solution chemistry 02

7/28/2019 Solution Chemistry 02

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SOLUTE + SOLVENT SOLUTION

Gas + Gas Air

Gas + Liquid Aerated water, coke

Gas + Solid H2 in Pd

Liquid + Gas Vapors in air

Liquid + Liquid Alcohol in water (vodka)

Liquid + Solid Hg in Na or Ag

Solid + Gas Carbon in air (black smoke)

Solid + Liquid Glucose in water

Solid + Solid Alloys e.g (14-karat gold)

Types of Solution:


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The amount of substance dissolved per unit volume of

solution or in a specific weight of solvent is termed as

concentration. It can be expressed in a number of ways,

I). Percentage composition

II). Molarity (M)

III). Molality (m)

IV). Normality (N)

V). Mole Fraction (x)

VI). Parts per Million (ppm)

Concentration expressions of solution:


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I) Percentage composition= Solute x 100

Solution

weight weight percentage or (w/w %) weight volume percentage or (w/v %)

volume weight percentage or (v/w %)

volume volume percentage or (v/v %)


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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. w.t of solute

1 gm mol. w.t 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 solute

Mass of solvent in kg

Units : Molal or moles/kg

Molarity changes with temperature but molality does not

change.

Molarity decreases as temperature increases.Why?


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IV) Normality (N) = NO. of gm equivalent of solute

Volume of solution in liter/dm3

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

1 gm eq. w.t of solute

1 gm eq. w.t of solute = Molecular w.t of solute

Acidity / Basicity / e- loss or gain

Units: Normal or gm equivalent per liter


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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 = nB

nA + nB

XA + XB = 1 , no unit

To obtain mole percent we simply multiply mole fraction by 100.


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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

ppm = Amount of solute in g

Volume of solution in ml

ppm = Molarity x Molecular w.t x 1000


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Molarity = NO. of molesVolume of solution in liter

NO. of moles = Amount in gMolecular w.t

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

Volume in liter NO. of moles

ppm = Amount in g x 1000Volume in liter

ppm = Amount in mg

Volume in liter


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ppm = Normality x Equivalent w.t x 1000

Normality = NO. of g eq. w.tVolume of solution in liter

NO. of g eq. w.t = Amount in gEquivalent w.t

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

ppm = Amount in g x 1000

Volume in liter

ppm = Amount in mg

Volume in liter


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Q. Which concentration is used when we want

temperature independent criteria?


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NUMERICAL

If 24.5 g of H2SO4 has been dissolved in 250 g of water resulted 300

ml of solution. Write down names of solute & solvent and calculateMolarity, Molality, Normality, Mole fraction of solute, w/w %, w/v %

and concentration of solution in ppm?


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Data:

Amount of H2SO4 = 24.5 g (solute)

Amount of water = 250 g = 0.25 kg (solvent)

Volume of solution = 300ml =0.3 liter

1 gm Mol w.t of H2SO4 = 98 g

1 gm Eq. w.t of H2SO4 = 98 g = 49 g

2


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Solution:

NO. of mole of H2SO4 = 24.5 = 0.25 moles

98

NO. of mole of water = 250 = 13.88 moles

18

Molarity= 0.25 moles = 0.83 M or Molar

0.3 liter

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


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Normality

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

49

Normality = 0.5 equivalent = 1.66 N or Normal

0.3 liter

Mole fraction of H2SO4

X H2SO4 = 0.25 = 0.017

0.25 + 13.88

w/w % = 24.5 x 100 = 8.92 w/w %

24.5 + 250 g

w/v % = 24.5 g x 100 = 8.16 w/v %

300 ml


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ppm

ppm = 24.5 g x 1000 = 81666 ppm

0.3 liter

ppm = 0.83 x 98 x 1000 = 81666 ppm

ppm = 1.66 x 49 x 1000 = 81666 ppm


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

meniscus surface.

Surface tension is the amount of work required to extend a liquid

surface.

Symbol = Units: In CGS system = dyne/cm

In SI system = N/m

SURFACE TENSION


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Cohesion

Water clings to polar

molecules throughhydrogen bonding

Cohesionrefers to

attraction to other water

molecules. responsible forsurface

tension

a measure of the force

necessary to stretch or

break the surface of aliquid


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Q. Compare the surface tension of hexane

water and mercury?


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Q. What will be the effect of temperature onsurface tension?


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Methods for determination of

Surface tension1) Capillary rise method.

2) Drop formation method.

(i) Drop weight method

(ii) Drop number method

3) Ring detachment method.

4) Maximum bubble pressure method.


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(i) Capillary rise Method:

acting along the inner circumference of the tubeexactly supports the weight of liquid column.

Upward force = Downward force

2 r = mg2 r = v d g

= r2h d g2 r

= r h d g2

t t f


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Capillary actionwater evaporates fr om

leaves = transp irat ion

adhesion,

cohesion and

capi l lary act io n

water taken up by

roots
http://rds.yahoo.com/S=96062883/K=capillary+action/v=2/SID=e/l=IVI/*-http://physicsed.buffalostate.edu/courses/00/spring/sci420/misc/students/bayham_zoe/Celery_files/cups.jpeg


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(ii) Drop formation method:

Drop supported by the upward force of surface tension acting

at outer circumference of the tube

Upward force = Downward force

2 r = mgThe apparatus employed is called stalagmometerordrop pipette.

A. Drop weight method:

For sample liquid 2 r = mgFor water 2 r w = mwg

Divide equations 2 r = mg2 rw mwg

= mw mw = m w

mw

w = 72.0 dynes/cm


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B. Drop number method:Volume of one drop of liquid = v

nMass of one drop of liquid = m = v d

n

= mw mw = (v/n) dw (v/nw)dw = nw dw n dw

= nwd wn dw


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Viscosity of a liquid is a measure of its frictional resistance.

It is internal frictional resistance force to flow

F A dV

dx

VISCOSITY


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Is the property of a liquid, by which a liquid

opposes relative motion between its

layers.

VISCOSITY


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F = A dVdx

= F dxA dV

Force of resistance per unit area, which will cause unit velocity difference b/wtwo adjacent layers 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 lengthlength

2

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

Ostwalds method

d t

= k d t w = k dwtw = k d t w k dwtw = d t

w dwtw

= d t wdwtw

w = 0.0101 poise


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Liquid Vaporization Gaseous

Condensation

At equilibrium,

Rate of vaporization = Rate of condensation

The pressure exerted by the vapours in dynamic equilibrium with

its liquid at a specified temperature called vapour pressure at

that temperature.

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

VAPOUR PRESSURE


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Vapour pressure 1

Intermolecular force

At 60 degree C Ethanol = 350 torr, water= 150 torr

Vapour pressure Temperature

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

Boiling point is the temperature at which vapour pressure of liquidbecomes equal to atmospheric pressure or surrounding pressure

So, Boiling point Atmospheric pressure

e.g high altitude areas, pressure cooker


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Q. Why at high altitude areas rate of cooking is slow?

Q. Why cooking is fast in pressure cooker?


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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 + C2H5COONa)Mechanism:

CH3COOH

CH3COONa

CH3COOH

H2O

CH3COO

-

+ H+partially dissociated

completely dissociatedCH3COO

-+ Na

+

Addition of OH-

Addition of H+

___________________________________________________________

___________________________________________________________

AcidicBuffer

BUFFER SOLUTION


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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.


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Henderson Equation for acid buffer:

Ka

CH3COOH CH3COO- + H+

Ka = [H+][CH3COO-]

[CH3COOH]

[H+] = Ka [CH3COOH][CH3COO-]

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

log [H+] = log Ka log [Acid]

[Salt]

pH = pKa log [Acid]

[Salt]


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pH [Salt][Acid]

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

[Acid]

So pH = pka

pH = pKa + log [Salt]

[Acid]


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2) Basic buffer : Mixture of weak base and its salt

e.g (NH4OH + NH4Cl)

Mechanism:

NH4Cl

NH4OH

NH4OH

H2O

NH4+

+ OH-partially dissociated

completely dissociated NH4

+ + Cl

-

Addition of H+

Addition of OH-

___________________________________________________________

___________________________________________________________

Basic

Buffer


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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. Ammoniumhydroxide is a weak base, which is very feebly ionized, and hencethe concentration of OH- ions almost remain same and therefore the

pH of the solution remains unaltered.


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Henderson equation for Basic Buffer:

KbNH4OH NH4+ + OH-

Kb = [NH4+] [OH-]

[NH4OH]

[OH-] = Kb [NH4OH]

[NH4+]

[OH-] = Kb [Base]

[Salt]

log [OH-] = log Kb log [Base]

[Salt]


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pOH = pKb log [Base]

[Salt]

pOH [Salt][Base]

If [Salt]=[Base] then log [Salt] = log 1= 0

[Base]

So pOH= pkb

pOH = pKb + log [Salt]

[Base]


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NUMERICAL

A chemist desire to prepared 300 ml of a buffer solution at pH = 9.

How many grams of NH4Cl have to be added to 0.20 M NH3 to

make such a buffer. pKb value of ammonia is 4.75 ?


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Solution:

pH = 9pOH = 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] = 0.356 M


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Amount = Molarity Mol w.t Volume (liter)

Amount of [NH4Cl] = 0.356 53.5 3001000

Amount of [NH4Cl] = 5.71 g


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NUMERICAL

Find out the pH of a 200 ml buffer solution containing 2.88 g of

CH3COONa and 1.8 g of CH3COOH While ka for acetic acid is

1.8 x 10 -5 ?


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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 mole

82 g/mole

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

B. Acetic Acid

Amount 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 M

0.2 liter


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pH = pka + log [Salt][Acid]

pH = pka + log [CH3COONa]

[CH3

COOH]

pH = - log(1.8 10-5) + log 0.175

0.15

pH = 4.74 + log 0.175 log 0.15pH = 4.74 0.69 + 0.82

pH = 4.87

solution chemistry 02

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