Chem Lab Manual Sem-II

8/6/2019 Chem Lab Manual Sem-II

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GAJANAN MAHARAJ SHIKSHAN PRASARAK MANDALS

Sharadchandra Pawar College of Engineering,

Dumbarwadi.

Department of Applied Science(Chemistry)

Lab Manual

SUBJECT :Applied Chemistry

Semester-II

20 -20

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Sharadchandra Pawar College of Engineering,

Dumbarwadi.

Department of Applied Science(Chemistry)

Certificate

This is to certify that Mr./Ms.

Roll No...of Second Semester of F.Ehas completed the term work satisfactorily in Applied

Chemistry for the academic year 20 . . . to 20 . . . as prescried

in the curriculum.

Place: Exam Seat No. : . . . . . . .

Date :

Subject Teacher H.O.D. Principal

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IMPORTANCE OF THE SUBJECT

Chemistry is a basic science subject.Its study is essential to all

engineering courses.It provides knowledge of various elements

and their large number of compounds that are available in

nature or prepared synthetically.The study of their materials

gives knowledge of their proparties and related applications in

the engineering field and making them ecofriendly.

Due to advancement in the engineering and technological

processes the use of different materials cause hazardous

effects on environment and living organisms.The study of

these effects will bring awareness in the learner about ill

effects.The study will further lead to understand the

precaution and prevention measures to be taken in such

situations and step ahead in the preservatiov of the nature.

The study of this subject will develop thinking about cause and

effect of various materials during their engineering

applications.

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Sharadchandra Pawar College of Engineering,Dumbarwadi.

Applied Chemistry Lab Manual Sem (II)

List of Practicals

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Sharadchandra Pawar College of Engineering, Dumbarwadi.

EXPERIMENT No.:

________________________________________________________

Date of performance : Date of submission :

Title:Determination of hardness of water sample.

Aim: To determine temporary and permanent hardness of water sample by EDTA

method.

Chemicals: EDTA solution, standard hard water solution or standard calcium

chloride solution, buffer solution of pH=10, Eriochrome Black - T(EBT)

indicator solution.

Apparatus: Burette, pipette, conical flask etc

Theory: Total hardness of water is due to the presence of water soluble bicarbonates ,

Sulphates, Chlorides and nitrates of calcium and magnesium. Total hardness is the sum

of temporary hardness and permanent hardness. Temporary hardness is due to presence

of bicarbonates of calcium and magnesium which easily get decomposed when heat and

form carbonates. Thus temporary hardness is also known as carbonate hardness.

Permanent hardness is due to the presence of sulphates, chlorides and nitrates of calcium

and magnesium.

To determine total hardness of a given water sample EDTA used. The disodium

salts of ethylene diamine tetra acetic acid are widely used in the analysis of water only

because it is a good reagent to form stable metal complexes. The efficiency of complex

formation with EDTA is affected by variation of the pH of the solution and is favorable

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in basic solution. Therefore to get better results, alkaline buffer of ammonium hydroxide

+ ammonium chloride is used.

This is very accurate method based upon the fact that when EBT(i.e. blue dye) is

added to hard water is alkaline medium (at about 10 pH) it gives a wine red colored

unstable complex Ca++ and MG++ ions. EDTA takes Ca++ and Mg++ions from the

unstable Ca++ chrome black-T, Mg++ chrome black T complex and forms a colorless stable

complex with the regeneration of blue colored dye (EBT).Thus the colour of dye changes

from wine red of its original blue colour at the end point.The chemical reactions taking

place can be shown as follows.

1) Ca++ + EBT Ca- chrome black T

(Unstable wine red coloured complex)

i.e. Metal+Indicator Metal Indicator Complex

2) Ca++ chrome black T + EDTA Ca-EDTA + EBT

(Colourless (blue dye)complex)

i.e. Metal-Indicator+EDTA Metal-EDTA+Indicator

Complex Complex

Structure of Metal-EDTA complex:

NaOOCH2C CH2COONa

N-H2C-CH2-N

OOCH2C CH2COO

M

M = Ca++ , Zn++ etc.

Note-Wash the apparatus with distilled water only and as far as possible use dry

apparatus for the titration to get better results.

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Procedure:

Pipette out 50ml of the hard water sample in to 250ml conical flask. Add 2 ml of

the buffer solution and 3 drops of Eriochrome Black-T indicator. Titrate the solution with

standard EDTA solution from the burette untill the colour changes from wine red to sky

blue at the end point.Repeat the titration untill constant reading is obtained.Note the

constant burette reading as Xml,which corresponds to total hardness.

Measure out 250ml of hard water sample in 500ml beaker,boil gently for half an

hour,filter the solution in to a 250ml volumetric flask.Make the solution up to the mark

with de-ionized water and shake thoroughly. Pipette out 50ml of this solution in to 250ml

conical flask add 2ml of the buffer solution and 3 drops of Eriochrome Black-Tindicator. Titrate the solution with standard EDTA solution from the burette until colour

changes from wine red to sky blue at the end point. Repeat the titration until constant

reading is obtained. Note the constant burette reading as Y ml which corresponds to

permanent hardness.

Observation Table:

Part A:-

1) In burette - 0.01 M EDTA

2) Pipette - Hard Water sample(50)

3) Indicator - EBT

4) End Point - Wine red to sky blue

Part B:-

1) In burette - 0.01 M EDTA

2) Pipette - Hard Water sample(boiled)(50)

3) Indicator - EBT

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4) End Point - Wine red to sky blue

Readings :-

Sr. No.

Volume of

water

sample taken

Volume of 0.01 M EDTA solution run-down

Total hardness expt.

(X ml)

Permanent hardness expt.

(Y ml)

1. 50 ml

2. 50 ml

3. 50 ml

Calculation :

1 ml of EDTA = 1 mole of Ca2+

1 ml of 1 M EDTA = 100 mg of CaCO3

Total Hardness:-

XX ml of 0.01 M EDTA = 100 x ------------ x 0.01 = X mg of CaCO3

1

-------- ml of 0.01 M EDTA = 100 x --------- x 0.01 = -------------mg of CaCO31

Hence the hardness present in 50 ml of the water sample taken =--------mg of CaCO3

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The hardness present in 1000 ml of the water sample.

1000

= X x ---------- = ------------mg of CaCO3

50

1000

= --------- x ---------- = ------------mg of CaCO350

The total hardness =------------mg/l =------------ppm

Permanent Hardness :-

Y

Y ml of 0.01 M EDTA = 100 x ------------ x 0.01 = Y mg of CaCO31

-------- ml of 0.01 M EDTA = 100 x --------- x 0.01 = -------------mg of CaCO31

Hence the hardness present in 50 ml of the water sample taken =--------mg of CaCO3

The hardness present in 1000 ml of the water sample.

1000= Y x ---------- = ------------mg of CaCO3

50

1000

= --------- x ---------- = ------------mg of CaCO350

Permanent hardness =------------mg/l =------------ppm

Temporary Hardness :-

Temporary Hardness = Total Hardness - Permanent Hardness

= .ppm -.ppm

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= ppm

Result:

1) Temporary Hardness of water sample=.ppm

2) Permanent Hardness of water sample=.ppm

Conclusion:

Precaution:

1. Avoid the contact of acid / harmful chemical with skin.

2. If acid comes in contact with the skin then wash it with plenty of water .

3. Handle the glass wares and apparatus very carefully.

4. Keep the working table always clean and neat.

5. Close the water taps and gas taps immediately after use.

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Sign. of Teacher



EXPERIMENT No.:

________________________________________________________

Date of performance : Date of submission :_________________________________________________________________

Title:Mohrs method

Aim: Determination of Chloride content in given sample of water by Mohrs Method.

Chemicals: 0.01N AgNo3, 5%K2CrO4, Sample of water etc.

Apparatus: Burette, Pipette, Conical flask, Beakers, etc.

Theory: Water contains Chloride in the form of dissolved salts such as CaCl2, MgCl2,NaCl, etc. These salts ionize and produce Chloride ion in water. Water which contains

Chloride is not useful for drinking, industrial or boiler purpose and also cause corrosion

in boiler because at higher temperature and pressure, CaCl2, MgCl2 react with water and

form HCL, hence water becomes acidic causes corrosion of boiler plate.

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PRINCIPLE: In this titration, sample of water is titrated with AgNO3 solution, using

K2CrO4 as an indicator. Silver ions react with Chloride as well as well as Chromate ions

to form silver salts.

AgNO3 Ag+ + NO3-

Ag+ + NO3- + Cl- AgCl + NO3

-

The solubility product of AgCl if greater at concentration of Ag+ ions than that of Silver

chromate (Ag2CrO4 ). Hence AgCl is precipitated first. When all th Cl+ ions are removed

I nth form of AgCl, then the extra drop of AgNO3 reacts with indicator (K2CrO4) forming

brick red precipitate.

Procedure:

1.Wash apparatus and rinse burette wih AgNO3.

2.Fill burette with 0.01N AgNO3 solution.

3. Rinse the pipette with given water sample and then pipette out 10ml of given water

sample in flask.

4. Add 2 drop of K2CrO4 indicator in titration flask. Flask solution turns to yellow.

5. Then add AgNO3 from Burette, slowly and drop wise, with constant shaking, till

permanent brick red precipitation is obtained.

6. Note the reading, repeat titration to get two constant readings.

Observation Table:1.Solution in Burette: 0.01N AgNO3

2.Solution by pipette : Water Sample(10 ml)

3.Indicator : K2CrO4

4.End Point : Yellow to brick red ppt.

Table for Constant Burette Reading:

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BuretteLevels

Burette Readings in ml Constant BuretteReadings (X ml)

Pilot 1 2 3

Initial

Final

Difference

Volume of AgNO3 (X ml)=ml

Calculation:

1 N 1000ml AgNO3 = 35.5 g Cl-

0.01 N 1ml AgNO3 = (0.355 g of Cl-)

(But X ml of 0.01 N AgNO3 = 10ml of H2O Sample)

10ml of H2O Sample = X x 0.355 mg of Cl-

x 0.355x0.010.01 N X ml AgNO3 =

0.01

= 0.355 x .

1000 x 0.355 x X

1000ml of H2O Sample =10

1000 x 0.355 x=

10

1000ml of H2O Sample contains = ppm of Cl-

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Result :

The Sample of water contains =ppm of Cl-

Conclusion:

Precaution:

1. Avoid the contact of acid / harmful chemicals with skin.

2. If acid comes in contact with the skin then wash it with plenty of water .



5. Close the water taps immediately after use.

Sign. of Teacher

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Sharadchandra Pawar College of Engineering, Dumbarwadi.__________________________________________________________________

EXPERIMENT No.Date of performance: Date of submission :

Title:Alkalinity of water.

Aim: To determine the alkalinity of given sample of water.

Theory:

1) Alkalinity of water

Alkalinity of water is due to presence of OH- ,CO3-- and HCO3

- ions in water.

Total alkalinity of water is the sum of total of OH- ,CO3-- and HCO3

- in ppm.

2) Alkaline ions

Waterbecome alkaline due to HCO3- ions only,CO3

-- ions only ,OH- ions only,

OH- and CO3-- ions together, CO3

- - and HCO3- ions together

Presence of OH- , and HCO3- ions together is ruled out because they react and produce

CO3- - ions.

OH- + HCO3- CO3

- - + H2O

e.g. NaOH + NaHCO3 Na2CO3 + H2O

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This reaction shows that OH- and HCO3- are converted into CO3

- - indicating that the

alkalinity is due to CO3- - only.

3) Neutralization Titration

H+ ions of hydrochloric acid neutralize OH- , CO3-- and HCO3- ions present in

given sample of water.

4) Suitability of alkaline water

When alkaline water is used in boiler to generate steam or when it comes in

contact with metallic structures creates problems like corrosion.

Principle and Chemical reaction:

The alkalinity of water is due to presence of either OH- ions, or CO3-- ions or

HCO3- ions or OH-, CO3

- - together or CO3-- and HCO3

- together.

The alkalinity due to these ions can be separated and estimated by titration against

the standard acid using phenolphthalein and methyl orange as an indicator. The

determination is based on the following reactions.

OH- + H+ H2O

CO3- - + H+ HCO3

-

HCO3- + H+ H2O + CO2

The volume of standard acid up to phenolphthalein end point [p] marks the completion of

reaction (i) and (ii) while the volume of acid run down after [p] corresponds to the

completion of equation (iii). The total volume of acid used from the beginning of the

experiments, i.e. [M] corresponds to total alkalinity.

Alkalinity is expressed in ppm or mg/lit.

Sr.No. Relation between

[P] and [M]

Hydroxyl

[OH]-

Carbonate

[CO3]--

Bicarbonate

[HCO3]-

1 [P] = O Nil Nil [M]

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2 [P] = [M] [P] or [M] Nil Nil

3 [P] = [M] Nil 2[P] or [M] Nil

4 [P] > [M] 2[P] - [M] 2[M] [P] Nil

5 [P] < [M] Nil 2[P] [M] 2[P]

P = Reading for phenolphthalein end point.

M = Reading for methyl orange end point

Chemicals:

HCl solution (O.5 N), alkaline water sample, phenolphthalein indicator and methyl

orange indicator.

Apparatus:Glassware:-

Burette (25ml), Pipette (25ml), conical flask (250 ml), 2 Beakers (200 ml) each.

Procedure:

1. Wash the burette and pipette with water.

2. Rinse and fill the burette with o.5 N HCl.

3. Remove air bubble if present in burette and adjust zero level correctly.

4. Rinse the pipette with given water sample.

5. Take 50 ml of sample water in conical flask with the help of pipette and add to it

2 -3 drops of phenolphthalein indicator.

6. Titrate the content of conical flask against 0.5 N HCl until pink colour just

disappears. Note the reading as phenolphthalein end point [P].

7. Then, add 2 to 3 drops of methyl orange indicator to the same solution of conical

flask.

8. Continue the titration till a sharp colour change from yellow to orange occurs.

9. Note the total volume of 0.5 N HCl from the beginning of the titration as methyl

orange end point [M].

Observation :

For Phenolphthalein End Point

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1. In burette: 0.5 N HCl solution

2. In flask : 50 ml water sample

3. Indicator: Phenolphthalein

4. End Point: Pink to colourless

For Methyl Orange End Point

1. In burette: 0.5 N HCl solution

2. In flask : same solution from phenolphthalein end point.

3. Indicator: Methyl Orange

4. End Point: Yellow to orange

Observation table for alkalinity of water

Sr. No. Volume of water

sample

Volume of 0.5N HCl for

phenolphthalein EndPoint[P]

Volume of 0.5N HCl for

Methyl orange EndPoint [M]

1 50 ml X=.. ml Y=..ml

2 50 ml X=.. ml Y=..ml

Now [P] = Xml =mland

[M] = Yml =ml

Calculation :

Step I:

Identification of ions (with the help of standard table)

If [P] < [M] then the sample contains only CO3-- andHCO3

-but it does not contains

OH-.

Step II:

Determination of volume of 0.5N HCl consumed by each alkaline, ion during titration.

According to given standard table 2[P] corresponds to CO3-- and[M] 2[P] corresponds

to HCO3- .

Therefore,

Volume of 0.5N HCl equivalent to CO3-- = 2P = 2 (X) ml

=

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= (A) ml.

Volume of 0.5N HCl equivalent to HCO3- = [M]-2[P]

= (Y)-2(X)

= (.. ) 2 () ml

= (B) ml.

Step III: Calculation for alkalinity

1. ForCO3-- ions

HCl = Water

N1V1 = N2V2

0.5 A = N2 50 ml

N2 = 0.5 A

--------------

50

0.5

N2 = --------------

50

N2 = .N

Strength in gm/liter = N2 Eq.wt.

= N2 30 (Equivalent Weight of CO3-- is 30)

=. 30

=gm/lit.

Alkalinity due to carbonate (CO3--) =Strength (gm/lit) 1000

=.. 1000

=..ppm or mg/lit.

2. For HCO3- ions

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Hydrochloric Acid (HCl) = Water (H2O)

N1V1 = N2V2

0.5 B = N2 50 ml

N2 = 0.5B

50N2 = 0.5.

50

N2 =..N

Strength (gm/lit ) = N2 Eq.wt

= N2 61 (Equivalent Weight of HCO3-

is 61)

=. 61

=gm/lit.

Alkalinity due to bicarbonate (HCO3-) = Strength (gm/lit) 1000

= 1000

=.ppm or mg/lit.

Result:

1. Alkalinity with respect to CO3-- =ppm

2. Alkalinity with respect to HCO3- =ppm

3. Total alkalinity [CO3- - + HCO3-] =ppm

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__________________________________________________________________

EXPERIMENT No.Date of performance: Date of submission :

Title:Study ofcorrosion of metals.

Aim: To study the corrosion of metals in medium of different pH.

Theory:

Any process of deterioration or destruction and consequent loss of solid material

through an unwanted chemical or electrochemical attack by its environment starting at its

surface is called corrosion. Thus corrosion is process reverse of extraction of metal. The

process of corrosion is slow and occurs only at surface of metals. The losses incurred due

to corrosion are enormous.Eg.Distruction of machine, equipment and different types of

metallic products. In general, life of equipment plants etc. is very much reduced due to

the corrosion. It is very difficult to predict exactly the losses incurred due to corrosion.

Corrosion of metals occurs by the attack of surrounding environment / medium on

the surface of metals. This attack occurs in two ways

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I) Direct chemical corrosion or dry corrosion

II) Electrochemical corrosion or wet corrosion or immersed corrosion.

Chemicals:

Acid solutions of different concentrations.

Apparatus:

Measuring cylinder, Beaker, Electronic Balance, Metal strips, Filter Papers, Forceps.

Procedure:

Weigh accurately the given metal strips and dip it in 50% and 20% acid solutions.Keep it in acid solutions for about half an hour. After half an hour remove the metal strips

and dry it completely by using filter paper.

Again weigh the metal strip. Calculate the loss in weight of metal strips due to

corrosion.

Observation :

1) Weight of metal strips no.-1 before corrosion = gm.

2) Weight of metal strips no.-2 before corrosion = gm.

3) Weight of metal strips no.-1 after corrosion = gm.

4) Weight of metal strips no.-2 after corrosion = gm.

Calculation:

1) Loss in weight of metal strips no.-1 due to corrosion = gm.

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2) Loss in weight of metal strips no.-2 due to corrosion = gm.

A) % weight loss by metal in 50% acid solution

= Loss in wt. of metal 100Wt. of metal strip before corrosion

B) % weight loss by metal in 20% acid solution

= Loss in wt. of metal 100

Wt. of metal strip before corrosion

Result :

1) % weight loss by metal in 50% acid solution =

2) % weight loss by metal in 20% acid solution=

Conclusion:

Precaution:

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1. Weigh accurately metal strips.

2. Avoid the contact of acid / harmful chemicals with skin.

3. If acid comes in contact with the skin then wash it with plenty of water.



6. Close the water taps immediately after use.

Sign. of Teacher:

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Chem Lab Manual Sem-II

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