King Saud UniversityCollege of SciencesDepartment 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:

BurettePipette

Conical Flask(Erlenmeyer Flask)Beakers

Volumetric FlaskFunnel

Conical flasks Beakers

Waste beaker

glass funnel

Indicator

Volumetric pipette

Bulb

Burette

Stand

Burette clamp

Distilled water

Valve (tap)

Main Tools (glassware) used:

Balance

How we use glassware ???

Burette :

1 2 3

Pipette:

Very..Very important

OK

You can reading the volume ?

.

.

.V = ………………ml

Equipments (glassware) used:

• 2 beakers (glass cups)• Burette• Graduated pipette• Plastic Funnel• StandInstruments(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 = m / V = d 6. Repeat the steps 3 to 5

* 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

m (g)

V (ml)

y2

y1

x2x1

y2 –y1

x2 –x1

Slope = = d

1 g/ml

Exp(2): * Introduction of Titrations(Reactions of Neutralization). and *Preparation of a standard solution of Sodium Carbonate (Na2CO3) (0.05 M).

• Titration: is method of quantitative chemical analysis • that is used to determine the concentration of an unknown

solution by a standard solution.

OR

• Titration : 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: . Equivalent(equivalence) point :is the volume of titrant required

to neutralize the sample (No. moles acid = No. moles base).

• End point :is the pH value at the equivalent point of a titration.

• Indicator :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)

• A standard Solution: is a solution of known concentration.

• Remember that: Acid + Base Salt + Water • Example: HCl + NaOH NaCl + H2O

Summary of Titration:

Titrant solution

(acid or base)

Known of Volume

Another solution

2 drops of indicator

Burette

Conical flask

Valve

Stand

Clamp

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/lS=m(gm) of solute / V(L)of

Solution

LecturePLUS Timberlake 19

pH Range

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Neutral

[H+]>[OH-] [H+] = [OH-] [OH-]>[H+]

Acidic Basic

pH Calculations (PH laws):

pH

pOH

[H+]

[OH-]

pH + pOH = 14

pH = -log[H+]

[H+] = 10 -pH

pOH = -log[OH-]

[OH-] = 10 -pOH

[H+] [OH-] = 1 x10-14

LecturePLUS Timberlake 21

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 ? It’s..................

Basic Laws:n = m / M.wt & M = n / V(L)

Where: n: No. moles of solute m: mass M.wt:molecular weight (molar mass) M: molarity V: volume of solution

m = M × V(ml) × M.wt / 1000This 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..

PH meterPH paperB A

Indicators Used :

Indicator Color in Acidic middle

Color in Basic middle

Range in PH

Phenolphthalein(ph.ph)

Colorless Pink 8 - 10

Methyl Orange(M.O)

Orange red yellow 3.1 - 4

The idea of the experiment:• To determine the suitable indicator we will make

2 Titrations : A) a strong acid with strong base .. B) 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.

Titration Curves:

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.

The objective of titration:IS to know the concentration of an unknown solution.

But ,

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.

Generally:This figure shows the titration curve for both cases.

15_330

Vol NaOH

Strong acid

pH

Weak acid

A)Titration between a strong acid(HCl) with a strong base(NaOH) HCl + NaOH NaCl + H2O

ph.ph

M.o

*The suitable indicator for this titration is………….......

OK

OK

B)Titration between weak acid(CH3COOH)with a strong base(NaOH) CH3COOH + NaOH CH3COONa + H2O

M.O

Ph.ph

*The suitable indicator for this titration is………….......

OK

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

pH endpoint equivalent point 7 X

equivalent point volume ( V NaOH (ml) ) 0 10 20 30 40 Volume of titrant added (mL)

EXP(4): Determination of Sodium Hydroxide(NaOH) Concentration BY Titrations With A Standard Solution of Hydrochloric Acid (HCl) .

NaOHM = ? V = average

HClM’ = 0.1 MV’ = 10 ml

+

2 drops of

ph.ph or

M.O

HCl + NaOH → NaCl + H2O

1) The titration by :

M × V /n = M’ × V’ /n’

▪ Volume of NaOH:1-2-3-

*Results:

*Calculations :

▪ Molarity (M) of NaOH :

▪ Strength of Solution (S) of NaOH : S = M × M.wt

2) The titration by :

*Reaction equation:

Ph.ph M.O

EXP(5):Determination of Acetic Acid (CH3COOH) Concentration BY Titrations With A Standard Solution of Sodium Hydroxide (NaOH) .

NaOH M = 0.1 M

V = average

CH3COOHM’ = ?

V = 10 ml

+2 drops of ph.ph

* The titration by (only) :

*Reaction equation: CH3COOH + NaOH CH3COONa + H2O

* Results:

… Volume of NaOH :1-2-3-

*Calculations :.. Molarity ( M ) of CH3COOH :

M × V /n = M’ × V’ /n’

.. Strength ( S )of CH3COOH :

S = M × M.wt

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) The titration by ph.ph: 2 HCl + Na2CO3 2 NaCl +CO2 + H2O HCl + Na2CO3 NaHCO3 + NaCl

*Results:..Volume of HCl1-2-3-

*Calculations:. Molarity of HCl : M × V /n = M’ × V’ /n’

. Strength of HCl: S = M × M.wt

*Results:..Volume of HCl 1-2-3-

*Calculations:. Molarity of HCl : M × V /n = M’ × V’ /n’

.Strength of HCl : S = M × M.wt

HClM =?

V=average

Na2CO3

M’ = 0.05M V’ = 10 ml

+2 drops of indicator

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 : ...........................................................

38

Molecular diffusion :

Diffusion: “gas molecules spreading out to fill a room are diffusion.”

Its not easy since an average gasmolecule at room temperature and

pressure will experience about10 billion collisions per second!

It only travels about 60 nm between collisions!

39

N2

He

Which balloon will lose pressure sooner (quickly)?

40

N2 He

NOW :

Which balloon will lose pressure sooner ?

The answer : ..............

Big molecules Small molecules

41

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 :

1

2

2

1

2

1

d

d

r

r

42

1

2

2

1

r

r

r = Diffusion rate of gas.M = Molar mass (Molecular Weight) of gas.

Example: Compare the rates of diffusion of He and N2 .

65.200.4

0.28

2

N

He

r

rHe diffuses 2.65 times as fast as N2

43

Dalton’s Law of Partial Pressures:

Ptotal = P1 + P2 + P3 +...

He H2 N2

EXP(8): Determination of Critical Solution Temperature (C . S . T )

• 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).

H2O

Principle of experiment:

Warning: toxic

…………………………………………………………

C .S .T : Is temperature at which be all compositions of mixture or solution completely miscible.

Above (C . S .T) be Homogeneous .

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

mixture

* Diagram between %phenol & miscibility temperature:

*

T (O C)

%Phenol

C.S.T =………OC

%Phenol at C.S.T=…....%

Homogeneous

Heterogeneous

% 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 ” .

AH1

B

H2

CH3

D

H4

F

W

E

L

H1 H2 H3 H4

48

Explanation of Hess’s Law:

Start FinishA State Function: Does not depend on path .

Both lines accomplished the same result, they went from start to finish.

Net result = same.

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

exothermic

Absorb heat

that

Liberate heat

H = +

H = ─

Procedure:Calorimeter:

Stirrer

Cork stopper

Thermometer

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 .

Rate = k [Na2S2O3]n [HCl ] m

Rate = [ S ]

t

Factors that effect on Reaction Rate :1- Nature of reactants .2- Presence of a catalyst .3- Temperature .4- Concentration of reactants .

2HCl + Na2S2O3 → 2NaCl + SO2 + H2O + S

Precipitate yellow glutinous (slimy)

OR

1) Order of Na2S2O3 n 2) Order of HCl m

Log 1/t = log k + ? log M’

Slope= y2- y1/ x2-x1= n

Log Na2S2O3

Log 1/t Log 1/t

Log HCl

Slope= y2- y1/ x2-x1= m

Order of total reaction = n + m

Log k Log k

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

EXP(11): Determination of the Molar Mass (M.wt) of An Organic Compound By The Depression of Its Freezing Point (ΔTf ) .

*********

ΔTf = kf . n2 . 1000 / m1(g)

m= n2 / m1(kg)

m= n2 . 1000 / m1(g)

ΔTf = kf . m2 . 1000

M2 . m1

= kf × m2 × 1000

ΔTf × m1

ΔTf = kf . m

molality

moles of solute

mass of solvent

Where:

M.wt

M

Molar mass

Procedure:

Split stopper

Thermometer

Stirrer

Ice & salt

Big beaker

(4.8 ~ 5) gm of

unknown

25 ml ofH2O

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 .

mA / mB = PA × ( M.wt)A / PB × (M.wt )B

(M.wt)B = PA × ( M.wt )A × m B / mA × PB

PTotal = PA + PB

For a mixture of two miscible liquids (A and B), the total vapor pressure is the sum of the individual vapor pressures:

Raoult’s law:

Suppose that : H2O = A Unknown liquid = B&

:The steam distillation apparatusح

Graduated Cylinder

100 ml of unknown compound (B) 50 ml of H2O (A)

+100 ml

50 ml

Thermometer

Procedure:

VB

VA

condenser

****************************************************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 = …… mlCalculations:

(M.wt)B = PA × ( M.wt )A × m B / mA × PB

dB × VB dA × VA

(M.wt)B = ....... gm / mol

End of the experiments

***************************************