Koç University
2010, Fall
Physical Chemistry Laboratory (Chem 301)
Experiments
1
Laboratory Safety Rules
ALWAYS:

Wear safety glasses

Wear protective clothing

Know the location and use of all safety equipment

Use proper techniques and procedures

Add acid to water

Be very cautious when testing for odors

Use hoods whenever poisonous or irritating fumes are evolved

Discard wastes properly - flush liquids down the sink with a large excess of water

Report any accident, however minor, to the instructor at once

All the times think about what you are doing

Be alert, serious, and responsible
NEVER:

Eat or drink in the lab

Perform unauthorized experiments

Leave anything unattended while it is being heated or is reacting rapidly

Aim the opening of a test tube or flask at yourself or at anyone else

Add water to acid

Insert droppers, pipettes and other laboratory equipment into reagent bottles – this is a
sure way of contaminating the contents

Return unused reagents to stock bottles

Clutter your work area

Take unnecessary risks

Enter chemical storage area
2
Experiment 1
Ideal Gas Laws
1. Purpose
The purpose of the experiment is to demonstrate the ideality of a real gas at given temperature
and atmospheric pressure.
2. Introduction
An ideal gas is a gas that conforms, in physical behavior, to a particular, idealized relation
between pressure, volume, and temperature called the ideal gas law. This law is a
generalization containing both Boyle's law and Charles's law as special cases and states that
for a specified quantity of gas, the product of the volume, V, and pressure, P, is proportional
to the absolute temperature T; i.e., in equation form, PV = kT, in which k is a constant. Such a
relation for a substance is called its equation of state and is sufficient to describe its
macroscopic behavior.
The ideal gas law can be derived from the kinetic theory of gases and relies on the
assumptions that (1) the gas consists of a large number of molecules, which are in random
motion and obey Newton's laws of motion; (2) the volume of the molecules is negligibly
small compared to the volume occupied by the gas; and (3) no forces act on the molecules
except during elastic collisions of negligible duration.
Although no gas has these properties, the behavior of real gases is described quite closely by
the ideal gas law at sufficiently high temperatures and low pressures, when relatively large
distances between molecules and their high speeds overcome any interaction. A gas does not
obey the equation when conditions are such that the gas, or any of the component gases in a
mixture, is near its condensation point.
The ideal gas law may be written in a form applicable to any gas, according to Avogadro's
law, if the constant specifying the quantity of gas is expressed in terms of the number of
molecules of gas. This is done by using as the mass unit the gram-mole; i.e., the molecular
weight expressed in grams. The equation of state of n gram-moles of a perfect gas can then be
written as PV/T = nR, in which R is called the universal gas constant. This constant has been
measured for various gases under nearly ideal conditions of high temperatures and low
pressures, and it is found to have the same value for all gases: R = 8.314 J/g mol K.
3
According to Boyle’s law, the pressure (P) of a gas varies inversely with the volume (V) of
the gas when the temperature (T) and quantity (n) are kept constant. Mathematically, this is
expressed as
P ~1/V (T, n constant) Boyles’s law
By rearranging it can be found that the product of pressure and volume of a specific amount
of gas is constant as long as the temperature does not change.
P1V1= constant
At the same temperature, the PV product of a gas remains constant. Therefore, it can be
written as equality of the initial and the final products of pressure and volume.
P1V1=P2V2
According to Boyle’s law, volume decreases when the pressure increases, and volume
increases when the pressure decreases.
When temperature rises, the kinetic energy of the molecules increases. To keep the pressure
constant, the volume must expand. According to Charles’ law, the volume of a gas changes
directly with the Kelvin temperature as long as the pressure and the number of the moles
remains constant.
V ~ T (P and n constant) or V/T= constant
Therefore Charles’ law can be written as
V1 V2

T 1 T2
2. Experimental procedure
2.1. Charles’ Law
Materials: 50 ml Erlenmeyer flask, 200 ml beaker, one-hole rubber stopper with a short piece
of glass tubing inserted and attached to a piece of rubber tubing, water containers,
thermometers, pinch clamps, burette clamp, hot plate, graduated cylinder, boiling chips, ice.
4
Procedure: Dry any moisture on the inside of the 50 ml Erlenmeyer flask. Place a one-hole
rubber stopper and tubing in the neck of the flask. Set a 200 ml beaker on a hot plate. Add a
few boiling chips to the bottom of the beaker. Attach the burette clamp to the neck of the flask
and lower the flask into a 200 ml beaker without touching the flask of the bottom. Fasten the
burette clamp to the ring stand. Pour the water into the beaker until it comes up to the neck of
the flask. Leave place at the top for the water to boil without boiling over (see figure 1).
Figure 1. Setup for heating an Erlenmeyer flask in boiling water bath
Begin heating and bring the water to boil. Boil gently for 10 min to bring the temperature of
the air in flask to that of the boiling water. Measure the temperature of the boiling water.
Convert from degree Celsius to the corresponding Kelvin temperature.
Place the pinch clamp on the rubber tubing. In the lab there will be 3 large containers with
water at different temperatures. Using the burette clamp as a holder, carefully lift the flask out
of the hot water and carry it over the one of the cool water containers. Keep the stopper end of
the flask (closed with pinch clamp) pointed downward, immerse the flask in the cool water
(see Figure 2). Keep the flask inverted and remove the pinch clamp. Water will enter into the
flask as the air sample cools and decreases in volume.
After the flask has been immersed into the cool water at least 10 minutes, measure the
temperature of the cool water bath. It is assumed that the temperature of the cooled air sample
in the flask is the same as the cool water outside the flask. Convert the degree Celsius
temperature to Kelvin.
5
Figure 2. Placement of the heated flask into a cool water bath
Keeping the flask inverted (upside down, stopper still pointed downward), raise or lower the
flask until the water level inside the flask is equal to the water level in the container (see
figure 3). While holding the flask at this level, reattach the pinch clamp to the rubber tubing.
Remove the flask from the water, and set it upright on the desk. Remove the clamp and use a
graduated cylinder to measure the volume of water that entered the flask as the air sample
cooled. Record the volume in ml.
Equalizing
water levels
Figure 3. Procedure to equalize the water level inside and outside the flask
Measure the total volume (ml) of the flask by filling it to the top with water, but leave space
for the volume occupied by the rubber stopper assembly. Record. Repeat the measurement 3
times!
List the temperatures (K) of the boiling water bath and the total gas volume (ml) obtained at
three different runs.
Calculate the average boiling water bath temperature (K) and the average for the total gas
volume of the flask.
Repeat the measurement in three different cool water baths, measure the temperature and the
volume of water that entered to the flasks.
6
2.2. Data Analysis
a. Use average temperature for the boiling water baths. Calculate the volume of the air in the
flask for each temperature by subtracting the amount of cool water that entered the flask from
the average total gas volume.
Volume of cool air = total flask volume - volume of water flask (average)
b. For each sample, calculate the V/T(K) value. These values should be constant, according to
the Charles’ law. Any value that is not similar will not be a good data point to use on the
graph.
c. Graph the volume-temperature relationship of the gas. (Absolute zero is the theoretical
value for the coldest temperature that matter can attain. The value of absolute zero is
predicted by extrapolating to the axis where the volume of the gas would decrease to 0 ml.)
d. Boyle’s law
In an experiment, the volume of a specific amount of gas is measured at different volumes,
while the temperature is kept constant. The results are summarized in a report sheet table.
Determine the P x V product by multiplying the pressure and the volume in each sample.
Round the product to give correct number of significant figures.
Mark the vertical axis in equal intervals of mm Hg of pressure. Divide the horizontal axis into
equal intervals of ml. For the lowest pressure value, use a pressure that is slightly below the
lowest value in the data: the highest value should be just above the highest measured value for
pressure. For example, the pressure scale might begin at 600 mm Hg and go up to 1000 mm
Hg. The volume values also should begin with a value near the smallest volume obtained. Plot
the pressure (mm Hg) of the gas in each reading against the volume (ml). Draw a smooth line
through the points obtained from the data. The data points will fall on a slight curve called a
hyperbola, not a straight line. Use the graph to discuss the meaning of Boyle’ law.
7
Pressure / mm Hg
Volume / ml
630
32.0
690
29.2
726
27.8
790
25.6
843
24.0
914
22.2
PxV
8
EXPERIMENT 2
THERMOCHEMISTRY
I.
PURPOSE
In this experiment we will investigate the principles of calorimetry and calculate the
enthalpy of reaction and neutralization as well as heat capacity.
II.
THEORY
Most chemical reactions involve the transfer of energy between the system (the
chemical reaction) and the surroundings. One form of energy exchanged is heat, given the
symbol q. In some reactions, heat is released when the products are formed (exothermic
reaction, q<0) and in other reactions, heat must be added to the system in order for the
reaction to proceed (endothermic reaction, q>0). When the reaction is carried out under
conditions of constant pressure, the heat exchanged is called the enthalpy of reaction, Hrxn.
In SI units, the enthalpy of reaction is expressed in kilojoules per mole of reactant.
The enthalpies of several acid-base reactions and the dissolution of a solid substance
will be determined. These enthalpy values will be used to calculate other thermodynamic
quantities and the enthalpies for related reactions.
Reactions will be carried out in a calorimeter (a styrofoam cup). A calorie is the
amount of energy required to raise the temperature of one gram of water one degree Celsius.
In this experiment, you will mix hot and cold water of known temperature and mass. Using
the definition of calorie, you will be able to determine the amount of the heat energy that is
transferred in bringing the hot and cold water to their final equilibrium temperature, and
thereby determine if heat energy is conserved in this process.
Changes in temperature can be measured, from which the energy change (quantity of
heat released/absorbed) can be calculated. The enthalpy of the reaction, Hrxn = qsystem, since
the experiment is carried out under conditions of constant pressure.
Hrxn = qsystem = -qsurroundings
9
Where; qsurroundings = Csurr (Tsurr) (msurr).
C = specific heat of the surroundings (J/g.C)
m = mass of the surroundings (g)
T = temperature change (C) = Tfinal - Tinitial
The “surroundings” consist of the calorimeter and its contents. If the calorimeter did
not absorb any heat, all of the energy given off by the reaction would raise the temperature of
the reaction mixture. However, since some of the heat will be absorbed by the calorimeter, a
calorimeter constant, W, must be calculated. Each calorimeter has its own calorimeter
constant dependent on its size and composition. This must be experimentally determined.
The reaction will be carried out in water, which has a specific heat capacity of 4.184
J/g.C. We will assume that the specific heat of the solutions is also 4.184 J/g.C.
Fig. 1 Experimental Calorimeter
Calculation of H for the neutralization reactions:
The heat evolved (q) is equal to the sum of the heat absorbed by the reaction mixture and
the heat absorbed by the calorimeter. The enthalpy of reaction H is a molar quantity
obtained by dividing the heat evolved, q, by the number of moles of water formed in the
neutralization reaction. For an exothermic reaction, H is negative because heat flows from
the system to the surroundings. Calculate H values for each of the three neutralization
reactions to the proper number of significant figures.
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Calculation of H for Related Reactions – Hess’s Law:
Hess’s Law states: The amount of heat generated or absorbed by a chemical reaction
is constant regardless of whether the reaction takes place in one or several steps. All chemical
reactions which start with the same reactants and end with the same products involve the same
net energy change, which is independent of the pathway by which the final state is reached.
Enthalpy is a state function.
To illustrate Hess’s Law, the following reactions can be considered:
a) Sodium hydroxide with hydrochloric acid
H1
b) Sodium hydroxide with acetic acid
H2
c) Sodium acetate with hydrochloric acid
H3
Specific heat:
The Specific Heat of a substance, usually indicated by the symbol (c), is the amount of
heat required to raise the temperature of one gram of substance by one degree Celsius. From
the definition of the calorie, it can be seen that the specific heat of water is 1.0 cal/gr .°C. If an
object with mass (m) is made of a substance with specific heat of (c), then the heat (H)
required to raise the temperature of that object by an amount (T) degrees is:
H = (m) x ( c ) x (T)
In this experiment, you will measure the specific heats of aluminum, copper, and lead.
III.
PROCEDURE
III. A. Determination of the calorimeter constant:
1. Weigh one of the beakers to be used in the calorimeter.
2. Put 30 g of water at room temperature into beaker by using the balance. Record the
exact mass of this cold water. Place the beaker in the calorimeter and cover with
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aluminum foil. Make a small hole in the foil and place a thermometer into the
calorimeter.
3. Put nearly 30 g of water into the other beaker and record its exact mass as the mass
of hot water. Place the beaker on the heater. Attach a thermometer to the stand
using clamps and dip it into water in the beaker. Make sure the thermometer does
not touch the bottom of the beaker which will be hotter than water.
4. Heat until the temperature reaches ~10 oC above room temperature.
5. Record the temperature in the calorimeter as Tcold and the temperature of hot water
as Thot and immediately mix the hot water into the cold water beaker in the
calorimeter.
6. Cover the calorimeter immediately and stir gently with the glass rod. The
temperature will stay constant after decreasing for a few minutes. Record the
maximum temperature reached as the final temperature Tfinal.
7. Repeat steps #2-6 three more times. Make sure each time you put 30.0  0.5 g of
water at room temperature into the calorimeter. The temperature of hot water
should be ~RT + 20 oC in the second trial, ~ RT + 30 oC in the third trial.
III.B Determination of the reaction enthalpy for acid/base pairs:
1. Put exactly 20.0 ml of the standard 1.00 M NaOH solution into a beaker and place in
the calorimeter. Pour 20.0 ml of standard 1.00 M HCl solution into the other beaker.
Measure and record the temperature of the acid. Rinse and dry the thermometer, then
measure and record the temperature of the base.
2. Quickly add the acid solution to the calorimeter, stir, and observe the temperature
continuously. Make sure the temperature has reached a maximum before you stop
measuring.
This should take 3-4 minutes.
Record the maximum temperature.
Finally, record room temperature and the weigh the calorimeter including the reaction
mixture before discarding. Use these data to calculate H1.
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3. Follow the above procedure for each of the following acid/base pairs. Obtain one (1)
set of data for each acid/base pair.
1.00 M CH3CO2H/1.00 M NaOH (CH3CO2H = acetic acid)
1.00 M HCl/1.00 M NaHCO3 (NaHCO3 = sodium hydrogencarbonate)
H2
H3
I II.C Determination of Specific Heat:
1. Measure Mcal, the mass of the calorimeter you will use ( it should be empty and dry).
2. Measure the masses of aluminum, copper and lead samples.
3. With the help of thread, suspend each of the samples in boiling water. Do not allow
the thread to fall into the water. Steam generator may show the setting 5 for a faster
boiling, but you should better adjust the setting to 3 after it boils.
4. Fill the calorimeter approximately ½ full of cold water; use enough water to easily
cover any one of the metal samples.
5. Measure Tcold; the temperature of the cold water.
6. Immediately following your temperature measurements, remove the metal sample
from the boiling water, then suspend it in the cold water in the calorimeter. (the
sample should be completely covered but should not touch the bottom of the
calorimeter.)
7. Stir the water with the glass rod and after waiting for some time measure T water as it
comes into thermal equilibrium with the metal sample.
8. Immediately after taking the temperature, measure and record Mtotal; the total mass of
the calorimeter, water, and the metal sample.
9. Repeat the steps above for all the mentioned metal samples.
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Materials needed:
Distilled water
Standard 1.00M NaOH soln, Standard 1.00M HCl soln., 1M HAc soln, 1M NaHCO3 soln.
Aluminum, copper, lead samples
IV.
TREATMENT OF THE EXPERIMENTAL DATA
IV.A Calculation of calorimeter constant:
1. Plot (mhot* Thot) as a function of Tcold. The data should look linear. Why?
2. Fit a line (by eye) to the data and calculate the slope of the line.
3. Calculate the average mass of the cold water you used in calibration measurements.
4. The heat capacity of the calorimeter, (mc)calorimeter is given by
(mc)calorimeter = slope – mcold-avg * cwater
where cwater = 1 cal/g.oC.
IV.B Calculation of H for the neutralization reactions:
The heat evolved (q) is equal to the sum of the heat absorbed by the reaction mixture
and the heat absorbed by the calorimeter. The enthalpy of reaction H is a molar quantity
obtained by dividing the heat evolved, q, by the number of moles of water formed in the
neutralization reaction. H = q / n . For an exothermic reaction, H is negative because heat
flows from the system to the surroundings.
1. Calculate H values for each of the three neutralization reactions to the proper
number of significant figures.
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For q mixture calculation assumes Cmix as Cwater.
2. Discuss how this approximation affects your results.
IV.C Calculation of H for Related Reactions
Question 1. Hess’s Law:
To illustrate Hess’s Law, consider the following reactions:
a) Sodium hydroxide with hydrochloric acid
H1
b) Sodium hydroxide with acetic acid
H2
c) Sodium acetate with hydrochloric acid
H3
1. Write a balanced chemical equation for each reaction.
2. Show that Eq. 3 can be obtained by an algebraic sum of Eq.1 and Eq.2.
Question 2.
Compare your experimental results: H1 (for reaction of sodium hydroxide with
hydrochloric acid) and H2 (for reaction of sodium hydroxide with acetic acid) with the
theoretical values. Account for any differences between the values.
IV. D How would the following affect the magnitude of the calculated H, i.e., too large,
too small, no effect? Include a short explanation for each case.
The actual concentration of HCl was more than 1.0 M in the reaction with 1.0 M
NaOH. The calorimeter constant is assumed to be zero. The thermometer reads all
temperatures 1.5 oC more than it should be.
15
A student thought he/she weighed out 2.0 g solid substance but in fact only weighed
out 1.7 g solid substance.
IV.E Calculation of the Specific Heats of the Metal Samples
1. For each metal sample calculate M water, the mass of the water used, T water, the
temperature change of water when it came into contact with each meal sample, and T
sample, the temperature change of the metal sample when it came into contact with the water.
2. Tabulate your results.
3. From the law of energy conservation, the heat lost by the metal sample must be
equal to the heat gained by water. Use this law to calculate the heat capacities of each metal
sample.
4. Compare your results with the literature values. Discuss the possible reasons for the
deviations.
5. You ignored the heat absorbed by the calorimeter; why can’t you use the mc
calorimeter that you found in previous experiment?
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EXPERIMENT 3
HOMOGENEOUS EQUILIBRIA
EQUILIBRIUM IN ORGANIC LIQUIDS
I. PURPOSE
This experiment illustrates the determination and calculation of apparent equilibrium
constants.
II. THEORY
An acid and an alcohol combine to form an ester and water; the ester hydrolyzes to
give the acid and the alcohol. The two opposing reactions proceed until equilibrium is reached
with all four compounds present in definite concentrations. The equilibrium constant K gives
a measure of relation between the activities of the products and the reacting materials. When
the equilibrium constant has been determined, it is possible to calculate the yield that may be
expected from any given concentration of reacting materials.
When concentrations are used in calculating the equilibrium constant, the equilibrium
constant is in error on account of the failure of the mass law, particularly at high
concentrations. An accurate value of the equilibrium constant is obtained by using activities
instead of analytically determined concentrations. Activities may be determined by vaporpressure measurements or by other methods. In many cases, however, the equilibrium
constant may be calculated with fair accuracy from the concentrations.
The equilibrium is sometimes reached very slowly, particularly in organic reactions,
and it may be necessary to raise the temperature or to use a catalyst.
The equilibrium between ethanol, acetic acid, ethyl acetate, and water is studied in this
experiment, and the reaction is catalyzed by the addition of hydrochloric acid. Its
concentration is great enough to change the character of the water and alter the numerical
value of the equilibrium constant, but the results are fairly constant for a given concentration
17
of hydrochloric acid. The hydrochloric acid is added merely as a catalyst to speed-up the
reaction, and it takes no part in the stoichiometric reaction.
III. EXPERIMENTAL
III. 1. Apparatus. Burettete ; 5-ml pipette; 2-ml pipette; 1-ml pipette; 50-ml glass-stoppered
jars; phenolphthalein, ethyl acetate, 0.5 N sodium hydroxide, 3 N hydrochloric acid, glacial
acetic acid, absolute ethanol.
III. 2 Procedure. Standardize the 3 N hydrochloric acid prepared with 0.5 N sodium
hydroxide; mix the standardized 3 N hydrochloric acid with ethyl acetate and other materials
in small jars, in the following proportions:
a) 5 ml 3 N HCl + 5 ml water
b) 5 ml 3 N HCl + 5 ml ethyl acetate
c) 5 ml 3 N HCl + 4 ml ethyl acetate + 1 ml water
d) 5 ml 3 N HCl + 2 ml ethyl acetate + 3 ml water
e) 5 ml 3 N HCl + 4 ml ethyl acetate + 1 ml ethanol
f) 5 ml 3 N HCl + 4 ml ethyl acetate + 1 ml acetic acid
g) 5 ml 3 N HCl + 4 ml ethanol + 1 ml acetic acid
Duplicate determinations should be made to understand the reproducibility of the
experiment. Each of the bottles is stoppered immediately and allowed to stand in a 40°C for at
least 48 hours and preferably for a week, with occasional shaking. It is necessary that stoppers
fit tightly to prevent evaporation. A thermostat is unnecessary because this equilibrium is
affected only slightly by temperature changes.
The weight of each material is determined by discharging the pipette directly into a
glass-stoppered weighing bottle and weighing. In this way the following weighings are made:
5 ml 3 N hydrochloric acid
5 ml, 4ml and 2 ml ethyl acetate
4 ml and 1 ml ethanol
18
1 ml acetic acid
5 ml, 3 ml, and 1 ml water
The drainage of the pipettes will not always be uniform, but the error involved in using
of a pipette is not serious for the present work.
After standing, each solution is titrated with the 0.5 M sodium hydroxide, using
phenolphthalein as an indicator.
III. 3. Evaluation of the data
Calculations. The original weight of ethyl acetate is obtained by weighing the liquid
ethyl acetate discharged from the pipette. The original weight of acetic acid and ethanol in (e),
(f), and (g) is obtained in the same way. The total weight of water is obtained by adding the
weight of pure water in the 3 N hydrochloric acid solution to the water contained, obtained
from titration, from the weight of the 5 ml of hydrochloric acid solution.
The amount of acetic acid at equilibrium is obtained by subtracting the number of
milliliters of sodium hydroxide used in the solution (a) from the number used for the final
equilibrium titration. The former is a measure of the hydrochloric acid, and the latter is a
measure of both the hydrochloric acid and the acetic acid in the equilibrium. The difference,
then, gives the amount of acetic acid at equilibrium. In (f) and (g) acetic acid is added to the
original solution, and this amount must be used in calculating the equilibrium amounts of the
other reactants. For every mole of acetic acid produced in the reaction, 1 mole of ethanol is
produced, 1 mole of water disappears, and 1 mole of ethyl acetate disappears.
If the number of moles of each of the four materials in the original mixture and the
number of moles of acetic acid produced in the reaction are known, the equilibrium constant
KN may be computed. It is defined as
KN = N (CH8COOC2H6 ) N ( H2O)
N( C2H6OH ) N (CH3COOH)
Where N represents mole fraction.
19
As indicated in the opening paragraphs, the value KN obtained in this way is not
particularly constant. It deviates more and more as the concentrations of the reactants and
products are increased. Thus, the values of KN which are obtained are apparent ones; true
equilibrium constants can be obtained by the use of activities in place of concentrations.
Practical Applications. In planning any chemical synthesis, it is desirable to know
what yield of material may be expected from a given concentration of reacting materials. Such
a calculation may be made when the value of the equilibrium constant is known, provided that
the reaction is fast enough to come to equilibrium in the time allowed.
Suggestions for Further Work.
Similar experiments may be carried out with other esters.
More significant results are obtained without the use of a catalyst, but under these
conditions it is necessary to heat the mixture to about 150
0
C in sealed tubes to effect an
equilibrium within a couple of days. If sufficient precautions are taken to avoid danger from
bursting tubes, the equilibrium constant may be determined by titrating mixtures that have
been weighed out, sealed off in small glass tubes, and heated. The original work of Berthelot
and St.Gilles may be repeated.
The equilibrium involved in the reaction between acetaldehyde and alcohol to give
acetal and water may be studied. A little hydrochloric acid is used as a catalyst, and the
equilibrium concentration of the acetaldehyde is determined volumetrically by the sulfite
method, with thymolphtalein as indicator.
20
EXPERIMENT 4
BOILING POINT DIAGRAM OF A BINARY MIXTURE
(Liquid-Vapor Equilibrium in Two Component Systems)
1. Refractive indices
The refractive index (n) is a physical constant that can be used to characterize liquids. It can
be used to determine the optical performance, purity, concentration and dispersion etc.
Therefore, refractometer is widely used as one of the indispensable tools in petrology, oil and
fat analysis, pharmaceutical, paint, food, chemical and sugar making industries as well as in
universities and industrial research centers for R&D and quality control (QC).
Refractive index is the ratio of the velocity of light in air to the velocity of light in the liquid.
It is also equal to the ratio of the sine of the angle of incidence (θ2) to the sine of the angle of
refraction (θ1).
Velocity in air
sin θ2
n = ——————— = ———
Velocity in liquid
sin θ1
Fig 1. Refraction of light
The angle of refraction is also a function of temperature and the wavelength of light. Because
the velocity of light in air is always greater than that through a liquid, the refractive index is a
number greater than 1.
21
The measurement is made on a refractometer using a few drops of liquid. Temperature
correction must be applied to the observed reading by adding 0.00045 for each degree above
20 °C:
n20D = ntD + 0.00045 (t-20 °C)
The superscript 20 indicates that the refractive index was measured at 20 °C and the subscript
D refers to the yellow D-line from a sodium vapor lamp, light with a wavelength of 589 nm.
PROCEDURE
There are many different designs of refractometers. The most common is the Abbé design
shown in Fig. 2.
Fig. 2. Abbé refractometer
To determine the refractive index, 2 -3 drops of the sample are placed on the open prism using
a Pasteur pipette (please be careful not to scratch the prism). The prism is closed and the light
is turned on and positioned for maximum brightness as seen through the eyepiece. The knob
is turned so that the line separating the dark and light areas is at the crosshairs, as in Fig. 3 (b).
Then the chromatic adjustment is turned until the demarcation line is sharp and colorless. At
this point the value of the refractive index can be read on the dial. Read the temperature on the
22
thermometer attached to the refractometer, and make the appropriate temperature correction
to the observed index of refraction.
Fig. 3. (a) View when index knob is not adjusted. (b) View into refractometer
when properly adjusted (c) View when chromatic adjustment is correct.
Fig. 4. Refractometer reading scheme
EXPERIMENTAL PROCEDURE:
Read and record the refractive indices of the following solutions and the temperature for each
measurement. Take at least 3 readings for each material. Clean the prisms of refractometers
after each reading as you are instructed before starting the experiment.
Read and record the refractive index of the mixtures of distilled water and IPA (2-propanol),
provided in the table.
Calculate the weight percents and mole percents of the solutions below and fill in the
appropriate cells in the chart below.
23
After every reading clean the prism by wiping gently with a piece of tissue!
Concentration
Sample
Refractive
Temperature
T
index
(ºC)
Corrected
Mol %
Wt %
Average -
IPA
IPA
by volume
IPA
1
100%
2
80%
3
50%
4
30%
5
0%
1
2
3
1
2
3
RI
DATA HANDLING
Using the data in Table above, plot two graphs on a graph paper.
1) Mole percent vs. Refractive index (IPA-Water composition)
2) Weight percent vs. Refractive index (IPA-Water composition)
2. Liquid-Vapor Equilibrium in Two Component Systems
I. PURPOSE
To study the liquid-vapor phase diagram (or temperature-composition diagram) of isopropyl
alcohol-water mixtures using refractometer.
24
II. THEORY
For two component systems one can construct “temperature-composition diagrams” (also
termed as phase diagrams) by plotting boiling points at constant pressure (usually atmospheric
pressure) against the composition of the liquid and vapor phases. One can also plot total
pressures at a given temperature against compositions and obtain “pressure-composition
diagrams”.
In general for a solution of two volatile liquids, the vapor pressure is the sum of the vapor
pressures of individual components. For ideal solutions, which follow Raoult’s Law, the
vapor pressure (Pi) of each component is simply the product of its mole fraction (x i) in the
liquid mixture and the vapor pressure of pure component (Pio) at that temperature, as shown
below.
P =  Pi =  Pio . xi
i
i
Most of the solutions, however, do not behave ideally (often resulting from intermolecular
interactions among different molecules) and hence show either positive or negative deviations
from Raoult’s Law.
For benzene-toluene system, which behaves ideally, the liquid-vapor equilibrium diagram (a
temperature-composition diagram) is given in Figure 1.
For benzene-toluene system the pressure-composition diagram at 25oC is given in
Figure 2.
25
Figure 1. Liquid-vapor equilibrium for benzene-toluene mixtures: normal boiling
temperature versus composition
Figure 2. Liquid vapor equilibrium for benzene-toluene mixtures at 25oC.
26
AZEOTROPES
An azeotrope or constant-boiling mixture, is a mixture of liquids which has a fixed
composition and characteristic boiling point. This term was introduced by Wade and
Merriman in 1911 to designate all binary and ternary mixtures that cannot be separated from
each other by distillation.
Not all liquids form ideal solutions. Ethanol and water are such liquids. Because of
molecular interaction, a mixture of 95.5% ethanol and 4.5% water ( by weight ), boils at
78.15oC, below the boiling point of ethanol. No matter how efficient the distilling apparatus,
100% ethanol cannot be obtained by distillation of a mixture of, say, 75% ethanol and 25%
water. This is because ethanol and water form a binary azeotrope.
The boiling-point-composition curve for the ethanol-water mixture is seen in fig.3.4.
To prepare 100% ethanol, the water can be removed chemically (reaction with calcium oxide)
or by removal of the water as an azeotrope with another liquid. Absolute alcohol is made by
addition of benzene to 95% alcohol and removal of the water in the volatile benzene-wateralcohol azeotrope.
A pure liquid has a constant boiling point. A change in boiling point during distillation
is an indication of impurity. The converse proposition is not always true, however, as,
azeotropes have constant boiling points that can be either above or below the boiling point of
the individual components.
27
Figure 3: Boiling-point-composition curve for ethanol water mixtures.
III. EXPERIMENTAL
III. 1. Apparatus and Chemicals
In this experiment you will be using a distillation set-up shown in Figure 3 and an Abbe
refractometer. The following chemicals and equipment are also needed:
Isopropyl alcohol
(b.p: 82.4 oC (760 mmHg), density: 0.785 g/cm³(20oC), MW: 60.09
g/mole)
Water
(b.p:100.0 oC (760 mmHg), density: 0.998 g/cm³ (20oC), MW:18.0153
g/mole)
Test tubes
Pasteur pipettes
Tube rack
Heating mantle
100 ml 2-neck round bottom flask
Distillation Setup
Thermometer
Figure 4. Distillation Setup
28
III. 2. Procedure
Before starting the experiment a calibration curve giving the chemical composition (mole
fractions) of alcohol-water mixtures versus their refractive indices must be constructed. For
distillation part, alcohol-water mixtures from 10% to 95% (concentration by volume) will be
studied. Please calculate the molar compositions and fill the respective columns in Table I,
before coming to the laboratory.
Table I: Preparation of stock solutions
Soln.
IPA
water
IPA
water
IPA
water
(V/V)
(ml)
(ml)
(g)
(g)
(mole)
(mole)
%10
5
45
%40
8
12
%50
10
10
%80
16
4
%90
18
2
%95
19
1
III. 2. 1. Collection of data
1. Pour about 20 ml of sample into the boiling flask of the distillation apparatus.
2. Add boiling chips.
3. Turn the heat on (setting on heater about 3-4).
4. Watch the system carefully until boiling.
5. When distillation starts record the boiling point.
6. Wait for about 2 minutes for the boiling temperature to stay constant, then replace the
collection flask with a clean a test tube and collect about 2 ml of a sample and cap it
tightly. Then, determine its refractive index (Vi) using the refractometer.
7. Continue the distillation for second sample (vapor phase fraction, 1V), record the
boiling temperature again (do not remove the test tube before the system cools down
completely).
8. Turn the heater off. Wait system to reach equilibrium.
29
9. Take a sample from the distillation flask (1L) and determine its refractive index.
10. Empty the contents of the distillation flask into the waste container.
Dry distillation flask and fill it with next sample. Repeat this procedure for all the samples.
Soln.
IPA
water
Vapor Initial (Vi)
Vapor (V)
Liquid (L)
no
(ml)
(ml)
Refractive
Refractive
Refractive
index @T(°C)
index @T(°C)
index @T(°C)
%10
5
45
%40
8
12
%50
10
10
%80
16
4
%90
18
2
%95
19
1
III. Data Handling
1. Construct calibration curve giving the chemical composition versus their refractive indices.
2. Discuss the difference between Vi and V.
3. Calculate mole fractions of liquid (L) and vapor phases (V) and construct a temperaturecomposition diagram.
4. Does this system obey Raoult’s Law? If not why? What kind of deviation is observed from
ideal systems?
5. On a separate graph, plot the boiling point diagram of the same system as predicted by
Raoult’s law (i. e. as if it were an ideal system).
In order to do this use the following equations (Antoine equation), which will give you the
vapor pressures of isopropylalcohol and water at selected temperatures(1).
ln P = A- (B/ T+ C) (Antoine equation)
ln P = 7.7402 - [1359.517/(T + 197.527)]
for IPA [T range 52.3 – 89.3 (°C)]
lnP = 8.10765 - [1750.286 /(T + 235.000)] for water [T range 0 – 60.0 (°C)]
30
lnP = 7.96681 - [ 1668.210 /(T + 228.000)] for water [T range 60.0-150.0 (°C)]
where (P) is the vapor pressure in mmHg and (T) is the temperature in degrees centigrade.
For ideal solutions at temperature (t),
Ptotal = P1 + P2 = x1 . P1o(t) + x2 . P2o(t)
= (1- x2 ) . P1o(t) + x2 . P2o(t)
= P1o(t) + x2 [P2o(t) - P1o(t)]
Assuming Ptotal = 760 mmHg, the mole fraction [x2(liq)] can be calculated. Do this calculation
for five different temperatures spaced between boiling points of water and isopropyl alcohol
and plot the points on your graph. Next draw the liquid composition curve through these
points. Corresponding points for the vapor (or distillate) curve are obtained from the
following relationship:
P2
x2 (liquid) . P2o(t)
y2 (vapor) = —— = ————
Ptotal
Ptotal
6. Plot these and draw the vapor curve similarly for the ideal solution as well.
IV. QUESTIONS
1. Determine compositions of the following by using calibration curve that you prepared.
i) 4 portions of your distillate ii) resulting mixture in your flask
2. Calculate mol fraction of liquid and vapor phases and construct a compositiontemperature graph like in your manual. ( Make your comments about this graph)
3. Which composition of IPA-water mixture forms azeotrope? Why?
4. Give the possible reasons of azeotrope formation.
5. Find other alcohol mixtures which can form azeotrope?
6. If we use simple distillation apparatus to separate IPA-water mixture, how will your
composition-temperature graph be affected?
7. Propose a proper way to separate azeotropic mixtures.
V. REFERENCES
1. Shoemaker D.P., Garland C.W., Nibler J.W., Experiments in Physical Chemistry, 5.
edition, 1989, McGraw Hill Inc.
31
EXPERIMENT 5
SOLUBILITY OF SOLIDS IN LIQUIDS
I. PURPOSE
Investigate the solubility of solutes at different temperatures.
II. THEORY
When a sufficiently large amount of solute is in contact with the limited amount of solvent,
dissolving of the solute starts taking place. After a certain period of time, the reverse process,
precipitation, starts to occur. When dissolving and precipitation of the solute takes place at
equal rates (i.e. equilibrium has been established), the solution is said to be saturated. After
this point the concentration of the solute in the solution remains unchanged. The
concentration of the saturated solution is defined as the solubility of the solute in the given
solvent. Solubility is generally a function of temperature.
For a pure solid A to be in equilibrium with a solution containing A, the chemical potential of
A, μA, must be the same in the solid (s) and liquid (l) phases; μAl = μAs. The chemical
potential of component A in an ideal solution is
μAl = μA0l + RT ln XA
(1)
Where μA0l is the chemical potential of pure liquid A, and X is the solubility. Hence, the
equilibrium condition can be written
μAs = μA0l + RT ln XA
(2)
Where μA0I and μAs are the molar free energies of pure liquid and pure solid A. Remember
that:
 G 

 i  
 ni  P ,T ,nJ
(3)
32
Hence
GA
 GA
RT
OS
Ol
 LnX A
(4)
And since
 G / T 
H
 2
T
T
(5)
Differentiation of equation (4) with respect to T yields (with ΔHf, the latent heat of fusion)
HA
 HA
RT 2
OS
Ol

H f
RT
2

dLnX A
dT
(6)
It is a good approximation to take ΔHf independent of T over moderate ranges of temperature.
Alternately, equation 6 can be written in the following form (see equation 7 below). As a
result, the linear variation of the solubility, XA, of pure solid in an ideal solution with
temperature is
LnX A 
H f  1 1 
  
R  T2 T1 
(7)
In determining the solubility of a solute, it is practically more correct to start off at a high
temperature, cool the solution down to the required temperature and wait for thermal
equilibrium to be reached. This approach is known as “reaching equilibrium from the
molecular rather than the crystal side.
The solubility of the solute (benzoic acid in this case) can be determined as a function of
temperature. As shown in equation (7), a plot of lnXA versus 1/T should give a straight line
with the slope being equal to heat of solution (ΔHf / R).
The heat of solution is the difference between the amount of energy associated with the
formation of a layer of solvent molecules around the solute and the amount of energy
33
associated with formation of the crystal packing in the solute. When the enthalpy is positive,
solid will absorb heat when it goes into solution. At high temps, more heat is available so
more solid will dissolve. A negative enthalpy means the solid gives off heat when it dissolves
and subsequently absorbs heat when it precipitates. At high temps then, precipitation is
favored. The heat of solution (enthalpy) at the saturation point determines the temperature
effect.
When a pure solid is mixed with a pure solvent at a given temperature and pressure, the free
energy of mixing will be given by:
G  H  TS
Being H , the change in enthalpy of mixing, T the absolute temperature in the process, and
S the change in entropy of mixing. According to the equation, and from the thermodynamic
point of view, the dissolution will only take place if G sign is negative. S is usually
positive, since in solution, the molecules display a more chaotic arrangement than in the solid
state, and on the other hand, the absolute temperature must be also positive. However, H
may be either positive or negative. According to Rosen (1982), a positive H indicates that
both the solid and the solvent are in their lower energy state, whereas a negative value
suggests the solution is in its lower energy state, showing interactions such as hydrogen
bonding, which are established between solid and solvent molecules.
Experimentally, the heat of solution measured is the heat absorbed when 1 mole of solid is
dissolved in an almost saturated solution. This value is different from the heat of solution
measured at “infinite dilution”, and the difference is equivalent to the heat of dilution.
III. PROCEDURE
1. 1 gram of benzoic acid will be provided in 150 ml beaker. Add 125 ml of distilled
water with graduated cylinder into it, place the beaker on heater stirrer, start heating while
stirring gently with a glass rod up to 70-750C. Caution, above 800C benzoic acid may
decompose. Record time, temperature, and any observations.
34
2. After all benzoic acid dissolves, distribute this solution carefully into 20 ml volume
marked and labeled 4 separate test tubes. There should be 20 ml sample solution in each test
tube.
3. Keep two of the tubes in water bath at room temperature, and place a glass rod into
one of them.
4. Place two of the tubes in a water bath at 350C or 450C and place a glass rod into one
of them. Maintain the temperature constant for 15-20 minutes.
5. Check for the formation of crystals. Record any observation, time.
6. Starting from the higher temperature sample test tubes, filter your each sample that
are waited at the same temperature into separate beakers. Then, take 5 ml sample from this
beaker with a graduated cylinder into a 50 ml. Weigh the Erlenmeyer flask and the solution.
7. Titrate the ‘weighed’ solution with the standardized 0.05 M NaOH solution, using
phenolphthalein as the indicator. Perform at least 2 titrations, until the results are in +/ - 0.1
ml range.
8. Repeat your “solubility” determinations for each sample. You should have duplicate
titration results for each solution that were kept either at room temperature, at 350C or at
450C.
Please remember to record the room temperature!
IV. TREATMENT OF THE EXPERIMENTAL DATA
1. Use titration results to determine the solubility, X, of benzoic acid. Define the solubility as
moles per 1000 g of solvent.
2. Plot lnX versus 1/T. The slope of the plot of lnXA vs. 1/T will result ΔHf /R where R=8.314
j/mol*K or
R=1.986 cal/mol*K
and T in Kelvin. You need to consider 1/T in 5
significant figures. Otherwise you will result with a linear curve
35
3. Make calculations for RT, 35°C and 45°C.Draw tangents to the curve for these
temperatures, and determine ΔHf from each slope.
4. Determine the deviations in each value obtained. Why do you think it differs for each
temperature?
36
Experiment 6
Electrochemistry
1. Introduction
Electrochemistry, branch of chemistry concerned with the relation between electricity and
chemical change. Electrochemistry studies chemical reactions which take place in a solution
at the interface of an electron conductor (a metal or a semiconductor) and an ionic
conductor (the electrolyte). These reactions involve electron transfer between the electrode
and the electrolyte or species in solution.
Many spontaneously occurring chemical reactions liberate electrical energy, and some of
these reactions are used in batteries and fuel cells to produce electric power. Conversely,
electric current can be utilized to bring about many chemical reactions that do not occur
spontaneously. In the process called electrolysis, electrical energy is converted directly into
chemical energy, which is stored in the products of the reaction. This process is applied in
refining metals, in electroplating, and in producing hydrogen and oxygen from water.
A galvanic cell or voltaic cell is a device in which a redox reaction spontaneously occurs and
produces an electric current. In order to transfer electrons in a redox reaction to produce an
electric current and be useful, the electrons are made to pass through an external electrically
conducting wire instead of being directly transferred between the oxidizing and reducing
agents. In a galvanic cell, two solutions, one containing the ions of the oxidation half-reaction
and the other containing the ions of the reduction half-reaction, are placed in separated
compartments called half-cells. For each half-cell, the metal, which is called an electrode, is
placed in the solution and connected to an external wire. A redox reaction results when an
oxidation and a reduction half-reaction are combined to complete a transfer of electrons as in
the following example:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
The electrode at which oxidation occurs is called the anode (Zn) and the electrode at which
reduction occurs is called the cathode (Cu). The two half-cells are connected by a salt-bridge
that allows a “current” of ions from one half-cell to the other to complete the circuit of
electron current in the external wires. When the two electrodes are connected to an electric
37
load (such as a light bulb or voltmeter) the circuit is completed, the oxidation-reduction
reaction occurs, and electrons move from the anode (−) to the cathode (+), producing an
electric current.
The cell potential, Ecell, which is a measure of the voltage that the battery can provide, is
calculated from the half-cell reduction potentials:
Ecell = Ecathode - Eanode
At standard conditions, indicated by the superscript o, the standard cell potential, Eocell, is
based upon the standard reduction potentials.
E°cell = E°cathode – E°anode
Based on the values for the standard reduction potentials for the two half-cells [–0.76 V for
zinc anode and +0.34 V for copper cathode], the standard cell potential, E°cell, for the
galvanic cell would be:
Eocell= +0.34 V – (–0.76 V) = +1.10 V
The positive voltage for Eocell indicates that at standard conditions the reaction is
o
o
spontaneous. Recall that ΔG = − nFEocell, so that a positive Eocell results in a negative ΔG .
Thus the redox reaction would produce an electric current when set up as a galvanic cell.
When conditions are not standard, the Nernst equation, equation (6), is used to calculate the
potential of a cell. In the Nernst equation, R is the universal gas constant with a value of
8.314 J/Kmol, T is the temperature in K, and n is the number of electrons transferred in the
redox reaction. Q is the reaction quotient for the ion products/ion reactants of the cell. The
solid electrodes have constant “concentrations” and so do not appear in Q. F is the Faraday
constant with a known value of 96500 J/Vmol.
 RT 
0
E cell  E cell

ln Q 
 nF 
For example, Zn(s) + Cu2+(aq) → Zn2+ (aq) + Cu(s), so Q and ln Q are:
Zn  and ln Q = ln [Zn2+] − ln [Cu2+]
Q
Cu 
2
2
Since n = 2, the Nernst equation for this redox reaction becomes:



 RT 
 RT 
0
2
2
E cell  Ecell

 ln Zn  
ln Cu
2
F
2
F




38

In a series of galvanic cells, in which [Zn2+] is kept constant while [Cu2+] is varied, Ecell can
be measured and it will be found to vary with ln[Cu2+]. A plot of the data obtained in which y
is Ecell and x is ln[Cu2+] will result in a straight line: y = mx + b. For Cu/Zn galvanic cell the
terms Eocell and −[RT/2F]ln [Zn2+] are constant and together they equal the intercept, b, of the
line. [RT/2F] will be the constant slope, m, provided the temperature is constant.
Thus a plot can be generated by measuring the cell potential for different values of [Cu 2+],
while keeping [Zn2+] constant (at 1.0 M). The equation of the line in this plot can then be used
to determine ln[Cu2+] in a solution of unknown concentration from its measured Ecell, as long
as [Zn2+] remains at 1.0 M and the temperature at 25 oC. This is the same principle used in a
pH meter for the determination of [H+]. Notice that as the concentration of Cu2+ ion reactant
increases (or ln[Cu2+] becomes less negative), the potential of the cell increases.
2. Galvanic cell (Nernst Equation for varying Cu2+ concentrations):
Purpose of the experiment
Galvanic cells with different known Cu2+ concentrations and a fixed Zn2+ concentration will
be prepared and their cell potentials measured. A plot of the Nernst equation, Ecell vs.
ln[Cu2+], constructed from this data will then be used to find [Cu2+] of an unknown solution.
2.1. Procedure
1. Prepare three dilute solutions of Cu(NO3)2 by serial dilution from a 1.0 M Cu(NO3)2 stock
solution. Transfer 0.5 ml of 1.0 M Cu(NO3)2 stock solution into a labeled (with tape) 50.00 ml
volumetric flask and dilute to the mark with de-ionized water to form a 0.01 M solution.
Rinse the pipette with a small volume of the solution to be transferred prior to use. Next take
the 0.01 M solution just formed and transfer 0.5 ml of it into a labeled 50.00 ml volumetric
flask. Dilute to the mark with de-ionized water to form the 0.0001 M solution. Repeat one
more time so that you have the following set of solutions in labeled flasks:
Solution A – 1 M Cu(NO3)2 (stock solution) Solution B – 0.01 M Cu(NO3)2
Solution C – 0.0001 M Cu(NO3)2 Solution D – 0.000001 M Cu(NO3)2
3. Clean and degrease the Cu and the Zn electrode by Al2O3 particles and detergent until the
surface is shiny. Rinse it carefully with deionized water.
4. Set the glass cell (U shaped) with a help of a clamp to vertical position as shown in
figure 1.
39
Figure 1. The experimental set up for Zn/Cu galvanic cell
5. Prepare a half-cell of Cu2+ by placing the copper solution D into one side of the U shape
glass cell. Place the Cu electrode with a plastic stopper into the cell.
6. Place 1.0 M Zn(NO3)2 in a cell next to the Cu2+ half-cell together the Zn electrode. Connect
the copper and zinc electrodes to the multimeter. Measure and record the cell potential in your
laboratory notebook.
7. Remove the Cu(NO3)2 solution with a disposable pipette and repeat the measurement of
Ecell (steps 4) for the remaining three copper solutions in order of increasing concentration.
8. Then, in the same way, measure the Ecell for the unknown Cu2+ solution. Record the
measured values in a table in your notebook.
Data analysis and report
1. Construct and attach to your report a graph of the measured values of Ecell versus ln[Cu2+]
from your data table.
2. Find the best fit line for the plot and equation of the line.
3. From the equation for the line, determine [Cu2+] of the unknown solution.
40
3. Electroless plating
3.1. Introduction
Electroless plating is a process to produce thin metallic coatings on objects without the
application of external current. The plating bath contains a dissolved salt of the metal and a
reducing agent. However, the reduction of the metal cation to metal occurs only on the surface
of the object to be coated due to the catalytic nature of the surface. The advantages of this
process over electroplating are the possibility to produce coatings on insulator materials, and
to produce uniform thickness coatings on geometrically complex surfaces.
The simplest of chemical plating is the so called metal-displacement reaction. For example
when Zn metal is immerse into copper sulphate solution, the zinc metal dissolves and
spontaneously replace by copper atoms from the solution. Chemical plating by displacement
yields deposits limited only few micrometer thickness, hence has few applications.
In order to continuously build thick deposit by chemical means without consuming the
substrate (the surface has to be coated), it is essential that a sustainable oxidation reaction be
employed as an alternative to the dissolution of the substrate. The deposition reaction must
occur initially and exclusively on the substrate and subsequently continue to deposit on the
initial deposit (autocatalytic reaction).
The chemical deposition of nickel metal by hypophosphite meets both the oxidation and redox
potential criteria without changing the mass of the substrate, moreover nonconductive
substrates can be coated with this method after activation of the surface.
Reduction: Ni2+ + 2 e- → Ni0
E0= -25 mV
Oxidation: H2PO2- + H2O → H2PO3- + 2H+
E0= +50 mV
Ni2+ H2PO2- + H2O → Ni0 + H2PO3- + 2H+
E0= +25 mV
This reaction does not represent the true electroless plating reaction, since electroless nickel
deposition is accompanied by hydrogen evolution. Electroless nickel plating is undoubtedly
the most important catalytic plating process used today. The chemical a physical properties of
the coating depends on its composition which in turn depends on the formulation and
operation conditions of the electroless plating bath. Typically, the constituents of an
electroless plating solution are:

source of nickel ions

reducing agent

suitable complexing agents
41

stabilizer/inhibitors.
3.2. Purpose of the experiment
To get familiar with the chemical deposition of metal from an aqueous solution containing the
desired metal as a salt. The process based on electrochemical mechanism, both oxidation and
reduction reactions involving the transfer of electrons between the reacting chemical species.
3.4. Procedure
Component
Concentration / M
NiCl2
0.126
NaH2PO2
0.114
CH3COONa
0.122
Solution pH=5
1. Calculate, and put the amount of salt required for 50 ml solution by weighing.
2. Dissolve the CH3COONa salt in ~25 ml deionized water in 50 ml volumetric flask, add the
nickel salt and dissolve in the same solution and finally add the reducing agent (NaH2PO2).
Fill up the volumetric flask to 50 ml.
3. Pour the solution into a beaker and set the pH to ~5 by adding few drop of HCl solution.
4. Heat the solution by a hot plate while continuous stirring to ~80 oC.
5. Clean and degrease the Cu plate by Al2O3 particles and detergent until the surface is shiny.
Rinse it carefully with deionized water. Dry it and measure the weight.
6. Immerse the copper plate into the hot solution and observe the surface.
7. After 15 min deposition remove the plate from the bath, stop the heating, rinse the plate by
deionized water and dry it on paper towel. Measure the weight of the plate and the coated
surface area.
Report
1. Report the observations: change in Cu surface, electrolyte bath color, odors.
2. Report the weight of the deposit!
2. Calculate the thickness of the coating from the surface area and mass! Assume, that pure
nickel was deposited without any impurity (density of nickel is 8.908 gcm-3).
42