1

2

Week Topic Page

Number

1 Preparation of solutions 2

2 Relative viscosity 5

3 Viscosity at different temp. 8

4 Catalytic decomposition of Hydrogen peroxide 10

5 Phase diagram of partially miscible liquids 12

6 Specific Heat of calorimeter 15

7 Heat of Solution 17

8 Rate of Reaction of Sodium Thiosulfate and

Hydrochloric Acid

19

9 Heat of Neutralization 21

10 Saponification of Ethyl Acetate 24

3

Experiment 1:

Part A. Preparation of Molar solution of Sodium Chloride salt

Molar solution (M) is a solution that contains 1 mole of solute in a liter of

solution. Molar (M) solutions are based on the number of moles of compound in 250

milliliter of solution. A mole consists of 6.02 x1023 molecules or atoms. Molecular

weight (MW) is the weight of one mole of a compound. Determine MW using a

periodic table by adding the atomic mass of each atom in the chemical formula.

Example: For the MW of CaCl2, add the atomic mass of Na (23) to that of Cl 35.45) to

get 58.5 g/mole. Therefore, a 1M solution of NaCl consists of 58.5 g of NaCl dissolved

in enough water to make one liter of solution.

Procedure:

1. To prepare 100 ml of 1 M CaCl2 solution, use the previous formula to find out how

many grams of NaCl you need.

2. Grams of NaCl = (1) x (58.5) x (100) ÷ (1000) = x g

3. Weigh the calculated amount of substance accurately using balance for 0.01.

4. Transfer weighed amount into glass beaker then dissolve it in in sufficient distilled

water.

5. Transfer the dissolved amount into chemical to a clean, dry 100ml volumetric flask

using a funnel.

6. Add distilled water carefully to volumetric flask using graduated cylinder, before

reaching the calibration mark add water slowly using Pasteur pipette until the

bottom of the meniscus touches the calibration mark.

Step 1 Step 2 Step 3 Step 4 Step 5

Material:

1. Graduated cylinder

2. Volumetric flask

3. Pasteur pipette

4. Analytical balance

Grams of substance = (Molarity of solution in mole/liter) x (MW of substance

in g/mole) x (Volume of solution in ml of solution) ÷ 1000 ml/liter

Stage 4

4

5. Beaker

6. Funnel

7. Sodium Chloride salt

Part B. Preparation of Diluted Solution of Sodium Chloride from stock

(concentrated solution)

Normal solution (N) is a solution that contains 1 gram equivalent weight of a solute in a

liter of solution. Prepare 100 ml of 0.05 N of NaCl solution, use the previous solution and

dilute it, calculate volume of NaCl you need to prepare the new solution:

Procedure:

1. Calculate the volume of solution needed to make 100ml of solution 0.05 N using

the above formula.

2. Take the amount of calculated solution needed using a pipette.

3. Transfer the volume to a clean, dry 100ml volumetric flask.

4. Slowly add distilled water to the volumetric flask. Keep adding water until you

reach the 100ml mark on the neck of the flask using Pasteur pipette.

5. Place the stopper in the flask and gently swirl the flask until all the chemical is

dissolved.

millis of NaCl = (N conc.) x (V conc.) ÷ (N dil.) = ــــــــ ml

Material:

1. Graduated cylinder

2. Volumetric flask

3. Pipette

4. Funnel

N X V (diluted) = N X V (concentrated)

Graduated cylinder Volumetric Flask Pasteur pipette

5

Experiment 2:

Determination of Relative viscosity of liquid

Viscosity describes how a liquid resists the laminar movement of two neighboring layers.

This resistance to flow can be seen with gases, liquids and even solids.

Stoke's law:

In laminar flow, a ball with radius R moving through a liquid with viscosity η has a speed

proportional to the frictional force.

Fr = 3 π η d v

Or:

where:

Fd is the frictional force acting on the interface between the fluid and the particle (in

N),

η is the fluid's viscosity (in [kg m-1 s-1]),

R is the radius of the spherical object (in m), and

v is the particle's velocity (in m/s).

Viscosity can be readily followed using an

Ostwald viscometer as shown in the figure.

Ostwald viscometer: This, essentially, consists of a capillary tube down which a known

volume of solution is allowed to flow under gravity. The time taken for this flow is

measured (t1 and also that of water (t0); the relative viscosity is then given by:

Where η1 is the viscosity of the liquid of density ρl and η0 the viscosity of water of density

ρ0. If the densities are taken to be the same then the expression = ( t1/t0 )

Relative viscosity:

ηrel = η1/η0 = ( t1/t0 ) . ( ρl/ρ0)

6

Material: 1. Ostwald viscometer

2. Syringe or Pipette

3. Distilled water

4. Glycerin.

5. Stop watch accurate to at least 0.1 s.

Ostwald Viscometer

Capillary

tube

Add certain volume

of liquid

Measure the

time as the

Liquid Passes

through

This length)

7

Procedure:

Viscosity is very sensitive to temperature, so all solutions and the viscometer must be kept at

room temperature. Always handle the viscometer by one limb only and never squeeze the two arms together.

1. Rinse the viscometer with water and place it in position by carefully clamping one limb. Check

that it is vertical using a plumb line

2. Transfer 15 ml of water into 50 ml beaker and weigh it ( weight of water = x g)

3. Introduce exactly 15 ml of water into the viscometer with a pipette.

4. Leave for 5 min to equilibrate, apply gentle suction to the other limb (II, that has capillary tube)

until the meniscus rises above the upper graduation mark B.

5. Release the pressure and measure the time (to the nearest 0.1 s) for the liquid to flow between the

two graduation marks A and B. Repeat the experiment until the flow times agree within 0.2 s

6. Calculate the average flow time for water.

7. Repeat steps from 2 to 5 using different viscous liquids (ex. Glycerin).

8. Tabulate all readings in a table as shown.

9. Calculate the relative viscosities (η1/η0 ) using the equation below.

η1/ηw = ( tl/tw ).(ρ1/ρw) Results and Calculations:

Lab. Temperature = ـــــــــــــــــــــــــ oC

Density of water at Lab temp. ρw= 1 g/cm3

Viscosity of Water at Lab. Temp. ηw = 1.00166 cp (cp = Centipoise)

Volume of liquid = ــــــــــــــــــــ ml

Weight of liquid = ـــــــــــــــــــ g

Density of liquid ρl = ـــــــــــــــ g/cm3

T1 T2 Average (t) Wt. of liquid

Density of Liquid ρl = w/v

Absolute Viscosity ηl =tl/tw dl ηw

Relative Viscosity ηl =tl/tw dl

Water

Glycerin

8

Experiment 3:

Effect of Temperature in the viscosity of a Pure Liquid

At certain temperature T, time of flow through viscometer for liquid t l and water tw

w

l

w

l

w

l

t

t

d

d

Multiplying both sides with

l

w

d

to obtain

ww

wl

l

l

td

t

d

, the value

l

l

d

is called experimental viscosity of

liquid .

w

ww

wl dtd

t Where dw =1, then

ww

wl

td

t

The relation between experimental viscosity and temperature is given through exponential function

T

B

Ae

Where; A, and B are constants, T is absolute temperature in kelvin.

T

B

Ae

T

BA lnln

T

BA log303.2log303.2

T

BA

303.2loglog

From the relation between log , andT

1, the equation of a straight line. Its slope is

303.2

B and the

intersection with the y-axis Alog .

Procedure:

1. Prepare thermostated water path at 40 oC.

2. Insert a viscometer in water path after filling it with glycerin as mentioned in the previous

experiment.

3. Leave it inside water path for 10 minutes then start to determine the flow time of glycerin t l.

4. Repeat steps2 and 3 with water instead of glycerin and determine tw.

5. Decrease the water path temperature to 30 oC.

6. Repeat previous steps 2, 3, and 4 at 30 oC.

Results and Calculations:

9

1. Lab.Temperature= ــ ـــــ oC ـــــــ

2. Water density at lab. temperature= 1 g/cm3

3. Viscosity of water at lab. Temp. = ـ ــ ـــــــــــــــــ

4. Viscosity of water at lab temperature= ـ ــــ ـــــ ـــــ mPa.K.s ــ

Temperature

ToC

Absolute Temp. =

T+ 273

1/T Flow time of liquid tl Experiment

al Viscosity

w

wl

t

t

log

30

40

5. Draw the relation between

T

1on x-axis, and log on y-axis, then calculate A and B constants. The

slope is 303.2

B and the intersection is Alog .

Values of water viscosity with different temperature are tabulated in table below.

Temperature oC

Water Viscosity, mPa.K.s = cp

20 1.00166

28 0.835052 30 0.800117

35 0.72202

40 0.655688 45 0.598494

50 0.548981 55 0.505851

60 0.468063 65 0.434778

70 0.405311 80 0.35569

90 0.315778

N.B. The relation between Centipoise and Pascal and Newton units of viscosity 1 cP = 1 mPa·s = 0.001 Pa·s = 0.001 N·s/m2.

10

Experiment 4:

Catalytic Decomposition of Hydrogen Peroxide

Rate of chemical reaction depends on several variables, concentration of reactant, temperature, and

presence of catalyst.

H2O2 → H2O + ½ O2

Rate of decomposition of H2O2 increases as catalyst is added, such as Manganese dioxide MnO2.The

original concentration of H2O2 is (a ), x : the amount of H2O2 has decomposed, so the new

concentration of H2O2 is (a-x), k is reaction rate constant. Thus equation (1) the differential equation of

rate law and expresses the first order rate law.

dx/dt= k(a-x) (1)

Equation has been integrated into equation (2).

ln 𝑎

𝑎−𝑥= 𝑘𝑡 (2)

Then t = (2.303

𝑘) log

𝑎

(𝑎−𝑥) (3)

(t) is time of reaction

Relationship between (t) and 𝑎

(𝑎−𝑥) is straight line with a slope

2.303

𝑘.

If x = a/2, t= t1/2, where t1/2 is the half life time of reaction (time required for reactant to reaches half its

original quantity). From equation (3), t1/2 can be deduced as in equation (4).

t1/2 = 0.693

𝑘 (4)

Materials:

1. H2O2 1.2%

2. KMnO4 0.1N

3. H2SO4 2N

4. MnO2 (solid) or Fe2O3 or PbO2

5. Burette

6. Pipette 5 ml

7. Graduated cylinder 10 ml

8. Stop Watch

Procedure:

1. Fill up the burette with KMNO4 to 50 ml.

2. Take 5 ml H2O2 into conical flask then add 10 ml H2SO4.

3. Titrate KMnO4 against H2SO4 till reach light pink, record the volume; it represents the original

concentration of H2O2 (a).

4. Add 0.01 g of MnO2 to the rest of H2O2, record the time of addition.

5. After 5 minutes from addition withdraw 5 ml into conical the add 10 ml H2SO4 and titrate

against KMnO4.

6. Repeat step 5 at different times, 10, 15, 20, 25 minutes.

7. Tabulate all results as shown below:

11

8. Draw the relation between (t) and ( log𝑎

(𝑎−𝑥)) and calculate the value of (k) and (t1/2).

Results and Calculations:

t (minute) a a-x 𝑎

(𝑎 − 𝑥) log

𝑎

(𝑎 − 𝑥)

0 5

10

15

20

25

12

Experiment 5:

Phase Diagram- Mutual Solubility Curve for Phenol and Water

Objective

To determine the solubility of two partially miscible liquids properties by using phenol in water

Introduction Few liquids are miscible with each other in all proportions while others have miscibility under certain

proportions only. A typical example for this is phenol and water. Under certain temperature and

concentration of phenol and water, the phenol and water maybe in one phase condition or maybe being

separated into 2 different phases. Generally, both liquids become more soluble with rising temperature

until the critical solution temperature is attained, and above that point, the liquids will turn into only one

phase.

Material:

1. Phenol

2. water

3. Thermometer

4. boiling tubes or Erlenmeyer flask 50ml

5. pipette

6. Graduated Cylinder

Procedure: 1. Transfer 1.6 ml of phenol in a dry clean boiling tube then add the calculated volume of

Water so as to obtain a mixture of 8% phenol and 92% water by weight.

2. Cover the tube with a cork stopper and insert thermometer then put the tube inside water path.

3. Prepare other four boiling tubes (or Erlenmeyer flask 50ml) have mixture of phenol and

water, where phenol was added in water in various percentages from 8%, 11%, 37%, 50%,

and 63% .

Erlenmeyer flask 50 ml

4. The total amount of two liquids in the boiling tubes was fixed to be 30ml and the boiling

tubes were labeled accordingly from 1 to 5.

5. Then, the boiling tube 1 was heated in hot water and the mixture was stirred. The temperature

at which the turbid liquid became clear was recorded.

6. The boiling tube 1 was then been cooled gradually and the temperature at which the liquids

became turbid again forming two separated layers was recorded. Then, boiling tube 1 was

heated again and the average temperature for heating and cooling was recorded.

1 2 3 4 5

13

7. Finally, steps (3-5) were repeated for boiling tubes 2 to 5.

8. A graph of temperature at complete miscibility against phenol composition in the different

mixtures was plotted.

Critical Temperature: the temperature above it the two liquids turn into one phase.

Results and Calculations:

Volume of water added:

When phenol percentage is 8.0 %

mlterVolumeofwa

mlx

4.186.120

6.120100

0.8

Volume of water added:

When phenol percentage is 11.0 %

terVolumeofwa ml ـــــــــــــ

And so on ……

Percentage

of phenol

(%)

Volume

of phenol

(mL)

Volume

of water

(mL)

Temperature (ºC) Average

temperature

(ºC) 1 2 3

0.0

0.0

20.0

25.0

25.0

25.0

25.0

8.0 1.6 50.0 56.0 50.0 52.0

11.0 2.2 58.0 64.0 56.0 59.3

37.0 7.4 68.0 68.0 68.0 68.0

50.0 10 65 65 65 65.0

63.0 12.6 64.0 63.5 57.0 61.5

14

15

Experiment 6:

Determination of the specific heat of the calorimeter

The heat (q) released by a reaction or process is absorbed by the calorimeter and any substances in

the calorimeter. If the only other substance in the calorimeter is water, the following energy

balance exists:

q = qcal + qw

Where qcal is the heat flow for the calorimeter and qw is the heat flow for the water. Both of these

individual heat flows can be related to the heat capacity and temperature change for the substance.

qcal = Ccal ΔT

qw = Cw ΔT

Where Ccal is the heat capacity of the calorimeter, Cw is the heat capacity of the water. Because the

water and calorimeter are in thermal equilibrium, they both have the same temperature and thus

ΔT is the same for both. The consequence is that the heat capacity of the entire system (C) is the

sum of the heat capacities for the individual components.

C = Ccal + Cw

The heat capacity is Heat required to change the temperature of a substance one degree, its units is

J oC-1. Specific heat capacity (s), Heat required to change the temperature of one gram of a

substance one degree.

sw =

Cw

mw

Where the weight of water (mw), sw = 1.00 calorie/g oC = 4.184 J oC-1 g-1

Overall one can write Q= Cw (T f - Ti)

Then Cw= sw. mw

Ccal = scal . mcal

Quantity of Heat lost by Boiling water (Q) = Quantity of heat absorbed by Calorimeter (Qcal) +

Quantity of heat absorbed by water (Qw)

Cw (100 - Tf ) = sw (Tf - Ti) + scal mcal (Tf - Ti)

scal = sw. mw (100 – Tf ) - sw. mw (Tf - Ti)/ mcal (Tf - Ti) (1)

s cal = ــــــــــــــــ calorie/g oC OR

16

s cal = ــــــــــــــــ J oC-1 g-1

Material:

1. Coffee cup calorimeter or 2 Styrofoam cups

2. Thermometer

3. Stirring rod

4. Analytical balance

5. Graduated Cylinder

Procedure:

1. Weigh the internal cup of calorimeter, and record it as mcal. 2. Transfer 50 ml of distilled water into internal cup of calorimeter, weigh water as m1 and

measure temperature= Ti.

3. Transfer 50 ml from boiling water into internal cup of calorimeter and Measure

temperature of the system at equilibrium as Tf.

4. Calculate the change in temperature for the system.

5. AFTER THE FINAL TEMPERATURE IS DETERMINED, WEIGH THE

CALORIMETER WITH BOILING WATER.

Results and Calculations:

1. Mass of empty Calorimeter = ـــــــــــــــــــg

2. Mass of 50 ml cold water = ـــــــــــــــــــg

3. Mass of 50 ml hot water = ـــــــــــــــــــg

4. Heat capacity of water, the specific heat capacity for water is

sw = 1.00 calorie/g oC or 4.184 Joules/g oC.

5. Calculate specific heat of the calorimeter using equation )1(.

scal = sw. mw (100 – Tf ) - sw. mw (Tf - Ti)/ mcal (Tf - Ti)

Ccal= scal. mcal

Coffee cup calorimeter 2 Styrofoam cups

17

Experiment 7:

Determination of the heat of solution (∆H)

Heat Solution is the heat evolved or absorbed when one mole of a substance dissolves completely. The

value is positive if heat is absorbed (endothermic) and negative if heat is released (exothermic).

Materials:

1. Coffee cup calorimeter or 2 Styrofoam cups

2. Thermometer

3. Stirring rod

4. Analytical balance

5. Graduated Cylinder

6. Solid Na2SO4

Procedure:

1. Weigh and record the mass of the clean, dry calorimeter W1. 2. Place about 50 mL of distilled water in the calorimeter and weigh m2.

3. Measure the temperature of the water (Repeat twice) to 0.01o C.

4. Weigh out about 5 grams of the solid Na2SO4 (or any solid substance). Add the solid to the

calorimeter.

5. Stirring the calorimeter, determine Temperature to 0.01oC.

6. The stable maximum or minimum temperature reached as the solid dissolves.

7. Check to make sure that the solid is completely dissolved. 8. A temperature change of at least five degrees should be obtained in this experiment.

9. AFTER THE FINAL TEMPERATURE IS DETERMINED, WEIGH THE CALORIMETER WITH

THE DISSOLVED SOLID.

When a substance undergoes a change in temperature, the quantity (Q) of heat lost or gained

can be calculated using the mass (m), specific heat (s), and change in temperature

( T = Tf - T i)

sw = 1.00 calorie/g oC or sw = 4.184 Joules/g oC.

If the system doesn’t exchange heat with the surroundings, then Q system=0, and

18

Q reaction = -Q solution = - (m s T) solution

Because the experiment is performed under constant pressure conditions, the heat flow is also enthalpy

change H for the reaction. H is generally reported as the heat of reaction per a fixed amount of the

reacting chemical, and we will calculate H per gram or per mole of the solute.

solute

rxnrxn

mass

QH (J/g)

Results and Calculations:

19

Experiment 8:

Rate of Reaction of Sodium Thiosulfate and Hydrochloric Acid

Introduction

Studying the effect of sodium thiosulfate concentration on the rate of reaction of sodium thiosulfate with hydrochloric acid. Sodium thiosulfate reacts with hydrochloric acid to form sulfur and sulfur

dioxide (Equation 1).

Na2S2O3 (aq) + 2HCl (aq) → S(s) + SO2 (g) + 2NaCl (aq) (Equation 1)

The reaction, will be followed by measuring the time needed for the reaction mixture to become turbid

(unclear). The results will be analyzed graphically to determine the order of reaction; the mathematical

relationship between the reactant concentration and the rate.

Materials:

1. Hydrochloric acid solution, HCl, 2 M, 25 mL

2. Graduated cylinders, 10-mL, 5

3. Sodium thiosulfate solution, Na2S2O3, 0.15 M, 150 mL

4. Distilled or deionized water Permanent marker

5. Beakers, 100-mL,

6. 5 Stirring rods

7. Graduated cylinders, 50- or 100-mL,

8. 2 Stopwatch or timer

Procedure:

9. Label five 100-mL beakers 1–5 and clean the bottom of each beaker.

10. Draw a large “X” across the bottom on the outside of each beaker in black color. Put the beaker

above a white paper to see the “X” letter easily.

11. Add the required amounts of 0.15 M sodium of thiosulfate and distilled water to each beaker

Using graduated cylinders.

12. Calculate the final concentration of sodium thiosulfate in each beaker 1–5.

13. Measure 5.0 ml of 2 M hydrochloric acid into each of five 10-ml graduated cylinders.

14. Starting with beaker #1, carefully add the HCl all in one pour to the sodium thiosulfate solution.

Stir the solution once with a stirring rod and immediately start timing.

15. Stop timing when the black “X” is no longer visible. Record the reaction time in seconds.

16. Repeat steps 5 and 6 with beakers 2–5.

20

Results and Calculations:

1. Beaker # 1: Concentration of Na2S2O3 = ـــــ Mــــــــــــــــ

2. Beaker # 2: Concentration of Na2S2O3 = ــــ Mـــــــــــــــــ

3. Beaker # 3: Concentration of Na2S2O3 = ــــ Mـــــــــــــــــ

4. Beaker # 4: Concentration of Na2S2O3 = ــــ Mـــــــــــــــــ

5. Beaker # 5: Concentration of Na2S2O3 = ــــ Mـــــــــــــــــ

6. Calculate 1/reaction time for each beaker= ــــــــــ sec–1

7. Plot concentration vs. time

8. Plot concentration vs. 1/time.

Beaker Volume of Na2S2O3 (mL)

Volume of H2O (mL)

[Na2S2O3], M Reaction time (sec)

Reaction rate (1/time, sec–1)

1 50 0 0.15 22.5

2 40 10 0.15 27.3 3 30 20 0.090 35.1

4 20 30 0.060 60.0

5 10 40 0.030 159.1

21

Experiment 9:

Heat of Neutralization

Heat of neutralization (ΔHn) is amount of heat releases when one equivalent of an acid neutralizes with

one equivalent of a base to form water and a salt.

The heat (Q) released during a reaction is:

Quantity of released Heat Q = Quantity of heat absorbed by Calorimeter (Qcal) +

Quantity of heat absorbed by solution (Qw).

Qreleased = Qsystem

Q= (m s ΔT) calorimeter + (m s ΔT) solution

Q= (scal mcal+ ssol . msol) T

Where m is the mass of the solution, s is the specific heat capacity of the solution, and ∆T is the

temperature change observed during the reaction. From this, the standard enthalpy change (∆H) is

obtained by division with the amount of substance (in moles) involved.

Materials:

1. Calorimeter

2. HCl 0.1M

3. NaOH 0.1M

4. 2 Graduated Cylinder 50 ml

5. Thermometer

6. Beaker 150 ml

Procedure:

Part A. Calibration of the Calorimeter; to measure Ccal = the calorimeter constant (i.e. Experiment # 6)

1. Weigh the internal cup of calorimeter, and record it as mcal. 2. Transfer 50 ml of distilled water into internal cup of calorimeter, weigh water as mw and

measure temperature as Ti.

3. Transfer 50 ml from boiling water into internal cup of calorimeter and Measure temperature of

the system at equilibrium as Tf.

4. Calculate the change in temperature for the system. After the final temperature is measured,

weigh the calorimeter with boiling water.

5. Calculate specific heat of the calorimeter using equation )1(. Ccal is called calorimeter constant,

it equals Ccal = scal. mcal

scal = sw. mw (100 – Tf ) - sw. mw (Tf - Ti)/ mcal (Tf - Ti)

Ccal= scal. mcal Equation )1(

22

Part B. Measurement of the Heat of Neutralization of NaOH (aq) and HCl (aq).

1. Prepare and weigh the empty calorimeter as in steps A-1 and A-2.

2. Using a 50 ml graduated cylinder, pour 50 ml of NaOH solution into calorimeter. Measure and

record the temperature of the base solution. Weigh the calorimeter containing NaOH and record

its mass.

3. Using a clean and dry 50 mL graduated cylinder measure 50 ml of acid solution and pour it into

the calorimeter. Measure and record the temperature of the acid solution. Weigh the calorimeter

containing the acid and record its mass.

4. If the temperature of the acid and the base are not within 0.5 oC of each other, either heat or cool

the base (use a hot plate or an ice bath) do not add water or ice directly to the base or you will

change both the mass and the concentration of the base.

5. Mix the base with the acid in the calorimeter in one pour record the time of mixing. Stir briefly

with a glass stirring rod. Measure and record the temperature of the mixture at 30 second intervals

for the next 12.5 minutes (a total of 15 minutes of measurements).

6. Discard the contents of the calorimeter into the sink. Rinse the calorimeter with tap water and dry

the calorimeter with a paper towel. Rinse and dry the thermometer.

7. Set up the calorimeter assembly again and repeat steps 1-6 with a second base-acid system.

Results and Calculations:

1. ssol = specific heat of the solution 1.0 M NaCl = 3.89 J/g K.

2. Ccal = ـــــــــــــــــــ J/ oC, where Ccal = the calorimeter constant.

3. Ti = ـــــــــــــ oC, where Ti = (Tacid + Tbase)/2 = average temperature of the acid and the base before

mixing.

4. Tmax sol = extrapolated maximum temperature of the reaction after mixing = ــــــــــــ oC.

5. ΔTsol = (Tmax sol – Ti) temperature change of the solution during the reaction= ـــــــ oC. 6. Calculate the heat, Q, released by the reaction is calculated from Equation 2

Q = - (msol ssol + Ccal) ΔTsol Equation 2

Where: msol = macid + mbase = mass of the solution in the calorimeter

7. Calculate the molar enthalpy of neutralization ΔH = Q/n,

Where: n = number of moles of reactant. n = C. V= 1.0 x 0.05= 0.05 moles

C = concentration in “M” = moles/L.

V = volume in liters.

23

24

Experiment 10:

Saponification of Ethyl Acetate

Studying kinetics of ethyl acetate in alkaline medium is second order reaction. The reaction proceeds

according to the following equation:

CH3COOCH2CH3 + NaOH → CH3COONa + CH3CH2OH

The rate of reaction for any reaction is defined as how fast a reaction takes place. For example, the

oxidation of iron is a slow reaction which can take many years, but the combustion of butane in a fire is

a reaction that takes place in a second. Consider a typical chemical reaction

aA + bB → pP + qQ

The lowercase letters (a, b, p, and q) represent stoichiometric coefficients, while the capital letters

represent the reactants (A and B) and the products (P and Q).

Rate of reaction depends on the concentration of both ethyl acetate and sodium hydroxide.

))(( xbxakdt

dx

Taking integration for both sides of above equation

)(

1

xaa

x

tk

With rearrangement the above equation,

)( xaa

xkt

Where k is known as the reaction rate coefficient or rate constant, a is the original concentration of

reactants, while x is the amount of consumed from reactants to give products, and (a-x) is the remaining

amount of ester and alkaline.

Materials:

1. Pipettes 10 ml

2. Burette 50 ml

3. Titrimetric flasks 250 ml

4. Stop watch

5. Solution of ethyl acetate (c = 0,05M),

6. Solution of sodium hydroxide (c = 0,05M),

7. Solution of hydrochloric acid (c = c = 0,025M),

8. Phenolphthalein

25

Procedure:

1. Transfer 50 ml of the solution of ethyl acetate (c = 0,05M) into the volumetric flask (V=50 ml) and

50 ml of the solution of sodium hydroxide (c = 0, 05 M) into another volumetric flask. Both flasks

cork down and put them in the thermostated bath (25°C) for 10 minutes.

2. Fill the burette with the solution of sodium hydroxide (c = 0, 05 M).

3. Take out the flasks with solutions from thermostated bath and pour the solution of ethyl acetate to

the solution of sodium hydroxide, put the mixture again into the thermostated bath and start the

stopwatch.

4. After 5 minutes, transfer 10 ml from the mixture (leave the flask in the bath) into a titrimetric flask

contains 100 ml distilled water.

5. Pipette 10 ml of solution of hydrochloric acid (c = 0,025M) into titrimetric flask to stop reaction then

add one drop of Phenolphthalein.

6. Titrate with the solution of sodium hydroxide. When the endpoint of titration has been reached, read

the used volume of NaOH from the burette.

7. Repeat the step 4, 5 and 6 every 5 minutes six times more (after10th, 15th, 20th, 25th, 30th and 35th min.

from the moment of mixing).

8. Write down to the Tab. used volume of NaOH for each titration as (x) ml and the original

concentration of reactants equals the volume of HCl as (a= 0.025 M)

Results and Calculations

Draw the relation between t andxa

a

, it gives a straight line passing through zero point its slope is k.

t, min. x ml, Volume of NaOH, xa

)( xaa

x

5 10

15

20

25 30