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Undergraduate Practical Chemistry
Physical Chemistry Lab
Experiment no: 01
Studies of the kinetics of alkaline hydrolysis of Ethyl acetate and determination of (a) the rate constant (b) activation energy.
Principle
The reaction is described by the following equation
EtOAc + OH- = EtOH + AcO- …………………….(1)
Here Et and Ac represent ethyl and acetyl radical is respectively.
There are two convenient methods for measuring the progress of the reaction. The consumption of the reactant OH- may be measured by taking aliquot portions out of the reaction mixture and measuring its concentration by titrating with standard acid solution.
The second and most convenient method of following progress of the above reaction is through the measurement of the reaction conductance of the reaction mixture. As the reaction shows highly conducting OH- ions are replaced by poorly conducting AcO- ions, and this reaction in a decrease in the overall conductance of the reaction mixture.
In order to test whether the reaction is second order, it is convenient to work with equal initial concentrations of both the reactants. Under this condition, the rate equation is
-d[A]d[t] = k[A]2
And the integrated form
1[A]-1Ao = kt ………………………….(2)
Where [A]0 is the initial concentration and [A]0 is the initial concentration and [A] is the concentration at t, and k is the second order rate constant.
The equation (2) can be rearranged into the form
k= 1tA0.(Ao-[A])[A] ……………………(3)
If C0, Ct, and C are the conductance of the reaction mixture at time 0, t, and infinite respectively, it can be shown that
(C0- Ct) [A]0
(Ct- C ) [A]
(C0- Ct) ([A]0- [A]
Putting these proportionality relations in equation (3) and rearranging, we get
Ct = 1Ao .(Co-Ct)t + C ………………(4)
Therefore a plot of Ct vs. C0- Ct/t should give a straight line with a slope equal to 1/[A]0.k
From the values of k at different temperatures the activation energy can be calculated by applying the Arhenious equation
K = Ae-Ea/RT …………………….(5)
Where Ea = activation energy, A = Frequency factor
Equation (5) can be written as
ln K = lnA – Ea/RT
A plot of lnK vs. 1/T should give a straight line whose slope is –Ea/R.
Procedure
Prepare 0.1 M standard oxalic acid solution.
Prepare 0.1 M NaOH solution and standardize it by titrating against standard oxalic acid solution using phenolphthalein indicator.
Prepare 0.02 M standard NaOH solution by proper dilution of std. 0.1 M NaOH solution.
Prepare exact 0.02 M Ethyl acetate solution.
Take 50 ml of NaOH solution and CH3COOC2H5 solution in two different boiling tube.
Suspend both of them in the thermostat maintain at room temperature.
Thoroughly rinse the conductance cell with distilled water. Measure the conductance of NaOH solution.
When the two solutions have attained the temperature of the thermostat, mix them together and measure the conductance.
Record the conductance of the solution at regular time intervals (1 minute- 3 minute), till there is no measurable changes.
Repeat steps (6) to (9) for four different temperatures, e.g. 30°C, 35°C, 40°C, 45°C, etc.
Calculation
MW of oxalic acid (C2O4H2.2H2O) = 126 g
Molarity, S = 1000 wMV or, weight, w = SMV1000 =0.1 ×126 ×1001000 gm = 1.26 gm
Weight 1.26 g oxalic acid and fill up to 100 ml in volumetric flux.
MW of NaOH = 40 gm
Weight 1gm NaOH and fill up to 250 ml in volumetric flux.
Sl. No.
Volume of oxalic acid(ml), V1
Conc. of oxalic acid(M), S1
Burette reading of NaOH (ml), V2
Initial
Final
Average
1
10
0.1
0.0
2
10
0.1
3
10
0.1
H2C2O4 (aq) + 2 NaOH (aq) --> Na2C2O4 (aq) + 2 H2O (l)
2V1S1 = V2S2 or, S2 = 2V1S1 /V2
Diluted by this equation V1S1 = V2S2
Concentration of ethyl acetate is 99%, Density is 0.9 g/ml
Molecular mass is 88.11 g/mol
1000 ml ethyl acetate contains 900x0.99=891 g
It can calculate in term of Molarity =891 (g)/ (88.11 (g/mol)) =10.11 M
So, C1V1=C2V2
or, 10.11(M) x V1=0.02 x 500ml or, V1= 0.989 ml
So, 0.989 ml ethyl acetate is added into water adjusted to 500 ml to obtain 0.02 M.
Treatment of Data
Prepare conductance-time data, along with (C0- Ct)/t in tabular form.
Time, t (min)
Initial cond., C0
(µS)
Conductance at time t, Ct (µS)
C0- Ct
C0- Ct/t
Draw conductance vs. time graph and calculate C0.
Draw Ct vs. C0- Ct/t, and evaluate the rate constant, k from the slope (1/[A]0k) of the line.
From the four k values at four different temperatures measure lnk and 1/T (derived data).
Plot lnk vs. 1/T to evaluate E, the activation energy.
Sample questions
What is conductivity?
Ans. Electrical conductivity is a measure of the ability of a solution to carry a current. Current flow in liquids differs from that in metal conductors in that electrons cannot flow freely, but must be carried by ions. Ions are formed when a solid such as salt is dissolved in a liquid to form electrical components having opposite electrical charges. For example, sodium chloride separates to form Na+ and Cl- ions. All ions present in the solutions contribute to the current flowing through the sensor and therefore, contribute to the conductivity measurement. Electrical conductivity can therefore be used as a measure of the concentration of ionizable solutes present in the sample.
Why OH- ion is more conducting than AcO- ion?
Ans. Because of OH- ion is very small size than AcO- ion.
How can you make different types of solutions?
Ans. See calculation.
What is rate constant?
Ans. In chemical kinetics a reaction rate constant, k quantifies the rate of a chemical reaction. For a chemical reaction where substance A and B are reacting to produce C, the reaction rate has the form: Reaction: A + B C
r= k(T)[A]m[B]n
k(T) is the reaction rate constant that depends on temperature.
What is the significance of Arrhenius equation?
Ans. Arrhenius' equation is a simple, but remarkably accurate, formula for the temperature dependence of reaction rate. By this equation we can easily calculate the activation energy, molar gas constant etc.
Experiment no: 02
Extraction of caffeine from tea
Principle
Tea and coffee have been popular beverages for centuries, primarily because they contain the refreshment caffeine. It stimulates respiration, the heart, and the central nervous system, and is a diuretic (promotes urination). It can cause nervousness and insomnia and, like many drugs, can be addictive, making it difficult to reduce the daily dose. A regular coffee drinker who consumes just four cups per day can experience headache, insomnia, and even nausea upon with drawl from the drug. On the other hand, it helps people to pay attention and can sharpen moderately complex mental skills as well as prolong the ability to exercise. During the course of the day, an average person may unwittingly consume up to a gram of this substance. The caffeine content in some common foods and drugs are below.
Table 1. Caffeine content of common foods
Beverage
Caffeine (mg) per cup
Coffee
80-125
Tea
30-75
Coffee, decaffeinated
2-4
Chocolate milk/cocoa
3-30
Soft drinks
20-50
Cocoa
5-40
Caffeine belongs to a large class of compounds known as alkaloids. The main source of it is plants, contains basic nitrogen, often has bitter taste and complex structure, and usually has physiological activity. Their names are usually end in "ine", many are quite familiar by name if not chemical structure-nicotine, cocaine, morphine, strychnine.
Figure-1: Structure of Caffeine (1,3,7-trimethylxanthine)
IUPAC Name: 1,3,7-trimethyl-1H-purine-2,6(3H,7H)-dione
Tea leaves contain tannins, which are acidic, as well as colored along small amount of un-decomposed chlorophyll (soluble in dichloromethane). To ensure that the acidic substances remain water soluble and that the caffeine will be present as the free base, sodium carbonate is used for the extraction. The solubility of caffeine in water is 2.2 mgmL-1at 250C, 180 mgmL-1 at 800 C, and 670 mgmL-1 at 1000 C. It is quite soluble in dichloromethane.
Caffeine has the chemical elements carbon, hydrogen, and oxygen and nitrogen in its structure. It has a bitter taste and soluble in water and organic solvents. It is a white crystalline xanthine alkaloid and psychoactive stimulant.
Table 2. Physical properties of caffeine
Physical parameters
Value
Molecular Formula
C8H10N4O2
Molecular Weight
194.19 g/mol
Melting Point
227-2280C (anhydrous);
234-2350C(monohydrate)
Boiling Point
1780 C (sublimation)
Procedure
In this experiment, caffeine will be isolated directly from tea bags. The chief problem with the isolation is that caffeine does not exist alone in tealeaves, but is accompanied by other natural substances from which it must be separated. Caffeine constitutes as much as 5% by weight of the leaf material in tea plants and is water-soluble. Unfortunately, Tannins in the tealeaf are also water-soluble. The Tannins, which steep from tealeaves, are in the form of Catechin and Flavonoid derivatives.
Fortunately, Caffeine is soluble in polar aprotic solvents whereas the Tannins are soluble in protic solvents with hydrogen bonding. Thus, we can carryout the isolation of caffeine from tealeaves in the following steps:
Extraction of the Caffeine and Tannins into hot water.
Extraction of the Caffeine into a non-polar organic solvent; dichloromethane.
Drying the dichloromethane of any remaining water.
Evaporation of the dichloromethane, leaving impure Caffeine.
Purification of the Caffeine by re-crystallization using a mixed-solvent system.
Caffeine was extracted in two steps: (1) extraction of crude caffeine and (2) purification of crude caffeine.
Crude Caffeine
Weigh 5 tea bags and dip in 200 mL water and mix 5g sodium carbonate.
Heat the mixture up to vigorous boil with a hot plate and stir to dissolve Na2CO3 completely.
Allow the mixture to cool for 5 min, and decant into another beaker.
Gently squeeze the tea bags to liberate the rest of the water.
Cool the aqueous solution to near room temperature by continue cooling in an ice bath. Because, tea must be cooled (about 20° C) before coming in contact with dichloromethane (boiling point = 40°C).
Extract the solution three times with 30-mL portions of dichloromethane (CH2Cl2). After 30mL addition, the solution shake for 15 min vigorously in the shaker machine to cause emulsion formation, hence if it is not shaken enough the caffeine will not be extracted into the organic layer.
Pour the solution in a separatory funnel and use a gentle rocking motion of the separatory funnel. Vent the Sep funnel periodically with the spigot pointing away to relieve gas pressure.
When the contents have been sufficiently shaken, put the Sep funnel back on the ring stand and let the two layers separate out.
Remove the stopper and drain the bottom layer into a flask. The steps were repeated two more times.
Dry the combined dichloromethane solutions and any emulsion layer with anhydrous Sodium Carbonate (Na2CO3). Add sufficient drying agent until it no longer clumps together on the bottom of the flask – about 1 teaspoon usually works well.
Carefully gravity filter the dichloromethane solution into a tarred (previously weighed) beaker.
Evaporate the CH2Cl2 solvent in a hot water bath in the hood. A wooden stick works better than boiling chips to promote smooth boiling because it is easily removed once the solvent is gone.
When all the solvent will be removed, it will be observed a residue of greenish-white crystalline caffeine.
Purification of crude caffeine
Re-crystallize the crude caffeine using a mixed solvent system that involved dissolving it with 10 mL hot acetone followed by the addition of hexane until the solution turned cloudy. The solution was cooled and the crystalline caffeine was collected by vacuum filtration. Repeat this extraction in several times.
Take UV of 250 ppm caffeine solution in dichloromethane solvent.
Calculation
Percentage of caffeine extraction
Table 3. Weight of crude caffeine
Weight of 5 tea bags
Weight of crude caffeine
Percentage of yield = Mass of crude caffeine / mass of tea bags x 100%
Weight of pure caffeine
Weight of filter paper
Weight of filter paper + caffeine
Weight of pure caffeine
Recovery percentage of pure caffeine
= Mass of purified caffeine / mass of crude caffeine x 100%
Sample questions
What is alkaloid?
Ans. Alkaloids are a group of naturally occurring chemical compound that contain mostly basic nitrogen atoms. This group also includes some related compounds with neutral and even weakly acidic properties. Some synthetic compounds of similar structure are also attributed to alkaloids.
Draw the structure of caffeine
Ans. See principle.
Why organic dichloromethane gives bottom layer?
Ans. Dichloromethane (1.33 g/cm3) is denser than water so it appears at the lower layer while upper layer is aqueous layer.
Why Na2CO3 is used?
Ans. Low molecular weight of tannin is soluble in organic solvent and it makes isolation of caffeine more difficult. Sodium carbonate was used to hydrolyze tannin to produce glucose and sodium salt of gallic acid in which they are not soluble in organic layer due to their high polarity. At the same time, the base also converted caffeine to a free base, which is more soluble in organic layer.
Experiment no: 03
Determination of the Dissociation constant of Phenolphthalein indicator colorimetrically.
Principles
Colorimetry is concerned with the determination of the concentration of a substance by measurement of the relative absorption light with respect to a known concentration of the substance. The variation of light absorption is due to the variation of the color of the system. This method can be consistently applied in the determination of the dissociation constant of an acid-base indicator when undissociated.
All indicators in general use are very weak organic acids or bases. If we consider the undissociated acid indicator as HIn, then the equilibrium in aqueous solution may be written as
HIn H+ + In-
Unionized color ionized color
Now, in acid solution, i.e., in the presence of excess of H+ ions, the ionization will be depressed (common ion effect) and the color will therefore be that of unionized form. If the medium is alkaline, the decrease of [H+] will result in the further ionization of the indicator becomes apparent. By applying the law of mass action, we obtain
Kin = H+[In-][HIn] ……………….(2)
Where Kin is the ionization constant of the indicator and the activity coefficients are assumed to be unity.
Rearranging equation (2)
[H+] = Kin [HIn][In-]
log[H+] = log Kin + log [HIn][In-]
pH = logKin + log [HIn][In-] ……………….(3)
It is evident from equation (3) that as the pH of the solution increases the value of log [HIn][In-] also increases (since pKin is a constant), i.e., there will be greater dissociation.
Phenolphthalein is an example of acid indicator which gives purple color in alkaline medium due to the color of the ionized species [In-]. Whereas in acid medium the solution is colorless because the equilibrium shifts to the left and exist as undissociated condition, HIn which is colorless. It is observed that at pH 10, phenolphthalein is particularly completely dissociated form. On the other hand at pH 7 or below it assumed to exist in the dissociated form. Using eq. (3) it is possible to determine the dissociation constant, Kin of phenolphthalein (std. value of pKin 9.1-9.6), colorimetrically.
Now let us consider the simplest form of the Beer-Lambert Law:
A = log I0/It = εcl ……………….(4)
Where, A= absorbance; I0 = intensity of the incident light
It = intensity of the transmitted light; ε = molar absorption coefficient
c = concentration of the colored solute expressed in moledm-3
l = thickness of the medium in cm
If this law is applied to an alkaline solution of phenolphthalein the equation (4) can be written as
A =ε [In-]l ........................(5)
Equation (5) implies that with the increase in the degree of dissociation, concentration of In- increases. Thus the solution absorbs more light and absorbance increases.
At pH 10, phenolphthalein is completely dissociated and [In-] = [HIn]
Then, equation (5) becomes
Amax = ε[HIn]l ……………...(6)
Where Amax corresponds to the light absorbed at complete dissociation.
Dividing equation (5) by equation (6) we get
AAmax =[In-][HIn] ……………………..(7)
Combining this equation with equation (3), one can easily get
pH = logA/Amax + pKin ………………………(8)
Equation (8) can be used to calculate the pKin and hence the dissociation constant, Kin of phenolphthalein indicator.
Procedure
Prepare the following three solutions in three different volumetric flask;
250 ml 0.2M NaOH solution (M.W. of NaOH = 40)
250 ml 0.2M Boric acid solution (M.W. of H3BO3 = 61.8)
50 ml of saturated solution of Na2CO3
Freshly prepared phenolphthalein indicator solution by dissolving 0.5 gm of the pure reagent in 50 cm3 of ethanol (C2H5OH) and then mix 50 cm3 of water with constant stirring.
Prepare the buffer solution of different pH by mixing different volume of NaOH and H3BO3 solution as follows:
Solution No.
Volume of 0.2M NaOH (ml)
Volume of 0.2M H3BO3 (ml)
1
2
18
2
3
17
3
4
16
4
4.5
15.5
5
5
15
6
6
14
7
7
13
8
8
12
Add few drops phenolphthalein solution to all solutions.
Measure the pH of each of the solution.
Note the pH of the saturated solution of Na2CO3 and measure the λmax using different filter of the colorimeter.
At λmax, measure the absorbance of each of the solution.
Note:
If the instrument gives the value of transmittance instead of absorbance, equation (5) to (8) should be changed according to the relation: A = log 1/T.
Calculation
Make NaOH and Boric acid solutions by this equation: w = SMV/1000
Treatment of Data
Calculate A/Amax and log A/Amax of each of the solution (or T/Tmin and logT/Tmin)
Solution No.
A
Amax
logA
logA/Amax
1
2
Plot pH vs. logA/Amax (or, logT/Tmin). It should give a straight line. From the slope of the line calculate the value of Kin, the dissociation constant of phenolphthalein.
Sample questions
What is colorimetry?
Ans. Colorimetry is the analysis of chemical samples to collect information about their concentration. It involves passing light through a sample and measuring how much is absorbed by the solution, using equipment like a spectrophotometer to measure as precisely as possible. The technique may be one of several tests performed on a sample to determine its composition, check for quality and clarity, or demonstrate scientific principles in a classroom. Manufacturers of scientific instruments produce a range of instruments for this purpose.
What is dissociation constant?
Ans. Dissociation constant is a constant whose numerical value depends on the equilibrium between the undissociated and dissociated forms of a molecule. A higher value indicates greater dissociation.
What is common ion effect? How does common ion effect occur in this experiment?
Ans. Common-ion effect describes the suppressing effect on ionization of an electrolyte when another electrolyte is added that shares a common ion. In the presence of excess of H+ ions, the ionization of phenolphthalein will be depressed (common ion effect) and the color will therefore be that of unionized form.
Write the reaction between NaOH and Boric acid?
Ans. 4H3BO3 + 2NaOH = Na2B4O7 + 7H2O and H3BO3 + 2NaOH = Na2BO3 + 2H2O
