excess nitrogen, which can't leave any other way.  Excess oxygen can be used up by normal metabolism.
 

It is easy to understand why the solubility of any gas in a liquid should increase with pressure.   The relevant equilibrium is
 
gas  +  solvent <====>  solution


Suppose we've established this equilibrium and then increase the pressure (by reducing the volume available to the gas, for example.  Le Chatelier's principle says that the system will attempt to change in order to counteract this "stress" and bring the pressure down somewhat.  This can happen if some molecules leave the pas phase and enter the solution.  There will be fewer molecules in the gas phase, and these fewer molecules exert less pressure.  The converse os this is also true.  By reducing the pressure on the gas above the solution, by increasing the volume,  the system will release dissolved gas from the solution to reach a new equilibrium. 
 

For gases that do not react with the solvent, there is a simple relationship between gas pressure and gas solubility.  The Pressure-Solubility law, often referred to as Henry's Law.

Pressure-Solubility Law (Henry's Law)
The concentration of a gas in a liquid at any given temperature is directly proportional to the partial pressure of the gas on the solution.
 

CgkgPg


where Cg = the concentration of the gas, kg is the proportionally constant for that gas and Pg is the partial pressure of the gas above the solution.  Constant temperature is assumed.  This equation  is true only at relatively low concentrations and pressure and for gases that do not do react with the solvent.

A more useful expression of Henry's Law is
 

C1  C2
P1        P2


In a calculation using Henry's Law you are usually given data on the initial concentration at some initial pressure and are asked to find what the concentration of the gas will be at some new pressure.  We can do this without first finding the proportionality constant because we know it will have the same value at both pressures. 

Example:  At 20oC the solubility of N2 in water is 0.0150 g/L when the partial pressure of nitrogen is 580 torr.  What will be the solubility of N2 in water at 20oC when its partial pressure is 800 torr?

Solution
Gather the data first:

C1 = 0.0150 g/L          C2 = ? 
P1 = 580 torr               P2 = 800 torr


Using the equation above for Henry's Law, we have:

           C2C1  X  P2
                          P1
                 =  0.0150 g/L X 800 torr
                                580 torr
                 =  0.0207 g/L

The solubility under the higher pressure is 0.0207 g/L

Example #2
How many grams of nitrogen and oxygen are dissolved in 100 g of water at 20oC when the water is saturated with air?  Above this solution, at a total pressure of 760 torr, the air is saturated with water vapor and the partial pressure of the nitrogen is 586 torr and the partial pressure is the oxygen is 156 torr.  The solubility of oxygen in water at a total pressure of 760 torr is 0.00430 g O2 / 100 g H2O.  The solubility of nitrogen in water at a total pressure of 760 torr is 0.00190 g N2 / 100g H2O.
 

Go to the Henry's Law Worksheet

Lowering of a Solution's Vapor Pressure with a Nonvolatile Solute
Raoult's Law
All liquid solutions of nonvolatile solutes have lower vapor pressures than their pure solvents.  The extend of how much the vapor pressure is lowered depends upon whether the solute is molecular (and non-ionizable) or ionic.  We will consider the molecular or non-electrolytic solutes first.   Consider a simple sugar.  It will dissolve in water because of its polarity but it will not ionize because of its strong covalent bonds.
There is a particularly simple law that correlates the solution's vapor pressure with the mole fraction of the solvent.  This is the vapor pressure - concentration law, or Raoult's Law.
Vapor Pressure - Concentration Law, or Raoult's Law
 
Psolution = Xsolvent   X   Posolvent

Psolution = is the vapor pressure of the solution.  Xsolvent = is the mole fraction of the solvent in this solution and Posolvent = is the vapor pressure of the pure solvent. 

A graphical plot of  Psolution  vs  Xsolvent would be linear for a for a solution that obeyed Raoult's Law at all concentrations.
 

The reason for Raoult's Law is easy to understand.  When the particles of solute are present, the opportunities for the solvent to escape into the vapor phase are reduced or lowered.  Therefore, the vapor pressure of the solution is less than the vapor pressure of the pure solvent at the same temperature.
The interference caused by the solute molecules and therefore the extent  to which the vapor pressure is lowered depends on the fraction of molecule at the surface which are the solute.  If 25% of these molecules are solute, then the vapor pressure is lowered by 25%.  Another way of looking at this, however , is that 75% of the molecules art the surface are those of the solvent, and the vapor pressure, having been lowered by 25%, is now 75% of  its value for the pure solvent.  This  is exactly what Raoult's law  says, because if 75%  of the molecules are those of the solvent, then the mole, fraction of the solvent must be 0.75 and 0.75 multiplied by the vapor pressure of pure solvent gives a value that is 75% of Posolvent
Example:  Carbon tetrachloride has a vapor pressure is 100 torr at 23oC.  This solvent can dissolve candle wax, which is essentially nonvolatile.  Although candle wax is a mixture, we can take its molecular formula to be C22H46 with a formula weight of 310.  What will be the vapor pressure at 23oC of a solution prepared by  dissolving 10.0 grams of wax in 40.0 g of CCl4?  The formula mass of CCl4 is 154.

Solution:
Find the mole fractions first, then use Raoult's Law.

For CCl4        n = g / mm = 40.0 g /  154 g/mol = 0.260 mol of CCl4

For C22H46   n = g / mm = 10.0 g / 310 g/mol = 0.0323 mol of C22H46

The total number of moles is 0.260 mol + 0.0232 mol = 0.292 mol.

The mole fraction of CCl4 is 0.260 mol = 0.890
                                               0.292 mol

The  mole fraction of CCl4, the solvent is 0.890.  The vapor pressure of the solution will be the fraction of the vapor pressure of pure CCl4 (100 torr) as calculated by Raoult's Law.
Psolution = 0.890 X 100 torr
              =  89.0 torr
 

The Effect of a Volatile Solute on the Vapor Pressure of a Solution
When two (or more) components of a liquid solution can evaporate, the vapor contains molecules of each.  Each volatile component contributes its own partial pressure to the total pressure and this partial pressure is directly proportional to the component's mole fractions in the solution.  This is reasonable because the rate of evaporation of each compound has to be lowered by molecules of the others in the same way that occurs with nonvolatile solutes.  So, using Dalton's Law. the total vapor pressure will be the sum of all the partial pressures.   To calculate these partial pressures, we use Raoult's Law for each component.  For component A, present in a mole fraction XA , its partial pressure PA, is found as a fraction of its vapor pressure when pure, PoX, thus,
 
PA = XAPoA


and the partial pressure of component B would be

PB = XBPoB


If you have more than two components than do similar calculations of them.

The total pressure of the solution of liquids A and B is then:

Ptotal = PA + PB


Example:  Acetone is a solvent for both water and molecular liquids that do not dissolve in water, like benzene. At 20oC  acetone has a vapor pressure of 162 torr.  The vapor pressure of water at 20oC is 17.5 torr.  What is the vapor pressure of a solution of acetone and water with 50.0 mole percent of each?

Solution: To find Ptotal we need to calculate the individual partial pressures and then add them.

Pacetone = 162 torr X 0.500 = 81.0  torr
   Pwater = 17.5 torr X 0.500 = 8.75 torr
                                     Ptotal = 89.8 torr

Thus the vapor pressure of the solution is much higher than that of pure water but much less than that of pure acetone. 


Go to the Raoult's Law Worksheet

Boiling Point Elevation
When the only volatile component of a solution is the solvent, then the vapor pressure of the solution is less than the vapor pressure tat the solvent has at any given temperature.   At the temperature at which the pure solvent normally boils, the vapor pressure of the solution is still not equal to the atmospheric pressure.  So to make the vapor pressure of the solution come up to atmospheric pressure, we have to increase the temperature of the solution further.  The presence of a nonvolatile solute thus elevates the boiling point of the solution.
The most common application of this property of solutions occurs in the use of permanent-type antifreezes.  These protect the liquid in a vehicle's cooling system not just from freezing but also from boiling over.  These products are based on either propylene glycol or ethylene glycol, which are high-boiling, nearly nonvolatile, water soluble liquids, and they elevate the boiling point as well as lower the freezing point when dissolved in water.

Molal Boiling Point Elevation Constants
The number of degrees of elevation of a boiling point caused by a nonvolatile solute is directly proportional to its molal concentration, m
For molecular compounds only:   delta t proportional to m
where Δt = (bpsolution - bpsolvent)  With a proportionality constant, this becomes
 
Δt = kbm


This equation works best for solutions with low solute concentrations.  The constant, kb, is called the molal boiling point elevation constant.  Each solvent has its own value of kb, and a few examples are given below.  The units of kb are Δt / m.
 

Molal Boiling Point Elevation Constants

BP (oC) kb
Water 100 0.51
Acetic Acid 118.3 3.07
Benzene   80.2 2.53
Chloroform   61.2 3.63
Cyclohexane   80.7 2.69

Example:  At what temperature will a 0.15 m solution of sugar in water boil when the pressure is 1 atm?

Solution:  First off the change in temperature will be

       Δt = 0.51 X 0.15
            = 0.0765
            = 0.077 oC

The final temperature will be  100oC + 0.077 oC = 100.077 oC

The resulting elevation will be  100.077oC for 0.15 m solution.
 

Formula Weights from Boiling Point Elevation Data
If the molality, kb and Δt are known we can work backwards to find the formula weight.

Example:  A solution made by dissolving 10.0 g of an unidentified compound in 100 g of water boiled at 100.45oC at 1 atm.   What is the formula weight of the compound?  (The solute is not an electrolyte)

Solution:  The kb for water is 0.51 kg-solvent/mol-solute.  The value of Δt is found first.

Δt = 100.45oC - 100.00o
            = 0.45oC

Now we use the equation from above:
 

     Δt = kbm
     0.45oC = 0.51 oC kg-solvent   X m
                                  mol-solute

therefore   m =          0.45oC
                            0.51 oC kg-solvent
                                         mol-solute 

                        = 0.88 mol-solute
                                    kg-solvent

From this molal concentration we can calculate the number of moles of solute.  The solution we know has 100 g or 0.100 kg of solvent, so it also contains:

      0.88  mol-solute   X   0.100 kg-solvent  =   0.088  mol-solute
                                                                                       kg-solvent

From this we can calculate the formula weight.
  mm = g-solute / mol-solute 
         =      10.0 g-solute 
              0.088 mol-solute 
          =  114 g/mol

Giving due consideration to significant figures, the formula weight is between 110 and 120.


Go to the Boiling Point Elevation Worksheet

Freezing Point Depression
A solution freezes at a temperature below the temperature at which the pure solvent freezes.  We say that the solute causes a freezing point depression.  Why this should be can be explained if we remember that pure crystals of one component form when something starts to crystallize.  As we cool a solution, solvent molecules eventually lose enough average kinetic energy to enable them to settle into lattices of crystals of pure solvent.  Although these solvent crystals do not incorporate molecules of then solute, as the  crystals grow, the solute molecules get in the way and interfere with the growth of the solvent crystals.  So to compensate more kinetic energy must be lost in order for the solvent crystals to form -- i.e.. the solution must become cooler.  Thus, the solute depresses the freezing point.
Molal Freezing Point Depression Constants
The number of degrees,  Δt, by which a solution's freezing point is depressed is proportional to the molality, m, at least at relatively low concentrations.

With a proportionality constant, the freezing point depression constant, kf, we convert this into an equation:
 

Δt  = kf  X m


where Δt is the change in freeing point from normal, kf is the freezing point depression constant and m is the molality of the solution.  Each solvent has its own value of kf.
Solvent MP oC kf
Water 0.00 1.86
Acetic Acid 16.6 3.57
Benzene 5.45 5.07
Camphor 178.4 37.7
Cyclohexene 6.5 20.0


Example:  At what temperature will of 0.250 m solution of sugar in water freeze?

Solution:  From the above table the kf of water is 1.86 oC kg-solvent / mol-solute. 

       Δt  = 1.86 okg-solvent  X  0.250       mol-solute      
                                                          mol-solute kg-solvent
             = 0.465 oC

The freezing point of water, 0.00 oC is lowered by 0.465 oC; so the solution itself will freeze at -0.465oC.

The depression of a freezing point by a solute is the principle behind the use of salt to melt ice on city streets as well as the use of antifreeze in vehicle cooling systems.


Formula Weights from Freezing Point Depression Data
The freezing point depression data can be used to calculate the formula weight of an unknown solute.

Example:  A solution made by dissolving 5.65 g of an unknown compound in 110 g of benzene from at 4.39oC.  What is the formula weight of the solute?

Solution:  The kf for benzene is 5.07 kg-solvent/mol-solute.  The value of Δt is found first.

Δt = 5.45oC - 4.39o
      = 1.06oC

Now we use the equation from above:
 

            Δt = kfm
     1.06oC = 5.07 oC kg-solvent   X m
                                 mol-solute

therefore   m =          1.06oC
                            5.07 oC kg-solvent
                                       mol-solute 

                        = 0.209 mol-solute 
                                     kg-solvent

From this molal concentration we can calculate the number of moles of solute.  The solution we know has 110 g or 0.110 kg of solvent, so it also contains:

      0.209  mol-solute   X   0.110 kg-solvent  =   0.0230  mol-solute
                kg-solvent

From this we can calculate the formula weight.
  mm = g-solute / mol-solute 
         =    5.65    g-solute 
              0.0230 mol-solute 
          =  246 g/mol

The formula weight is 246. 


Go to the Freezing Point Depression Worksheet

Osmosis and Dialysis
When two solutions of different concentrations are separated by the right kind of membrane, their concentrations change in the direction of becoming equal.

In living things, membranes of various kinds keep mixtures organized and separated.  These membranes are semipermeable, which means that they let water molecules as well as small ions and other small molecules pass through.  However they do not allow the passage of very large molecules, like those of proteins.  This phenomenon, the selective passage of small ions and molecules through a membrane, is called dialysis, and the membrane is a dialyzing membrane.

In the limiting case of dialysis, osmosis, only solvent molecules can get through the membrane, now called an osmotic membrane.   Such membranes are rare, but they can be made.
 

One theory that can be used to explain osmosis is that an osmotic membrane separates two solutions from each other, one solution being more concentrated in solute than the other.   The solvent in each solution has its own "escaping tendency", which is just like the vapor pressure of a solution.  The solution with the lower solute concentration has, of course, the higher mole fraction of solvent.  Therefore, solvent molecules in the less concentrated solution (or the pure solvent) have a higher escaping tendency than those in the more concentrated solution.  So solvent molecules escape through the membrane more frequently from the less concentrated side and pass into the more concentrated solution.

Some solvent molecules come the other way, of course, but not as frequently.  The effect is a net shift of solvent through the membrane from the less concentrated side to the more concentrated side.  This is osmosis, and it will continue, in principle, until the concentrations of the solutions become equal.   Only then would the escaping tendencies of the solvent on both sides of the membrane become equal.   In principle, of course, osmosis could not continue indefinitely if one liquid were pure solvent, instead of more dilute solution.  The more concentrated solution, no matter how much it became diluted, would always have a higher concentration of solute than the pure solvent.  What sometimes stops osmosis, however is the development of a back pressure.
 

Osmotic Pressure
Osmosis has all the appearance of there being something that creates a pressure that "forces" the solvent through the membrane.  But there is no "something," only unequal concentrations separated by an osmotic membrane plus nature's driving force to get everything as mixed up as possible.  (If the membrane wasn't there the two solutions would intermingle and eventually become uniform everywhere in concentration.)

As solvent moves into the more concentrated solution a column of solution rises.  This creates a back pressure caused by the weight of the liquid in this column.  We could actually prevent osmosis, and even force it to reverse, by putting an opposing pressure in place.  The exact back-pressure needed to prevent any osmotic flow when one of the liquids is pure solvent is called the osmotic pressure of the solution.  Its symbol is  Π .

The word pressure here is being used in a somewhat different way than we have done before.  Pure water alone, for example, does not "have" a special pressure called osmotic pressure.  What it does have is an escaping tendency that nature exploits as osmosis in a very special situation - when the water is separated from a solution by an osmotic membrane.  Only because the escaping tendency of water from a solution is less than from the pure water can there be a net flow of water into the solution.

It is not hard to imagine that the more concentrated the solution, the more water will transfer to it.  Thus the osmotic pressure of a solution is proportional to the relative concentrations of its molecules of solute and solvent.  Molality served well to express this ratio, but in any dilute aqueous solution the molarity and molality are virtually identical, and we will use molarity.   Therefore we can say that the osmotic pressure of a dilute solution is proportional to its molarity.
 

Π V = nRT      or         Π = MRT


Osmotic pressures can be very high even in dilute solutions.

Example:  A very dilute solution, 0.0010 M sugar in water, is separated from pure water by an osmotic membrane.  What osmotic pressure developed at 25oC or 298 K?  The gas constant R = 0.0821 L atm / mol K.

Solution: Apply  the molarity equation from above:
 
Π
 = MRT

     = 0.0010 mol  X 0.0821  atm  X 298 K
                      L                     molK

     = 0.024 atm 

The osmotic pressure is 0.024 atm.

Mercury has a density of 13.59 g/mL.  Since the density of sugar water in this solution is basically equivalent to pure water we can convert  0.024 atm into the the equivilant expressed in inches of water.   0.024 atm will support a colulm of water 10 inches high.  A 0.1 M sugar solution, still relatively dilute, could support a column 100 times as high or 1000 inches, or about 83 ft high.  Thus osmosis is a part of the explanation for the rise of sap in trees.
 

Dialysis
Although membranes in living systems are not true osmotic membranes, they do hold back large colloid-sized molecules and ions, such as those of proteins.  When a colloidal dispersion is separated from water by a dialyzing membrane, therefore, the dispersion will receive water molecules from the other side faster then it lets them escape.  This is just like osmosis, but it is also called dialysis because if other solutes consisting of small ions and molecules are also present, they will pass back and forth through the membrane.  Only the presence of the large particles evokes the osmosis-like behavior.  In this situation, there is an associated colloidal osmotic pressure.

One practical use of dialysis is the separation of solutes in true solution from colloidal substances.  Cellophane makes an acceptable dialyzing membrane.  When, for example, an aqueous solution of sodium chloride that contains collodially dispersed starch is placed in a cellophane bag and this is immersed in water, the ions of the salt dialyze out leaving just the starch in the bag.  Water moves in both directions, but more goes into the bag than leaves because of the colloidal osmotic pressure created by the starch. 

If the liquid surrounding the cellophane bag also contained NaCl at the same concentration as the system inside, then sodium and chloride ions would dialyze through the bag at equal rates.   No net removal of salt from the contents of the bag occur.  Hemodialysis is a special application of these principles that is used to cleanse the blood.

Our kidneys are organs that cleanse the blood of many of the wastes of metabolism.  If the kidneys quit working, these wastes will cause death.

The artificial kidney provides a way to take over a person's kidneys should they develop problems.   It works by dialyzing, and the procedure is called hemodialysis.  The blood is diverted from the body into a long coiled dialyzing membrane, often cellophane.  A solution called the dialysate is constantly circulated around the outside of this tube. 

The dialysate contains in solution all of the solutes that must remain in the blood at the concentrations that they have in the blood - NaCl, for example, plus traces of other ionic and molecular compounds.  The dialysate will not initially contain any of the wastes to be removed. 

As the blood flows within the dialyzing membrane, its sodium ions move into the dialysate.  But they return at the same rate because the concentrations inside and outside are the same.  Impurities in the blood, such as urea, however, dialyze through the membrane faster than they can return. Thus the blood experiences a loss of the impurity.  Proteins, other large molecules, and blood cells cannot, of course, pass through the membrane, so they stay in the blood.  Eventually the blood re-enters the body.

Go to the Osmosis and Dialysis Worksheet


Colligative Properties of Solutions of Electrolytes
A 1 m solution of NaCl, an ionic compound, freezes at -3.37oC, instead of  -1.86oC, the expected freezing point of a 1 m molecular compound dissolved in water.  This much greater depression of the freezing point by the salt - almost twice as much - is not hard to understand if we remember that colligative properties depend upon the cocnentrations of particles, not their chemical identities.

In a 1 m NaCl solution, the solute is dissociated.
 

NaCl(s) --------->   Na+(aq)  +  Cl-(aq)

So, 1 m NaCl actually has a 2 m concentration in dissolved solute particles.  Theoretically, then 1 m NaCl should freeze at 2 X (-1.86oC) = -3.72oC.

If we made up a solution of 1 m H2SO4, we would have to consider the following dissociation.
 

H2SO4(l) --------->  2 H+(aq)  +  SO42-(aq)
 
Thus, 1 mol H2SO4 can give a total of 3 mol of ions -- 2 mol of H+ ions and 1 mol of SO42- ions.   A 1 m solution of this acid has a calculated freezing point of -5.58oC (3 X -1.86oC).  It actually freezes at -4.22oC.

Example: Estimate the freezing point of a 0.106 m aqueous MgCl2, assuming that it ionizes completely.

Solution:  When MgCl2 dissolves in water it breaks up as follows:
 

MgCl2(s) ------>  Mg2+(aq)  +  2 Cl-(aq)
 
Since 1 mol of MgCl2 gives 3 mol of ions, the true molality of what we have called 0.106 m MgCl2 is three times as great.

effective molality =  3 X (0.106) = 0.318   mol-solute
                                                                    kg-solvent

Since, Δt = kfm
           Δt = 1.86 oC kg-solvent   X   0.318 mol-solute
                           mol-solute                    kg-solvent 

                 = 0.519oC

Thus the freezing point is depressed below 0.000oC by 0.519 oC, so we calculate that this solution freezes at -0.519oC.  It actually freezes at -0.517oC.
 

Ions Interact with Each Other in Aqueous Solutions
Neither of the 1 m solutions above actually freezes quite as low as calculated. What is wrong, or course, is not nature; it is our assumption of 100% dissociation of the ions.  This assumption is not quite correct.  To be sure, the ions in all these aqueous solutions are hydrated, so the solvent helps shield them from each other, but this shielding is not perfect.  The ions are not really 100% free of each other so the solutes do not behave as if they were 100% dissociated.  As solutions of electrolytes are made more and more dilute, however, the observed and calculated freezing points come closer and closer together.  At greater dilutions, ions interact with each other less and less, so the solutes behave more and more as if they were really 100% separated into their ions.

Chemists compare the degree of dissociation of electrolytes at different dilutions by a quantity called the van't Hoff factori.     It is a ratio of the observed freezing point depression to the calculated value on the assumption that the solute dissociates as non-ionized molecules.

i  =                  (Δtf)measured_________________ 
                  (Δtf)calculated as a non-electroltye

The van't Hoff factors for several electrolytes at different dilutions are given below.  Notice that with decreasing concentration the experimental van't Hoff factors agree better and better with their corresponding hypothetical van't Hoff factors -- calculated on the assumption of 100% disscociation.
 
van't Hoff Factor
                       Concentration 
                     (mol salt / kg water)
If 100 %
Dissociation
Salt 0.1 0.01 0.001 Occurred
NaCl 1.87 1.94 1.97 2
KCl 1.85 1.94 1.98 2
K2SO4 2.32 2.70 2.84 3
MgSO4 1.21 1.53 1.82 3


Go to the Colligative Properties of Electrolytes Worksheet