Soln Chem Clr
-
Upload
ahmed-khan -
Category
Documents
-
view
254 -
download
0
Transcript of Soln Chem Clr
-
8/4/2019 Soln Chem Clr
1/13
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
-
8/4/2019 Soln Chem Clr
2/13
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
-
8/4/2019 Soln Chem Clr
3/13
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
-
8/4/2019 Soln Chem Clr
4/13
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
-
8/4/2019 Soln Chem Clr
5/13
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
-
8/4/2019 Soln Chem Clr
6/13
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
-
8/4/2019 Soln Chem Clr
7/13
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
-
8/4/2019 Soln Chem Clr
8/13
Measurement of viscosity by Oswalds method
dt
= kdt
w = kdwtw = k d tw kdwtw
= dtw dwtw
= dt wdwtw
w = 0.0101poise
-
8/4/2019 Soln Chem Clr
9/13
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.
-
8/4/2019 Soln Chem Clr
10/13
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
-
8/4/2019 Soln Chem Clr
11/13
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
-
8/4/2019 Soln Chem Clr
12/13
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]
-
8/4/2019 Soln Chem Clr
13/13
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