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SOLUTION

Solution is a homogenous mixture or uniform physical combination with out chemical changesof two or more substancesSolute: Substance which is smaller in amount in a solution i.e. minor component.Solvent: Substance which is larger in amount in a solution i.e. major component /

dominated component.

Types of Solution:

SOLUTE + SOLVENT SOLUTION

(i) Gas + Gas Air(ii) Gas + Liquid Aerated water(iii) Gas + Solid H2 in Pd(iv) Liquid + Gas Vapors in air(v) Liquid + Liquid Alcohol in water

(vi) Liquid + Solid Hg in Na or Ag(vii) Solid + Gas Carbon in air (black smoke)(viii) Solid + Liquid Glucose in water(ix) Solid + Solid Alloys

Concentration expressions of solution:

I). Percentage composition= Solute x 100Solution

i. weightweightpercentage or (w/w%)ii. weightvolumepercentage or(w/v%)iii. volumeweightpercentage or(v/w%)iv. volumevolumepercentage or (v/v%)

II). Molarity (M) = No. of Moles of soluteVolume of solution in liter

No. of moles of solute = Amount of solute in gm

1 gm mol. Wt. of solute

1 gm mol. Wt. of solute = Sum of Atomic mass of all atoms present in molecule

Units:Molar or moles /liter

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III). Molality (m) = NO. of moles of soluteMass of solvent in kg

Units : Molal or moles/kgMolarity changes with temperature but molality does not change.

IV). Normality (N) = NO. of gm equivalent of soluteVolume of solution in liter/dm3

No. of gm equivalent of solute = Amount of solute in gm1 gm eq. wt. of solute

1 gm eq. wt. of solute = Molecular w.t of soluteAcidity / Basicity / e- loss or gain

Units:Normal or gm equivalent per liter

V). Mole Fraction (x)Mole ratio of a component with whole solution.e.g. A solution composed of solute A & solvent B.

Mole fraction of solute XA = nAnA + nB

Mole fraction of solvent XB = nBnA + nB

XA+ XB = 1 , no unit

VI). Parts per Million : (ppm)It is number of parts of solute which are present in a million parts of solution.

ppm = Amount of solute in mgVolume of solution in liter

Or

ppm = Amount of solute in gVolume of solution in ml

ppm = Molarity x Molecular w.t x 1000

Molarity = NO. of molesVolume of solution in liter

NO. of moles = Amount in gMolecular w.t

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ppm = NO. of moles x Amount in g x 1000Volume in liter NO. of moles

ppm = Amount in g x 1000Volume in liter

ppm = Amount in mgVolume in liter

ppm = Normality x Equivalent wt. x 1000

Normality = NO. of g eq. wt.Volume of solution in liter

NO. of g eq. wt. = Amount in gEquivalent wt.

ppm = NO. of g eq. wt. x Amount in g x 1000Volume in liter NO. of g eq. wt.

ppm = Amount in g x 1000Volume in liter

ppm = Amount in mgVolume in liter

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NUMERICALIt 24.5 g of H2SO4 has been dissolved in 250 g of water resulted 300 ml of solution. Write downnames of solute & solvent and calculate Molarity, Molality, Normality, Mole fraction of solutew/w %, w/v % and concentration of solution in ppm.

Data:

Amount of H2SO4 = 24.5 g (solute)Amount of water = 250 g = 0.25 kg (solvent)Volume of solution = 300ml =0.3 liter1 gm Mol wt. of H2SO4 = 98 g1 gm Eq. wt. of H2SO4 = 98 g = 49 g

2Solution:

NO. of mole of H2SO4 = 24.5 = 0.25 moles98

NO. of mole of water = 250 = 13.88 moles18

i. Molarity= 0.25 moles = 0.83 M or Molar0.3 liter

ii. Molality= 0.25 moles = 1 m or Molal0.25 kg

iii. Normality

NO. of Eq w.t of H2SO4 = 24.5 = 0.5 equivalent49

Normality = 0.5 equivalent = 1.66 N or Normal0.3 liter

iv. Mole fraction of H2SO4

X H2SO4 = 0.25 = 0.0170.25 + 13.88

v. w/w % = 24.5 x 100 = 8.92 w/w %24.5 + 250 g

vi. w/v % = 24.5 g x 100 = 8.16 w/v %300 ml

vii. ppm

ppm = 24.5 g x 1000 = 81666 ppm0.3 liter

ppm = 0.83 x 98 x 1000 = 81666 ppm

ppm = 1.66 x 49 x 1000 = 81666 ppm

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Physical Properties

Properties of a substance depend on the intermolecular forces, which depends upon internastructure. These properties do not involve chemical changes.1. Additive property: When a property of a molecule is equal to sum of that property of

constituent atoms. e.g. molecular mass.

2. Constitutive property: When a property of a molecule depends on the arrangement ofatoms and bond structure in molecule. e.g. M.P, B.P, Optical activity.

3. Additive & constitutive property: An additive property which also depend upon theintermolecular structure e.g. Surface tension, Viscosity and vapour pressure

SURFACE TENSION

It is the measure of force per unit length acting at a tangent to the meniscus surface.

Symbol = Units: In CGS system dyne/cm

In SI system N/m

Methods for determination of Surface tension1. Capillary rise Method:

acting along the inner circumference of the tube exactly supports the weight of liquidcolumn.

Upward force = Downward force

2 r = mg2 r = vdg

= r2hdg2 r

= rhdg2

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2. Drop formation method:Drop supported by the upward force of surface tension acting at outer circumference of thetube

Upward force = Downward force

2 r = mgA. Drop weight method:

For sample liquid 2 r = mgFor water 2 r w = mwgDivide equations 2 r = mg

2 rw mwg = mw mw = m w

mw

w = 72.0dynes/cm

B. Drop number method:

Volume of one drop of liquid = vn

Mass of one drop of liquid = m = v dn

= mw mw = (v/n) dw (v/nw)dw = nwdw ndw = nwd w

ndw

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VISCOSITY

It is internal frictional resistance force to flow

F A dV

dx

F = A dVdx

= F dxA dV

Force of resistance per unit area, which will cause unit velocity difference b/w two adjacentlayers of a liquid at a unit distance from each other.

Reciprocal of viscosity is called fluidity() = 1

Units :

= F dxA dV

= mass length x time -2 lengthlength2 length time-1

= mass x length-1 x time-1 gCm-1S-1 = poise Kgm-1S-1

CGS System SI System

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Measurement of viscosity by Oswalds method

dt

= kdt

w = kdwtw = k d tw kdwtw

= dtw dwtw

= dt wdwtw

w = 0.0101poise

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VAPOUR PRESSURE

Liquid Vaporization GaseousCondensation

At equilibrium,Rate of vaporization = Rate of condensation

The pressure exerted by the vapours in equilibrium with its liquid at a specifiedtemperature called vapour pressure at that temperature.

It is the measure of tendency of the tendency of a substance to evaporate.

Vapour pressure 1

Intermolecular forceAt 60 0C Ethanol = 350 torr, water= 150 torr

Vapour pressure Temperature

At 80 0C Ethanol = 730 torr, water = 410 torr

Boiling point is the temperature at which vapour pressure of liquid becomes equal toatmospheric pressure or surrounding pressure

So, Boiling point Atmospheric pressure

e.g. high altitude areas, pressure cooker

GlycerinB.P is 290 0C at 760 torr but it decompose

B.P is 2100

C at 50 torr.

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BUFFER SOLUTION

A solution which has tendency to reserve its pH

1) Acid buffer: Mixture of weak acid and its salts e.g. (CH3COOH + CH3COONa),(HCOOH + HCOONa), ( C2H5COOH + CH5COONa)

Mechanism:

CH3COOH

CH3COONa

CH3COOH

H2O

CH3COO- + H+partially dissociated

completely dissociatedCH3COO

-+ Na

+

Addition of OH-

Addition of H+

___________________________________________________________

___________________________________________________________

Acidic

Buffer

Case I : Addition of small amount of base (OH-)

The OH-ions added will react with H

+to form H2O. CH3COOH will ionize to compensate deficiency of

H+

ions and hence pH of the solution remains constant.

Case II : Addition of small amount of acid (H+)

The H+

ion added will combine with CH3COO-to form CH3COOH. CH3COOH is a weak acid, which is

very feebly ionized, and hence the concentration of H+

ion almost remain same and therefore the pH of

the solution remains unaltered.

Henderson Equation for acid buffer:

CH3COOH CH3COO-

+ H+

Ka = [H+][CH3COO

-]

[CH3COOH]

[H+] = Ka[CH3COOH][CH3COO

-

]

[H+] = Ka [Acid][Salt]

log[H+] = logKa log[Acid][Salt]

pH = pKa log [Acid][Salt]

Ka

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pH [Salt]

[Acid]

If[salt] = [Acid] thenlog[salt] = log 1 = 0[Acid]

So pH = pka

2) Basic buffer :Mixture of weak base and its salte.g. (NH4OH + NH4Cl)

Mechanism:

NH4Cl

NH4OH

NH4OH

H2O

NH4+

+ OH-partially dissociated

completely dissociated NH4

+ + Cl

-

Addition of H+

Addition of OH-

___________________________________________________________

___________________________________________________________

Basic

Buffer

Case I : Addition of small amount of acid (H+)

The H+

ion added will combine with OH-

to form H2O molecule. . NH4OH will ionize to compensate

deficiency of OH-

ions and hence pH of the solution remains constant.

Case II : Addition of small amount of base (OH-)

The OH-ions added will react with NH4

+to form NH4OH. Ammonium hydroxide is a weak base, which

is very feebly ionized, and hence the concentration of OH-

ions almost remains same and therefore thepH of the solution remains unaltered.

Henderson equation for Basic Buffer

NH4OH NH4+

+ OH-

Kb = [NH4+][OH

-]

[NH4OH]

[OH-] = Kb [NH4OH]

[NH4+]

[OH-] = Kb [Base]

[Salt]

pH = pKa + log[Salt][Acid]

Kb

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log[OH-] = logKblog[Base][Salt]

pOH = pKb log [Base][Salt]

pOH [Salt]

[Base]

If [Salt] = [Base] thenlog[Salt] = log1 = 0[Base]

So pOH = PKb

NUMERICAL

A chemist desire to prepared 300 ml of a buffer solution at pH = 9. How many grams of NH4Chave to be added to 0.20 M NH3 to make such a buffer. pkb value of ammonia is 4.75 .

Solution :pH = 9

pOH = 14 pH = 14 9 = 5

pOH = pKb + log [Salt][Base]

5 = 4.75 + log[NH4Cl][NH3]

5 4.75 = log[NH4Cl] log[NH3]

0.25 = log[NH4Cl] log 0.20

log[NH4Cl] = 0.25 0.698

log[NH4Cl] =0.448

[NH4Cl] = Antilog ( 0.448)

[NH4Cl] = 0.356M

Amount = MolarityMol w.tVolume (liter)

pOH = pKb + log[Salt][Base]

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Amount of [NH4Cl] = 0.35653.53001000

Amount of [NH4Cl] = 5.71 g

NUMERICALFind out the pH of a 200 ml buffer solution containing 2.88 g of CH3COONa and 1.8 g ofCH3COOH While ka for acetic acid is 1.8 x 10

-5

Solution:

Volume of solution = 200 ml = 0.2 liter

A. Sodium AcetateAmount of CH3COONa = 2.88 g

NO. of mole of CH3COONa = 2.88 g = 0.035 mole82 g/mole

Molarity of CH3COONa = 0.035 mole = 0.175 M0.2 liter

B. Acetic AcidAmount of CH3COOH = 1.8 g

NO. of mole of CH3COOH = 1.8 g = 0.03 mole60 g/mole

Molarity of CH3COOH = 0.03 mole = 0.15 M0.2 liter

pH = pka + log [Salt][Acid]

pH = pka + log [CH3COONa][CH3COOH]

pH = - log(1.8 10-5) + log 0.1750.15

pH = 4.74 + log0.175 log0.15

pH = 4.74 0.69 + 0.82

pH = 4.87