This document provides an overview of physical chemistry topics related to solutions, including ideal and real solutions, solubility of gases in liquids, and colligative properties. It defines key terms like solute, solvent, and solution. It describes the three main types of solutions and factors that influence gas solubility. The document also explains concepts like Raoult's law, ideal and non-ideal solutions, Henry's law, and how temperature and pressure impact gas solubility. Finally, it summarizes four colligative properties - osmotic pressure, vapor pressure lowering, boiling point elevation, and freezing point depression - and provides examples of calculations for each.

1. PHYSICAL CHEMISTRY I
TU6: IDEAL AND REAL SOLUTIONS, SOLUTIONS OF
GASES IN LIQUIDS, COLLIGATIVE PROPERTIES, PH:
DETERMINATION OF PH BUFFERS, THEORY OF BUFFERS.

2. GENERAL INTRODUCTION
Solute-Is a substance that can be dissolved in a
solvent to produce a solution.
Solvent-It is the part that the solute is dissolved
in.
Solution-Is a homogeneous mixture of one or
more solutes dissolved in a solvent.

3. Solutions
-Can be divided into three types: gaseous, liquid and solid solution
1. Gaseous solution-solutions in which solvent is present in gaseous state.
This can be divided into three types based on the phases of solute and
solvent.
a. Gas-gas solution-Solutions in which solute and solvent are both gases. For
example-solution (mixture) of nitrogen and oxygen.
b. Liquid-gas solution-Solutions in which solute is in liquid state and solvent is
in the gaseous state. For example, solution of chloroform in nitrogen gas.
c. Solid-gas solution-Solutions in which solute is in its solid state and solvent
is in gaseous state. For example, solution of camphor in nitrogen gas.

4. 2. Liquid solution
Solutions in which solvent is present in liquid state.
a. Gas-liquid solution-Solutions in which solute is in in gaseous state
and solvent in liquid state. For example-solution (mixture) of oxygen in
water.
b. Liquid-liquid solution-Solutions in which solute and solvent are both in
the liquid state. For example, solution of ethanoic acid and water
(vinegar).
c. Solid-liquid solution-Solutions in which solute is in its solid state and
solvent is in liquid state. For example, solution of glucose in water.

5. 3. SOLID SOLUTION
Solutions in which solvent is present in solid state.
a. Gas-solid solution-Solutions in which solvent is in solid state and
solute in gaseous state. For example-solution (mixture) of hydrogen and
palladum.
b. Liquid-solid solution-Solutions in which solvent is in liquid state and
solute is in the liquid state. For example, solution of amalgum of
mercury with sodium..
c. Solid-solid solution-Solutions in which solute and solvent are both in
the solid state. For example, solution of gold and copper.

6. RAOULT’S Law
In 1986, a French chemist (Francois Marte Raoult)
Proposed a relationship between partial pressure and mole fraction of
volatile liquids.
According to the law, ‘the mole fraction of the solute component is
directly proportional to its partial pressure’. (PV=nRT)
On the basis of this law, liquid solutions can be of two types.
-1. Ideal solutions
-2. Non-ideal solutions

7. Ideal solutions
Solution which obey Raoult’s law under all standard temperature and concentration.
No heat is evolved or absorbed during the mixing process. (i.e. It satisfies the
∆𝑉𝑚𝑖𝑥𝑖𝑛𝑔 = 0 & ∆𝐻𝑚𝑖𝑥𝑖𝑛𝑔 = 0)
The solution of two components A and B in which the A---B interactions are of same
magnitude as A---A and B---B interaction.
Only solutions with low concentration of solute behave ideally.
Example: benzene+toluene, chlorobenzene+bromobenzene, n-hexane+n-heptane
 When the intermolecular forces of attraction between A—A, B—B and A—B are
nearly equal.

8. Non-ideal solutions
The solution which do not follow Raoult’s law.
∆𝑉𝑚𝑖𝑥𝑖𝑛𝑔 ≠ 0 𝑎𝑛𝑑 ∆Hmixing≠0
It is the solution in which solute and solvent molecules interact with one
another with a different force than the forces of interaction between the
molecules of the pure compounds.
Ex. Sulphuric acid (solute) and water (solvent) the amount of heat is evolved
is large and thus change in volume is also seen.
Non-ideal solutions are of two types: non ideal positive and negative
deviations

9. Deviations
Positive deviations
In mixtures showing a positive deviation from Raoult’s law, the vapour pressure
of the mixture is always higher than you would expect from an ideal mixture.
This means that molecules are breaking away more easily than they do in the
pure liquids.
Negative deviations
Here the vapour pressure of the mixture is less than would be expected by the
Raoult’s law.
This means that molecules break away from the mixture less easily than they
do from the pure liquids.

10. SOLUBILITY OF GASES IN LIQUIDS
Solubility of gases in liquids is the concentration of dissolved gas in the
liquid when it is in equilibrium with the pure gas above the solution.
Examples include: effervescent preparations containing dissolved CO2,
ammonia and hydrochloride gas.
NH3 (aq)+H2O (l) NH4+(aq) + OH-(aq)
Aerosol products containing nitrogen or carbon dioxide as propellant
 The gas solubility is greatly affected by temperature and pressure as
well as the nature of the solute and solvent.

11. Factors that influence solubility generally include
Temperature
Nature of solvent
Pressure
pH
Particle size
Crystal structure
Molecular structure
Solute-solvent interactions
Melting and boiling points
Addition of substituent
Solubilizing agents
The two major factors that influence solubility of gases in liquid are:
Pressure and temperature

12. 1. Effect of Pressure
The solubility of a gas in a liquid is directly proportional to the partial pressure of the
gas present above the surface of the liquid or solution.
Henry’s law- the partial pressure of a gas above a solution is proportional to the mole
fraction of the gas in the solution.
P=KH x, where
P=partial pressure of the gas
X=mole fraction of the gas in solution
KH =Henry’s law constant
Effect of pressure-increase in pressureincrease in solubility

13. 2. Effect of temperature
Temperature is a measure of the average kinetic energy.
Effect of temperature-increase in temperaturedecrease in solubility
Why?
As temperature increases, kinetic energy increases. This greater kinetic
energy results in greater molecular motion of the gas particles. As a
result, the gas particles dissolved in the liquid are more likely to escape to
the gas phase and the existing gas particles are less likely to be
dissolved.

14. Physical properties of substances
Classified as additive, constitutive and colligative
A. Additive properties-depend on the total contribution of the
atoms in the molecule or on the sum of the properties of the
constituents in a solution.
- An example of additive property of a compound is the molecular
weight, i.e. the sum of the masses of the constituent atoms.
- The masses of the components of a solution are also additive.
The total mass of the solution being the sum of the masses of the
individual components.

15. B. Constitutive properties
 Depend on the arrangement and to a lesser extent on the
number and kind of atoms within a molecule.
Many physical properties may be partly additive and partly
constitutive.
The refraction of light, electric properties, surface and
interfacial characteristics, and the solubility of drugs are at least
in part constitutive and in part additive properties.

16. C. Colligative properties:
depend ONLY on the number of dissolved particles (molecules or ions, small or large) in
solution and not on their identity.
The properties include:
(i) Osmotic pressure
(ii) Vapour pressure lowering
(iii) Boiling point elevation
(IV) Freezing point depression

17. (I) OSMOTIC PRESSURE
 Diffusion in liquids-Substances tend to move or diffuse from regions of
higher concentration to region of lower concentration so the differences
in concentration disappear.
 Osmosis is the passage of the solvent molecules from pure solvent into
a solution through a semipermeable membrane. Solvent molecules move
through a semipermeable membrane (SPM) from a dilute solution to a
concentrated solution.
The flow of solvent towards the solution side across the SPM can be
stopped by applying extra pressure on the solution.

18. Osmotic pressure
Is the external pressure that must be applied to the solution in order to
prevent it being diluted by the entry of solvent via osmosis.
Osmotic pressure depends on the concentration of the solution.
Depends on the number of solute molecules and NOT their identity.
The osmotic pressure is proportional to the reduction in vapour pressure
brought about by the concentration of solute present.

19. Van’t Hoff’s equation:
𝜋𝑉 = 𝑛𝑅𝑇
𝜋 is the osmotic pressure in atm, V is the volume of the solution in liters, n is the number of moles pf solute,
R is the gas constant, equal to 0.082 liter atm/mole degree, T is the absolute temperature (Kelvin)
Example: 1 g of sucrose, molecular weight=342 g/mol, is dissolved in 100 mL of solution at 25 ⁰C. What is the
osmotic pressure of the solution?
Solution:
Step 1: calculate the moles of sucrose
Moles of sucrose=
1 𝑔
342 𝑔/𝑚𝑜𝑙
= 0.0029 mol
Step 2: Using Van’t Hoff’s equation, 𝜋𝑉 = 𝑛𝑅𝑇
𝜋𝑥100
𝑚𝐿
1000
𝑚𝐿 = 0.0029 𝑚𝑜𝑙𝑥 0.082 𝐿.
𝑎𝑡𝑚
𝑚𝑜𝑙.𝐾
𝑋 (25 + 273𝑘)=0.71 atm

20. Osmotic pressure of non-electrolytes (not ionized)
Osmotic pressure∝ 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏. This means twice concentration, twice osmotic pressure
𝑶𝒔𝒎𝒐𝒕𝒊𝒄 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 ∝number of molecules. This means osmotic pressure of two solutions having
the same molal concentration are identical.
Example: Explain why solution contains 34.2 g sucrose (mol wt 342) in 1000 g water has the same
osmotic pressure as dextrose solution (mol wt 180) contains 18 g/1000 g water?
Solution:
No of moles of sucrose=wt/mol wt (=34.2/342=0.1 molal)
- No of moles of dextrose
=18/180=0.1 molal.
Therefore, the two solutions are iso-osmotic.

21. Important terms to note
Isotonic solutions-Solutions that have the same osmotic
pressure at a given temperature.
Hypotonic solutions-Solutions that have a lower osmotic
pressure than other solutions.
Hypertonic solutions- Solutions that have higher osmotic
pressure than other solutions.

22. (II) Vapour pressure lowering
 When a non-volatile solute is dissolved in solvent, the
vapour pressure of the solvent is lowered.
Solvent molecules on the surface which can escape into
vapour is replaced by solute molecules and have little (if
any) vapour pressure.
The lowering of vapour pressure is ∆𝑝 = 0.018𝜌𝑖°𝑚

23. (III) BOILING POINT (BP) ELEVATION
The boiling point of a liquid is defined as the temperature at which the
vapour pressure of that liquid equals the atmospheric pressure (760
mmHg).
 For a solution, the vapour pressure of the solvent is lower at any given
temperature. Therefore, a higher temperature is required to boil the
solution than the pure solvent.
Elevation of the BP- The bp of a solution of a non-volatile solute is
higher than that of a pure solvent. This is due to the fact that the solute
lowers the vapor pressure of the solvent.
The more of the solute that is dissolved, the greater is the effect.

24.  Addition of solute will decrease the vapour pressure and so
will increase the boiling point.
The elevation of the boiling point is given as ∆𝑇𝑏 = 𝑇 − 𝑇0
∆= 𝑇𝑏 = 𝐾𝑏𝑚
m-molality(molality is the moles of solute per kilograms of
solvent)
Kb: boiling point elevation constant that depends on the
particular solvent being used. (Kb water=0.51 ⁰C/m)

25. (IV) FREEZING POINT DEPRESSION
Freezing (melting) point is the temperature at which solid and liquid are in equilibrium under 1 atm.
Addition of solute will decrease the vapour pressure and so will decrease the freezing point.
For a liquid to freeze, It MUST achieve a very ordered state that results in the formation of a crystal.
Depression of the freezing point
-If a solute is dissolved in the liquid at the triple point, the escaping tendency or vapour pressure of the
liquid solvent is lowered below that of the pure solid solvent.
-The temperature must drop to reestablish equilibrium between the liquid and the solid.
-Due to this, the freezing point of a solution is always lower than that of the pure solvent.
-The more concentrated the solution, the farther apart are the solvent and the greater is the freezing
point depression. ∆𝑇𝑓 = 𝑖𝐾𝑓𝑚

26. As per the law, the freezing point for a given dilute
solution stays directly proportional to molality of the
solute, same as the boiling elevation point
∆Tf = iKfm
∆Tf = change in temperature
𝑖 = 𝑡ℎ𝑒 𝑣𝑎𝑛′𝑡 𝐻𝑜𝑓𝑓 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑖𝑛𝑡𝑜 𝑤ℎ𝑖𝑐ℎ
𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑠
Kf = the molal freezing point constant for water is − 1.86
m = the molality, which is the moles of solute per kilograms of solvent

27. Example: What is the freezing point of a solution containing 3.42 g of sucrose and 500 g
of water? (The molecular weight of sucrose is 342. In this relatively dilute solution, Kf is
approximately equal to -1.86 ⁰C/m)
Solution:
∆𝑇𝑓 = 𝐾𝑓𝑚 = 𝐾𝑓
𝑤2𝑚1
𝑤1𝑚2
=−1.86 𝑥
1000𝑥3.42
500𝑥342
∆𝑇𝑓=- 0.037 ⁰C
Therefore the freezing point of the aqueous solution is
-0.037 ⁰C

28. Buffer solutions
A buffer solution is a solution that can resist changes in pH only slightly when small
amounts of strong acid or strong bases are added.
A buffer contains significant concentrations of both
-weak acid and its conjugate base
-Weak base and its conjugate acid
-Examples of buffers: Sodium acetate buffer-this buffer relies on the dissociation
reaction of acetic acid.
-CH3COOH ch3coo- + H+

29. Buffer capacity and range
- There is a limit to the ability of a buffer solution to neutralize added acid or base.
The buffer capacity is reached before either buffer component is consumed.
As a rule, a buffer is most effective if the concentrations of the buffer acid and its
conjugate base are equal or nearly so

30. PH: determination of PH buffers, theory of buffers.
The pH of a buffer is determined by two factors:
1. The equilibrium constant Ka of the weak acid
Different weak acids have different equilibrium constants (Ka). Ka tells us what proportion of HA will be
dissociated into H+ and A – in solution. The more H+ ions that are created, the more acidic and lower
the pH of the resulting solution.
2. The ratio of weak base [A-] and weak acid [HA] in solution.
The ratio of weak base [A-] to weak acid [HA] in a buffer also affects the pH.
-If a buffer has more base than acid, more OH- ions are likely to be present and the pH will rise.
-If a buffer has more acid than base, more H+ ions are present and the pH will fall.
-When the concentrations of A- and HA are equal, the concentration of H+ is equal to Ka, (or
equivalently pH=pKa)

31. Handerson-Hasselbalch equation
In predicting or determining the pH of a buffer solution, the HH equation is used given below
pH=pKa+log(
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
)
Since pKa is a property of the weak acid used in selecting our buffer, we can control the pH by
manipulating the proportion of weak base (A-) and weak acid (HA) in solution.
When [𝐴−]
[𝐻𝐴] remains close to 1, the pH remains close to pKa.
As the [𝐴−]
[𝐻𝐴] goes from 1/10 to 10, the pH changes from pKa+1 to pKa-1