# 105 Chem.الجزء الثاني

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```					                         Stoichiometry
Chemical Arithmatic

Two aspects of chemistry

Qualitative                                Quantitative
‫كيفي أو نوعي‬                                     ‫كمي‬

Stoichiometry in the quantitative aspect of chemical reaction
and composition.

H2O
- Is composed of hydrogen and oxygen. (qualitatively).
- Is composed of two hydrogen atoms and one oxygen atom.
(Quantitatively)

H2 + 1/2 O2              H2O
Hydrogen + Oxygen                Water (Qualitatively)

One mole hydrogen + half mole oxygen                one mole water
(quantitatively).

1
Basic SI units:

Physical Quantity            Unit                Symbol
Mass                 Kilogram                kg
Length                  Meter                  m
Time                  Second                  s
Temperature               Kelvin                  K
Amount of substance           mole                  mol

The Mole:
Is formally defined as the amount of substance of a system that
contains as many dementary entities as there are atoms in 0.012
kg of 12C.

Or: It consists of Avogadro's number of objects.
i.e. 6.02 x 1023 object

mole = atomic weight (mass) in grams if the substance consists
of atoms.
1 mole of C = 12.01 g C

2
1 mole = Molecular weight (mass) in grams if the substance is
composed of molecules.

1 mole = Formula weight (mass) in grams if the substance is
composed of ions or molecules.

1 mole of CCl 4
Contains Avogadro's number of CCl4 molecules.
Contains Avogadro's number of C atoms.
Contains 4 moles of Cl atoms.
We need two moles of Cl2 molecules to make 1 mole of CCl4.

Ex: How many gram in 2.25 mol Cu. Cu = 63.5 U amu

1 mol Cu = 63.5 g Cu

moles of Cu = 2.25 mol Cu x 63.5g = 142.88 g
1mol cu

3
E.g: How many moles of Aluminum are required to react with
28.5g of Fluorine to form AlF3
F = 19

1 mole of Al atoms   ~     3 moles of F atoms

Number of mol of F atoms = 28.5g F atoms x 1 mol Fatoms
19 g F atoms

= 1.5 mol F

No. of moles of Al = 1.5 mol F x 1 mol Al
3 mol F

= 0.5 mol Al

Molecular Weight (mass):
Sum of atomic weights of the atoms in a molecule is called
molecular weight or mass.
CH4       C =12,      H=1
MW = ( 1 x 12 ) + ( 4 x 1 ) = 16

Formula Weight:
The sum of atomic weights of atoms in a formula unit is called
formula weight or mass.
Ca Cl2      Ca = 40 ,        Cl = 35.5
F.W = ( 1 x 40 ) + ( 2 x 35.5 ) = 111

4
E.x: Calculate the number of carbon atoms and the number of
hydrogen atoms in 600g of propane, C3 H8 C = 12 , H = 1.

MW = ( 3 x 12 ) + ( 8 x 1 ) = 44 amu

Number of moles of propane = 600 g x 1 mol = 13.63 mol
44 g

1 mole of C3 H8 contains 3 mol of C atoms and contains 8 mol
of H atoms.
It contains also Avogadro's number of C3 H8 molecules.

Number of moles of
C atoms = 13.63 mol C3 H8 x 3 mol of C atoms
1 mol C3 H8
= 40.89 mol of C

Number of C atoms = 40.89 mol C atoms x 6.02 x 1023 atoms
1 mol of C atoms
25
= 2.46 x 10 C atoms

Number of mol of H atoms = 13.63 mol C3 H8 x 8 mol H atoms
1 mol C3 H8
= 109.04 mol H atoms

Number of H atoms = 109.04 mol H atoms x 6.02 x 1023 atoms
1 mol of H atoms
25
= 6.56 x 10 H atoms

5
Sun 24/12/1424

Molar mass = atomic weight or molecular weight in grams.

Percentage Composition:
Percentage by mass contributed by each element present in the
sample.

E.x: What is the percentage composition of CO2.
C = 12   ,      C = 16

MW = ( 1 x 12 ) + ( 2 x 16 ) = 44
1 mol = 44g CO2

C% = 12g x 100 = 27.3%
44g
O% = 32g x 100 = 72.7%
44g

Chemical formulas:
1. Empirical formula gives the simplest whole number ratio
between atoms present in the compound.
2. Molecular formula states the actual number of atoms of
each element present in the molecule.

6
E.F.                M.F.
CH2O               C2H4O             Acetic acid
H2O                 H2O                Water
CH2O               C6H12O6            Glucose

Derivation of empirical formula:

The E.F doesn't only give the simplest ratio between number of
atoms but also the simplest ratio between moles of atoms.

We can, therefore, find the empirical formula by determining the
number of moles of atoms from their masses present in the
sample. Then divide the number of moles of atoms each by the
smallest number of moles present.

C 0.25      H 1.5     O 0.5        CH6 O2
0.25       0.25      0.25

PO2.5       P2 O5

C1.33 H2      C4 H6

7
Tues 26/12/1424

E.X. What is the empirical formula of a sulfur oxygen
compound, a sample of which contains 2.1g oxygen and
1.4g sulfur. S = 32 , O = 16.

No. of moles of S = 1.4 g S x 1 mol = 0.0437 mol S
32 g S

No. of moles of O = 2.1 gO x 1 mol = 0.131 mol O
16 g O
S0.0437 O0.131

S0.0437    O0.131          S1O3         SO3
0.0437    .0437

Remember:
E.F is a ratio between atoms or moles of atoms, so we use
atomic weight and not molecular weight to calculate number of
moles of atoms.
Empirical formula from percentage composition :

Ex.: The same compound contains 40% S and 60% oxygen.
Determine the empirical formula.
Suppose we have 100g sample, then calculate number of moles.

S = 40 = 1.25 mol S , O = 60 = 3.75 mol O
32                    16
S1.25 O3.75     S1.25 O3.75
1.25    1.25
SO3

8
Derivation of Molecular Formula:
Since molecular formula is an integral multiple of empirical
formula, we divide MW by EFW to get molecular formula.

Ex: Find the molecular formula for a compound whose
empirical formula is CH2 and MW = 84.
C = 12 , H = 1

E.F.W = (1 x 12) + (2 x 1) = 14
Number of times of E.F occur in M.F. = 8_4 = 6
14
Molecular formula = C6 H12

Balancing chemical equations:
1. Write all reactants and products with correct formula.
2. Balance by adjusting coefficients that preceed formulas.

e.g. C4 H10 + O2        CO2 + H2O
C4 H10 + O2        4CO2 + 5H2O
C4 H10 + 13/2 O2     4CO2 + 5H2O
2C4 H10 + 13O2       8CO2 + 10H2O

9
E.X: NH3 + O2        NO + H2O
2NH3 + O2        2NO + 3H2O
2NH3 + 5/2O2       2NO + 3H2O

4NH3 + 5O2         4NO + 6H2O

E.g: Calculations based on chemical equations:

C + O2        CO2

1 mol of C atoms reacts 1 mol of O2 molecules to give 1 mol of
molecules
(requires)

1 atom + 1 molecule       1 molecule
12 g + 32 g      44 g

Ex.: Calculate the mass of carbon required to produce 1 g of
CO2

12g      44 g
Mass of carbon       1  g
1 x 12 = 3 g of C
44  11

10
Limiting Reactant Calculations

If one of the reactants is used in excess of mole ratio
than the other indicated by a balanced equation,
reactant which is not in excess will be used up
completely before the one present in excess.

2 H2 + O 2        2H2O

The amount of product is determined by the reactant
which disappears first. The reactant which is
consumed first is called limiting reactant.

11
Sun 2/1/1425

Ex.: Freon gas, CCl2F2 is prepared by the following reaction:

3CCl4 + 2SbF3             3CCl2F2 + 2SbCl3

If 150g of CCl4 and 100g of SbF3 were used.
a) How many grams of CCl2F2 can be formed.
b) How many grams of which reactant will remain.

C = 12, F = 19, Cl = 35.5, Sb = 122

a)
molar mass of CCl4 = (1 x 12) + (4 x 35.5) = 154 g
molar mass of SbF3 = (1 x 122) + (3 x 19) = 179g

no. of moles of CCl4 = 150 gCCl4 x 1mol CCl4 = 0.974mol SbF3
154g CCl4

no. of moles of SbF3 = 100g SbF3 x 1mol SbF3 = 0.559mol SbF3
179g SbF3

no. of moles of SbF3 required to react completely with CCl4 =
0.974 mol CCl4 x
2mol SbF3 = 0.649mol
3mol CCl4      SbF3

.          0.649 mol is more than 0.559 mol which is available
of SbF3

CCl4 is in excess or SbF3 is the limiting reactant.

12
Another method to determine limiting reactant divide no. of
moles of each reactant by its coefficient. The smallest value is
for the limiting reactant.

0.559 mol SbF3 = 0.279        0.974 mol CCl4 = 0.325
2                             3

SbF3 is limiting reactant.

no. of moles of CCl2F2 produced=0.559mol SbF3 x 3mol CCl2F2
2mol SbF3

= 0.839 mol CCl2F2

Mass of CCl2F2= 0.839mol CCl2F2 x 121g Freon=101.5g Freon
1mol Freon
121g is molar mass of CCl2F2
b)
no. of moles of CCl4 consumed = 0.559 x 3 = 0.839 mol CCl4
2

Mass of CCl4 consumed = 0.839 x 154g = 129g CCl4

mass of CCl4 remaining = 150 – 129 = 21g

13
Theoretical Yield

Theoretical yield of a given product is the maximum yield that
can be obtained if the reaction gave only that product.

Percentage Yield

Is a measure of the efficiency of the reaction

% Yield = actual yield x 100
Theoretical Yield

E.g: Calculate the percentage yield of the previous
reaction (in the previous example) if the actual yield
is 76g.

% yield = actual yield x 100
Theoretical yield

= 76g x 100 = 75%
101.5g

E.g: Calculate the actual yield if the percentage yield of
the previous reaction is 80%
actual yield = 101.5g x 80 = 81g
100

14
Molarity

To express the amount of solute in solution, we use
concentration units. Most important unit is molarity. Molarity is
the number of moles of solute in 1L (dm3) solution.

Molarity = no. of mole of solute
Volume solution in litres

Unit: mol L -1
Or: mol dm-3

no. of moles of solute = molarity x volume of solution in litres

Example.: Calculate the molarity of a solution containing
4g of NaOH in 50 ml solution.
Na= 23, H = 1, O = 16

Formula wt = (1 x 23) + (1 x 1) + (1 x 16) = 40 amu
1 mol = 40g

no. of moles of NaOH = 4 g x 1 mol = 0.1 mol
40g

Molrity    =     0.1mol__ = 2.0 M
0.05 L

15
Dilution

On dilution, the number of moles of solute do not change.

No. of moles before dilution = no. of moles after dilution

M1 . V1 = M2 . V2

E.g: How many ml of 1 M HCl must be added to 50 ml of
0.5 M HCl to get a solution whose concentration is 0.6 M.

no. of moles before mixing = no. of moles after mixing

( Y x1) + (50 x 0.5) = (Y + 50) x 0.6
1000   1000             1000

Y + 25 = 0.6Y + 30
Y – 0.6Y = 30 – 25
0.4Y = 5
Y= 5
0.4

Y = 12.5 ml

16
E.g: Calculate the volume of 0.2M H2SO4 required to react
completely with 500ml of 0.1M NaOH.

The equation:
H2SO4 + 2NaOH         Na2SO4 + 2H2O

no. of moles of NaOH = 500 x 0.1 = 0.05 mol NaOH
1000

no. of moles of H2SO4 required = 0.05mol NaOH x 1mol H2SO4
2mol NaOH
= 0.025 = volume x 0.2

Volume = 0.025 = 0.125 L
0.2

Volume = 125 ml

17
Sun 9/1/1425

Oxidation – Reduction Reactions
(Redox Reactions)

Oxidation number is defined as the charge which would the
atom have if electrons of covalent bonds were assigned to the
more electronegative atoms.

-2
O           H – Cl                    O=C=O
+1 -1                     -2 +4 -2
H            H
+1           +1

Rules for Assigning Oxidation Number

1. Oxidation number of atoms in the elemental form equals
zero, regardless of the number of atoms in the molecules.

He, Ne, Ar                      Na, Cu, K, Fe
H2, F2, N2, O2, Cl2             P4
S8

2. Oxidation Number of atoms in simple ions (monatomic
ions) equals charge of the ion.

M2+, X3-, M3+
+2 3- 3+

3. Sum of oxidation number of all atoms in a molecule =
Zero and for polyatomic ion = charge on the ion.

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PO4-3                   S2O3-2                 Mn O4-
y + (4 x -2) = -3       2y + (3 x -2) = -2      y + (4 x -2) = -1
Y – 8 = -3              2y – 6 = -2             y – 8 = -1
y = +5                  2y = 4                  y = +7
y = +2

S4O6-2
(4 x y) + (6 x -2) = -2
4y = +10
y = 2.5

4. Fluorine in its compounds always has -1 Oxidation
number
MF       MF2      MF3
+1-1          +2-1        +3-1

CF4 PF6-
+4-1   +5-1

5. Metals of group IA (Li, Na, K, Rb, Cs, Fr) in their
compounds always have +1 oxidation number.

6. Alkaline earth metals group IIA (Be, Mg, Ca, Sr, Ba, Ra)
in their compounds always have +2 O.N.

19
7. Oxygen in its compounds almost always has -2 O. N.

Exceptions:
Peroxides          Na2O2 ,            H2O2 ,             BaO2
-1                 -1               -1

Super Oxide        KO2
+1 -1/2

Fluorine Oxide F2O
+2

8. Hydrogen almost always +1 O.N.
Exceptions: metal hydrides Na H, Ca H2, Al H3
+1 -1     +2 -1         -1

KH , LiA H4, Na BH4
-1          -1

9. Halogens in binary halides (Metal + halogen) have -1
O.N.
Fe Cl3
-1

10. For familiar ions the oxidation number can be considered
the charge of the ions.

PO4-3           Na3PO4                         Ca3(PO4)2
CO3-2       NO3-
SO4-2                      NH4+
-2          -1                 -2                 +1

20
Oxidation: An atom, ion or molecule loses electrons or
its oxidation number increases.

Reduction: An atom, ion or molecule accept electrons or
its oxidation number decreases.

Reducing agent: An atom, ion or molecule providing
(losing) electrons.

Oxidizing agent: An atom, ion or molecule accepting
(receiving) electrons.

E.g: Zn + Cu+2      Zn+2 + Cu    it is Redox reaction
O    +2       +2     O

Zn       is   Oxidized
Cu+2     is   Reduced
Zn       is   Reducing agent
Cu+2     is   Oxidizing agent

21
E.g: Which of the following are redox reaction:

H2 + 1/2 O2         H2O
O      O          +1 -2

H+ + OH-          H2O            x    neutralization
+1    -1         +1 -2

H+ + H2O           H3O+          x    acid-base reaction
+1 +1-2             H -2

Ca CO3 heat Ca O + CO2                x
decomposition
+2 +1 -2      +2 -2 +4 -2

H2 + F2         2HF
O     O         +1 -1

Ag+ + Cl-         Ag Cl          x    precipitation
+1 -1           +1 -1

Cu+2 + 4Cl-         Cu Cl4       x    complex formation
+2 -1             (Lewis acid – base
Reaction)

Fe+3 + 6 CN-         Fe (CN)6-3 x     complex formation
+3     -1          (Lewis acid – base
Reaction)

22
Balancing of Redox Equations by ion electron method:

We have to remember that the equations
must be balanced in mass and charge.
In the final equation, their should be no
electrons.

This method is convenient for ionic equations. Steps:

1. Divide the equation into two halves (Reduction half and
oxidation half).

2. Balance each half separately for mass and charge.

3. Balance the charge by adding electrons to the more
positive (+ve) side or less negative (-ve) side.

4. Multiply each half by a suitable factor to eliminate
electrons.

 in acid solution add a number of H+ to the side deficient in
hydrogen, and to balance oxygen atoms add number of
H2O molecules and to other side add 2H+ for each H2O

23
Balance in Acid medium

Cl2         Cl- + Cl O-3

Divide then carry out balancing

Cl2         Cl-

Cl2         Cl O-3

I Cl              I- + I O-3 + Cl-

I Cl              I- + Cl-

I Cl              IO-3 + Cl-

Then carry on as usual

Balance in acid medium

CN- + As O-34           As O-2 + CN O-

CN-          CNO-       ,      As O-34         As O-2

CN- + H2O            CNO- + 2H+ + 2e-

As O-34 + 4H+ + 2e-            As O-2 + 2H2O

CN- + As O-34 + 2H+            CNO- + As O-2 + H2O

24
E.g: Balance in acid medium:

Zn + NO-3            Zn+2 + NH+4

Zn          Zn+2 + 2e- ………….. (1) x4

NO-3        NH4+

NO-3 + 10H+         NH4+ + 3H2O

NO-3 + 10H+ + 8e-         NH4++ 3H2O ……….. (2) x 1

4Zn + NO-3 + 10H+ + 8e-          4Zn+2 + 8e- + NH4+ + 3H2O

4Zn + NO-3 + 10H+          4Zn+2 + 8e- + NH4+ + 3H2O

25
SUN 16/1/1425

Balancing in Basic medium

The best method is to balance first in acid medium, then carry
out the following steps:

1. Add number of OH- to eliminate H+ to both side.
2. Combine H+ and OH- into H2O.
3. Cancel water if necessary.

Ex.: Balancing in Basic medium:

MnO-4 + C2O4-2          MnO2 + CO3-2

MnO-4           MnO2

MnO-4 + 4H+ + 3e-           MnO2 + 2H2O ………(1) x           2

C2O-24              CO-23

C2O-24 + 2H2O          2CO-23 + 4H+ + 2e- ………(2)       x   3

2MnO-4 + 8H+ + 6e- +3C2O-24 + 6H2O            2MnO2 + 4H2O +
6CO-23 + 12H+ + 6e-

2MnO-4 + 3C2O-24 + 2H2O            2MnO2 + 6CO-23 + 4H+

The equation is now balanced in acid medium.
2MnO-4 + 3C2O-24 + 2H2O + 4OH-           2MnO2 + 6CO-23 +
4H+ + 4OH-
2MnO-4 + 3C2O-24 + 4OH-          2MnO2 + 6CO-23 + 2H2O
The equation is now balanced in basic medium.

26
Gases:

The gas occupies the entire volume in which it exists.
Volume of gas = volume of container (vessel)

If you have a mixture of gases, the volume of each
Gas = volume of the container.

Pressure in all directions is equal

Pressure = force    . (Nm-2) unit of pressure. It is called Pascal.
area

Atmospheric pressure:

Measured by barometer.
Pressure exerted by air. It equals the weight of a column of air.

It equals the height of mercury in barometer.

Standard atmosphere:

1 atm = 760 mm Hg = 760 torr = 76 cm Hg = 101325 Nm-2 =
101325 Pascal = 101.325 kpa.

27
E.g: 0.8 atm in torrs

0.8         ?               608 torrs
1           760

0.8 atm in Pascal

1.0          101325
0.8          ?
81060kPa

0.8 atm in KPa = 81.06 kPa

Boyel’s law:

For a fixed amount of a gas, at constant temperature, the volume
is inversely proportional to pressure.

Vα 1       ,     V = constant
P                   P

PV = constant

P1 V1 = P2 V2 = P3 V3

Graphically

28
Tues 18/1/1425

At low pressure, the behavior of the gas approaches ideal
behavior

Charle’s law:

At constant pressure, the volume of a given amount of gas is
directly proportional to absolute temperature.

Vα T             T absolute temperature
(Kelvin)

V = constant X T

The volume of a gas changes linearly with Celsius degrees if
you extrapolate lines for gases to volume zero, they will meet at
-273°C. This is the absolute zero or it is zero at Kelvin scale.

29
At high temperature a real gas approaches ideal behaviors.

Real gas approaches ideal behavior at low pressures and high
temperatures.

Amonton’s law (Gay – Lussac’s law):

The pressure of a given quantity of gas is directly proportional
to absolute temperature if the volume is kept constant.

PαT              P = constant
T
P = constant X T or P1 = P2 = ……
T1 T2

Combined gas law:

Pi Vi = Pf Vf         i = initial
Ti      Tf           f = final

At constant T the law reduces to Boyle’s law:
Pi Vi = Pf Vf

At constant pressure  Vi = Vf reduces to Charle’s law
Ti Tf

At constant volume  Pi = Pf reduces to gayLussac’s law
Ti   Ti w

STP standard temperature and pressure:

1 atm           101325 Pa (Nm-2) 760 torr

O°C 273 K

30
Dalton’s law of partial pressure:

The total pressure exerted by a mixture of gases, that do not
interact, equals sum of partial pressures of all gases, if each gas
occupies the container on its own.
P total = P1 + P2 + P3 + …………

Note: if you collect gas above water, the gas will be saturated
with water vapor total pressure.
Pt = Pgas + P H2O                P total for wet gas
Pg for dry gas

Chemical reactions between gases:

Gay-Lussac’s law of combining volumes: the volume of
gaseous reactants and products are in a simple whole number
ratio, if the volumes are measured under the same conditional of
temperature and pressure.

2H2 (g) + O2 (g)     2H2O (g)
2 volumes + 1 volume     2 volumes
2     :     1   :   2

H2 (g) + Cl2 (g)          2HCl (g)
1     :    1     :         2

31
Avogadros hypothesis: Equal volumes of gases contain the
same number of molecules if the volumes were measured
under the same conditions of temperature and pressure.

Since equal number of molecules implies equal number of
moles, then

Vαn        where (n) is the number of moles

It follows that the volume of one mole of any gas is the same
under the same conditions of temperature and pressure. This
volume is called Molar volume.

Molar volume of any gas at STP = 22.4 L

E.g: The molar volume of a gas at temperatures higher that
STP and pressures lower than STP, equal:
a) 22.4 L.
b) Less than 22.4 L.
c) more than 22.4 L. 
d) we can’t tell.

32
The ideal gas law:

Vα 1               V α nT
P                   P
VαT
Vαn                V = R nT          R: universal gas constant
P

PV = nRT          Ideal gas law or equation of
state for an ideal gas.

Obeyed by ideal gases and by real gases at normal laboratory
conditions.

Value of R:
For one mole of gas at STP (101325 Pa and 273 K), it occupies
22.4 L.
SI unit

R = PV = 101325 Nm-2 x 0.0224 m3 = 8.314 J mol-1 K-1
nT        1 mol x 273 K

for 1 mol P= 101.325 kPa , V = 22.4L

R = 101325 kPa x 22.4L = 8.314 kPa L mol-1 K-1
1 mol x 273 K

1 mol      1 atm       22.4 L

R = 1 atm x 22.4 L = 0.0821 atm L mol-1 K-1
1 mol x 273K

33
E.g: Calculate the volume of 4g methane, CH4, at 25°C and 80
kPa. C = 12     ,    H=1

nCH4 = 4g x 1 mol = 0.25 mol
16g

V = nRT = 0.25 mol x 8.314 Nm mol-1 K-1 x 298K
P               80000 Nm-2
= 7.074 x 10-3 m3
J = N.m

Or

V = 0.25 mol x 8.314 kPa mol-1 K-1 x 298K
80 kPa
= 7.74 L

Or

V = 0.25 mol x 0.0821 atm mol-1 K-1 x 298K
80      atm
101.325
= 7.74 L

For a mixture of gases
P1 V = n1 RT
Divide     Pt V = nt RT

P1 = n 1 = X1         mole fraction
Pt   nt

P1 = Pt x X1

34
E.g: Calculate total pressure exerted by a mixture of 6g helium
and 40 g oxygen in 10L vessal at 25°C He = 4 , O = 16

T = 273 + 25 = 298°K

nHe = 6g x 1 = 1.5 mo.                    Inert gases have
49                             monoatomic molecules

nO2 = 40g x 1 = 1.25 mol
32g

PHe = nHe RT , PO2 = no2 RT
V              V

P = 3.67 atm , PO2 = 3.06 atm
He
P total = PHe + Po2 = 3.7 + 3.06 = 6.73 atm

Or
Pt = nt RT
V

Pt = Po2 + PH2o  Wet oxygen (Gas) ‫ارا ركر السؤال حنسة معه ضغط املاء‬
PO2 only  dry oxygen (gas): ‫أما‬

35
Density of a gas:

PV = n RT                          n= m      mass of gas
M      molar mass

P = n RT
V

P = m RT              m =d         density
M V               V

P = d RT
M

PM = dRT

E.g: calculate the density of CO2 at 86.7 kPa and 25°C
C = 12      ,    O = 16

86.7 kPa x 44 = d x 8.314 x 298

d = 1.536 gL-1

or you can convert 86.7 into atmospheres and then use R =
0.0821 atm L mol-1 K-1

36
Real Gases:
Gas laws apply perfectly to ideal gases, but real gases deviate
from ideal behavior.

Example: PV = n RT        ‫للغاز احلقيقي‬
‫خاصح عنذ درجاخ منخفضح وضغط عايل‬

There are two reasons for deviation of real gases:
1. Molecules of a real gas have a volume. Therefore, the
volume we measure for a real gas is bigger than the
volume in which molecules are free to move.

2. There are attractive forces between molecules of a real gas,
whereas molecules of an ideal gas have no attractive
forces.

Molecules of an ideal gas would be cooled absolute zero without
condensing (Because no attractive forces) in to a liquid.

37
The measured pressure of a real gas is less than that of an
ideal gas because molecules which are about to collide
with the wall are attracted towards interior by other
molecules.

Van der Waals Equation for real gases.

(P + n2a) (V – nb) = nRT              for n mole
V2

(P + a ) (V – b) = RT             for 1 mole
V2

P = measured pressure
V = measured volume
R = Gas constant
T = Absolute temp.
n = number of moles

a and b constants characteristic for each gas.

a related to attractive forces

b related to molecules sizes

38
E.g: Oxygen gas generated by the reaction:

KClO3           KCl + O2 (un balanced)

was collected over water at 30°C in 150 mL vessel until the total
pressure 600 torr.

a. How many grams of O2 were produced?

b. How many grams of KClO3 were consumed?

PH2O=31.8 torr at 300C

S: a)        2KClO3          2KCl + 3O2

PO2 = Pt – PH2O = 600 – 31.8 = 568.2 torr

568.2 x 0.150
nO2 = PV = 760
RT 0.0521 x 303              = 4.5 x 10-3 mol

Mass of O2 = 4.5 x 10-3 x 32 = 0.144g

b) No of moles of KClO3 consumed = 4.5 x 10-3 x 2 = 3 x 10-3 mol
3
-1
Molar mass of KClO3 =122.5g mol
Mass of KClO3 consumed = 3 x 10-3 x 122.5 = 0.369g

39
Change direction of Chemical Thermodynamics
Thermochemistry:

The study of heats of reactions is called thermochemistry.
Virtually every chemical reaction is associated with absorption
or release of energy how this comes out and why?

Bonds are formed and broken. The differences in potential
energies associated with formed and broken. Bonds are the
source of the energy evolved.

Hess’s law of constant heat summation:
The heat change for a process, ΔH, is the same whether change
is carried out in one step or in stepwise manner.

A           C                  change in energy is the same

B
Two Steps

It is an application of law of conservation of energy.

For example:
C(graphits) + O2(g)     CO2(g)       ; ΔH = -393.5kJ

This is called a thermochemical equation

ΔH = enthalpy or heat change = H products – H reactents

H = heat content or enthalpy

40
41
If ΔH is engative, the reaction is exothermic.

If ΔH is positive, the reaction is endothermic.

H is called a state function. The change in its value depends only
on intial and final state.
In two steps:

C(graphite) + 1/2O2(g)                CO(g) ;ΔH2 = -110.5kJ

CO(gas) + 1/2O2(g)               CO2(g)     , ΔH3 = -283KJ

C(graphite) + O2(g)              CO2(g)     , ΔH = -393.5KJ

Thermochemical equation can be treated algebraically.

They can be added or subtracted. ΔH is also treated
algebraically.

Enthalpy diagram to illustrate Hess’s law:

C(graphite) + O2(g)
ΔH1 = ΔH2 + ΔH3
ΔH1         ΔH2                       We can calculate
ΔH of a reaction from
CO(g) + 1/2O2 (g)         known heats of other
reactions
ΔH3

CO2(g)

42
ΔH is the heat gained or lost by a system under constant
pressure the only work done being due to volume change.

ΔH = qp

System: any thing we focus our study on it. Anything else is
called surroundings. If ΔH for a system is + ve, ΔH for the
surrounding is – ve.

Manipulating thermochemical equations:

How to calculate ΔH of a reaction from known ΔH for other
reactions:

In order to get the desired equation we multiply or divide by a
suitable factor or change direction of reaction if necessary. Do
the same thing for ΔH.

Ex: Given the following:

2H2(g) + O2(g)        2H2O(l) ΔH° = -571.5kJ

N2O5(g) + H2O(l)                      2HNO3(l)   ΔH° = 76.6kJ

½ N(g) + 3/2 O2(g)          HNO3(l), ΔH° = -174kJ

Calculate ΔH for the reaction:

2N2(g) + 5O2(g)        2N2O5 (g)

43
1) Multiply equation (3) by 4, multiply equation (2) by 2 and
change direction. And change direction of equation 1.

Do the same thing for ΔH and add.
2N2(g) + 5O2 (g) + 2H2 (g)      4HNO3(l), ΔH° = -696KJ

4HNO3(l)            2N2O5 (g) + 2H2O(l) ,              ΔH° = +153.2
KJ

2H2O(l)             2H2(g) + O2(g)                ΔH° = +571.5 kJ

2N2(g) + 5O2 (g)             2N2O5(g) ,           ΔH° = + 28.7kJ

Heats or enthalpy formation:
Heats or enthalpy change when one mole of substance is formed
from its elements under stated conditions. ΔH° = standard
enthalpy or heat formation, is the enthalpy change when one
mole of pure substance is formed from its elements in their most
stable forms under standard state conditions.

Standard state conditions (°) at 25°C (298K) and 1 atm.

Ex: which of the following has ΔH°f for H2SO4(l) :

SO3(g) = H2O(l)          H2SO4(l)

SO2(g) = ½ O2(g) + H2O(l)              H2SO4(l)

S(s) + H2(g) + 2O2(g)        H2SO4(l)

S(g) + H2(g) + 2O2(g)          H2SO4(l) because S(g) is not most
stable

H2(s) + Br2(l)          2HBr(g)

½ H2(g)            H(g) ΔH°f

44
Example: H2(g) + ½ O2(g)                 H2O(l)        ΔH°f (1)

C(graphite) + 2H2 (g)             CH4 (g)        ΔH°(2)

1     C(graphite) + ½ O2                CO(g)          ΔH°f (3)

Reactions like (2) and (3) are difficult to measure. So, we use
calculations to know ΔH°f
E.x: C(graphite) + O2(g)           CO2 (g) ΔH° = -393.5kJ (1)

H2(g) + ½ O2(g)           H2O(l)            ΔH° = -285.9kJ (2)

CH4(g) + 2O2(g)           CO2(g)+2H2O(l)ΔH° = -890.4kJ
(3)

Calculate ΔH°f for CH4.
C(graphite) + 2H2(g)             CO2(g) (2), multiply (2) by 2
C(graphite) + O2(g)                CO2(g) ΔH° = -393.5kJ

2H2(g) + O2(g)          2H2O(l)             ΔH° = -571.8kJ

CO2(g) + 2H2O(l)                  CH4(g)+2O2(g)ΔH°=
+890.4kJ

C(graphite) + 2H2(g)           CH4(g)       ΔH° = -74.9kJ mol

C (graphite) + 2O2(g) + 2H2(g)     ΔH1 CO2(g) + 2H2O(l)

ΔHf                    ΔH2

ΔH1 = ΔHf + ΔH2         CH4 + 2O2

ΔH depends on status of reactants and products.

45
Ex:        H2(g) + ½ O2         H2O(l)       ΔH° = -286kJ (1)
H2(g) + ½ O2         H2O(g)        ΔH° = -242 kJ (2)

Calculate ΔH vap
H2O(l)             H2O(g)          ΔH vap

Change direction of (1) and add to (2)
H2O(l)          H2 + ½ O2           ΔH° = 286 kJ
H 2 + ½ O2          H2O(g)          ΔH° = - 242 kJ

H2O(g)           H2O(g)       ΔH° vap = 44 kJ

H2(g) + 1/2 O2(g)        ΔH1            H2O(g)

ΔH2                      ΔH vap

ΔH1 = ΔH2 + ΔHvap           H2O(l)

ΔHf for many substances are tabulated
ΔH reaction = Σ ΔHf (products) - Σ ΔHf (reactants).
ΔHf for elements in their standard state = o

Ex: CH4(g) +       2O2(g)            CO2(g)       +   2H2O(l)
ΔH = ?

ΔHf (CO2) = -393.5 KJ mol-1

ΔHf (H2O) = -285.9 KJ mol-1

ΔHf (CH4) = -74.9 KJ mol-1

ΔH reaction = Σ ΔHf (products) - Σ ΔHf (reactants)
= Σ (-393.5 + 2 x -285.9) - Σ (-74.9) = -890.9 kJ

46
Heat of combustion:

Heat change when one mole of a substance is burned completely
in oxygen.

Bond energy or Bond enthalpy:
Energy required to break a bond into neutral fragements.

H2 + energy              2H

Atomization Energy:
Energy required to reduce a gaseous complex molecule into
neutral gaseous atoms. It is the sum of bond energies.

E.x: The structure of propane is             H H   H

H–C–C=C

H     H

ΔH°f (C3H6) = +8 kJ mol-1

Calculate C = C bond energy, you are given the following data:
ΔH°f C (atoms) = 715 kJ mol-1
ΔH°f H (atoms) = 218 kJ mol-1
C – C bond energy = 348 kJ mol-1
C – H bond energy = 418 kJ mol-1

ΔHf = ΔH1 + ΔH2
ΔH1 = 3 x 715 kJ + 6 x 218 kJ = 3453 kJ      ΔH2 is formation
 8 kJ = 3453 kJ + ΔH2                       of bonds from
ΔH2 = -3445 kJ                     atoms.
ΔH atom = +3445 kJ                      it is the reverse of
+ 3445 = 6 x 418 + 348 + C = C bond energy        atomization
C = C bond energy = 607 kJ

47
The first law of thermodynamics:
This is the law of conservation of energy. Change in internal
energy, ΔE, equals heat added to the system plus work done on
the system by surroundings.

ΔE = ΔE final – ΔE initial        , ΔE = q + w
w = -p ΔV = - p (Vfinal - V initial)

E.x: if a system absorbs 442 kj of heat expands from 100L to
445L against 1 atm pressure. Calculate the change in internal
energy.

(445 – 100) L
-2
ΔE =442 kJ – 101325 Nm x 1000L / m3
1000J / kJ
= 42 kJ – 34.96 kJ
= 7.05 kJ

Measurement of ΔEL:

Bomb calorimeter

48
Heat capacity:

Heat required or released to raise the temperature of the system
1ºC.

Heat released     = heat compacity x Δt
= heat capcity x (t2 – t1)

Then calculate for one mole, ΔE.
ΔE = qv heat at constant volume when ΔV = 0
For pure substances = M x Y x Δt   Y: specific heat, M: mass
Heat change

The relation ship between ΔH and ΔE:

H = E = PV              P ΔV = Δn RT
ΔH = ΔE+ PΔV            Δn = number of moles of gases produced
ΔH = ΔE + nRT           - number of moles of gases consumed
Neglect Δn change in solid and liquids.

E.x: Relation ship between ΔH and ΔE:
1. C(s) + O2(g)                 CO2(g)
ΔH = ΔE         because Δn( gas) = O

2. CO(g) +  1/2O2(g)                       CO2(g)
ΔH = ΔE – 1/2 RT         because Δn( gas) = -½

3. Zn(s) +       2HCl(aq)                  Zn Cl2(aq) +   H2(g)

ΔH = ΔE + RT             because Δn(gas) = 1

49
E.x: if ΔH for the above reaction = -151.5 kJ mol
Calculate ΔE at 27ºC , R = 8.314 J mol-1 K-1
ΔH = ΔE + RT
-151.5 = ΔE + 8.314 x 300
1000

ΔE = -154.44 kJ mol-1

Chemical Bonding

Atoms combine together to form molecules or compounds. The
force of attraction that holds atoms together is the chemical
bond.

Lewis symbols (Lewis structure):
1A 2A 3A 4A 5A            6A 7A 8A
Li Be B         C    N    O      F   Ne
.     .     .    .    ..    ..   ..
X . X . X. . X . . X . . X . ..X . ..X ..
.       .  ..   ..     ..   ..
Valence electrons.

The Ionic Bond:

50
Metals tend to react with non-metals to give ionic compounds.
Ionic or electrovalent bonds occurs between positive ions or
cations (atoms losing electrons) and negative ions or anions
(atoms accepting electrons) i.e. it results from attraction between
oppositely charged ions.

Example: Reaction between sodium and fluorine?

11Na                 9F
2        2     6      1
11Na   1s       2s      2p     3s               9F   1s2   2s2   2p5
.    .                         :
Na +: F:                +
[Na ] + [: F- : ]
:

51
Noble gas structure is reached. This corresponds, except for
helium, to eight electrons in the valence shell. This is the basis
of octet rule "atoms accept or lose or share electrons until there
are eight electrons in the valence shell". Not all ions obey the
octet rule, transition metal ion and post transition metal ions do
not obey octet rule:

26Fe   1s2 2s2 2p6 3s2 3p6 4s2 3d6           26Fe
+2
[Ar] 366
[Ar]

24Cr    [Ar] 4s1 3d5         24Cr
+3
[Ar] 3d3

30Zn    [Ar] 4s2 3d10     30Zn
+2
[Ar] 3d10 pseudo noble gas
structure
The ratio between is not always 1 : 1
.       ..
Ca .+ 2 :Cl:              [Ca]+2 + 2 [Cl]- ratio 1 : 2
.
.
2 Li +: O .        2 [Li]+ + [O]-2       ratio 2 : 1
:

E.x: deducing formula from charges:

Ca+2   (PO4)-3        Ca3 (PO4)2

Al+3 SO4-2            Al2 (SO4)3

Mg+2 O-2          Mg2 O2           Mg O

52
* Foctors influencing the formation of ionic compounds :-
Q : why compounds are formed ?
A : To reach stability. The system becomes of lower energy.
Lower energy          More Stability .

In the reaction : Li(s) +      1/2 F2(g)                   Li+F-(s)
1 mole       1/2 mole                     1mole

We can analyze the factors that contribute to the energy change of
this reaction:

- Born Haper cycle :
ΔHf
Li(s) + 1/2 F2 (g)                       Li+F- (s)
(1)               (2)                                (5)
F(g) (4) + 1e-          F- (g)
Li(g)       (3) - 1e-           Li+ (g)

Hf = H1 + H2 + H3 + H4 + H5

Step (1) : Vaporization or sublimation. Li(s)         Li(g); H=155 kJ
endothermic
Step (2) : dissociation 1/2 F2 (g)             F(g) ; H= 79 kJ
endothermic
Step (3) : ionization Li(g)        Li+ (g) + 1e- ; H = 520 kJ
endothermic
Step (4) : electron affinity
F(g) + 1e-      F-(g) ; H = -330 kJ exothermic
Step (5) :Lattice energy: Li+(g)+ F -(g)    Li+F-(s) ; H =- 1016kJ
highly exothermic

53
‫طاقة البناء الشبكي البلوري‬
* Lattice energy : is primarily responsible for the formation and
stability for ionic compounds.

HF0 = 155 + 79 + 520 + (-330) + (-1016) =-592 kJ mol-1

generally ionic compounds are formed from metals and non-metals
particularly group IA and IIA( low ionization energy) with VIIA and
VIA (high electronic affinity).

* The covalent Bond :
When conditions are not favorable for the formation of ionic bond,
a covalent bond is formed.
A covalent bond is formed as a result of sharing a pair of electrons
between the two atoms. Bonding results from attraction between this
pair and both nuclei.

- Formation of H2 molecule :-

Energy diagram of H2 formation

Energy                                  Bond
energy

bond distance
or bond length

Inter nuclear distance

54
As the two atoms comes close to each other the energy decreases to
a minimum. At minimum Energy, the distance between nuclei is
called bond length. The depth of this minimum is the bond energy.
Both electrons in bond have opposite spins.

. .
H+H             H : H or H – H         or H2

Some atoms require to form more than a covalent bond.
H
.     .             ..
.C. + 4H        H..C.. H
.                  ..
H

..                ..Lone pair or free fair or
N + 3H           H..N.. H                non- bonding pair
H
O + 2H              .. O.. H
H

F+H                 F.. H

Not all atoms obey octet rule

F                                      F
.x                                  F :P: F
C| .x Be.x C|        xB x                  PF5             F:
F    F                                    F
F
F: : F
S
F: F
F

55
atoms from third period and beyond can expand their valence shell
to include empty d subshell and thus exceed octet rule.

Some atoms make more than covalent bond :

O=C=O             N≡N

- Drawing Lewis structure for molecules :
After deciding the central atoms, follow the following steps :
1- Count all electrons plus or minus charge.
2- Put a pair of electrons in each bond.
3- Complete 8 electrons to the surrounding atoms.
4- Put any remaining pairs on central atoms.
5- If the central atom didn't reach 8 electrons, make
multiple bond.

* Ex Draw Lewis structure for PF3         (15P – 9F)
1s2 2s2 2p6 / 3s2 3p3
1s2 / 2s2 2s5

1- (1x5) + (3x7) = 26 e-
2- F:P:F
F                       4- F:P:F:
F
3- :F:P:F:
:F:

56
NO3-           (9N, 8 O)
-
1- (1x5) + (3 x 6) = 24e
o
2- o:N:o
o
3- o:N:o
4- X :o:

5- :o: :N:
:o:

____________________________________

CO2
1)   4 + (2 x 6) = 16 e-
2)   O:C:O
3)   o:c:o
4)   X
5)   O:C:O

57
Bond order and Bond properties :

Bond length and bond energy are two characteristic properties
of the covalent bond.

H–H+E                    2H.
Bond energy

* Bond order : number of covalent bonds between two atoms.

H         H
H- C-C- H                    C=C        -C≡C-
H              H
Bond order
between carbon           1                 2             3
atoms

Increase in bond energy
Increase of bond order
decrease in bond length

- Vibrational frequency :-
Two atoms vibrate towards and away each other along the axis
joining the two atoms.
The factors affecting vibrational frequencies :-
Increase of mass of atoms decreases vibrational frequency.
Increase of bond order increases vibrational frequency.

58
between

- State of mater and Intermolecular forces :

Intra bonding(intra means inside) is
stronger than intermolecular forces.
a) Dipole - dipole Interaction :

it occurs between polar molecules. It amounts to 1% strength
of the covalent bond.
In solids > liquids > gases

b) hydrogen bonding :
Occurs between molecules when hydrogen is covalently
bonded to one of the three small most electronegative atoms which
are of F, O, N. example : H2O
H
O                    H-O
H         H

OO             hydrogen
O              bond
H        H
OO                      OO
O                       O
H        H            H     H
Its strength amounts to 5-10% the strength of the covalent bond. It is
responsible for exceptionally high boiling point and heat of
vaporization : H2O>H2S
HF >HCl
* Ex: Is there hydrogen bonding in the following :
O        x        H3CF x            HF         CH3OH
H3C       CH3
H           11O
O               H
N  CH3 – CO-CH3 x CH3- CH x                   H - N-H 

59
OH    H
CH3             CH3

C) London Forces :
It plays a significant role in attraction between non-polar
molecules and uncombined atoms.
It results from instantaneous random temporary polarization
due to continuous movement of electrons.

(+)       (-)    (+) (-)     Polarization in A induces polarization
in molecule B and attraction results
A               B

* It occurs between all particles polar, non polar and ionic.

Example : H9,          I2   , AR, Na+CL-

* London force increase with atomic size (atomic number)
I2 > Br2 > d2 > F2

* London force increases with increasing chain length or molecular
weight.

- c- c- c- c- ,    - c- c- c -c- c-   ,    - c- c- c- c- c- c-

B.P       10                     35c                     69c

Number of polarization sites in come with chain length
increases resulting in more attraction force.

60
Hydrogen bonding strength between H…F>H…O>H…N because of
electro negativity.

H2O        HF               NH3           Because water make
o         o                 o          more hydrogen bonds
B.P        100C       19C              -35C

O                  F        F                   N
H           H       H       H                    H       H
H

H2O               CH3O-H                         CH3 CH2 O-H
o                  o                               o
B.P 100C               68 C                             79C
..
Water makes more hydrogen bonds than CH3-O-H.
..
CH3 CH2 OH has stronger London Force

* Non polar molecules and compounds :
O2, N2, F2, d2, H2 ,…..
F
o
120
B
non polar CO2 O = C = O       BF3 F            F
compound

BF3, Bd3, CS2, S = C = S, CC/4, CF4, CBr4, CI4,
PCl5, PF5, SF6, hydrocarbons
Only London forces           Non polar
CH2 Cl2 Polar London forces and dipole – dipole
Cl- I, Br – I : Dipole dipole and London force

61
Definitions of acids
and bases

* Arrhenius definition in modern concepts :
An acid is a substance that increases concentration of
H3O+ in aqueous solution

Hcl + H2O                 H3O+ + Cl-
CO2 + H2O                 H2CO3
H2CO3+H2O                 H3O+ + HCO3-

- Non metal oxides : CO2, N2 O5, P2O5, SO2, SO3
are acid anhydrides

A base : substance that increase concentration of OH-
in aqueous solution
NaOH(s) + H2O           Na+aq + OH-aq aq= aqueous
NH3 + H2O                 NH4+ OH-

- metals oxides : CaO, Na2O , BaO, K2O, M9O
are basic anhydrides
CaO + H2 O              Ca (OH)2

Neutralization : H3O+ + OH-              2H2O

or simply :      H+ + OH-               H2O

62
* Bronsted – Lowry Definition :
An acid is a proton donor, a base is a proton acceptor

Hcl + H2O                H3O+ + Cl-
acid base                acid base

There are two acids and bases. An acid and abase that differ by only
one proton are called acid-base conjugate pairs.

* E.x :- H2S , S-2   x        H2O , OH- 
-3
H2S , HS-            H3PO4, PO4 x
+
H3O, OH- x
+
H3 O, H2O 

+     -
NH3 + H2O                NH4 + OH
base  acid                acid base

conjugated pairs

* water behaves as an acid and a base, it is amphitricha or
amphiprotic

NH3, CH3 COOH

+    -
H2O + H2O              H3O + OH
+     -
NH3 + NH3              NH4 + NH2

63
+        -
CH3COOH + CH3COOH                  CH3 COOH2 + CH3 COO
3+
Al (H2O) 6 hydrated ion is acidic
+
+3
Al (H2O)6 + H2O              H3O+ [A13+ (H2O)5OH-]+2
* Bronsted – Lowry definition is not restricted to water.
HCl + NH3        NH4+ + Cl-

Relative strength of acids and bases :
HCl + H2O                H3O+ + Cl-       100%
+     -
HF + H2O          H3O + F           only 3%
common
Base

HCl is more able to donate its proton than HF to the same base..:
HCl is stronger acid than HF.
F- is able to accept 97% protons from the same acid.
: F- is stronger base than Cl-.

a strong acid has a weak conjugate base.
as weak acid has a strong conjugate base.

-
NH2 + H2O         NH3+ OH-         100%
+    -
NH3 + H2O         NH4 OH           0.4%
-
with the same reference acid NH2 is much stronger base than NH3.

with the same reference base, NH4+ is much stronger acid than NH3.

64
Acids                      bases
strength    HClO4                              ClO4-           Strength
of acid     HCl                                 Cl-            of base
decreases                                                decreases

H2O                             OH-

- Relative strength oxy acids :
CClO4 > HClO3 > HClO2 > HClO
H2SO4 > H2SO3                 If you increases no of oxygen
HNO3 > HNO2                   atom, acid strength increases.
HClO4 > H2SO4 >H3PO4                increases of acid strength
Increases with electro negativity.

P S Cl
Electro negativity

HClO4 > HBrO4 >HIO4              increases of electro negativity,
Increases in acid strength.

- Binary acid :
HI > HBr > Hd > HF
Although polarity of H-F is highest, the strength of the covalent
bond is dominant factor.

NH3 < H2O > HF
Polarity of the bond is dominate factor.

65
Electro negativity
Increases

C     N     O        F
Cl     decreases
Br
I

- Lewis Definition :-
Lewis acid is a substance that accepts a pair of electrons to
form covalent band.
Lewis base is a substance that donates a pair of electrons to
form covalent bond.
H
+ -                   1
H + O–H                O–H
Acid base

+
+     ..                      H
H + H-N-H                  H- N- H
1                       1
H                        H

+
+      ..                     H
H + .. O-H                    1
1                     : O-H
H                        1
H

H                Cl               H          d
1                1                1          1
H - N ..    +        B-CL         H - N          B-Cl
1                1                1          1
H                Cl               H          d

66
Acidity and basicity is associated with a particular electronic
configuration.
‫ترتية الكتروين معني‬

An acid has an incomplete valence shell
A base has electron pairs to donate.

Acidity and basicity is not restricted to any element. Proton in an
acid.
+2

NH3

2+                                             Complex ion
Cu    + 4 NH3          H3N     Cu     NH3            complex compound
Lewis   Lewis
Acid    base
NH3

H2O    OH2    OH2

Al
3+                   2
Al + 6H2O              HO            OH2
Lewis Lewis
Acid  base                    OH2

HO

H2O + SO3                HO - S - O
Lewis Lewis
base  acid                      O
O
2-
O        +   S – O3   O- S–O              with rearrangement

O

67
-
O                          O
-        11                         C
HO    +    C                   HO         O
with rearrangement
bi coronate HCO3-

HCl + H2O -------       H3O+ + Cl-

H2O stronger base than Cl- and displace it.

Lewis base is also a Bronsted base.

68
Solutions

Homogeneous mixture of two of more components in one phase
composition is uniform (constant).

1) gaseous; such as air
gas in liquid; CO2 in water
2) Liquid         liquid in liquid; acetone in water
solid in liquid; sugar in water

3) Solid           alloys

- Concentration units :

na
mole fraction Xa = ____________
na+nb+nc+…

no of moles of one component over total number of moles of all
components. sum of mole fractions =1

mole percent = mole fraction X 100

mass of one component
mass fraction = ___________________
total mass of solution

mass % = mass fraction X 100

69
- molarity : no. of moles of solute in 1L solution.
- molality : no. of moles of solute in 1kg of solvent.

 Only molarity changes with change in temperature.

* Ex. A solution of 121g of Zn (NO3)2 in 1L solution has a density of
1.107gmL-1. calculate :
a) mass percent         b) mole fraction
c) molarity             d) molality of Zn (NO3)2
Zn = 65.4 , N=14       ,      O=16 ,           H=1

molar mass of Zn (NO3)2 = 65.4 + (2X14) + (16X16) =189.4 gmoL-1
a) mass of 1L solution = 1000 mL X 1.107 gmL-1 = 1107g

121
mass percent =____ X 100 = 10.93%
1107

b) mass of water = 1107 – l2l = 986g
no.of moles of Zn (NO3)2 = l2l g x 1mol = 0.639 mol
189.4g

no.of moles of H2O = 986g x 1mol = 54.78 mol
18g

mole fraction =    0.639    . = 0.0115
0.639+54.78

c) molarity =     0.639 mol = 0.639 M or (molL-1)
IL

D) molality =     0.639 mol = 0.648 molkg-1 or m
0.986 kg

70
- Conversions amongst concentration units :
to convert mass percent to or from mol fraction or to molality
we need to know molecular weight. To convert these into molarity
we must know also density of solution.

Ex :An aqueous solution of 1 M H2SO4 has density of 1.05g mL-1
.
calculate its molality.
H=1 ; O = 16 ; S = 32

mass of 1L = 1000 ml X 1.05 gmL-1 = 1050 g
molar mass of H2SO4 = 98 g mol-1
mass of water = 1050 – 98 = 952 g
molality = 1mol = 1.05 mol kg-1
0.952 kg

OH
1
E.X A solution of CH3-CH-CH3 in water has a mole fraction of
alcohol equals 0.25; what is the mass percent of alcohol in solution.

C = 12 ; O = 16 , H = 1

Assume we have 1 mol of solution
no. of moles of alcohol = 0.25 mol
no. of moles of water = 1-0.25 = 0.75 mol

mass of alcohol = 0.25 mol X 60g = 15 g
1mol

mass of water = 0.75 mol X 18g = 13.5 g
1mol

mass of percent of alcohol = 15g x 100= 52.6%
15+135

71
E.X Calculate molality of the same solution above.

Molality = 0.25 mol = 18.5 mol kg-1
0.0135 kg

- Liquid Solution :
* The solution process in liquid solution :
The ease of distribution of solute particles in solvent depends on
relative forces between :
(1) Solvent molecules
(2) Solute particles
(3) Solute – solvent attraction forces

CCl4        +       O

relatively        relatively
weak London        weak London
force             force

These two liquids are miscible. They dissolve in each other
(similar attraction forces).

CCl4       +      H2O
relatively        relatively
weak London        strong hydrogen
force             bonding

they are immiscible. They form two layers.

72
C2H5OH          + H2O
relatively         relatively
strong hydrogen strong hydrogen
force               bonding
They are miscible in all propotions
Between miscibility and immiscibility, these is partial
miscibility . If you increase length of the hydrocarbon chain,
solubility decreases.

* Solids in liquids :
Attractive forces between particles of a solid is a maximum
because they are closely arranged.
The solute – solvent attractive forces must be relatively high to
get a solution of solid in liquid.
molecular solid dissolve in non polar solvent.

I2(s) dissolve in CCl4

ionic solid lkike NaCl does not dissolve in CCl4.
weak attractive forces can't overcome strong ionic forces between
ions.
ionic solid dissolves in very polar solvents like water.
strong attractive forces between H2O and ions is enough to tear ionic
lattice.

like dissolves like
polar dissolves polar
non polar dissolves non polar

73
- Heat of solution, H soln :-

Hsoln = Hsolution – Hcomponents

If H soln is – Ve, energy is released (exothermic)
If H soln is +Ve, energy is absorbed (endothermic)

The magnitude of solution depends on relative forces between
various particles that make a solution.

- Energetics of liquid in liquid solution :-
There are 3 steps, each of which involves heat change :
1- Separation of solvent molecules (endothermic).
2- Separation of solute molecules (endothermic).
3- Bringing solvent and solute molecules together
(exothermic)

H solution is the sum of the energies.

1st case energy diagrams

(2)                                 CCl4 + benzene
to expand solute           (3) solvent-   Hsdn = O
solute
together
Energy required                           solvent – solvent,
(1)                                  solute – solute =
To expand solvent                         solvent – solute
Solvent + solute                          attractive forces
A-A, B-B=A-B

74
2nd case
acetone / H2O
H-Ve (exothermic)
(2)                      solute-solute, solvent-solvent <
Solute-solvent

volume of solution < volume of
solvent + volume of solute
(1)                        (3)

Hsolution

3rd case
Hexane+ ethanol
H+Ve (endothermic)
(2)                (3) solvent – solvent, solute-solute >
Solute-solvent attractive force
volume of solution > volume of
(1)                         solvent + volume of solute

Hsolution

75
- Energetic of solution of ionic compounds in water :-
Two processes
K+I-(s)           K+(g) + I-(g)   endothermic equals lattice energy
+    -
+      -
K (g) + I (g) + xH2O            Kaq + Iaq hydration energy (exothermic)

K+I-(s)+K+(g) + I-(g) + xH2O       K+(g)+I-(g)+K+aq+I-(aq), Hsoln
+    -
+ -
K I (S) + XH2O            Kaq + Iaq ; Hsoln

If lattice energy > hydration energy
Hsoln + Ve (endothermic)

If lattice energy < hydration energy
Hsoln – Ve (exothermic)

Energy diagram

K+(g)     +   I-(a)

Lattice               hydration energy
+      -
Energy                Kaq + Iaq

KI(s)             Hsdn

endothermic
for LiCl, it is exothermic

76
increasing charge of the ion, increase both lattice energy and hydration
energy.

Decreasing the size of the ion, increases both lattice energy and hydration
energy.
‫التنبؤ‬
It is difficult to predict in advance which phenomenon predominates.

- Solubility and Temperature :-
If the process is exothermic, solubility decreases with rising temp.
If the process is endothermic, solubility increases with increasing tempt.

Solvent + gas          solution always exothermic
Solubility of a gas decreases with increasing temperature.
That is why gaseous beverages and preferred to be drunk while cold.

- The effect of pressure on solubility :-
Changing pressure has no effect on solubility of liquid in liquid or solid in
liquid.
But increasing pressure increases solubility of gas in liquid.

77
gas (solute) + Solvent        solution

Two applications :-
1- carbonated beverages are canned or bottled under high pressure of
CO2.
2- The same phenomenon is responsible for decompression sickness
(bends).

- Henry's Law :-
The concentration of a gaseous solute in solution is directly proportional to
the partial pressure of the gas above the solution.

Cg = Kg x Pg             Partial
Conc. of gas                             Pressure
of the gas
constant
characteristic
for the gas and
liquid

Henry's law applies for relatively low concentrations and pressures, and for
gases that do not interact extensively with solvent.

78
- Vapor pressure of solutions :
When a solute dissolve in a solvent, the vapor pressure of the solvent
in solution decreases. When a non volatile solute dissolves in
a solvent, the vapor pressure of the solution (also the solvent) is
given by Rauole's law.
O
V.P. of solvent         PA = XA. PA          V.P of pure solvent
in solution
mole fraction
of solvent
*
O

PA

PA

0,0          XA            1

* E.X:- A solution prepared from 96.0g of non volatile, non
dissociating solute in 5.25 mol toluene O CH3 has a vapor

pressure of 16.31 kPa at 60C. what is the molecular mass of the
solute. The V.P. of pure toluene at 60 C is 18.63 kPa.

o
PA = XA. PA

16.31 kPa = XA. 18.63 kPa

79
XA = 0.875
nA
0.875 =          __________
nsolute+nA

5.25 mol
0.875 =          __________
nsolute+5.25

n   solute   = 0.754 mol

1mol
0.754 mol = 96 x _______
M

M = 127.3 gmol-1
__________________
* A Solution of two volatile components : In this case the vapor
pressure of solution equals sum of V.P of each components in
solution.
o
PA = XA. P A applying Rauol's law

o
PB = XB. P B
o            o
Ptotal= PA+ P B = XAPA+XB PB

Graphically

V.P. of solution
O
PA
O
B
P

O          XA           1
1          XO           0

80
Volume of solution
<volume of A+
Volume of B
O        - Ve deviation A-B>A-A+B-B
PA          exothermic H - Ve
O                                            rise temp.
B
P

O          XA            1
1          XO            0

O        + Ve deviation endothermic
PA        H + Ve decreased in temp
O                                           volume of solution > volume of
PB                                              A + volume of B
A-B< A-A+B-B

O          XA            1
1          XO            0

* E.X Heptance, C7H16, has a vapor pressure of 791 torr at 100c. At
this temp., octane C8H18 has a vapor pressure of 352 torr. What
will vapor pressure in torr of a mixture of 25 g heptane and 35 g
octane assume ideal solution behavior.

Psolution = Pheptane + Poctane

Xheptanex Pleptans +XoctanexPoctane

Xhep =           25/100          , Xoctane =         35/114
25/100+35/114                        35/114+25/100

Psolution = 791 X xhep + 352x Xoctane
Psolution = 355.16 torr + 193.9 torr = 549.06 torr

81
‫الصفات التجمعية‬
- Colligative Properties :
These properties depend on the number of solute particle in
solution and not on its nature. These are :
1- Lowering of vapor pressure.
2- Elevation of boiling point and depression (lowering) of
freezing point.
3- Osmotic pressure.

Phase Diagram of Water

Water liquid
for solution

1atm     ice                     vapor

Pressure                           tb

0C        100C
tf at freezing  t   boiling point
point
___________________________________________

Normal boiling point : the temperature at which vapor pressure of
liquids = 1 atm.
Normal freezing point : the temperature at which 1 atmosphere line
cross solid liquid equilibrium line.
As a result of solution, vapor pressure decreases and causes rise in
boiling point and loweing of freezing point.

82
These properties are used for determination of molecular masses.

tb = Kb molality

Boiling point constant
Characteristic for each liquid

tf = Kf molality

Freezing point constant
Characteristic for each liquid

* For water : Bp = 100 + tb
fp = 0 - tf

* E.X : What is the molecular mass and molecular formula of a non-
dissociating compound whose empirical formula is C4H2N
if 3.84 g of the compound in 500 g benzene, C6H6, gives a
freezing point dep. 0.307C.
Kf (benzene) = 5.12Cm-1.

tf = Kf. molality
0.307C = 5.12 Cm-1 X molality
Molality = 0.06m or mol Kg-1

0.06 mol Kg-1 nsolute
0.5 Kg

nsolute = 0.03 mol

83
3.845
molar mass = ________ = 128 g mol-1
0.03 mol

E.F. Wt = (4X12) + (2X1) + (1X14) = 64

128
no. of time E.F. in M.F = _____ = 2
64

Molecular formula = C8 H4 N2

3- osmotic pressure :
Osmosis is a process whereby solvent molecules pass through a semi
permeable membrane from dilute solution to a more concentrated
one.
Dialysis : occurs at cell wall permits water and small particles but
restricts large molecules.
There is a net transfer of solvent from dilute to conc. solutions.

11      h        hydrostatic
pressure =
osmotic pressure

II & C. T     absolute temp.
osmotic                                   dilute conc.
pressure    molarity

II=RCT

Universal gas constant

II = CRT                   IIV = nRT        Van't Hoff
II = n/v RT                                 equation

volume

84
This phenomenon is used for determination molecular masses
particularly high molecular mass like proteins or plastics (polymers).

* E.X : The osmotic pressure found for solution of 5g horse
hemoglobin in IL solution is 1.8X10-3 atm. At 25C, what
is the molar mass of hemoglobin.

IIV = nRT
1.8X10-3 X 1 = n X 0.0821 X 298 K
n = 7.4 X 10-5 mol

5g
-5
7.4 X 10        = _____________
molecular mass

5g
Molar mass =___________ = 68000 g mol-1
7.4X10-5mol

85
-   Solution of electrolytes :
Electrolytes dissociate in solution. The effect of 0.1 M NaCl in
solution is twice the effect 0.1M glucose, because the former
gives twice as many particles as the latter. Because colligative
properties depends on the number of particles of solute in
solution.
Ca Cl2 has 3 times the effect of non electrolytes.
Al2 (SO4)3 has 5 times the effect of non electrolytes.

A weak electrolyte has an effect between a strong electrolyte and
non electrolyte. In case of association like benzoic acid in
benzene, the effect is halved because two particles act as one
particle.

O……H - O

O   -C                 C-       O

O – H ….O

- Inter ionic attraction :
These are attractive force between ions in solution, which increase
with increasing concentration. This makes the solute behaves as if
it is not ionizing completely.

L : Vant Hoff factor

tf (measured)
I =_____________ _______________
tf (calculated as non electrolyte

86
It can be used to calculate degree of dissociation.

- Chemical Equilibrium :-
forward reaction
C+D             E+F
back word reaction

At equilibrium, forward rate = backward rate.
The concentrations of reactants and products remain constant at
equilibrium.

Rate                  forward

time

There is relationship between concentrations of reactants and
products at equilibrium called equilibrium law or law of mass action.

[E] . [F]                 [ ] = molarity
Kc = _______
[C] . [D]

Kc : equilibrium constant. It is constant at constant temperature. In
general for :-

87
cC + dD          eE + fF

[E]e . [F]f
Kc = _________
[C]c . [D]d

for gases we can use also Kp.

PEe . PFf                   P : Partial pressure
Kp = _______
PCc . PDd

If you have large value Kc,
H2 + Cl2          2HCl               Kc = 5 X 1032
The reaction goes to large extent to the right.
For small value of Kc

2H2O             2H2+ + O2      Kc = 1.2 x 10-82

The equilibrium lies to the left.

This is dynamic equilibrium. The reaction does not stop but the
forward rate equals backward rate.

Some important relationship :

H2(g) + 3H2(g)         2NH3 (g)         Kc
2NH3 (g)               N2 (g) + 3H2 (g) K`c

1
Kc = _____
K`c

88
1/2 N2 (g) + 3/2 H2 (g)           NH3 (g)      K"c

K"c = Kc       = Kc 1/2

2NO(g) + O2 (g)           2NO2 (g)       K1

N2 (g) + O2 (g)     2NO (g)     K2
___________________________________

N2 (g) + 2O2 (g)          2NO2           K3

K3 = K1 x K2

[NO2]2       [NO]2         [NO2]2
__________ X __________ = __________
[NO]2 [O2]    [N]2 [O2]    [N2] [O2] 2

K 1 x K2 = K3

- Lechatellier's Principle :
If a stress is applied to a system at equilibrium, the equilibrium
will shift to reduce or cancel this stress.

N2 (g) + 3H2 (g)          2NH3 (g)

* Change of conc. or partial pressure :
Increasing conc. of reactant(s) shifts eq. to the right (more products
are formed). Increasing conc. of product(s) shifts eq. to the left (more
reactants are formed).

89
Change of total pressure (opposite to change in volume) for previous
reaction (NH3 formation) :-
Increasing total pressure, the reaction shifts to products
to decrease total pressure by decreasing number of moles of gases.
For 2NO2(g) +             2NO (g) + O2(g)

Increasing total pressure, the reaction shifts to the left.

H2(g) + I2 (g)             2HI (g)

Change in total pressure or volume have no effect on equilibrium.
Effect of temperature change

N2(g) + 3H2 (g)            2NH3 (g) + heat or H – Ve

We treat heat of reaction as a reactant (endothermic) or products
(exothermic).
Increasing of temperature for endothermic reaction shift equilibrium
to products.
In this case Kc value increases. For exothermic reaction increasing
temperature reduces value of Kc.

90
4- Addition of Catalyst : has no effect on equilibrium constant or
equilibrium position.

5- Addition of inert gas : (any gas not present in the equation) has no
effect.

2NO2(g)          N2O4(g)
_____________________
- Reaction Quotient :
cC + dD         Ee + Ff

[E]e . [F]f
Q = _________           at any time before after or at eq
[C]c . [D]d

If Q > Kc the reaction goes to left (to reactants)
If Q< Kc the reaction goes to right (to products)
If Q = Kc the reaction is at equilibrium
_________________

- Relationship between Kp and Kc :-

PDV = nDRT

PD = __nD___ RT = [ D ] R T
V

PDd = [ D ]d (RT)d

91
PEe . PFf   [E]e [RT]e. [F]f [RT]f
Kp = _________ = ___________________
PCc. P.Dd   [C]c [RT]c. [D]d [RT]d

[E]e . [F]f   (e+f) – (c+d)
= _________ = (RT)
[C]c. [D]d

KP = KC. (RT)n                       An = no. of moles (coefficient)
of gases in products – no. of
moles of gases in reactants

* E.x. For H2 (g) + I2(g)              2HI(g)

In 10L vessel. at eq. there are : 0.1 mol H2
0.1 mol I 2
0.74 mol HI

If we insert 05 mol of HI after equilibrium has been reached calculate
the new equilibrium concentrations.

[HI]2         (0.74)2
KC = _________ = _____________ = 54.76
[H2] [I2]  (0.1) X (0.1)
10       10

H2(g) + I2(g)           2HI

H2(g) + I2(g)              2HI
no. of moles at start   0.1      0.1              0.74+0.5
change in no, of moles +X        +X               -2X
no. of moles at eq.     0.1+X 0.1+X               1.24-2X
conc.                   0.1+X 0.1+X               1.24-2X
10      10                 10
2
[HI]                (1.24 – 2x)2
KC = ________ = 54.76 = _____________
[H2][I2]          (0.1+X) (0.1+X)
10      10

92
1.24 – 2X
take square root 7.4 = _____________
0.1 + X
X = 0.053 mol

0.1 + 0.053  0.153
[H2] = _________ = _______ = 0.0153
10        10

[I2] = 0.0153

1.24 + 0.106
[HI] = __________ = 0.1134
10
* E.x. Calculate Kp for the reaction :

H2 (g) + I2(g)            2HI(g)
at 27 C if Kc = 7.5

Kp = Kc. (RT)n
n = 0
Kp = Kc = 7.5

heterogeneous Equilibrium

2NaHCO3(s)               Na2CO3(g)+ H2O(g) + CO2(g)

[Na2CO3(s)] [H2O(g)] [CO2(g)]
KC = _________________________________
[NaH CO3(s)]2

concentrations of pure solids and liquids constant and do not change

[NaHCO3(S)]2
__________ __        = constant              Kc = [H2O] [CO2]
[Na2 CO3(S)]

93
Kp = PH2O x PCO2

KP = Kc(RT)2

H2O(L) = H2O (g)         [H2O] = constant
Kp = PH2O (g)            = 3.17 kPa at 25oC
Kc = [H2O(g)]
Kp = 3.17 kpa
Kc = Kp = 3.17 kPa                      = 1.28 X 10-3 M
RT 8.31kPaLmol-1K-1x298K

94
95
* Acid – Base Equilibrium in aqueous solution :
Ionic equilibrium for acids and bases.

Ionization of water : H2O + H2O               H3O+ + OH-

[H3O+] [OH]
KC = __________ = 0.1134
[H2O]2

KC [H2O]2 = [H3O+] [OH-]
[H2O] = 55.5 M constant

KW [H3O+] [OH-] = 1x10-14              at 25 C

KW has same value in neutral acidic or basic solutions. It is an
equilibrium constant. It is called ionization constant or dissociation
constant or ionic product of water.

in pure water or neutral solution :
[H+] = [OH-] = 1X10-7 M
acid makes [H+] > 1 X 10-7 M
base makes [H+] < 1 X 10-7 or [OH-] > 1 X 10-7

96
in acidic solution [H+] is considered to come safely from acid
for two reasons :
1- Ionization of water is very small.
2- The presence of H+ from acid makes ionization still lower. n
basic solution. OH- is considered safely to come from base for
the same reasons in basic solutions.

Strong acids is considered to ionize completely.
Strong base is considered to ionize completely.

H2O         H+ + OH-

* E.x. Calculate [H+] and [OH-] for 0.1 M HCl (aq).
HCl(aq) is a strong acid.

H2O
HCl(aq)           H3O+ + Cl-

o.1 M             0.1 M    0.1 M

[HCl] = [H+] = 0.1 M

Kw      1x10-14
[OH-] =_____ = _______ = 1 X 10-13
[H+]     0.1

97
* E.x. Calculate [OH-] and [H+] for 0.1 M Ba (OH)2
which is a strong base.

Ba(OH)2                 Ba+2 + 2OH-
0.1 M 0.2 M

[OH-] = 0.2 M

Kw      1x10-14
[H+] =_____ = _______ = 5 X 10-14
[OH-]     0.2

* PH :-

1
+
PH = -log [H3O ] = Log ________
[H3O+]

1
-
POH = -log [OH ] = Log ________
[OH-]

KW = 1 X 10-14 = [H+] [OH-]

PKW = -log 1x10-14 = 14 = - log [H+] + (-log[OH-]
= PH + POH

For solution neutral, basic or acidic

PH + POH = 14

98
* Ex : Calculate PH and POH for 0.1 M HCl (strong acid)
[H+]= 0.1 M

PH = log [H+] – log 1x10-1 = 1
POH = 14 – 1 = 13

* Ex : Calculate POH and PH for 0.1 M Ba (OH)2
[OH-]= 0.2 M
POH = log 2xb-1 = 1-1092 = 0.7
PH = 14 – 0.7 = 13.3

- Note :

PH        POH

Neutral        7           7
solution
Acidic        <7         >7
solution
Basic         >7         <7
solution

* If PH changes from 7 to 5

10-5
_____ = 102                       ‫يتغري 110 مرج‬
10-7

* If PH changes from 5.5 to 7

- 5.5 shift log  3.16X6-6
=_____ _______ = ________31.6         ‫يتغري‬
-7 shift log    1X10-7

99
- Dissociation of weak acids and bases :-

HA+H2O                A- + H3O+

[H3O+] [A-]
Keq = ___________
[HA] [H2O]

[H3O+] [A-]
Keq [H2O] =___________ = Ka
[HA]

conc. At start    HA               M+ + A-
change            C                O    O
at eq            -X               +X   +X
C-X=C             X    X

X2      [H+]2
Ka = ____ = ________
C        C

Ka. C = [H+]2

[H+] =     Ka.C

100
* Ex : Calculate PH and POH for 0.1 M HC2 H3 O2
(actitic acid) Ka = 1.8 X 10-5

[H+] =    Ka. C =    1.8 x 10-5 X 0.1 = 1.34 X 10-3

PH = -log 1.34 X 10-3 = 2.87

POH = 14-287 = 11.13

- for weak bases like NH3 or organic amines,
NH3 + H2O             NH+4 + OH-
in the same way :

[OH-] =      Kb.C

* Ex : Calculate POH and PH for 0.1 M NH3, Kb=1.8 X 10-5

[OH-] =    Kb.C   = 1.34 X 10-3

POH = 2.87
PH = 14 – 2.87 = 1.13

101
- Buffers :
It is a solution which undergoes slight change in PH on
addition of small quantity of strong acid or strong base.

acidic buffer PH < 7
weak acid + salt of same acid + strong base
HC2 H3 O2 / Na C2 H3 O2

Basic buffer PH > 7
weak base + salt of same base + strong base
NH3 / NH4+

* How buffer works : for acidic buffers :
H+ + C2 H3 O2-          HC2 H3 O2            H+ consumed
from salt

b) addition of strong base :
OH- + HC2 H3 O2              H2O + C2 H3 O2-    OH- consumed

* if you have basic buffer :
a) H+ + NH3             NH4+             H+ is consumed

b) OH- + NH4+           H2O+ NH3       OH- is consumed

102
In acidic buffer :

HC2 H3 O2                   H+ + C2 H3 O2- common lon
salt ionize Na+ C2H3O2                  Na+ + C2 H3 O2-
completely

The presence of acetate from salt causes decrease in he
ionization of the acid, because of common ion effect. The
concentration of acetate can be considered safety to come from salt.

[H+] [C2H3O2-]
Ka =________________
[HC2H3O2]

[acid]
+
[H ] = Ka _______
[salt]

In the same way for basic buffer :

[base]
-
[OH ] = Kb _______
[salt]

Log [H+] = log ka + log [acid] – log [salt]         ‫تالضرب يف سالة‬

[salt]
pH = pKa +log ________                  pKa ‫يصثح احلمض ضعيف كلما زادخ قيمح‬
[acid]

This is Herdersn Hasselbakh equation

[salt]
pOH = pKb + log _______
[base]

103
(anti) in calculator = shift then log

* Ex : for a buffer solution 0.1 M acetic acid and 0.1 M sodium
acetate, calculate [H+] and pH. Ka = 1.8X10-5

[acid]             0.1 M
+                      -5
[H ]= Ka. _______ = 1.8X10 x _______ = 1.8x10-5
[salt]            0.1 M

pH = log 1.8X10-5 = 4.74
____________________

Note [acid] = [salt]    pH = pKa

* Ex : What ratio of acetic acid / sodium acetate concentration is
needed to form a buffer whose

pH = 5 , Ka = 1.8x10-5

[H+] = anti log – pH = 1x10-5

[acid]
-5
1X10        = Ka _______
[salt]

[acid]
-5           -5
1X10        = 1.8x10 X _______
[salt]

1    [acid]     55
____=______ or ____                    as long as ratio of
18 [salt]      100                    conc. is 55%, the pH =5

104
Dilution of the buffer has no effect on PH of the buffer.

To have effective buffer :
1) Use concentrated solution.
2) Ph
3) pH around pKa, [acid] = [salt].

105
Calculation on Equilibrium
13.54 At a certain temperature Kc = 7.5 for the reaction
2NO2         N2O4
If 2.0 mol of NO2 are placed in 2.0 dm3 container and permitted to reach, what will be
the concentrations of NO2 and N2O4 at equilibrium ? what will be the equilibrium
concentrations if the size of the container is doubled ? Does this conform to what you
would expect from the chatelier's principle ?
2NO2                    N2O4
conc. at start   2.0                     O
moldn-3          2
change in conc. -X                        X
2
[N2O4]     X/2
7.5 = _______ = ______
[NO2]2    (1-X)2
X
15 = _______
1-2X+X2
15-30X+15X2 = X
15-31X+15X2 = O , X = -b+ b2 – 4ac
2a
X= 31+ 961 – 900 = 31 +78=1.29 not acceptable X
30                      30
X = 31- 961 -900            = 31-7.8 = 0.773
30                     30
conc. of NO2 = 1M – 0.773 M = 0.227 M
conc. of N2O4 = 0.773 M = 0.387 M
2
_________________________________
Volume becomes 4dm3
concentration 0.5 X         X
eq.                         2
7.5 = X/2        , 15 =      X    .
2                  2
(0.5-X)           (0.25-X-X )
2
3.75 – 15 X + 15 X = X, 3-75-16X+15 X2 = O

X = 16 - 256 – 225 = 16-5.57 = 0.348 = 0.35M
30          30
conc. of N2O4 = 0.175 M
conc. of NO2 = 0.15 M
Total number moles of gas at V=2nd m3 = 0.46+0.78=1.24 mol
Total number moles of gas at V=4dm3 = 0.60+0.68=1.28 mol
Increasing the volume increase the total number of moles gas.

106
Draw Born-Haber Cycle for the formation of CaCl2 Given the
following data, calculate the lattice energy of CaCl2 in kilojoules per
mole. Energy needed to vapourize 1 mol of Ca(s) = 192 KJ; first
ionization energy of Ca=590KJ mol; second ionization energy of Ca
= 1146 KJ mol; electron affinity of Cl =-350 KJ mol-1, bond energy
of Cl2 = 238 KJ mol-1 of Cl-Cl bonds, energy change for the reaction,
Ca(s)+ Cl2(g)      Call2(s), - 795 KJ mol-1 of CaCl2(s)
formed
HF
Ca(s) +      Cl2(g)              CaCl2 (s)
(2)                       (5)
(1)       2Cl(g)           (4)   2Cl(g)

Ca(g)                 (3)           Ca2+(g)

Energy change for the above reaction is the sum of energy changes of
step (1) through step (5) as shown in the following :

Hf = H1 + H2 + H3 + H4 + H5

-795KJ = 192 KJ + 238 KJ + (590 KJ + 1146 KJ) + (2x(-350KJ)+ Hg

Hg = -2261 KJ mol-1
We have multiplied electron affinity by 2 because we have two chloride
cons.
If we have CaO, we must multiply bond energy by 1/2 because we need
only one mol of oxygen atoms. In addition we have to use 1 st electron
affinity an d 2nd electron affinity in order to get O2-.
In case of KB, we use also 1/2 bond energy of Br2, one ionization energy,
and one electron affinity.
Using fi2O, We have to multiply step (7) by 2 and step (3) by 2.

107
100 C the equilibrium constant, KC, for the reaction.
CO(g) + Cl2(g)          COCl2(g)
has a value of 4.6X109 if O.2 mol of COll2 is placed into a 100 dm3 Hask
at 100 C, what will be the concentration of all species at
equilibrium?
CO (g) + Cl2 (g)                     COCl2 (g)
Constant     O :       O                        (0.02-X)M      From 0.2 mol
10dm3
Constant      X         :       X               (0.02-X)M
[COCl2]
Kc = _________
[CO] (Cl2]
O.O2 - X
9
4-6X10           = __________
x.x
since the eq. constant has very high value we neglect X from
numerator, 4.6 X 109 = 0.02
x2
X2 = 0.00435 x 10-8 M2
X = 0.0208x10-4 M
[CO] = 2.1X10-6M
[Cl2] = 2.1X10-6M
[Clc2]= O.O2 M
13.49 Sodium bicarbonate (baking soela) has many useful protection.
Among them in the ability to serve a fire extinguisher because of
thermal decomposition to produce CO2, which some there's the fire.

2NaHCOS(s)                    Na2 CoS(s) + CO2 (g) + H2O (g)
It 125 C the value of Kp is 2.6X103 Kpa2.
What are the partial pressures of CO2(g) and H2O (g)…
This suplem at equilibrium ? Car yen explain why Na HCO 3 used in
banking ? This is a heterogeneous equilibrium.

Kp = CO2(g)           H2O(g)
26 X103 KP2c = P2CO2
PCO2(g) = 5.1 X 10 kPa
PHLO (g) = 5.1 X 10 kPa

108

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