ENGINEERING CHEMISTRY LAB – I B.TECH

MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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ENGINEERING CHEMISTRY LAB – I B.TECH

MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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ENGINEERING CHEMISTRY LAB – I B.TECH

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LIST OF ENGINEERING CHEMISTRY EXPERIMENTS INDEX S.NO

NAME OF THE EXPERIMENT

1

Preparation of Aspirin

2

Estimation of hardness of water by EDTA method

3

Determination Of viscosity by Ostwald viscometer

4

Conductometric titrations of a strong acid vs strong Base

5

Conductometric titrations of a mixture of acids vs strong base

6

Determination of percentage purity of Pyrolusite

7

Potentiometric titrations of a strong acid vs strong base

8

Potentiometric titrations of a weak acid vs strong base

9

Determination of % of copper in brass

10

Preparation of Thiokol rubber

11

Adsorption of Oxalic acid on charcoal

12

Estimation of ferrous iron by Dichrometry.

13

Determination of surface tension using Stalagmometer

PAGE NO.

DATE CONDUCTED

SIGNATURE

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SAFETY NORMS TO BE FOLLOWED. DO’S AND DONT’S IN CHEMISTRY LABORATORY.  NEVER SNIFF, TASTE OR TOUCH THE CHEMICALS.  ALWAYS POUR ACIDS INTO WATER. NEVER ADD WATER TO ACID. IF POUR WATER INTO ACID , THE HEAT OF REACTION WILL CAUSE WATER TO EXPLODE INTO STEAM, SOME TIMES VIOLENTLY, AND THE ACID WILL SPLATTER.  KNOW WHAT CHEMICALS YOU ARE USING. READ THE LABEL TWICE BEFORE YOU USE THEM.  LEARN WHERE THE SAFETY AND FIRST AID EQUIPMENT IS LOCATED, LIKE FIRE EXTINGUISHERS, EYE WASH STATIONS.  CONSIDER ALL CHEMICALS HAZARDOUS UNLESS YOU ARE INSTRUCTED OTHERWISE.  NEVER EXPERIMENT ON YOUR OWN.  DO NOT EAT /DRINK IN LAB AT ANY TIME.  DO NOT WEAR CONTACT LENS IN THE LAB.  NEVER LEAVE THE LAB WITHOUT WASHING YOUR HANDS.  DO NOT PIPETTE BY MOUTH.  DON’T CASUALLY DISPOSE OF CHEMICALS DOWN THE DRAIN.  MANY COMMON REAGENTS FOR EXAMPLE ALCOHOL AND ACETONE ARE HIGHLY INFLAMMABLE. DO NOT USE THEM ANYWHERE NEAR OPEN FLAMES.  IF CHEMICALS COME IN CONTACT WITH YOUR SKIN OR EYES, FLUSH IMMEDIATELY WITH WATER FOR ATLEAST 10 MIN. AND CONSULT YOUR INSTRUCTOR.

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No. 1

PREPARATION OF ASPIRIN AIM:To prepare aspirin from salicylic acid and acetic anhydride in presence of conc.H 2SO4 CHEMICALS REQUIRED: Salicylic acid = 2.5 gm or 5 gm Acetic anhydride Con. H2SO4 APPARATUS : Conical flask Beaker Funnel Glass rod PROCEDURE: Take 2.5 g of salicylic acid 5 ml of acetic anhydride in 100 ml conical flask. Add 3 drop conc. H2SO4 and shake well. Heat the flask on water bath keeping temp below 60-70oC for about 20 min Cool the mixture in an ice bath with stirring then add about 20 ml of ice cold water to decompose excess of acetic anhydride. Pour the content of the conical flask into a 250 ml beaker containing 50 g of crushed ice. Stir it until white solid appears. Scratch the side of the flask with a glass rod or stainless steel spatula. If white solid does not appear, filter through funnel and dry by pressing into the folds of filter paper. Recrystallise the crude acetylsalicylic acid. RECRYSTALLISATION: Dissolve the solid in minimum amount of ethanol and dissolve it until clear solution is obtained. Clear solution is allowed to cool slowly to get needle shaped crystals of Aspirin. Calculations: Theoretical yield = 5 gm X M .Wt of product M .Wt of starting material

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% yield of Aspirin =

Experimental value X 100 Theoretical value

RESULT:

The amount of crude Aspirin found

=

The amount of pure Aspirin found

=

VIVA QUESTIONS : 1) What is the chemical name of aspirin?

2) What is the medicinal importance of aspirin?

3) What are the chemicals required for the preparation of aspirin?

4) what is the process involved in the preparation of aspirin?

5) what is the function of acetic anhydride and sulphuric acid?

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SPACE FOR OBSERVATIONS AND CALCULATIONS

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SPACE FOR OBSERVATIONS AND CALCULATIONS

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Experiment No. 2

HARDNESS OF WATER BY EDTA METHOD AIM:To determine the total, permanent and temporary hardness of water by EDTA method. CHEMICALS REQUIRED: Standard ZnSO4 solution, EDTA Solution, Ammonia– Ammonium chloride buffer water sample EBT APPARATUS: Burette Conical flask Pipette INDICATOR:

Eriochrome black –T

END POINT: The end point is noted when the colour of the solution changes from wine red to blue PRINCIPLE: The hardness of water is due to the presence of salts of Ca 2+ and Mg2+. The bicarbonates of Ca2+ and Mg2+ impart temporary hardness to water which can be removed by boiling. The amount of Ca2+ and Mg2+ is estimated by complexometric method using EDTA. The obtained value gives the total hardness. The permanent hardness is determined first by precipitating the bicarbonates of Ca2+ and Mg2+ by heating and filtering off and the filtrate is titrated with EDTA. The temporary hardness is obtained by subtracting permanent hardness from total hardness.

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EDTA Titrations (or complexometric titrations): The titrations involving EDTA as a complexing agent are known as “EDTA Titrations” (or) “Complex Formation Titrations”. EDTA stands for ‘Ethylene Di-amineTetra Acetic acid”. The structure of EDTA is given by

CH2COOH

CH2COOH N - CH2 – CH2 – N

CH2COOH

CH2COOH EDTA forms complexes with metal ions like Ca2+, Mg2+, Ba2+, and

Sr2+ in aqueous solution. Complexo metric titration is the titration of metal ion with a reagent, usually EDTA, which forms chelated complex with a metal. The common indicator used is Erichrome Black-T Indicator (EBT), at the end point the colour generally changes from red to blue.

CH2COO-Na+

HOOCH2C N - CH2 – CH2 – N +

Na-OOCH2C

CH2COOH Disodium salt of EDTA

Ethylene di amine tetra acetic acid [EDTA] forms colorless stable complex with Ca+2 and Mg+2 ions present in water at pH 9-10. To maintain p H of the solution at 9-10 NH3-NH4Cl buffer solution is used. Erio chrome black –T (EBT) is used as indicator. The hard water when treated with (EBT) in presence of buffer solution forms unstable wine red color complex with Ca +2 and Mg+2 ions present in water. The stability of a metal indicator complex is less than that of metal EDTA complex. Hence during the titration EDTA extracts the metal ions from the metal ion EBT indicator complex and forms stable colorless metal EDTA complex releasing free

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indicator. Hence the end point of the titration is the color change from wine red to blue color.

PROCEDURE: PART A: STANDARDISATION OF EDTA SOLUTION: 20ml of standard Zn2+ solution is pipette out into a well cleaned conical flask. To it 2- 3 drops of EBT indicator and 5 ml of buffer are added. Then the solution is titrated against EDTA until wine red color changes from red to blue. This is taken as end point. Titrations are repeated until two successively concurrent values are obtained. The morality of EDTA is calculated by using the formula V1M1=V2M2 PART B: DETERMINATION OF TOTAL HARDNESS OF TAP WATER: 20ml of tap water is taken into a well cleaned conical flask. To it 5ml of buffer and 2-3 drops of EBT indicator are added. The color of resulting solution is wine red. The contents of the conical flask are titrated against EDTA which is taken in the

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burette until the colour of the solution changes from wine red to blue. This is the end point. Repeat the titration until two successively concurrent values are obtained.

The molarity of tap water is calculated by using the formula M2V2=M3V3 and M1V1=M2V2 M1 = Molarity of standard Zn2 + solution V1= Volume of standard Zn2 + solution V2=Volume of EDTA rundown M2 = Molarity of DTA Solution M2 = M1V1/V2 OBSERVATION:

PART A: STANDARDISATION OF EDTA SOLUTION: Burette readings

Volume of S.NO

standard Zn+2 solution (ml)

Initial

1.

20

0

2.

20

0

3.

20

0

Final

Volume of EDTA run down ‘x’ ml

V1M1=V2M2 M1= Molarity of tap water V1= Volume of tap water M2= Molarity of EDTA solution V2= Volume of EDTA rundown M2=M1V1/V2.

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PART B: DETERMINATION OF TOTAL HARDNESS OF TAP WATER: Burette readings (ml) S.NO

Volume of tap water (ml)

Initial

1.

20

0

2.

20

0

3.

20

0

Final

Volume of EDTA run down ‘x’ ml

M2V2=M3V3 M2= Molarity of EDTA solution V2= Volume of EDTA rundown M3= Molarity of tap water V3= Volume of tap water M3=M2V2/V3. PART C: DETERMINATION OF PERMANENT HARDNESS OF WATER: Place 100ml of the sample of water in a beaker and boil gently for 1020 minutes. Cool and filter, collecting the filtrate into a 100ml standerd flask. Make up the solution to the mark by adding distilled water and shake the solution well. Pipette out 50ml of this made up solution into a clean conical flask, which has been rinsed wth distilled water. Add 10ml of pH = 10 buffer solution and 2 to 3 drops of EBT indicator. Titrate this solution against the standard EDTA solution until the colour changes from wine red to blue. Note

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down the reading, repeat the process to get at least two equal titer values. Calculate the permanent and then temporary hardness as parts per million of CaCO 3.

EDTA

BOILED WATER

M2 = calculated in 2nd step

M4 =?

V2 = burette reading

V4 = 50 ml

n2 = 1

n4 = 1 M2V2

M4V4

n2

M4

=

=

n4

M2V2 V4

CALCULATIONS: Hardness = conc.x 105 ppm. Total hardness = M3 x 105 ppm Permanent hardness = M4 x 105 ppm

RESULT:

Amount of total hardness of the given sample water = ----------ppm Amount of permanent hardness of the given sample water = --------ppm Amount of temporary hardness of the given sample water = --------ppm

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VIVA QUESTIONS :

1) what is EDTA and write its structure?

2) what is buffer solution and mention its function?

3) which hardness can be removed by boiling?

4) what is hardness?

5) what is the indicator used and what is the colour change in this experiment?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No . 3

DETERMINATION OF VISCOSITY OF OIL BY OSTWALD VISCOMETER AIM: To determine the viscosity of given oil. APPARATUS: Ostwald Viscometer stop watch. Rubber bulb Specific gravity bottle Beaker CHEMICALS: Given oil (test liquid) water.(std.liquid) THEORY: In the laboratory the viscometer commonly used for comparing viscosities of liquids is the Ostwald viscometer. In its operation different liquids are taken in exactly the same volume. This is essential so that the height of the liquid columns creating the pressure heads are equal and then the pressure heads will be directly proportional to the densities of the respective liquids. The liquids are sucked up in to the upper bulb A. initially enough liquid volume is taken in bulb B such that when the liquid stands above the mark X in bulb A, a little is still left in the bend and the bulb B. the time of flow (t) is measured for each liquid for its flow from X to Y. the driving pressure P at all stages of the flow of the liquid is given by h*d*g, where g is the acceleration due to gravity; d is density of liquids; h is the difference between the heights of the liquids in the upper and lower bulbs (same for all liquids between the same points x and y) For any given capillary, radius, length being constant and volume fixed,   kPt where k includes all constant terms. This equation leads to an easy comparison between the viscosities of different liquids.

1 kP1t1 P1t1    2 kP2t 2 P 2t 2

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This equation can be simplified into

1 hd 1gt1 d 1t1   hd 2 gt 2 2 d 2t 2 PROCEDURE: Take a clean and dry Ostwald viscometer and set it vertically on a stand Fill the water through the larger bulb. Suck the liquid up into the bulb A through a rubber cock attached to the end D to a level the mark X Allow the liquid to flow freely through the capillary and note the time t 1 for the liquid to flow from X to Y. Repeat steps 3 to 4 to obtain the average time for the liquid to flow from X to Y. OBSERVATION TABLE: S.No

Liquid taken

Density (gm/cc)

Time (Sec)

Viscosity of liquid(milli poise)

 1 =8.90

1

Water

d1 =

t1 =

2

Test liq 1

d2 =

t2 =

2 =

3

Test liq 2

d3 =

t3 =

η3=

CALCULATIONS:

1 d 1t1   2 d 2t 2

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Y v i s c o s i t y

X Temp. 0C

RESULT: The viscosity of the given liquid = ………….. milli poise

VIVA QUESTIONS : 1) what is viscosity of a liquid?

2) what are the units of viscosity?

3) what are the effects of temperature and density on viscosity?

4) what are the apparatus used to determine the viscosity of a liquid?

5) what is the viscosity of water?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No : 4

CONDUCTOMETRIC TITRATION AIM:To determine the neutralization point in an acid base titration using conductivity meter. APPARATUS : Conductivity bridge Conductivity Cell Distilled water 250 ml beaker Stirrer CHEMICALS REQUIRED: 0.1M HCl 1.0 M NaOH. PRINCIPLE: As the alkali is added from the burette into the cell containing acid, the concentration of H+ ions change in a graphical manner, which leads to a considerable change in the electrical conductance of the solution, which is measured using a conductivity meter. Then from the plot of conductance versus volume of alkali, the precise neutralization point is determined. The titration of strong acid vz strong base involves the following equation. HCl + NaOH ― NaCl + H2O In the titration of HCl against NaOH, initially the conductance of HCl solution is maximum due to complete ionization. As the alkali is added, the conductance of solution decreases and after the neutralization point the conductance starts increasing. This is because of the addition of the alkali. The fast moving H+ ions are replaced by Na+ ions (slow moving) once the neutralization point is reached, addition of alkali introduces fast moving OH - ions, there by increasing the conductivity of the solution.

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PROCEDURE: Pipette out 20ml of 0.1M HCl into a clean 250ml beaker. Fill the burette with 1.0M NaOH solution. Dip a conductivity cell in the 0.1 M HCl solution and connect it to the conductivity meter and note the initial conductance of HCl solution. Run down the NaOH solution from the burette in small volumes of 1ml with stirring till the end point to be expected in to the cell and note the conductance. OBSERVATION: Conductance S.NO.

Volume of NaOH (ml) (Ohm-1 or mhos)

1

0 ml

2.

1ml

3.

2ml

4.

3ml

5.

4ml

6.

5ml

7.

6ml

8.

7ml

9.

8ml

10.

9ml

11

10ml

12

11ml

13

12ml

14

13ml

15

14ml

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Y

Conductance

EP

X Volume of NaOH

M1V1=M2V2

RESULT: From the graph the volume of 1.0 M NaOH required for the neutralization (Point of the graph) of 0.1M HCl is ------------

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VIVA QUESTIONS : 1)what is neutralization point?

2)why we don’t require an indicator in this experiment?

3)why the conductance decreases when we add NaOH to HCl solution?

4)After neutralization point why there is a rapid increase in the conductance?

5)what are the units of conductance?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No : 5

CONDUCTOMETRIC TITRATION OF MIXTURE OF ACIDS VS STRONG BASE AIM: To determine the neutralization point in an mixture of acids-base titration using conductivity meter. APPARATUS : Conductivity bridge Conductivity Cell Distilled water 250 ml beaker Stirrer CHEMICALS REQUIRED: 0.1M HCl 0.1M CH3COOH 1.0 M NaOH. PRINCIPLE: when a mixture containing CH3COOH & HCl is titrated against an alkali ,strong acid will b neutralized first, the neutralization of week acid [CH3COOH] commences only after the complete neutralization of strong acid[HCl]. Then conductance titration curve will be marked by two breaks. The first one corresponds to neutralization of HCl and second one two that of CH3COOH.Let V1 & V2 ml be volume of alkali corresponding to first and second breaks, respectively. V1 ml of NaOH = HCl

(V1-V2) ml of NaOH=CH3COOH

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PROCEDURE: Take a beaker and add 20ml of CH3COOH(0.1M) and 20ml of HCl(0.1M) . Fill the burette with 0.1M NaOH solution. Place the conductivity cell in distilled water.. Now measure initial conductance of mixture of solution. Then 1 ml of NaOH is added every time from burette into the solution and stirred well each time. Note down the conductance values till the conductance values decreases and increase considerably. OBSERVATION:

S.NO.

Volume of NaOH (ml)

1

0 ml

2.

1ml

3.

2ml

4.

3ml

5

4ml

6

5ml

7

6ml

8

7ml

9

8ml

10

9ml

11

10ml

12

11ml

13

12ml

14

13ml

15

14ml

Conductance (µ or mhos)

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GRAPH:

Y

Conductance

v2 v1 X Vol. of NaOH

M1V1=M2V2

RESULT: From the graph the volume of 1.0 M NaOH required for the neutralization (Point of the graph) of 0.1M HCl and 0.1M CH3COOH is---------

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VIVA QUESTIONS : 1) How many neutralization points are there in this experiment?

2) which acid reacts first with NaOH?

3) write the reaction between CH3COOH and NaOH?

4) what is the basic principle involved in this experiment?

5) How to determine the end points?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No . 6

ESTIMATION OF MANGANESE DIOXIDE IN PYROLUSITE AIM: To estimate the amount of MnO2 present in the given sample of pyrolusite APPARATUS: Standard flask Pipette Burette Conical flask. CHEMICALS REQUIRED: Pyrolusite Oxalic acid Dil H2SO4 KMnO4 PRINCIPLE : Manganese dioxide occurs in nature in the form of pyrolusite. The percentage of MnO2 is determined by treatment with excess of acidified solution of a reducing agent thus the MnO2present in pyrolusite sample is reduced to a known excess of standard sodium oxalate as acid medium and the unreacted sodium oxalate is titrates against standard solution of KMnO4. MnO2 + H2SO4 + H2C2O4

2CO2+2H2O+MnSO4

The ore is graded on the basis of its available oxygen content rather than on its percentage of Manganese. PROCEDURE : STEP I : STANDARDIZATION OF KMnO4 SOLUTION : Pipette out 20 ml of oxalic acid in a clean conical flask add 20 ml of 3M H 2SO4 and heat the conical flask to 700c . Fill the burette with KMnO4 solution and place the hot conical flask under the burette and rundown KMnO4 in drop wise manner .At the end point solution turns from colorless to pale pink . Note the burette reading and repeat the experiment for concurrent values .

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STEP II : ESTIMATION OF MnO2 IN PYROLUSITE : Weigh accurately 0.2 gm of finely powdered dry pyrolusite in to 250 ml conical flask and add 50 ml of N/10 oxalic acid .Add 50 ml of diluteH2SO4 (10%) and place a short funnel over the conical flask .Boil the contents of the flask gently until no black particles are visible in the flask . Allow it to cool about 700c and titrate the excess oxalate with standard KMnO4 solution until a permanent pink color appears .Note the burette reading and repeat the experiment for concurrent values . CALCULATIONS : S.No

Volume of oxalic acid

1

20ml

2

20ml

3

20ml

Initial

M1V1 --------

Burette reading (ml)

M 2=

Final

M2V2 =

--------

n1 oxalic acid (n1)=5

Volume of KMnO4(ml)

n2 KMnO4 (n2)=2 M1 * 20 * 2 V2 * 5

Molarity of KMnO4

=

-----------------------M

Molarity * M.Wt=Normality * equivalent wt .

Normality= Molarity * 158 31.6 =-------------------------N

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STEP 3: Weight ore taken =0.2 gm Volume of KMnO4 required for titration(V)= ----------------Volume of KMnO4 consumed = 50-V 1 ml of 1 N KMnO4 = 0.04346gm of MnO2 1 ml of --------------- N KMnO4 = ? 0.04346 * N of KMnO4 ---------------------------

= Y gm of MnO2

1N

CALCULATIONS: 1 ml of ----------------N KMnO4 = Y gm of MnO2 (50-V) ml of ---------- N KMnO4 = ? (50-V) * Y = A gm of MnO2 0.2 gm of Pyrolusite ore contains A gm of MnO2 100 gm of Pyrolusite ore contains --------------------100 * A = B gm 0. 2 Therefore % of MnO2 in pyrolusite = B %

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RESULT: Molarity of oxalic acid = ----------------Molarity of KMnO4

= -----------------

The percentage of MnO2 present in the given Pyrolusite Sample = B %

VIVA QUESTIONS : 1) what is pyrolusite?

2) what is self indicator in this experiment?

3) what is the equivalent weight of KMnO4 ?

4) what is molarity and normality?

5) what is the basic principle of pyrolusite experiment?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No . 7

POTENTIOMETRIC TITRATION OF STRONG ACID VS STRONG BASE AIM: To determine the neutralization point in the potentiometric titration of strong acid vs strong base. APPARATUS: Potentiometer bridge saturated calomel electrode platinum electrode Beaker Burette Stirrer CHEMICALS REQUIRED: N/10HCl N/10 NaOH Distilled water. PRINCIPLE: In the titration of 0.1N HCl with 0.1N NaOH on addition of the alkali, there is variation in the concentration of H+ ions, there variations in the concentration of H+ ions i.e. PH is followed potentiometrically using Quinhydrone (reversible to hydrogen ion) as, the indicator electrode in the HCl solution and coupling it with saturated calomel electrode (reference electrode) since the potential of the lather remains constant, the emf of the cell will vary only with PH of HCl solution. Therefore by measuring this emf act each stage of the titration and plotting it against the volume of base, we can deduce the equivalence point from the plot. At the end point, the emf increases at once which is clearly detected in the of graph. Cell reaction is: Hg / Hg2Cl2 (s).KCl(saturated solution)(s) // H+, QH2/Pt. MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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PROCEDURE: Pipette out 20ml of 0.1N HCl solution into beaker and saturate it with quinhydrone and dip the indicator electrode (platinum electrode) connect the indicator electrode and saturated calomel electrode (reference electrode) to the potentiometer. The two half cells are connected by means of a salt bridge. The potentiometer is standardized and used for measuring the emf directly. Take 20 ml of HCl solution in the beaker and add 1 ml of 0.1N NaOH from the burette every time. Shake well after each addition and measure the cell emf. From this rough titration find out the approximate volume needed for the end points. Now a fair titration is repeated by adding volumes of 1ml alkali. Subsequent additions

are made in

steps of 1or 2 ml of alkali. Then plot a graph between the measured emf on Y-axis and volume of alkali on X-axis. OBSERVATION: Vol. of alkali (ml)

EMF (mv)

1ml 2ml 3ml 4ml 5ml 6ml 7ml 8ml 9ml 10ml

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CALCULATIONS: Where

N1V1 = N2V2 N1 = Strength of acid(0.1N) V1 = Volume of acid taken in the beaker N2 = Strength of the Base(0.1N) V2 = ? (Volume of alkali from the graph)

RESULT: From the graph the equivalence point for the acid base titration, is potentiometrically determined to be ……ml

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ENGINEERING CHEMISTRY LAB – I B.TECH VIVA QUESTIONS :

1) what is a potentiometric titration?

2) what are the electrodes used in the experiment?

3) what is the indicator electrode?

4) what is neutralization point?

5) How can you determine the strength of acid from neutralization point?

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SPACE FOR OBSERVATIONS AND CALCULATIONS

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No . 8

POTENTIOMETRIC TITRATION OF WEAK ACID VS STRONG BASE AIM: To determine the neutralization point in the potentiometric titration of weak acid versus strong base. APPARATUS: Potentiometer bridge saturated calomel electrode platinum electrode Beaker Burette Stirrer CHEMICALS REQUIRED: N/10 CH3COOH N/10 NaOH Distilled water. PRINCIPLE: In the titration of 0.1N CH3COOH with 0.1N NaOH on addition of the alkali, there is variation in the concentration of H+ ions, there variations in the concentration of H+ ions i.e. PH is followed potentiometrically using Quinhydrone (reversible to hydrogen ion) as, the indicator electrode in the CH3COOH solution and coupling it with saturated calomel electrode (reference electrode) since the potential of the lather remains constant, the emf of the cell will vary only with PH of CH3COOH solution. Therefore by measuring this emf act each stage of the titration and plotting it against the volume of base, we can deduce the equivalence point from the plot. At the end point, the emf increases at once which is clearly detected in the of graph.

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Cell reaction is Hg/Hg2Cl2 (s).KCl(saturated solution)(s) // H+, QH2/Pt. PROCEDURE: Pipette out 20ml of 0.1N CH3 COOH solution into beaker and saturate it with quinhydrone and dip the indicator electrode (platinum electrode) connect the indicator electrode and saturated calomel electrode (reference electrode) to the potentiometer. The two half cells are connected by means of a salt bridge. The potentiometer is standardized and used for measuring the emf directly. Take 0.1N NaOH in the burette and add 1 ml of CH3COOH solution in the beaker everytime. Shake well after each addition and measure the cell emf. From this rough titration find out the approximate volume needed for the end points. Now a fair titration is repeated by adding volumes of 1ml alkali. Subsequent additions are made in steps of 1or 2 ml of alkali. Then plot a graph between the measured emf on Y-axis and volume of alkali on X-axis. OBSERVATION: Vol. of alkali (ml)

EMF (MV)

1ml 2ml 3ml 4ml 5ml 6ml 7ml 8ml

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GRAPH: Y EP

EMF

X Vol.of NaOH CALCULATIONS: N1V1 = N2V2 Where N1 = Strength of the Acid V1 = Volume of acid taken in the beaker V2 = ? (Volume of alkali from the graph) N2 = Strength of the base

RESULT: From the graph the equivalence point for the acid base titration, is potentiometrically determined to be ……ml

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VIVA QUESTIONS : 1) what is a potentiometric titration?

2) what are the electrodes used in the experiment?

3) what is the indicator electrode?

4) what is neutralization point?

5) How can you determine the strength of acid from neutralization point?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment no . 9

DETERMINATION OF PERCENTAGE OF COPPER IN BRASS AIM: To determine the Percentage of Copper in the given sample of brass APPARATUS:

Conical flask 250 ml standard flask Test tube Burette Pipette Beaker

CHEMICALS: Brass alloy Nitric acid Urea Ammonium hydroxide Hypo Starch PRINCIPLE: Brass is an alloy of copper and zinc. When dissolved in concentrated nitric acid, both the metals get converted to their nitrates. Cu + 4HNO3

Cu(NO3)2 + 2H2O +2NO2

When brass solution is treated with excess of KI solution, Cu+2 ions oxidize to liberate equivalent quantity of I2. 2 Cu(NO3)2 + 4KI

Cu2I2 + 4KNO3 + I2

The liberated I2 is titrated with hypo solution using starch as indicator. Na2S2O3 + I2

Na2S4O6 + 2NaI

The oxides of nitrogen present in brass solution are destroyed by adding urea. The presence of nitrogen oxides will be responsible for the liberation of extra I2 from KI, as they are also good oxidizing agents. The nitric acid present in brass solution is neutralized by adding NH4OH till a pale blue ppt of Cu(OH)2 is obtained. Otherwise HNO3 may also liberate MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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I2 from KI. The Cu(OH)2 ppt is dissolved in dilute acetic acid. Other mineral acids are not preferable as they lower the PH to a very low value and at which liberation of I 2 from KI by Cu+2 is not quantitative .Starch reacts with I2 to form a blue colored complex. At the end point, when free I2 is exhausted in the solution, added quantity of hypo dissociates starch-I2 and liberates starch, thereby discharging the blue color. End point is the color change from BLUE to COLORLESS. During the liberation of I2 from KI, Cu+2 gets reduced to Cu+1.Hence the equivalent weight of Cu = atomic weight of Cu= 63.54 PROCEDURE: Weigh accurately about 1 gm brass alloy. Transfer into a clean beaker. Dissolve the alloy in ¼ test tube conc. HNO3. Then add about 1 gm of Urea and 1 test tube of distilled water. Boil for 2 minutes. Cool the products and transfer quantitatively into a 250 ml standard flask. Make up the solution to the mark and shake well .Pipette out 25 ml of Brass solution into a conical flask. Add NH4OH solution drop wise until a bluish white ppt persists. Add dilute acetic acid solution drop wise just to dissolve the ppt and about 5 ml excess, followed by the addition of 10 ml of 10% KI solution .Titrate this solution against hypo solution taken in the burette until the color changes to pale yellow. At this stage add starch solution (1 ml) and continue the titration till the blue color changes to white .Repeat the titration for concordant values. OBSERVATIONS AND CALCULATIONS: Weight of brass = …………gm In burette = hypo solution In conical flask = 25 ml of brass solution, NH4OH, Acetic acid,10 ml of 10% KI Indicator = starch solution Color change = Blue To Colourless

S.No

Volume of Brass solution

1

25 ml

2

25 ml

3

25 ml

Initial reading on the burette

Final reading

Volume of hypo used V1 ml

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ENGINEERING CHEMISTRY LAB – I B.TECH

N1= Normality of Hypo V1 = Volume of Hypo N1 × V1 = N2 × V2

N2

=

N1 X V1

=

N1 X V1

V2

25

Weight of Copper in 1litre of solution = N2 X Equivalent weight of copper

= N2 X 63.54 = …………gm

N2 X 63.54 Wt of Copper in 250ml is = 4

N2 X 63.54 x 100 % of Copper in Brass

= 4 X Wt of Brass

RESULT: Percentage of Copper in the given sample of brass is …… …………….

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VIVA QUESTIONS :

1) What is brass?

2) What is the colour formed when starch reacts with iodine?

3) What is the role of KI in the experiment?

4) What is the indicator used?

5) What is the basic principal involved in this experiment?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No . 10

PREPARATION OF THIOKOL RUBBER AIM:To Synthesize Thiokol rubber using sodium poly sulphide with 1,2 dichloro ethane. APPARATUS: Beakers Glass rod. CHEMICALS: NaOH, Powdered Sulphur Ethylene chloride (1,2 dichloride ethane ) Benzene 5% Sulphuric acid Nitric acid. THEORY: It is a rubbery white substance and is obtained by treating sodium polysulfide with 1, 2 – dichloro ethane. S S

S S

|| ||

|| ||

nCl – CH2 – CH2 – Cl + n Na+ - S- - S- - Na+ → (- CH2 – CH2 - S – S - )n + (2n – 1 ) NaCl

PROCEDURE: In a 100 ml beaker dissolve 2 g NaOH in 50 -60 ml warm water.Boil the solution and to this add in small lots with constant stirring 4 g of powdered sulfur. During addition and stirring the yellow solution turns deep – red.Cool it to 60 – 700C and add 10 ml of 1,2 – dichloromethane (ethylene chloride)with stirring . Stir for an additional period of 20 minutes while rubber polymer separated out as a lump.Pour out the liquid from the beaker in the sink to obtain Thiokol rubber. Wash it with water under the tap.Dry in the fold of filter papers. The yield is about 1.5 g. Determine the solubility of the polymer in benzene, acetone, 5% H2SO4 and nitric acid. MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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ENGINEERING CHEMISTRY LAB – I B.TECH

NOTE: If some sulfur remains un dissolved filter the solution.

RESULT: Yield obtained = --------------g.

VIVA QUESTIONS :

1) what are the chemicals used in the preparation of Thiokol?

2) what are the different monomers are used?

3) what is the End point colour?

4) Is it natural rubber or synthetic rubber?

5) which rubber is not vulcanized with sulphur?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment No . 11

STUDY OF ADSORPTION OF OXALIC ACID ON ACTIVATED CHARCOAL AIM:

Study of adsorption of oxalic acid from solution on activated charcoal

CHEMICALS : Oxalic acid NaOH Phenolphthalein indicator Charcoal APPARATUS:

Reagent bottles Funnel Filter paper Glass rod

BASIC PRINCIPLE: Physical adsorption ,chemical adsorption and adsorption isotherms. Adsorption :- This is a surface phenomena Absorption :- This is a bulk phenomena Adsorbent :- The Solid substance on which adsorption takes place. Adsorbate :- The substance which undergo adsorption . PROCEDURE: 1. PREAPARATION OF 0.25M OXALIC ACID: Weigh out exactly 3.15gm of oxalic acid into a 1000 ml standard flask and make up the solution to the mark with distilled water after dissolving the salt in little distilled water. Shake the flask well for uniform concentration.

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2. PREPARATION OF 0.1M NaOH SOLUTION: Dissolve 4 gms of NaOH in 1000 ml of water and stir the solution well for uniform concentration. 3. PHENOLPHTHALEIN INDICATOR: Dissolve 1 gm of phenolphthalein indicator in100ml of ethanol. Take well cleaned and dried 6 stoppered reagent bottles and label them. With the help of two burettes transfer and oxalic acid and distilled water into these 6 bottles as shown below: Reagent bottle no.

1

2

3

4

5

6

Oxalic acid

50

40

30

20

10

00

50

60

70

80

90

100

(0.25M)(ml) Distilled water(ml)

Mix the solutions well and keep all the six bottles in a water bath for some time to acquire the temperature of the water bath. Weigh exactly 2 gms of activated charcoal on 6 glazed papers and add the same to each bottle. Shake the bottle well and keep in the water bath for half an hour time while shaking from time to time. Filter the contents of all the 6 bottles into 6 different labeled dried conical flasks. Titrate the 6 samples with M/10 NaOH taken in the burette using phenolphthalein indicator till pale pink colored end point is obtained. Let the titre values be V3,V4,V5,V6,V7,V8respectivelyfor bottle no.s 1,2,3,4,5,6. Take 20 ml of the stock solution into a clean 250 ml conical flask, add 2 drops of phenolphthalein indicator and titrate against 0.1M NaOH solution till pale yellow color is obtained as end point. Repeat the titration concurrent values. Let the titre value be x ml. CALCULATIONS: Molarity of oxalic acid solution prepared (M1) = Wt. of oxalic acid X 10 /126 Molarity of NaOH solution (M1) = V1M1/n1 = V2M2/n2 V1 = Volume of oxalic acid (20ml) M1 = Molarity of oxalic acid n1

= no. of moles of oxalic acid = 1

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V2 = Volume of NaOH (titre value x ml) n2 = no. of moles of NaOH reacted = 2 M2 = V1M1/n1 X n2/ V2

The initial concentration of the oxalic acid solutions in 1 to 6 conical flasks Bottle no.

Initial concentration(C)

1

50 X M2 /100 = a

2

40 X M2 /100 = b

3

30 X M2 /100 = c

4

20 X M2 /100 = d

5

10 X M2 /100 = e

6

00 X M2 /100 = zero

Concentration of oxalic acid solutions after adsorption: Bottle no. 1 = M (oxalic acid)

= 2 X M (NaOH) X V3 / 50 = p

Bottle no. 2 = M (oxalic acid)

= 2 X M (NaOH) X V3 / 40 = q

Bottle no. 3 = M (oxalic acid)

= 2 X M (NaOH) X V3 / 30 = r

Bottle no. 4 = M (oxalic acid)

= 2 X M (NaOH) X V3 / 20 = s

Bottle no. 5 = M (oxalic acid)

= 2 X M (NaOH) X V3 / 10 = t

Amount of Oxalic acid adsorbed x = (C1-C2) X100/1000 = (C1-C2)/10 Where, C1 = initial concentration C2 = conc. after adsorption MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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ENGINEERING CHEMISTRY LAB – I B.TECH

Bottle no.

(C1) initial concentration

(C2) concen. after adsorption

Amount of oxalic acid adsorbed (C1-C2)/10 = x

1

a

p

a-p/10

2

b

q

b-q/10

3

c

r

c-r/10

4

d

s

d-s/10

5

e

t

e-t/10

RESULT: Amount of Oxalic acid adsorbed x = ………….

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VIVA QUESTIONS :

1) what is adsorption?

2) what is adsorbate in the experiment?

3) what is adsorbent in the experiment?

4) what is the indicator used and what is the colour change?

5) Is adsorption bulk phenomena or surface phenomena?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment : 12

ESTIMATION OF FERROUS IRON BY DICHROMETRY. AIM: To estimate the amount of ferrous iron present in the given solution using standard Potassium Dichromate. APPARATUS: Burette Pipette Conical flask Burette stand standard flask weighing box weighing bottle. CHEMICALS: Potassium Dichromate Iron solution sulphuric acid phosphoric acid Diphenylamine indicator. PRINCIPLE: Potassium Dichromate is an oxidizing agent. It liberates nascent oxygen atoms in presence of dilute sulphuric acid. K2Cr2O7 + 4H2SO4 K2SO4 + Cr2 (SO4)3 + 4H2O + 3(O) The liberated nascent oxygen oxidizes ferrous salt to ferric salt in cold and in presence of acid. 6FeSO4 + 3H2SO4 + 3(O)  3Fe2(SO4)3 + 3H2O K2Cr2O7 +7H2SO4 + 6FeSO4  K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O

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PROCEDURE: Estimation of ferrous iron involves two steps: 1) Preparation of standard solution of potassium dichromate, 2)Estimation of ferrous iron.

STEP-I PREPARATION OF STANDARD POTASSIUM DICHROMATE Weight of empty weighing bottle

(w1) =

Weight of the weighing bottle + salt

(w2) =

Weight of K2Cr2O7

(W3) =

w2-w1

=

gm

Normality of potassium dichromate= Weight of K2Cr2O7

× 1000

Gram equivalent weight

V ml

STEP-I PREPARATION OF STANDARD SOLUTION OF POTASSIUM DICHROMATE: About 0.5 g of potassium dichromate crystals are taken in a clean dry weighing bottle and its weight is found accurately in analytical balance .Then it is transferred into 100ml standard flask through the funnel. The weight of the weighing bottle is again found out accurately. The difference between two weights gives the weight of potassium dichromate. Then the substance is dissolved and made up to the mark with distilled water, shaken well to get uniform concentration and kept aside for use. The Normality of potassium dichromate is calculated as follows:

Normality of potassium dichromate=Weight of K2 Cr2O7

×

Gram equivalent weight

1000 v ml

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STEP-II ESTIMATION OF FERROUS IRON: The given ferrous solution is made up to the mark with distilled water and shaken well for uniform concentration. The burette is rinsed with distilled water and then it is filled with potassium dichromate and initial reading is noted. Pipette is cleaned and rinsed with given iron solution. A 20ml of iron solution is pipette out in a clean conical flask and 5ml of acid mixture and few drops of diphenylamine indicator are added. It is titrated by adding potassium dichromate from burette until violet blue is observed. This is the end point. Then final burette reading is noted and the experiment is repeated to get concurrent readings. Then estimate the amount of ferrous iron present in the whole of given solution. ESTIMATION OF FERROUS IRON: S. No

Burette Reading(ml)

Volume of Iron (V2) ml

Initial

Volume of K2Cr2O7 (V2) ml

Final

1.

2.

3.

NIVI = N2V2

Volume of K2Cr2O7 Normality of K2Cr2O7

(V1) (N1)

Volume of ferrous iron solution (V2) Normality of ferrous iron solution(N2)

= = = 20ml = ?

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ENGINEERING CHEMISTRY LAB – I B.TECH

Normality of Ferrous solution (N2)

=

N1V1 V2

= Amount of ferrous solution present in the given solution

N. =

Normality of Iron (N2) ×Eq.Wt of Iron × 100 1000

RESULT: The amount of Fe2+ present in the given solution is ________gm.

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VIVA QUESTIONS :

1) what is an oxidizing agent?

2) what is the indicator used in the experiment?

3) which principal is involved in this experiment?

4) what is the end point colour of this experiment?

5) Define Normality and Molarity?

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ENGINEERING CHEMISTRY LAB – I B.TECH Experiment : 13

DETERMINATION OF SURFACE TENSION AIM: To determine the effect of soap and detergent on the surface tension of water using a Stalagmometer. APPARATUS:

Stalagmometer Specific gravity bottle Rubber cork Analytical balance

CHEMICALS:

Soap solution. Methanol (or) Ethanol

PRINCIPLE: Surface tension is the characteristic properly of every liquid and it is due to intermolecular attraction among molecules of liquid. A molecule in the interior part of the liquid is attracted by the surrounding molecules on the surface of the liquid are attracted only towards the interior i.e., sides and the bottom of a constant tension due to the downwards flow of the molecules in bulk, this tension at the surface is known as surface tension. It is defined as the force in dynes acting on a surface at right angles to any line of unit length. It is denoted by ”γ” (gamma) and its units are dynes/cm.

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Table given below lists the surface tension of several liquids at 20 0C. Liquid

Surface tension

Liquid

(dynes/cm)

Surface tension (dynes/cm)

Water

72.8

Ethylene glycol

47.7

Benzene

28.9

Glycerol

63.4

Toluene

28.4

Carbon tetrachloride

27.0

Acetone

23.7

Ethyl iodide

29.9

Methyl alcohol

22.6

Ethyl bromide

24.2

Ethyl alcohol

22.3

Nitrobenzene

41.8

PROCEDURE The determination of surface tension of a liquid involves the following 2 steps. 1.Determination of density of a liquid Density of a liquid is mass/unit volume. It is found by using specific gravity bottles. The specific gravity bottle is first washed with distilled water and finally with alcohol and dried. The weight of the empty specific gravity bottle is found by using analytical balance. Let it be W1gm.Then it is filled with the distilled water and its weight is accurately determined. Let it be W2 gm. Now the given unknown liquid is then filled in the specific gravity bottles and the weight us found. Let this weight be W3 gm the density of the unknown liquid (d2) is going to be calculated as:

Density of liquid (d2) =

Weight of liquid X Density of water Weight of water

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2.

DETERMINATION OF SURFACE TENSION OF A GIVEN LIQUID: The stalagmometer is first washed thoroughly with distilled water and finally with

little alcohol and dried. A clean rubber tube is attached to the upper end of the stalagmometer. A screw pitch cork is fixed on the rubber tube to regulate the flow of liquid by lifting the influx of air The stalagmometer is dipped in a beaker of water and suck the water till it rises above the upper mark (X). In the stalagmometer the water level is carefully brought to the mark (X). The pitch cork is opened in such a way that it allows the flow of 12-18 drops per minute. Now the stalagmometer is adjusted. Then water is sucked above the mark and the no of drops of water that flows from the mark (X) to the mark (Y) are noted. This is repeated 3 to 4 times with water. Now the stalagmometer is rinsed with the liquid whose surface tension is to be determined. The liquid is then filled in the stalagmometer as above and no. of drops that flow from the mark(X) to mark (Y) is noted. This is repeated 3 or 4 times with the liquid and the experimental results are noted in the tabular for

CALC ULATIONS: Calculation of density of the liquid Room temperature = 27oC Weight of empty specific gravity bottle (W1) = Weight of specific gravity bottle + water (W2) = Weight of specific gravity bottle + given liquid (W3) = Weight of water = (W2-W1) = Weight of given liquid = (W3-W1) = Density of water (d1) = 1 gm/cc

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Density of liquid (d2) =

Weight of liquid X Density of water Weight of water Number of drops Surface

Volume of S. No. liquid

1

Std.liquid

2

Test liquid-1

3

Test liquid-2

Exp.1

Exp. 2

Average

tension

(N)

(dynes/cm)

Therefore the surface tension of given liquid (γ2) is determined busing the following equation. n1d2 γ2 =

xγ1 n2d2

Where γ 1= Surface tension of water = 72.8 dynes/cm n1= no. of drops of water d1= density of water γ 2= surface tension of given liquid n2= no. of drops of given liquid d2= density of given liquid

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RESULT:

The surface tension of the given test liquid γ2 = …………… dyne/cm

VIVA QUESTIONS : 1) Define surface tension?

2) what are the units of surface tension?

3) what is the apparatus used to find out surface tension?

4) what are the forces that causes surface tension?

5) what is the value of surface tension of water?

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ENGINEERING CHEMISTRY LAB – I B.TECH

SPACE FOR OBSERVATIONS AND CALCULATIONS

MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.

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