King Saud University
College of Sciences
Department of Chemistry
Practical General Chemistry
101 chem & 104 chem
Nabil Al-Sahly .
Introduction:
• Reference Book: Practical General Chemistry
By : Dr. Ahmad Al-Owais
Prof. Abdulaziz Al- wassil
• Reports File :Present In building No. 5
Distribution of Marks:
1. Theoretical Part: In Class of
(70 Mark)
2. Practical Part: In Laboratory of (30 Mark)
a) 6 or 5 Experiments then exam 1 of (10 Marks)
b) 6 Experiments then exam 2 of (10 Marks)
c) Attendance on time & Reports
(10 Marks)
What must on the student?
•
•
•
•
•
Attendance at the time.
Lab Coat.
Reports File.
Pin – Calculator
Pencil & ruler in case Drawing (Graph)
Exp(1):Determination of a liquid Density.
• Density: is mass per volume unit.
d= m(g) /
V(ml)
m: is mass
V: is volume
 1 ml = 1 cm3
 Density is a physical property of a substance.
 In this experiment we will measure density of water
 Density of Water (H2O)= 1 g/ml
*****************
Before starting the experiment...you should
know What
are tools(equipments) that we will
use it for experiments?.. and How we use it?
* Important of Tools(glassware) Used:
Beakers
Conical Flask(Erlenmeyer Flask)
Volumetric Flask
Funnel
Pipette
Burette
Main Tools (glassware) used:
Burette clamp
Burette
Valve (tap)
Conical flasks
Beakers
Waste beaker
Stand
Indicator
Volumetric pipette
Bulb
glass funnel
Balance
Distilled
water
Burette :
1
Pipette:
2
3
Very..Very important
OK
You can reading the volume ?
.
.
.
V = ………………ml
Equipments (glassware) used:
• 2 beakers (glass cups)
• Burette
• Graduated pipette
• Plastic Funnel
• Stand
Instruments(apparatuses) used:
• Digital (Electronic) Balance .
Procedure:
1. Clean all of your glassware with water.
2. Find mass of the empty beaker in g = m1
3. Add the given volume of water(H2O) to the empty beaker
and find mass of the beaker and H2O in g = m2
4. Calculate mass of H2O in g (m2 – m1 ) = m
5 . Calculate density of H2O in g / ml
6. Repeat the steps 3 to 5
= m/V=
d
* Calculation of the density (d) by graph:
• Plot the relationship between the mass of H2O(m) on the Y-axis
versus its volume of H2O(V) on the X-axis , and find density(d) of
H2O from the slope = y2 – y1 / x2 – x1 = d
y2
Slope =
m (g)
y2 –y1
=
x2 –x1
d
y1
1 g/ml
x1
x2
V (ml)
Exp(2): * Introduction of Titrations(Reactions of Neutralization).
and
*Preparation of a standard solution of
Sodium Carbonate (Na2CO3) (0.05 M).
: is method of quantitative chemical analysis
• that is used to determine the concentration of an unknown
solution by a standard solution.
OR
: is a process transfer of a solution from a burette
(called the titrant) into a measured volume of another
solution for determining an unknown solution concentration.
Titration Terminology:
:is the volume of titrant required
to neutralize the sample (No. moles acid = No. moles base).
:is the pH value at the equivalent point of a titration.
:is a chemical which is added to the sample that
changes color at the equivalent point of a titration.
(Indicator is used to show when neutralization is occurs)
: is a solution of known concentration.
Acid + Base

• Example:
HCl + NaOH 
Salt + Water
NaCl + H2O
Clamp
Burette
Titrant solution
(acid or base)
Valve
Stand
Conical flask
Another solution
Known of Volume
2 drops of
indicator
The Concentration :
There are several ways of expressing
concentration and but we will focus on :
 Molarity(M):
Is number of moles of
solute dissolved in one
liter of solution.
M= n(mol) /
V(L)
 In case of the titration :
M × V /n = M’ × V’ /n’
This is law very important.
 Strength of Solution(S):
Is number of grams of solute
dissolved in one liter of
solution.
Relation between M & S :
S = M × M.wt
S=………. g/l
S=m(gm) of solute / V(L)of Solution
pH Range
0
1
2
3 4 5
6
7 8 9 10 11 12 13 14
Basic
Acidic
Neutral
[H+]>[OH-]
[H+] = [OH-]
LecturePLUS Timberlake
[OH-]>[H+]
19
pH Calculations (PH laws):
pH
pH = -log[H+]
[H+] = 10 -pH
[H+] [OH-] = 1 x10-14
pH + pOH = 14
pOH
[H+]
pOH = -log[OH-]
[OH-] = 10 -pOH
[OH-]
Example:
The [H+] of lemon juice is 1.0 x 10-3 M . What is the [OH-] , PH and POH?
[OH- ] = 1.0 x 10 -14
=
1.0 x 10-11 M
1.0 x 10 -3
PH= - log [H+] = -log(0.001) = 3
POH = - log [OH- ] = -log(1.0 x 10-11 ) = 11
OR
POH + PH =14   POH = 14 – 3 = 11
IS the solution acidic or alkaline(basic) or neutral ?
LecturePLUS Timberlake
21
It’s..................
Basic Laws:
Where: n: No. moles of solute
m: mass
M.wt:molecular weight
(molar mass)
M: molarity
V: volume of solution
This is law used for Preparation of Solution from
Solid Substance (material).
 To Preparation of Solution of Sodium Carbonate
(Na2CO3) in( 100 ml) & (0.05 M )
We apply the previous law:
m= 0.05 ×100 ×106 / 1000
=0.53 gm of Na2CO3
# Equipments and Material used:
 Sodium Carbonate (Na2CO3).
 Volumetric flask (100 ml).
 Electronic Balance .
…
Procedure:
1-Weigh out about(0.53 gm)of Na2CO3
.
2- Put it in volumetric flask(100 ml) .
3- Dissolve it with little of water(H2O).
4-Complete with water until the mark.
100 ml
Solution of Na2CO3(0.05M)
Exp(3)Determination of Organic Indicators for Acid base Titrations
At first we will know on methods of measuring pH, where there are two
ways …. A) Using pH meter. B) Using pH paper, a special type of paper.
.
B
PH paper
A
PH meter
Indicators Used :
Indicator
Phenolphthalein
(ph.ph)
Methyl Orange
(M.O)
Color in
Color in
Acidic middle Basic middle
Range in
PH
Colorless
Pink
8 - 10
Orange red
yellow
3.1 - 4
The idea of ​the experiment:
• To determine the suitable indicator we will make
Titrations :
a strong acid with strong base
..
a weak acid with a strong base
.. and draw a graphic relationship between the
volume of base added and the pH value .. where
we get Curve called (Titration Curve) ... through
which we can know on the suitable indicator.
A titration curve is a graph of the pH changes that occur
during an acid-base titration versus the volume of acid or
base added.
The objective (purpose) of study of titration curve:
Is to know the suitable Indicator for acid-base titration.
But ,
The objective of titration:
IS to know the concentration of an unknown solution.
Procedure:
• Where we will be adding different volumes of the
burette filled with a strong base(NaOH) to beaker
by the presence of known amount of strong
acid(HCl) or weak acid(CH3COOH), then follow the
change in pH value.
15_330
pH
W eak acid
Strong acid
Generally:
Vol NaOH
This figure shows the titration curve for both cases.
A)Titration between a strong acid(HCl) with a strong base(NaOH)
HCl + NaOH 
NaCl + H2O
ph.ph
OK
M.o
OK
*The suitable indicator for this titration is………….......
B)Titration between weak acid(CH3COOH)with a strong base(NaOH)
CH3COOH + NaOH 
Ph.ph
CH3COONa + H2O
OK
M.O
*The suitable indicator for this titration is………….......
Results from the drawing (graph):
• After that draw a relationship between the volume of base
added and the change in pH value.
• What determine from the drawing (graph)?
1- PH range at the equivalent point (….. to…..).
2- Draw two parallel lines for each indicator based on range
of each indicator, of which we can know the suitable
indicator (……………….) .
3- Find the volume of base at the equivalent point, through
dropping line from the region that occur then the sudden
change of the curve on the axis of the volume of
base(NaOH) added ( VNaOH = …… ml).
Explanation for pH curve:
This curve is similar to the previous curve, but the additive is acid
and not the base
14
endpoint
7
X
equivalent point
equivalent point volume ( V NaOH (ml) )
0
10
20
30
40
EXP(4): Determination of Sodium Hydroxide(NaOH)
Concentration BY Titrations With A Standard Solution of
Hydrochloric Acid (HCl) .
1) The titration by Ph.ph :
*Reaction equation:
2) The titration by M.O
:
HCl + NaOH → NaCl + H2O
*Results:
▪ Volume of NaOH:
123-
*Calculations :
▪ Molarity (M) of NaOH :
M × V /n = M’ × V’ /n’
▪ Strength of Solution (S) of NaOH :
S = M × M.wt
NaOH
M=?
V = average
HCl
M’ = 0.1 M
V’ = 10 ml
+
2 drops of
ph.ph
or
EXP(5):Determination of Acetic Acid (CH3COOH)
Concentration BY Titrations With A Standard Solution
of Sodium Hydroxide (NaOH) .
* The titration by Ph.ph
(only) :
*Reaction equation: CH3COOH + NaOH
CH3COONa + H2O
* Results:
… Volume of NaOH :
123-
NaOH
M = 0.1 M
V = average
*Calculations :
.. Molarity ( M ) of CH3COOH :
M × V /n = M’ × V’ /n’
.. Strength ( S )of CH3COOH :
S = M × M.wt
CH3COOH
M’ = ?
V = 10 ml
+
2 drops of ph.ph
EXP(6) : Determination of Hydrochloric Acid (HCl) Concentration By Titrations
With A Standard Solution of Sodium Carbonate (Na2CO3).
1)The titration by M.O:
2 HCl + Na2CO3
2 NaCl +CO2 + H2O
*Results:
..Volume of HCl
123*Calculations:
. Molarity of HCl :
M × V /n = M’ × V’ /n’
. Strength of HCl:
S = M × M.wt
2) The titration by ph.ph:
HCl + Na2CO3
HCl
M =?
V=average
Na2CO3
M’ = 0.05M
V’ = 10 ml
+
2 drops of
indicator
NaHCO3 + NaCl
*Results:
..Volume of HCl
123*Calculations:
. Molarity of HCl :
M × V /n = M’ × V’ /n’
.Strength of HCl :
S = M × M.wt
Is the molarity in two cases equal (same)
or
??
EXP(7): Measurement of Gas Diffusion
(Graham’s Law of Diffusion).
• In this experiment we will measure diffusion rate
to 2 gases :
NH3 (g)
&
HCl (g)
Which is faster diffusion you think ? And Why?
The answer : ……………faster than ………………
Because : ...........................................................
Molecular diffusion :
Diffusion:
“gas molecules spreading out to fill a room are diffusion.”
Its not easy since an average gas
molecule at room temperature and
pressure will experience about
10 billion collisions per second!
It only travels about 60 nm
between collisions!
38
He
N2
Which balloon will lose pressure sooner (quickly)?
39
N2
Big molecules
He
Small molecules
NOW :
Which balloon will lose pressure sooner ?
The answer : ..............
40
Graham’s Law of diffusion: diffusion rate is inversely the
proportional to square root of its molar mass or its density.
In case 2 gases :
r1 d1
2


r2 d 2
1
41
r1
2

r2
1
r = Diffusion rate of gas.
M = Molar mass (Molecular Weight) of gas.
the rates of diffusion of He and N2 .
rHe
28.0

 2.65
rN 2
4.00
He diffuses
times as fast as N2
42
Dalton’s Law of Partial Pressures:
He
H2
N2
Ptotal = P1 + P2 + P3 +...
43
EXP(8): Determination of Critical Solution Temperature
(C . S . T )
Principle of
experiment:
Warning: toxic
H2O
• Liquid water and phenol show
limited(partially) miscibility below( C.S.T). In
this experiment, miscibility temperatures of
several water-phenol mixtures of known
composition will be measured. Then
determination of the critical solution
temperature (C.S.T).
…………………………………………………………
C .S .T : Is temperature at which
be all compositions of
mixture or solution completely miscible.
Above (C . S .T) be
Homogeneous .
mixture
Under (C .S .T) be
Heterogeneous .
Calculation of percentage:
% X in a sample (compound or mixture or solution) :
= mass of X / mass of sample × 100
* Diagram between %phenol & miscibility temperature:
*
C.S.T =………OC
T (O C)
Homogeneous
Heterogeneous
%Phenol at C.S.T=…....%
%Phenol
% H2O at C.S.T = 100 ─ % phenol =….....OC
EXP(9):
Hess’s law
• Hess’s law states that “the change in enthalpy H
for any chemical reaction is constant, whether the
reaction occurs in one step or in several steps ” .
H1
A
B
F
H2
L
H4
W
E
D
C
H3
H1
H2
H3
H4
Explanation of Hess’s Law:
Start
Finish
A State Function: Does not depend on path .
Both lines accomplished the same result, they went
from start to finish.
Net result = same.
48
In this experiment you will measure:
the change in enthalpy(H ) of through knowing
amount of heat absorbed(q) , where:
#
q = m . P . t
q: amount of the heat
m: mass
P : Specific heat (constant)
t :The difference in temperature(t2 – t1)
# The Change in Enthalpy (H) =
Q/n
Q : Total (q1 + q2)
n : No. moles of NaOH
Specific heat & Heat capacity:
.Specific heat : Is the amount of heat (energy)
required to raise the temperature of (1 g ) of
a substance by (1 C).
. Heat capacity: Is the amount of heat (energy)
required to raise the temperature of something of
a substance by (1 C).
H
endothermic
Absorb heat
H = +
Liberate heat
H = ─
that
exothermic
Procedure:
Calorimeter:
Thermometer
Stirrer
Cork stopper
Cork cover(insulator)
EXP(10): Effect of Concentration on Reaction Rate:
(Determination of the order of the sodium thiosulphate and hydrochloric acid reaction)
In this experiment We will determine the order of Na2S2O3 and HCl , and thus
the order of total reaction .
2HCl + Na2S2O3 → 2NaCl + SO2 + H2O + S
OR
Rate = k [Na2S2O3]n [HCl ] m
Precipitate yellow
glutinous (slimy)
Rate = [ S ]
t
Factors that effect on Reaction Rate :
1- Nature of reactants .
2- Presence of a catalyst .
3- Temperature .
4- Concentration of reactants .
Log 1/t = log k + ? log M’
1) Order of Na2S2O3
Slope= y2- y1/ x2-x1=
Log 1/t
n
2) Order of HCl
n
m
Slope= y2- y1/ x2-x1=
Log 1/t
Log k
Log k
Log Na2S2O3
Order of total reaction =
n+m
Log HCl
m
EXP(11): Determination of the Molar Mass (M.wt) of An Organic
Compound By The Depression of Its Freezing Point (ΔTf ) .

If a nonvolatile solute is added to a liquid, a number of physical properties of the pure
substance change, including
vapor pressure depression, freezing point depression, and boiling point elevation. These
alterations are collectively
known as colligative properties of solutions.
* Freezing point depression: Solution freeze at lower temperature than pure liquid(solvent).
The amount that the freezing point is lowered called the freezing point depression (ΔTf) .
molality (m) = moles of solute / mass of solvent(kg)
* Freezing point ( T f ) : Is the temperature at which a liquid changes to a solid .
The change in the freezing point (ΔTf) in °C for a nonvolatile organic solvent
can be determined using the following
equation, where kf is characteristic for the solvent used:
ΔTf = kf . m
is molality don’t mass
*********
ΔTf = kf . m
molality
ΔTf = kf . n2 . 1000 / m1(g)
Where:
moles of solute
mass of solvent
m= n2 / m1(kg)
m= n2 . 1000 / m1(g)
ΔTf
=
kf . m2 . 1000
M2 . m 1
M
Molar mass
M.wt
= kf × m2 × 1000
ΔTf × m1
Procedure:
Split
stopper
Thermometer
(4.8 ~ 5) gm
of
unknown
25 ml of
H2O
Stirrer
Big
beaker
EXP(12):
Determination of the Molar Mass(M.wt) of An Organic Compound By
The Steam Distillation .
When mixing two liquid do not mix (immiscible) the boiling point of the
mix, will become less than boiling point of any of them .
Boiling Point(Tb ): It is the temperature at which internal vapor pressure
of the liquid is equal to the pressure exerted by its surroundings .
Suppose that :
H2O = A
m A / mB =
&
Unknown liquid = B
PA × ( M.wt)A / PB × (M.wt )B
(M.wt)B = PA × ( M.wt )A × m B / mA × PB
Raoult’s law:
For a mixture of two miscible liquids (A and B), the
total vapor pressure is the sum of the individual vapor
pressures:
PTotal = PA + PB
Procedure:
‫ح‬
The steam distillation apparatus:
Thermometer
condenser
100
100
mlml of unknown compound (B)
VA
+
50 50
mlml of H2O (A)
VB
Graduated Cylinder
****************************************************
Results:
1- Boiling point(Tb ) =………OC
PA = ……..
PB = PTOTAL ─ PA = …….
2- Volume of H2O (A) after distillation = ……. ml
3- Volume of unknown solution (B) after distillation = …… ml
Calculations:
d B × VB
dA × VA
(M.wt)B = PA × ( M.wt )A × m B / mA × PB
(M.wt)B = ....... gm / mol
End of the experiments
***************************************