Aqueous Solutions

1
Aqueous Solutions

• Dr. Afshin Fassihi
• Department of
• Medicinal Chemistry

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Solution

• A solution is a homogeneous mixture of a solute
  dissolved in a solvent.The solvent is generally
  in excess.
• Types of solutions are
• Gaseous (mixture of gases, ruled out by Daltons
  partial pressure law)
• Solid (metal alloys)
• Liquid.

3
Definitions

• SOLUTE The part of a solution that is being
  dissolved (usually the lesser amount)
• SOLVENT The part of a solution that dissolves
  the solute (usually the greater amount)
• Solute Solvent Solution
• Solubility The maximum amount of the solute that
  will dissolve in a definite amount of the solvent
  and produce a stable system at a specific
  temperature.
• Concentration The amount of solute dissolved in
  a given amount of solvent or dissolved in a given
  amount of solution at a given temperature.

4
Definitions

• Saturated solution contains the maximum quantity
  of solute that dissolves at that temperature, in
  other words contains the solute at its solubility
  concentration at the given temperature.

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Definitions

• Unsaturated solution contains less than the
  maximum amount of solute that can dissolve at a
  particular temperature
• Supersaturated solution contains more solute than
  is possible to be dissolved. Supersaturated
  solutions are unstable. The supersaturation is
  only temporary, and usually accomplished in one
  of two ways
• Warm the solvent so that it will dissolve more,
  then cool the solution
• Evaporate some of the solvent carefully so that
  the solute does not solidify and come out of
  solution.

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Solubility curve
Definitions
Supersaturated
Unsaturated
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Solubility curve
Definitions

• Any point on a line represents a saturated
  solution.
• In a saturated solution, the solvent contains the
  maximum amount of solute.
• Example
• At 90oC, 40 g of NaCl(s) in 100g H2O(l) represent
  a saturated solution.

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Solubility curve

• Any point below a line represents an unsaturated
  solution.
• In an unsaturated solution, the solvent contains
  less than the maximum amount of solute.
• Example
• At 90oC, 30 g of NaCl(s) in 100g H2O(l) represent
  an unsaturated solution. 10 g of NaCl(s) have to
  be added to make the solution saturated.

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Solubility curve

• Any point above a line represents a
  supersaturated solution.
• In a supersaturated solution, the solvent
  contains more than the maximum amount of solute.
  A supersaturated solution is very unstable and
  the amount in excess can precipitate or
  crystallize.
• Example
• At 90oC, 50 g of NaCl(s) in 100g H2O(l) represent
  a supersaturated solution. Eventually, 10 g of
  NaCl(s) will precipitate.

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Solubility curve
Definitions

• Any solution can be made saturated, unsaturated,
  or supersaturated by changing the temperature.

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The Solution Process

• A polar solute will dissolve in a polar solvent
  but not in a nonpolar solvent
• Like Dissolves Like"
• Example
• Water (polar substance) does not dissolve carbon
  tetrachloride (nonpolar substance)
• Carbon tetrachloride dissolves iodine.
• Alcohol (polar substance) dissolves in water
  (polar substance)

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The Solution Process

• Solids will dissolve in water if the attractive
  force between the water molecules and the solute
  molecule is stronger than the attractive force of
  the crystal.
• If not, the solids are insoluble.

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Dissolution of Ionic solids in water
15
Ionic solids
Dissolution of Ionic solids in water
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Enthalpy of Hydration

• The energy (KJ/mol) released by hypothetical
  process in which hydrated ions are formed from
  gaseous ions.
• K (g) Cl- (g) K (aq) Cl- (aq)
  ?H -684.1 KJ
• It depends upon the concentration of the final
  solution.
• Usually it is assumed that the enthalpy change
  pertains to the process of hydration of ions to
  the greatest possible extend enthalpies of
  hydration at infinite dilution.

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Enthalpy of Solution

• The enthalpy change (KJ/mol) associated with the
  process in which a solute dissolves in a solvent.
• It depends upon the concentration too and it is
  reported for infinitely dilute solutions, unless
  otherwise noted.
• It is a net result of
• The energy required to break apart solute-solute
  and solvent-solvent bonds or attractions.
• The energy released by the formation of solute
  solvent attractions.

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Enthalpy of Solution of KCl

• a) The energy required to break apart the KCl
  crystal structure
• KCl (s) K (g) Cl- (g) ?H
  701.2 KJ
• b) The enthalpy of hydration of KCl
• K (g) Cl- (g) K (aq) Cl- (aq)
  ?H -684.1 KJ
• The enthalpy of hydration is the sum of the
  energy required to break the hydrogen bonds
  between some of the water molecules and the
  energy released when these water molecules
  hydrate the ions.
• The overall process is endothermic
• KCl (s) K (aq) Cl- (aq)
• ?H 701.2 (-684.1) 17.1 KJ

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Enthalpy of Solution of KCl

• Some enthalpies of solution are negative because
  more energy is liberated by the hydration of the
  ions of the solute than is required to break
  apart the crystal structure.
• AgF (s) Ag (g) F- (g) ?H
  910.9 KJ
• Ag (g) F- (g) Ag (aq) F- (aq)
  ?H -931.4 KJ
• AgF (s) Ag (aq) F- (aq)
• ?H 910.9 (-931.4) -20.5 KJ
• The overall process is exothermic.

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Enthalpy of Solution

• The factors that lead to a large positive value
  for the first step (high ionic charges and small
  ions) also lead to a large negative value for the
  second step.
• So ignoring signs, the values for the steps are
  usually numerically close and
• The enthalpy of solution itself is a much smaller
  value than either of the values that goes into
  it.

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Enthalpy of Solution of Nonionic Materials

• The forces holding molecular crystals together
  are not so strong as those holding ionic crystals
  together. So the energy of the first step is low.
• Since the forces between the molecules of
  nonionic materials and the molecules of solvent
  are not strong the enthalpies of solvation for
  such materials are also low.
• the enthalpy of solution is endothermic and has
  about the same magnitude as the enthalpy of
  fusion of the solid, the enthalpy of the first
  step in the solution process.

22
Effect of Temperature on Solubility

• The solubility of solutes is dependent on
  temperature.
• When a solid dissolves in a liquid, a change in
  the physical state of the solid analogous to
  melting takes place. Heat is required to break
  the bonds holding the molecules in the solid
  together.
• At the same time, heat is given off during the
  formation of new solute - solvent bonds.

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Increase in solubility with temperature
Effect of Temperature on Solubility

• If the the net dissolving reaction is endothermic
  (energy required) the addition of more heat
  facilitates the dissolving reaction by providing
  energy to break bonds in the solid.
• The use of first-aid instant cold packs is an
  application of this solubility principle. A salt
  such as ammonium nitrate is dissolved in water
  after a sharp blow breaks the containers for
  each. The dissolving reaction is endothermic -
  requires heat. Therefore the heat is drawn from
  the surroundings, the pack feels cold.

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Decrease in solubility with temperature
Effect of Temperature on Solubility

• If the net dissolving reaction is exothermic
  (energy given off) the addition of more heat
  inhibits the dissolving reaction since excess
  heat is already being produced by the reaction.
• This situation is not very common.

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Solubility of Gases vs. Temperature

• As the temperature
• increases, the solubility of a gas decrease.

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Solubility of Gases vs. Temperature

• The solubility of gases in solvents is usually
  exothermic, because this is only the second step
  of dissolution process thats determines the net
  change in enthalpy during the process of
  dissolution of a gas in a solvent.

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Effect of pressure on the Solubility of Gases

• Henrys Law The amount of gas that dissolves in
  a given quantity of liquid at constant
  temperature is directly proportional to the
  partial pressure of the gas above the solution.
• This is only true for dilute solutions at
  relatively low temperatures.
• Gases that are extremely soluble generally react
  chemically with the solvent. For example HCl gas
  in water reacts to produce hydrochloric acid.
  They dont follow Henrys law.

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Effect of pressure on the Solubility of Gases

• Deep sea divers may experience a condition called
  the "bends" if they do not readjust slowly to the
  lower pressure at the surface.
• As a result of breathing compressed air and being
  subjected to high pressures caused by water
  depth, the amount of nitrogen dissolved in blood
  and other tissues increases.
• If the diver returns to the surface too rapidly,
  the nitrogen forms bubbles in the blood as it
  becomes less soluble due to a decrease in
  pressure. The nitrogen bubbles can cause great
  pain and possibly death.
• To alleviate this problem somewhat, artificial
  breathing mixtures of oxygen and helium are used.
  Helium is only one-fifth as soluble in blood as
  nitrogen. As a result, there is less dissolved
  gas to form bubbles.

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Concentrations of Solutions

• For example a 20 w/w aqueous solution of NaOH
  consists of 20 g NaOH and 80 g Water.
• 5 grams of a 20 w/w aqueous solution of NaOH
  consists of 1 g of NaOH and 4g of Water.
• A solution contains 15 g Na2CO3 and 235 g of
  H2O? What is the mass of the solution?
• 1) 15 Na2CO3
• 2) 6.4 Na2CO3
• 3) 6.0 Na2CO3

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Mole Fraction
Concentrations of Solutions

• For example, 1 mol ethylene glycol in 9 mol water
  gives a mole fraction for the ethylene glycol of
  1/10 0.10.
• An aqueous solution contains 39 w/w ethyl
  alcohol.
• What is the mole fraction of ethyl alcohol in
  this solution? Mw of ethyl alcohol is 46 g/mol
• Answer 0.200
• In a solution of naphthalene (Mw 128.2 g/mol) in
  toluene (Mw 92 g/mol) the molar fraction of
  naphthalene is 0.200. What is the of
  naphthalene in this solution?
• Answer 25.8 w/w

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Molarity
Concentrations of Solutions

• How many grams of an aqueous solution of 70 W/W
  HNO3 are required to prepare 250 mL of a 2.00 M
  aqueous solution? MW of HNO3 is 63 g/mol Answer
  45.0 g
• What is the molarity of concentrated HCl if the
  solution contains 37 HCl by mass and if the
  density of the solution is 1.18 g/mL.
• Answer 12 M (mol/L)

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Molality
Concentrations of Solutions

• For example, 0.20 mol of ethylene glycol
  dissolved in 2.0 x 103 g ( 2.0 kg) of water has
  a molality of
• What is the molality of a 12.5 solution of
  glucose in water? Mw of glucose is 180 g/mol.
• Answer 0.794

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Concentrations of Solutions Molality

• Final volume is not important. A 1m solutions of
  different solutes, each containing 1000 g of
  water, will have different volumes. All these
  solutions will have the same mole fractions of
  solute and solvent.
• Molality is used for the solutions other than
  aqueous solutions.
• A solution of a solute in toluene has the
  concentration of 1.00m. What is the mole fraction
  of the solute in this solution? Mw of Toluene 92
  g/mol Answer X 0.0842
• Molar fraction of all of the 1m solutions in
  toluene is 0.0842
• Molality doesnt change with temperature.

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Normality
Concentrations of Solutions

• The definition of equivalent, which is the amount
  of the reactant, depends upon the type of the
  reaction neutralization and oxidation-reduction.
  
• One equivalent of a given reactant will react
  with exactly one equivalent of another.
• The mass of one equivalent of a compound is
  called an equivalent weight. In general

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Concentrations of Solutions Normality

• The value of a depends upon the type of the
  reaction.
• For neutralization reactions, equivalent weights
  are based on the fact that one H (aq) ion reacts
  with one OH- (aq) ion.
• H (aq) OH- (aq) H2O
• One eq. weight of an acid is the amount of acid
  that supplies one mole of H (aq) ions for a
  neutralization reaction with one mole of OH- (aq)
  ions and vice versa.
• The value of a, therefore is the number moles of
  H (aq) supplied by one mole of the acid or the
  number of moles of OH- (aq) supplied by one mole
  of the base for the reaction being considered.

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Concentrations of Solutions Normality

• For oxidation-reduction reactions, equivalent
  weights are based either on the number of moles
  of electrons exchanged or oxidation-number
  changes.
• The number of moles of electrons lost by an
  oxidation (or the increase in oxidation number)
  must equal the number of moles of electrons
  gained by the reduction (or the decrease in
  oxidation number).
• Hence a is the number of moles of electrons lost
  or gained by a mole of the reactant, or the
  change in oxidation number.

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Concentrations of Solutions Normality

• The number of equivalents of A in a sample of A,
  eA, can be obtained by multiplying the volume of
  the sample, VA, (in liters), by the normality of
  the solution, NA, (which is the number of
  equivalents of A in one liter of solution)
• eA NAVA
• The number of equivalents of B in a sample of B,
  eB,
• eB NBVB
• Since one equivqlent of A reacts with one
  equivalent of B eA eB
  
• and therefore
• NA VA NBVB

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Vapor Pressure of Solutions Ideal Mixtures

• In a pure liquid, some of the more energetic
  molecules have enough energy to overcome the
  intermolecular attractions and escape from the
  surface to form a vapour.
• The smaller the intermolecular forces, the more
  molecules will be able to escape at any
  particular temperature.

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Vapor Pressure of Solutions Ideal Mixtures

• If you have a second liquid, the same thing is
  true. At any particular temperature a certain
  proportion of the molecules will have enough
  energy to leave the surface.

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Vapor Pressure of Solutions Ideal Mixtures

• In an ideal mixture of these two liquids, the
  tendency of the two different sorts of molecules
  to escape is unchanged.

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Vapor Pressure of Solutions Ideal Mixtures

• You might think that the diagram shows only half
  as many of each molecule escaping - but the
  proportion of each escaping is still the same.
  The diagram is for a 50/50 mixture of the two
  liquids.
• If the red molecules still have the same tendency
  to escape as before, that must mean that the
  intermolecular forces between two red molecules
  must be exactly the same as the intermolecular
  forces between a red and a blue molecule.
• Exactly the same thing is true of the forces
  between two blue molecules and the forces between
  a blue and a red.

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Vapor Pressure of Solutions Ideal Mixtures

• This is why mixtures like hexane and heptane get
  close to ideal behaviour. They are similarly
  sized molecules and so have similarly sized van
  der Waals attractions between them. However, they
  obviously aren't identical - and so although they
  get close to being ideal, they aren't actually
  ideal.
• For the purposes of this topic, getting close to
  ideal is good enough!
• Common examples include
• hexane and heptane
• benzene and methylbenzene
• propan-1-ol and propan-2-ol

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Raoult's Law

• The partial vapour pressure of a component in a
  mixture is equal to the vapour pressure of the
  pure component at that temperature multiplied by
  its mole fraction in the mixture.
• Raoult's Law only works for ideal mixtures.
• For a mixture of liquids A and B, this reads

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Raoult's Law

• The P0 values are the vapour pressures of A and B
  if they were on their own as pure liquids.
• XA and XB are the mole fractions of A and B. That
  is exactly what it says it is - the fraction of
  the total number of moles present which is A or
  B.

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Raoult's Law

• Suppose you had a mixture of 2 moles of methanol
  and 1 mole of ethanol at a particular
  temperature. The vapour pressure of pure methanol
  at this temperature is 81 kPa, and the vapour
  pressure of pure ethanol is 45 kPa. What are the
  partial pressures of each?

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Raoult's Law

• In this equation, PA and PB are the partial
  vapour pressures of the components A and B.
• In any mixture of gases, each gas exerts its own
  pressure. This is called its partial pressure and
  is independent of the other gases present.
• The total vapour pressure of the mixture is equal
  to the sum of the individual partial pressures.
• Total vapor pressure for the previous xample

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Vapour pressure / composition diagrams

• Imagine an ideal mixture of two liquids A and B.
  Each of A and B is making its own contribution to
  the overall vapour pressure of the mixture - as
  we've seen above.
• Let's focus on one of these liquids - A, for
  example.
• If you double the mole fraction of A in the
  mixture at a constant temperature according to
  Raoult's Law, its partial vapour pressure will
  become double. If you triple the mole fraction,
  its partial vapour pressure will triple - and so
  on.
• In other words, the partial vapour pressure of A
  at a particular temperature is proportional to
  its mole fraction.

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Vapour pressure / composition diagrams

• If you plot a graph of the partial vapour
  pressure of A against its mole fraction, you will
  get a straight line.

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Vapour pressure / composition diagrams

• If you do the same thing for B and plot it on the
  same set of axes you will see that mole fraction
  of B falls as A increases so the line will slope
  down rather than up. As the mole fraction of B
  falls, its vapour pressure will fall at the same
  rate.

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Vapour pressure / composition diagrams

• To get the total vapour pressure of the mixture,
  you need to add the values for A and B together
  at each composition. The net effect of that is to
  give you a straight line as shown below.

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Vapour pressure / composition diagrams

• For two liquids at the same temperature the one
  with the higher vapour pressure is the one with
  the lower boiling point.
• If we start with the boiling points of pure A and
  B. B has the higher vapour pressure. That means
  that it will have the lower boiling point.
• If we boil a liquid mixture of A and B, we wil
  find that the more volatile substance escape to
  form a vapour more easily than the less volatile
  one.

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Vapour pressure / composition diagrams

• That means that in the case we will find a higher
  proportion of B (the more volatile component) in
  the vapour than in the liquid. We can discover
  this composition by condensing the vapour and
  analysing it.
• If we condense it and boil it again we will have
  again a vapor with a higher proportion of the
  more volatile substance.
• If we repeat this process again and again we can
  have the nearly pure B in the vapor.
• We will have a liquid if we condense the vapor.
  This is what is happening during distillation.

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Non-Ideal Mixtures of Liquids

• Raoult's Law only works for ideal mixtures. In
  these, the forces between the particles in the
  mixture are exactly the same as those in the pure
  liquids. The tendency for the particles to escape
  is the same in the mixture and in the pure
  liquids.
• That's not true in non-ideal mixtures.
• Non-ideal mixtures show one of the two types of
  deviations from Raoult's Law
• Positive deviation
• Negative deviation

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Positive deviations from Raoult's Law

• In mixtures showing a positive deviation from
  Raoult's Law, the vapour pressure of the mixture
  is always higher than it would be expected from
  an ideal mixture.
• The deviation can be small - in which case, the
  straight line in the last graph turns into a
  slight curve.

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• But some liquid mixtures have very large positive
  deviations from Raoult's Law, and in these cases,
  the curve becomes very distorted.
• Notice that mixtures over a range of compositions
  have higher vapour pressures than either pure
  liquid. The maximum vapour pressure is no longer
  that of one of the pure liquids.

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Explaining the Deviation

• The fact that the vapour pressure is higher than
  ideal in these mixtures means that molecules are
  breaking away more easily than they do in the
  pure liquids.That is because the intermolecular
  forces between molecules of A and B are less than
  they are in the pure liquids.
• We can see this when we mix the liquids. Less
  heat is evolved when the new attractions are set
  up than was absorbed to break the original ones.
  Heat will therefore be absorbed when the liquids
  mix.
• The classic example of a mixture of this kind is
  ethanol and water. This produces a highly
  distorted curve with a maximum vapour pressure
  for a mixture containing 95.6 of ethanol by mass.

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Negative Deviations from Raoult's Law

• In exactly the same way, we can have mixtures
  with vapour pressures which are less than would
  be expected by Raoult's Law. In some cases, the
  deviations are small, but in others they are much
  greater giving a minimum value for vapour
  pressure lower than that of either pure
  component.

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Explaining the deviations

• These are cases where the molecules break away
  from the mixture less easily than they do from
  the pure liquids. New stronger forces must exist
  in the mixture than in the original liquids.
• You can recognise this happening because heat is
  evolved when you mix the liquids - more heat is
  given out when the new stronger bonds are made
  than was used in breaking the original weaker
  ones.
• Many (although not all) examples of this involve
  actual reaction between the two liquids. An
  example of a major negative deviation is a
  mixture of nitric acid and water. These two
  covalent molecules react to give hydroxonium ions
  and nitrate ions. We now have strong ionic
  attractions involved

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Boiling point / composition diagrams for
non-ideal mixtures

• A positive deviation from Raoult's Law ethanol
  and water mixtures
• As we learned a large positive deviation from
  Raoult's Law produces a vapour pressure curve
  with a maximum value at some composition other
  than pure A or B.
• If a mixture has a high vapour pressure it means
  that it will have a low boiling point. The
  molecules are escaping easily and you won't have
  to heat the mixture much to overcome the
  intermolecular attractions completely.
• The boiling point / composition curve will have a
  minimum value lower than the boiling points of
  either A or B.
• In the case of mixtures of ethanol and water,
  this minimum occurs with 95.6 by mass of ethanol
  in the mixture. The boiling point of this mixture
  is 78.2C, compared with the boiling point of
  pure ethanol at 78.5C, and water at 100C.

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Boiling point / composition diagrams for
non-ideal mixtures

• Note  This diagram is drawn grossly distorted.
  For example, at the temperature scale and the
  position of the 95.6 value on the composition
  scale. The shapes of the two separate areas
  between the curves are also exaggerated. This is
  to make it easier to see what is going on in what
  comes next.

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Boiling point / composition diagrams for
non-ideal mixtures

• Suppose we are going to distil a mixture of
  ethanol and water with composition C1 as shown on
  the next diagram. It will boil at a temperature
  given by the liquid curve and produce a vapour
  with composition C2.

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Boiling point / composition diagrams for
non-ideal mixtures

• When that vapour condenses it will, of course,
  still have the composition C2. If we reboil that,
  it will produce a new vapour with composition C3.

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Boiling point / composition diagrams for
non-ideal mixtures

• If we carry on with this boiling-condensing-reboil
  ing sequence, we would eventually end up with a
  vapour with a composition of 95.6 ethanol. If we
  condense that you obviously get a liquid with
  95.6 ethanol.
• What happens if we reboil that liquid?
• The liquid curve and the vapour curve meet at
  that point. The vapour produced will have that
  same composition of 95.6 ethanol. If we condense
  it again, it will still have that same
  composition.
• It is impossible to get pure ethanol by distiling
  any mixture of ethanol and water containing less
  than 95.6 of ethanol.
• This particular mixture of ethanol and water
  boils as if it were a pure liquid. It has a
  constant boiling point, and the vapour
  composition is exactly the same as the liquid.

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Boiling point / composition diagrams for
non-ideal mixtures

• It is known as a constant boiling mixture or an
  azeotropic mixture or an azeotrope.
• Distilling a mixture of ethanol containing less
  than 95.6 of ethanol by mass lets you collect
• a distillate containing 95.6 of ethanol in the
  collecting flask
• pure water in the boiling flask.
• What if you distil a mixture containing more than
  95.6 ethanol?
• Work it out for yourself using the phase diagram,
  and starting with a composition to the right of
  the azeotropic mixture. You should find that you
  get
• a distillate containing 95.6 of ethanol in the
  collecting flask (provided you are careful with
  the temperature control, and the fractionating
  column is long enough)
• pure ethanol in the boiling flask

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Boiling point / composition diagrams for
non-ideal mixtures

• Mixtures of nitric acid and water have vapour
  pressures which are less than would be expected
  by Raoult's Law.
• Mixtures of nitric acid and water have boiling
  points higher than either of the pure liquids.
• For these mixtures there is a maximum boiling
  point of 120.5C when the mixture contains 68 by
  mass of nitric acid. That compares with the
  boiling point of pure nitric acid at 86C, and
  water at 100C.
• Notice the much bigger difference this time due
  to the presence of the new ionic interactions.

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Boiling point / composition diagrams for
non-ideal mixtures
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Boiling point / composition diagrams for
non-ideal mixtures Distilling Dilute Nitric Acid

• Suppose we start with a dilute solution of nitric
  acid with a composition of C1 and trace through
  what happens.

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Boiling point / composition diagrams for
non-ideal mixtures Distilling Dilute Nitric Acid

• The vapour produced is richer in water than the
  original acid. If you condense the vapour and
  reboil it, the new vapour is even richer in
  water. Fractional distillation of dilute nitric
  acid will enable you to collect pure water from
  the top of the fractionating column.
• As the acid loses water, it becomes more
  concentrated. Its concentration gradually
  increases until it gets to 68 by mass of nitric
  acid. At that point, the vapour produced has
  exactly the same concentration as the liquid,
  because the two curves meet.
• You produce a constant boiling mixture (or
  azeotropic mixture or azeotrope). If you distil
  dilute nitric acid, that's what you will
  eventually be left with in the distillation
  flask. You can't produce pure nitric acid from
  the dilute acid by distilling it.

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Distilling nitric acid more concentrated than
68 by mass
Boiling point / composition diagrams for
non-ideal mixtures

• This time you are starting with a concentration
  C2 to the right of the azeotropic mixture.

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• The vapour formed is richer in nitric acid. If
  you condense and reboil this, you will get a
  still richer vapour. If you continue to do this
  all the way up the fractionating column, you can
  get pure nitric acid out of the top.
• As far as the liquid in the distillation flask is
  concerned, it is gradually losing nitric acid.
  Its concentration drifts down towards the
  azeotropic composition. Once it reaches that,
  there can't be any further change, because it
  then boils to give a vapour with the same
  composition as the liquid.
• Distilling a nitric acid / water mixture
  containing more than 68 by mass of nitric acid
  gives you pure nitric acid from the top of the
  fractionating column and the azeotropic mixture
  left in the distillation flask.

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Boiling Points and Melting Points of Solutions

• Boiling points and melting points of solutions of
  a non-volatile solute are different from those of
  pure solvent.
• The reason for this is the reduced vapor pressure
  of a solution.
• In 1886, Francois Marie Raoult observed that the
  vapor pressure of a solution depended on the mole
  fraction of the solute.

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Plot of vapor pressure solutions showing
Raoults law.
Boiling Points and Melting Points of Solutions
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Phase diagram showing the effect of nonvolatile
solute on freezing point and boiling point.
Boiling Points and Melting Points of Solutions
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Boiling Point Elevation

• The boiling-point elevation, ?Tb, is found to be
  proportional to the molal concentration, cm, of
  the solution.
• The constant of proportionality, Kb ,called the
  molal boiling-point-elevation constant (C/m)
• depends only on the solvent.

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Change in Freezing Point

• The freezing point of a solution is LOWER than
  that of the pure solvent

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Freezing Point Depression

• Freezing-point depression, ?Tf is also
  proportional to the molal concentration, cm , of
  the solute.
• where Kf ,the molal freezing-point-depression
  constant (C/m) depends only on the solvent.

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Colligative Properties
81
Colligative Properties

• An aqueous solution is 0.0222 m in glucose. What
  are the boiling point and freezing point for this
  solution?
• Kb and Kf for water are 0.512 oC/m and 1.86 oC/m,
  respectively. Therefore,
• The boiling point of the solution is 100.011oC
  and the freezing point is 0.041oC.

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Colligative Properties

• On adding a solute to a solvent, the properties
  of the solvent are modified.
• Vapor pressure decreases
• Melting point decreases
• Boiling point increases
• Osmosis is possible (osmotic pressure)
• These changes are called COLLIGATIVE PROPERTIES.
• They depend only on the NUMBER of solute
  particles relative to solvent particles, not on
  the KIND of solute particles.

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Colligative Properties of Ionic Solutions
Colligative Properties

• The colligative properties of solutions depend on
  the total concentration of solute particles.
• Consequently, ionic solutes that dissociate in
  solution provide higher effective solute
  concentration than nonelectrolytes.
• For example, when NaCl dissolves, each formula
  unit provides two solute particles.

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Colligative Properties of Ionic Solutions

• For solutes that are electrolytes, we must
  rewrite the formulas for boiling-point elevation
  and freezing-point depression.
• Here i is the number of ions resulting from each
  formula unit of the solute.

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Colligative Properties of Ionic Solutions

• Estimate the freezing point of a 0.010 m aqueous
  solution of aluminum sulfate, Al2(SO4)3. Assume
  the value of i is based on the formula.
• When aluminum sulfate dissolves in water, it
  dissociates into five ions.
• Therefore, you assume i 5.
• The estimated freezing point is 0.093oC.

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The vant Hoff factor, i
Colligative Properties of Ionic Solutions

• Ion pairing or association of ions prevents the
  effect from being exactly equal to the number of
  dissociated ions

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Colligative Properties of Ionic Solutions The
vant Hoff factor, i

• The vant Hoff factor, symbol i, is used to
  introduce this effect into the calculations.
• i is a measure of the extent of ionization or
  dissociation of the electrolyte in the solution.

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Colligative Properties of Ionic Solutions The
Vant Hof f Factor

• i has an ideal value of 2 for 11 electrolytes
  like NaCl, KI, LiBr, etc.
• i has an ideal value of 3 for 21 electrolytes
  like K2SO4, CaCl2, SrI2, etc.
• The freezing point of 0.0100 m NaCl solution is
  -0.0360oC. Calculate the vant Hoff factor in
  this aqueous solution.

89
Colligative Properties of Ionic Solutions The
Vant Hoff Factor

• meffective total number of moles of solute
  particles/kg solvent
• We can calculate the i factor.

90
Osmosis
Colligative Properties of Ionic Solutions

• Certain membranes allow passage of solvent
  molecules but not solute particles.
• Such a membrane is called semipermeable.
• Osmosis is the phenomenon of solvent flow through
  a semipermeable membrane to equalize solute
  concentrations on both sides of the membrane.
• Osmotic pressure is a colligative property of a
  solution equal to the pressure that, when applied
  to the solution, just stops osmosis.

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Colligative Properties of Ionic Solutions Osmosis
92
Colligative Properties of Ionic Solutions Osmosis
93
Colligative Properties of Ionic Solutions Osmosis
94
Colligative Properties of Ionic Solutions Osmosis

• The osmotic pressure, p, of a solution is related
  to the molar concentration of the solute.
• Similarities exist between the behavior of water
  molecules in osmosis and the gas molecules in
  diffusion. In both processes gas molecules
  diffuse from regions of high concentrations to
  regions of low concentration.
• ?V nRT
• ? n/V RT
• Here R is the ideal gas constant
• ( 0,08206 L.atm/K.mol)
• and T is the absolute temperature.