CBSE NOTES CLASS 12 CHEMISTRY CHAPTER 2

SOLUTIONS

Solution

Solution is a homogeneous mixture of two or more substances in same or different physical phases.

The substances forming the solution are called components of the solution.

A solution of two components is called binary solution.

Solute and solvent

In a binary solution, solvent is the component which is present in large quantity while the other component is known as solute.

Classification of solutions

(A) Following types of solutions are seen on the basis of physical state of solute and solvent.

 Type of Solution Solute Solvent Example Gaseous solution Gas Gas Mixture of oxygen and nitrogen gases Liquid Gas Chloroform mixed with nitrogen gas Solid Gas Camphor in nitrogen gas Liquid solutions Gas Liquid Oxygen dissolved in water Liquid Liquid Ethanol dissolved in water Solid Liquid Glucose dissolved in water Solid solutions Gas Solid Solution of hydrogen in palladium Liquid Solid Amalgam of mercury with sodium Solid Solid Copper dissolved in gold

Aqueous Solution

If water is used as a solvent, the solution is called aqueous solution and if not, the solution is called non-aqueous solution.

(B) Depending upon the amount of solute dissolved in a solvent we have the following types of solutions:

(i) Unsaturated solution

A solution in which more solute can be dissolved without raising temperature is called an unsaturated solution.

(ii) Saturated solution

A solution in which no solute can be dissolved further at a given temperature is called a saturated solution.

(iii) Supersaturated solution

A solution which contains more solute than that would be necessary to saturate it at a given temperature is called a supersaturated solution.

Solubility

The maximum amount of a solute that can be dissolved in a given amount of solvent (generally 100g) at a given temperature is termed as its solubility at that temperature.

• Polar solutes dissolve in polar solvents and non polar solutes in nonpolar solvents.

Like dissolves like.

Concentration of Solutions

The concentration of a solution is defined as the relative amount of solute present in a solution. On the basis of concentration of solution there are two types of solutions.

(i) Dilute solution (ii) Concentrated solution.

Methods of Expressing Concentration of Solutions :

(i) Percentage by mass (w/w %): It is defined as the amount of solute present in 100 g of solution.

(ii) Volume percentage (V/V %) is defined as:

(iii) Mass by volume (w/V) is defined as the weight of solute present in 100 mL of solution.

(iv) Parts per million (ppm) is defined as the parts of a component per million parts (106) of the solution. It is widely used when a solute is present in trace quantities.

(v) Mole fraction (χ) is defined as the ratio of the number of moles of a component to the total number of moles of all the components. For a binary solution, if the number of moles of A and B are nA and nB respectively, the mole fraction of A will be

(vi) Molarity (M) is the number of moles of solute present in 1L (dm3) of the solution.

Also,

• Molarity varies with temperature due to change in volume of solution.

When molarity of a solution is 1 M, it is called a molar solution. 0.1 M solution is called a decimolar solution while 0.5 M solution is known as semi molar solution

(vii) Molality (m) is the number of moles of solute per kilogram of the solvent.

• Molality is independent of temperature.

When solvent used is water, a molar (1 M) solution is more concentrated than a molal (1 m) solution. (Why?)

(viii) Normality (N) is the number of gram equivalents of solute present in 1 L of solution.

Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is, the mass of a given substance which will

• combine or displace directly or indirectly with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine – these values correspond to the atomic weight divided by the usual valence; or

• supply or react with one mole of hydrogen cations (H+) in an acid–base reaction; or

• supply or react with one mole of electrons (e) in a redox reaction.

• Like molarity, normality relates the amount of solute to the total volume of solution; however, normality is specifically used for acids and bases.

The mole equivalents of an acid or base are calculated by determining the number of H+ or OH- ions per molecule:

Relationship between molarity and normality

N = n × M (where n is an integer)

For an acid solution, n is the number of H+ ions provided by a formula unit of acid (i.e. basicity).

For a basic solution, n is the number of OH+ ions provided by a formula unit of base (i.e. acidity).

Also,

N × Eq. mass = M × mol. Mass

(ix) Formality (F) is the number of formula weights of solute present per litre of the solution.

(Mass fraction of any component in the solution is the mass of that component divided by the total mass of the solution.

• Molality, mole fraction and mass fraction are preferred over molarity, normality, etc., because former involve weights which do not change with temperature, whereas later involves volumes, which are temperature dependent.

Dilution Law

• For dilution from a solution of molarity M1 and volume V1 to molarity M2 and volume V2

M1V1 = M2V2

If a solution with volume V1 and normality N1 is diluted to volume V2 with normality N2, then,

N1 × V1 = N2 × V2

• For a balanced chemical equation, if n1 moles of reactant reacts with n2 moles of reactant 2, then,

Where n1 and n2 are stoichiometric coefficient in balanced equation

• If two solutions of the same solute having volumes and molarities V1, M1 and V2, M2 are mixed, the molarity of the resulting solution is,

$\mathrm{M}=\frac{{\mathrm{M}}_{1}{\mathrm{V}}_{1}+{\mathrm{M}}_{2}{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}+{\mathrm{V}}_{2}}$

And

$\mathrm{N}=\frac{{\mathrm{N}}_{1}{\mathrm{V}}_{1}+{\mathrm{N}}_{2}{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}+{\mathrm{V}}_{2}}$

• To dilute V1 mL of a solution having molarity M1 to molarity M2 up to the final volume V2 mL, the volume of water added is

${\mathrm{V}}_{2}-{\mathrm{V}}_{1}=\left(\frac{{\mathrm{M}}_{1}-{\mathrm{M}}_{2}}{{\mathrm{M}}_{2}}\right){\mathrm{V}}_{1}=\left(\frac{{\mathrm{N}}_{1}-{\mathrm{N}}_{2}}{{\mathrm{N}}_{2}}\right){\mathrm{V}}_{1}$

• For a chemical reaction involving an acid and a base,

na × Ma × Vb = nb × Mb × Vb

where,

na = Basicity of acid

nb = Acidity of base

Va = Volume of acid

Vb = Volume of base

Ma = Molarity of acid

Mb = Molarity of base

Solubility of a Solid in a Liquid

Effect of temperature

The solution being in dynamic equilibrium, must follow Le Chateliers Principle.

• If in a nearly saturated solution, the dissolution process is endothermic (ΔsolH > 0), the solubility should increase with rise in temperature

• If it is exothermic (ΔsolH < 0), the solubility should decrease.

Effect of pressure

Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

Solubility of a Gas in a Liquid

Effect of Pressure – Henry’s Law

The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.

χ ∝ p

The most commonly used form of Henry’s law states

“the partial pressure (p) of the gas in vapour phase is proportional to the mole fraction (χ) of the gas in the solution”

and is expressed as

p = KH χ .

Where, KH. is a proportionality constant called Henery’ constant.

• Greater the value of KH, lower the solubility of the gas.

• KH is characteristic of gas-liquid pair.

• The value of KH increases with increase in the temperature.

• SI unit of KH is bar

Applications of Henry’s Law

1. In manufacture of soft drinks and soda water, CO2 is passed at high pressure to increase its solubility.

2. In case of deep sea diving, increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life.

To minimise the painful effects (bends) accompanying the decompression of deep sea divers, O2 is diluted with less soluble He gas as breathing gas.

3. At high altitudes, the partial pressure of O2 is less than that at the ground level. This leads to low concentrations of O2 in the blood of climbers which causes anoxia.

Effect of Temperature

Dissolution of gases in liquids is an exothermic process. Hence the solubility of gases in liquids decreases with rise in temperature. Thus, aquatic species are more comfortable in cold water [more dissolved O2] rather than warm water.

Vapour Pressure of Liquid-Liquid Solutions – Raoult’s Law

For a solution of two volatile liquids, the vapour pressure of each liquid in the solution is less than the respective vapour pressure of the pure liquids and the equilibrium partial vapour pressure of the liquid is directly proportional to its mole fraction.

In case of a binary solution, containing two liquids 1 and 2 with mole fractions χ1 and χ2, for component 1 and 2 respectively,

p1χ1 p1 = k χ1

The proportionality constant is obtained by considering the pure liquid.

When χ1 = 1 then k = po1, where po1 is the vapour pressure of pure component 1 at the same temperature.

Hence, p1 = po1 χ1

Similarly, for component 2

P2χ2 p2 = po2 χ2

where po2 is the vapour pressure of pure component 2 at the same temperature.

According to Dalton’s law of partial pressures, the total pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution. Or,

ptotal = p1 + p2

Substituting the values of p1 and p2, we get

ptotal = po1 χ1 + po2 χ2

= (1-χ2) po1 + po2 χ2

= po1 + (po2 - po12

Konowaloff Rule

At any fixed temperature, the vapour phase is always richer in the more volatile component as compared to the solution phase.

In other words, mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase.

The composition of vapour phase in equilibrium with the solution is determined by the partial pressure of components. If y1 and y2 are the mole fractions of the component 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressure, then

p1 = y1 × ptotal

p2 = y2 × ptotal

Raoult’s Law and Henry’s Law

Raoult’s Law - The vapour pressure of a volatile component in a given solution is given by,

p1 = po1 χ1

[Raoult’s law gives the partial vapour pressures of components in terms of solutbility of components]

Henry’s Law - Solubility of gas is given by,

p = KHχ.

Hence Raoult’s law is a special case of Henry’s law in which KH becomes po1.

Raoult’s Law for Solutions of Solids in Liquids

Raoult’s law in its general form can be stated as,

For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

If component 1 is the solvent then,

p1 = po1 χ1

Since the other component is solid p2 = 0

Ideal solutions

The solutions which satisfy the following conditions called ideal solutions.

1. Solution must obey Raoult‟s law, i.e.,

p1 = po1 χ1 and p2 = po2 χ2

2. ΔmixH = 0 (No energy evolved or absorbed)

3. ΔmixV = 0 (No expansion or contraction on mixing)

Some solutions behave like nearly ideal solutions are,

• benzene + toluene

• n-hexane + n-heptane

• ethyl iodide + ethyl bromide

• chlorobenzene + bromobenzene.

Non-ideal solutions

Those solutions which show deviation from Raoult’s law, are called non-ideal solution.

For such solutions, ΔmixH ≠ 0 and ΔmixV ≠ 0

The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law.

If it is higher, the solution exhibits positive deviation if it is lower, it exhibits negative deviation from Raoult’s law.

Non-ideal solutions showing positive deviation

In such a case, the A – B interactions are weaker than A – A or B – B interactions and the observed vapour pressure of each component and the total vapour pressure are greater than that predicted by Raoult‟s law.

For such solutions

p1 > po1 χ1

p2 > po2 χ2

ptotal > po1 χ1 + po2 χ2

ΔHmix > 0 and ΔVmix > 0

Examples

• Ethnol + Water,

• CS2 + Acetone,

• CCL4 + C6H6,

• CCL4 + C6H5CH3,

• Ethnol + Cyclohexane,

• CCL4 + CHCl3

Explanation

• Mixtures of ethanol and acetone behave in this manner.

In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them. Due to weakening of interactions, more vapour is formed and the solution shows positive deviation from Raoult’s law.

• In a solution formed by adding carbon disulphide to acetone, the dipolar interactions between solute-solvent molecules are weaker than the respective interactions among the solute-solute and solvent-solvent molecules.

Non-ideal solution showing negative deviation

In such a case, the A – B interactions are stronger than A – A or B – B interactions and the observed vapour pressure of each component and the total vapour pressure are lesser than that predicted by Raoult’s law.

For such solutions

p1 < po1 χ1

p2 < po2 χ2

ptotal <po1 χ1 + po2 χ2

ΔHmix < 0 and ΔVmix < 0

Examples

• CHCl3 + CH3COCH3

• CHCl3 + C6H6

• H2O + HCl

• H2O + HNO3

• Methanol + acetic acid

Explanation

• An example of this type is a mixture of phenol and aniline.

In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules.

• Similarly, a mixture of chloroform and acetone forms a solution with negative deviation from Raoult’s law, because chloroform molecule is able to form hydrogen bond with acetone molecule.

This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting in negative deviation from Raoult’s law.

Azeotropic mixture

The binary mixtures which are having the same composition in liquid and vapour phase and boil at a constant temperature are called Azeotropic mixtures.

In such cases, it is not possible to separate the components by fractional distillation.

1. Minimum boiling azeotropes are formed by those liquid pairs which show large positive deviation from ideal behaviour. Such azeotropes have boiling points lower than either of the components, e.g., C2H5OH (95.57%) + H2O (4.43%) by mass.

2. Maximum boiling azeotropes are formed by those liquid pairs; which show negative deviation from ideal behaviour. Such azeotropes have boiling points higher than either of the components. e.g., H2O (20.22%) + HCl (79.78%) and HNO3 (68%) + H2O (32%) by mass.

Colligative properties

(colligative: Latin - co means together, ligare means to bind)

Colligative properties are those properties of solution which depend only on the number of solute particles in a solution irrespective of their nature.

Explanation

The vapour pressure of solution decreases when a non-volatile solute is added to a volatile solvent.

We know that evaporation is a surface phenomenon. The fraction of molecules changing from liquid phase to gaseous phase depends on the number of molecules of the solvent at the surface.

When a non-volatile solid is added to a solvent, the concentration of solvent molecules, at the surface, decreases. As a result, the number of molecules changing from liquid phase to gas phase decreases and vapour pressure decreases.

Some of the important colligative properties of solutions

1. Relative lowering of vapour pressure of the solvent

2. Depression of freezing point of the solvent

3. Elevation of boiling point of the solvent and

4. Osmotic pressure of the solution.

Relative lowering of vapour pressure

It is the ratio of lowering in vapour pressure to vapour pressure of pure solvent. The relative lowering in vapour pressure of solution containing a nonvolatile solute is equal to the mole fraction of solute in the solution.

For a solute (2) in solvent (1), we have,

p1 = po1 χ1

Hence

po1 - p1 = (po1 - po1 χ1)

= po1 (1- χ1) = po1 χ2

For dilute solutions n2 << n1

Therefore,

Or,

Where, W2 and W1 = mass of solute and solvent respectively. M2 and M1 = molecular weight of solute and solvent respectively.

From the above expression we can find the molecular weight of an unknown solute dissolved in a given solvent.

Elevation in boiling point (ΔTb)

Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure. As the vapour pressure of a solution containing a nonvolatile solute is lower than that of the pure solvent, its boiling point will be higher than that of the pure solvent.

The increase in boiling point is known as elevation in boiling point and is given by,

ΔTb = Tb – T°b

It has been found experimentally that for dilute solutions the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of the solute in a solution.

ΔTb = Kb m

where; m = molality of the solution.

Kb is called the molal elevation constant or ebullioscopic constant. The unit of Kb is K kg mol-1

Calculation of Molecular Mass of Solute

If W2 gram of solute of molar mass M2 is dissolved in W1 gram of solvent, then molality, m of the solution is given by

For water, Kb = 0.52 K kg mol-1

Depression in Freezing Point (ΔTf)

Freezing point of a liquid is the temperature at which vapour pressure of the solvent in its liquid and solid phase become equal. As the vapour pressure of solution containing non-volatile solute is lower than that of pure solvent, solid form gets separated out at a lower temperature.

This decrease in freezing point of a liquid is known as depression in freezing point.

ΔTf = T°f – Tf

• It has been found experimentally that for dilute solutions the elevation of boiling point (ΔTf) is directly proportional to the molal concentration of the solute in a solution.

ΔTf = Kf m

where; m = molality of solution

Kf is called the molal depression constant or Cryoscopic Constant constant. The unit of Kf is K kg mol-1

Calculation of molecular mass using depression in freezing point

For water, Kf = 1.86 K kg mol-1

${\mathrm{K}}_{\mathrm{f}}=\frac{\mathrm{R}×{\mathrm{M}}_{1}×{{\mathrm{T}}_{\mathrm{f}}}^{2}}{{\mathrm{\Delta }}_{\mathrm{f}\mathrm{u}\mathrm{s}}\mathrm{H}×1000}$

${\mathrm{K}}_{\mathrm{b}}=\frac{\mathrm{R}×{\mathrm{M}}_{1}×{{\mathrm{T}}_{\mathrm{b}}}^{2}}{{\mathrm{\Delta }}_{\mathrm{v}\mathrm{a}\mathrm{p}}\mathrm{H}×1000}$

Where ${\mathrm{\Delta }}_{\mathrm{f}\mathrm{u}\mathrm{s}}\mathrm{H}$ and are molar enthalpies of fusion and vapourisation of the solvent, respectively.

Ethylene glycol is usually added to water in the radiator to lower its freezing point. It is called antifreeze solution.

[Common salt (NaCl) and anhydrous CaCl2 are used to clear snow on the roads because they depress the freezing point of water.]

Osmotic pressure (π)

Osmosis is the phenomenon of spontaneous flow of the solvent molecules through a semipermeable membrane from pure solvent to solution or from a dilute solution to concentrated solution.

Some natural semipermeable membranes are animal bladder, cell membrane etc. Cu2[Fe(CN)6] is an artificial semipermeable membrane which does not work in non-aqueous solutions as it dissolves in them.

Osmosis may be

1. Exosmosis - outward flow of water or solvent from a through semipermeable membrane.

2. Endosmosis - inward flow of water or solvent from a through a semipermeable membrane.

The hydrostatic pressure developed on the solution which just prevents the osmosis of pure solvent into the solution through a semipermeable membrane is called osmotic pressure.

For dilute solutions, it has been found experimentally that osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Or,

π = C R T

Here π is the osmotic pressure and R is the gas constant.

Here V is volume of a solution in litres containing n2 moles of solute.

If W2 grams of solute, of molar mass, M2 is present in the solution, then

Hence,

where R = gas constant, T = temperature

Note: Osmotic pressure method is the best method for determining the molecular masses of polymers since observed value of any other colligative property is too small to be measured with reasonable accuracy.

Types of solutions based on osmosis

On the basis of osmotic pressure, the solution can be

1. Hypertonic solution

A solution is called hypertonic if its osmotic pressure is higher than that of the solution from which it is separated by a semipermeable membrane.

When a plant cell is placed in a hypertonic solution, the fluid from the plant cell comes out and cell shrinks, this phenomenon is called plasmolysis.

(ii) Hypotonic solution

A solution is called hypotonic if its osmotic pressure is lower than that of the solution from which it is separated by a semipermeable membrane.

(iii) Isotonic solution

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. These solutions have same molar concentration. 0.91% solution of pure NaCl is isotonic with human RBC‟s.

Isotonic solutions have same molar concentration, e.g., if x % solution of X is isotonic with y % solution of Y, this means molar concentration of X = Molar concentration of Y,

$\frac{\mathrm{x}}{100}×\frac{1000}{{\mathrm{M}}_{\mathrm{x}}}=\frac{\mathrm{y}}{100}×\frac{1000}{{\mathrm{M}}_{\mathrm{y}}}$

In case the solution is not dilute,

Reverse osmosis

When the external pressure applied on the solution is more than osmotic pressure, the solvent flows from the solution to the pure solvent, which is called reverse osmosis. Desalination of sea water is done by reverse Osmosis.

Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water.

Abnormal molecular masses

When there is dissociation of solute into ions, the experimentally determined molar mass is always lower than the true value.

Similarly, using freezing point and boiling point method also, the experimentally determined molar mass is again lower than the true value.

van’t Hoff factor (i)

These observed values are corrected by multiplying the experimentally determined values by van‟t Hoff factor (i).

It is the ratio of observed value of colligative property to the calculated value of colligative property.

or

or

So to correct the observed value of molar mass, van‟t Hoff factor (i) must be included in different expressions for colligative properties.

ΔTb = i Kb m

ΔTf = i Kf m

π = i C R T

Degree of dissociation (α) and van’t Hoff factor (i)

1. If one molecule of a substance A gets dissociated into n particles or molecules

A → nP

And if α is the degree of dissociation then

 Total number of moles A P Initially 1 0 At equilibrium 1 - α n α

Therefore

Degree of association (α) and van’t Hoff factor (i)

If n molecules of a substance A associate to form An

And if α is the degree of association then

 Total number of moles A An Initially 1 0 At equilibrium 1 - α $\frac{\mathrm{\alpha }}{\mathrm{n}}$

Therefore

Note:

• van’t Hoff factor i > 1 for solutes undergoing dissociation

• it is < 1 for solutes undergoing association.