Lab Manual - dbit.ac.in of alkalinity in the given water sample. 2. Determination of temporary and...

Lab Manual

B.Tech 1st Year

Chemistry Lab

Dev Bhoomi Institute of Technology Dehradun www.dbit.ac.in

Affiliated to

Uttrakhand Technical University, Dehradun www.uktech.in

http://www.dbit.ac.in/

http://www.uktech.in/

LIST OF EXPERIMENTS

1. Determination of alkalinity in the given water sample.

2. Determination of temporary and permanent hardness in given water sample

by using EDTA as standard solution.

3. Determination of amount of dissolved oxygen in given water sample.

4. Determination of available chlorine in bleaching powder.

5. Preparation of iodoform from acetone.

6. Preparation of sodium cobaltinitrite salt.

7. Preparation of phenol formaldehyde resin. (Bakelite)

8. Separation of metal ions by paper chromatography.

9. Determination of heat of neutralization of hydrochloric acid (HCl) and

sodium hydroxide (NaOH).

10. Determination of chloride content in the given water sample by using

Mohr’s method or argentometric method.

AIM: To determine the alkalinity of a given water sample.

CHEMICALS USED: N/10 HCl, phenolphthalein and methyl orange indicators, water sample.

Guidelines for preparation of solutions:

1. Methy1 orange indicator: Dissolve 20mg of methyl orange in 100 ml of hot water. Allow to cool and

filter if necessary.

2. Phenolphthalein indicator: Dissolve 1g of Phenolphthalein in 100ml of 95% alcohol.

3. 0.1N HCl: Measure 10mL of Conc HCl (12N) in to one liter measuring flask. Pour 30mL distilled

water and shake. Add more distilled water to make the solution up to the mark. Stopper the flask and

shake vigorously.

APPARATUS REQUIRED: Burette, pipette, conical flask, beakers, funnel, and dropper.

PRINCIPLE: Alkalinity of water means the total content of those substances in it which causes an

increased OH‐ ion concentration up on dissociation or due to hydrolysis. The alkalinity of water is

attributed to the presence of

(i) Caustic alkalinity (Due to OH‐ and CO3

2‐)

(ii) Temporary hardness (Due to HCO3‐)

Alkalinity is a measure of ability of water to neutralize the acids

Determination of alkalinity

OH‐, CO3

2‐ and HCO3

‐ can be estimated separately by titration against standard acid using

phenolphthalein and methyl orange as indicators

The determination is based on the following reactions

(i) OH- + H

+H2O P

(ii) Co32-

+ H+ HCO3

- M

(iii) HCO3- + H

+ H2O + CO2

The titration of water sample against a standard acid up to phenolphthalein end point (P) marks the

completion of reaction (i) and (ii) only. This amount of acid used thus corresponds to OH‐ plus one half of

the normal CO32‐

persent.

DEV BHOOMI INSTITUTE OF TECHNOLOGY

Department of Engineering chemistry

LABORATORY

MANUAL

PRACTICAL INSTRUCTION SHEET

EXPERIMENT NO. 1 ISSUE NO. : ISSUE DATE:

REV. NO. : REV. DATE :

PAGE:

LABORATORY NAME & CODE: Chemistry Lab (PCY-101/201) SEMESTER: 1st & 2

nd

On the other hand, titration of the water sample against a standard acid to methyl orange end point (M)

marks the completion of reaction (i), (ii) and (iii). Hence the total Amount of acid used represents the total

alkalinity. Thus,

P = OH-+ ½ CO3

2-

M = OH-+ CO3

2-+ HCO3

-

The table 1 below shows the type and amount of alkalinity in water

S.No Result of titration OH- CO3- HCO3-

1 P=0 nil nil M

2 P=1 P or M nil nil

3 P=1/2M(V1=V2) nil 2P nil

4 P>1/2M(V1>V2) 2P-M 2(M-P) nil

5 P<1/2M(V<V2) nil 2P M-2P

PROCEDURE: ‐

1. Pipette out 20 ml of water sample into a conical flask. Add 1‐2 drops of Phenolphthalein.

2. Rinse and fill the burette with N/10 HCl.

3. Titrate the water sample in conical flask with N/10 HCl till the pink color just disappear.

4. Note down the reading and repeat to get concordant readings.

5. Again take 20 ml of water sample in conical flask and add methyl orange indicator to it.

6. Titrate the water sample in conical flask with N/10 HCl till the yellow orange color changes to orange

red.

7. Note down the reading and repeat to get concordant readings.

OBSERVATION TABLE:

a) Using Phenolphthalein (Table 2)

Normality of the acid used: N/10

S.No Volume of the solution taken

in the titration flask (in mL)

Burette reading Volume of the Titrant

used

Say V1ml =P Initial

reading Final reading

1 20

2 20

3 20

b) Using Methyl orange (Table 3)

Normality of the acid used: N/10

S.No

Volume of the solution

taken in the titration flask

(in mL)

Burette reading Volume of the

titrant used

Say V2mL=M Initial reading Final reading

1

2

3

RESULT: ‐ Alkalinity of the given water sample is______ ppm.

PRECAUTIONS: ‐

1. All glass wares should be properly washed.

2. Before use, rinse burette and pipette properly

3. Find the correct end point

GENERAL CALCULATIONS:

Alkalinity due to OH

Acid Water

N1V1 = N2V2 Where, V2= 20mL

𝟏

𝟏𝟎 𝟐 × 𝑽𝟏 − 𝑽𝟐 = 𝑵𝟐 × 𝟐𝟎

𝑵𝟐 = 𝟐 𝑽𝟏 −𝑽𝟐

𝟏𝟎×𝟐𝟎

Strength = N2 X Eq. Wt of CaCO3 g/lt =2 𝑉! −𝑉2

200× 50 × 103 mg/l or ppm

In the same way the strength of other alkalinity causing ions can be calculated

VIVA QUESTIONS:

1. Write structural formula of phenolphthalein. In what forms does it exist in acidic and alkaline form.

2. What is alkalinity of water? How is it estimated?

3. Why OH‐ and HCO

3‐ ions cannot exist together in water?

4. What are the draw backs of using highly alkaline water?

5. What are the disadvantages of scale formation?

6. Name the salts that create hardness in water.

AIM: Determination of Ca

2+and Mg

2+hardness of water sample by using EDTA solution.

CHEMICALS USED: 0.02N CaCO3 solution (standard hard water), Approx. 0.2N EDTA solution,

Eriochrome Black-T indicator (EBT), NH4OH- NH4Cl Buffer solution (pH~ 9-10), Water sample

(Unknown Hardness)

Guidelines for preparation of solutions:

Determination of Ca2+and Mg2+hardness of water sample by using EDTA solution.

1. Preparation of 0.1M (0.2N) EDTA solution

Dissolve 37.225g of pure and dry EDTA disodium salt (M.W. 372.25) in distilled water and make up the

volume to 1 litre.

2. Preparation of 0.01M (0.02N) CaCO3 solution (SHW) Weigh 1g of CaCO3 (M.W. 100) and transfer it to one litre measuring flask. Add dil. HCl drop by drop

till there is effervescence and the salt is completely dissolved. Add more distilled water to make up the

volume to one litre

3. Preparation of Ammonia buffer solution

Dissolve 70g of NH4Cl in 570mL of Conc. NH3 and dilute it to one litre with distilled water.

4. Preparation of EBT solution Dissolve 0.5g of indicator in 100mL methanol.

APPARATUS REQUIRED: Conical flask, burette, pipette, measuring cylinder, funnel, filter paper,

dropper, beaker (100mL)

PRINCIPLE: Hardness is defined as the soap consuming capacity of water. Bicarbonates of Ca and Mg

cause temporary hardness while chlorides, sulphates and carbonates of Ca and Mg contribute for

permanent hardness. Temporary hardness can be removed by simple physical process like boiling but for

removing the permanent hardness chemical treatment is required. Hard water may have a number of

disadvantages thus softening of water is very essential for industrial and domestic uses.

Boiling

2HCO3- CO2

+ H2O + O2

Ca and Mg ions form complexes easily with EDTA (Ethylene diaminetetraacetic acid), which is the basic

principle of this experiment. The Ca+ and Mg

+2 ions present in water are titrated with EDTA solution

using Eriochrome Black T (EBT) as indicator. Since the action of the indicator and the formation of the

metal EDTA complex is governed by pH. Hence pH of the solution is kept nearly constant by adding a

basic buffer NH4OH- NH4Cl Buffer solution (pH~ 9-10).



LABORATORY

MANUAL




PAGE:


nd

Estimation of hardness by EDTA is based on following principle:

1. The indicator EBT which is blue in color forms less stable wine-red colored complex with Ca2+

and

Mg2+

Ions in hard water at pH of (9-10).

2. As this solution is titrated against EDTA the free Ca2+

band Mg2+

ions in water formed stable metal-ion

EDTA complex (colorless)

3. On addition of more amount of EDTA solution the metal ions from less stable {M-EBT} complex set

free and form {M-EDTA} which is more stable. The solution turns blue due to free EBT.

pH~9-10

M2+

(M=Ca, Mg) + EB {M-EBT} (less stable Wine - red complex)

pH~9-10

M2+

+ EDTA {M-EDTA} (More stable colorless complex)

pH~9-10

{M-EBT} + EDTA {M-EDTA} + EBT (Blue color)

PROCEDURE: -

1. Prepare standard hard water by dissolving 1 g of CaCO3 in one litre of distilled water.

2. Take 25 mL of standard hard water in to a conical flask.

3. Add 2 mL of buffer solution and 2-3 drops of EBT indicator to it.

4. The color of the solution becomes wine red, titrate it against standard EDTA solution till the color

changes from wine red to blue at the end point.

5. Repeat the titration to get concordant values

6. Now take unknown hard water sample and repeat the same process.

OBSERVATION TABLE: -

(i) Standardization of EDTA solution using standard hard water

Volume of standard hard water (SHW) taken for each titration = 25 ml.

Volume of 0.2N EDTA solution used = V1 ml.

S.No

Volume of the

solution taken in

the titration flask

Burette reading

Volume of the

titrant used Initial reading Final reading

1

2

3

SHW EDTA

𝑵𝟏𝑽𝟏 = 𝑵𝟐𝑽𝟐

𝟎. 𝟎𝟐 × 𝟐𝟓 = 𝑵𝟐𝑽𝟏 𝑵𝟐 =𝟎.𝟎𝟐×𝟐𝟓

𝑽𝟏

(ii) Determination of hardness of unknown hard water sample

Volume of unknown hard water (HW) taken for each titration = 25 ml.

Volume of 0.2N EDTA solution used = V2 ml.

S.No

Volume of the

solution taken in

the titration flask

Burette reading

Volume of the

titrant used Initial reading Final reading

1

2

3

HW EDTA

𝑵𝟏𝑽𝟐 = 𝑵𝟐𝑽𝟐

Where, 𝑵𝟐 =𝟎.𝟎𝟐×𝟐𝟓

𝑽𝟏

𝑵𝟏 × 𝟐𝟓 = 𝑵𝟐𝑽𝟐 𝑵𝟏 =𝟎.𝟎𝟐×𝟐𝟓

𝑽𝟏×

𝑽𝟐

𝟐𝟓 = 𝟎. 𝟎𝟐 ×

𝑽𝟐

𝑽𝟏

Strength = Normality X Eq.wt X 103 ppm (mg/lt)

RESULT: Ca+2

and Mg+2

Hardness of given water sample is ____ ppm.

PRECAUTIONS:

1. Use distilled water for washing and rinsing of glass apparatus.

2. Prepare EDTA solution in double distilled water.

3. Add same amount of indicator each time.

4. Maintain pH from 9-10 during the titration by adding buffer.

5. Correctly observe the end point.

Safety Instructions

1. NH3 (Conc.): It can cause serious burns to skin and eyes and its fumes irritate the eyes and respiratory

system. Wear gloves, use the solution under hood and do not breathe its gas.

2. HCl : It is poisonous and corrosive. Contact or inhalation can cause severe damage to the eyes, skin and

respiratory tract. Wear gloves and dispense under a hood, avoid contact and do not breathe the gas.

3. CH3OH: It is flammable and harmful if ingested, inhaled or absorbed through skin. Avoid contact with

the liquid and do not breathe its gas.

VIVA QUESTIONS:

1. Define hardness of water?

2. What do you mean by temporary and permanent hardness?

3. What is the principle of EDTA titration?

4. How will you find out permanent hardness of water?

5. Why is it necessary to add buffer sol. to the water sample in determination of hardness by EDTA

method?

6. Full form of EDTA?

AIM: To determine the amount of dissolved oxygen in the given water sample by Winkler’s

method, a standard solution of K2Cr2O7 of strength 0.01N is given.

CHEMICALS USED: Sodium thiosulphate, (N/40), MnSO4 KI, Starch, Conc. Sulphuric acid.

APPARATUS REQUIRED: Burette, pipette, conical flask, beaker etc.

PRINCIPLE: Determination of dissolved oxygen (D.O.) is important for Industrial purposes.Dissolved

oxygen is usually determined by Winkler’s method. It is based on the fact that dissolved oxygen oxidized

potassium iodide (KI) to iodine. The liberated iodine is titrated against standard sodium thiosulphate

solution using starch indicator. Since dissolved oxygen in water is in molecular state. It as such cannot

oxidize KI. Hence Manganese Hydroxide is used as an oxygen carrier to bring about the reaction between

KI and Oxygen. Manganese hydroxide, in turn, is obtained by the action of NaOH on MnSO4.

MnSO4 + 2NaOH → Mn (OH)2 + Na2 SO4

2 Mn(OH )2 +O2 → 2 MnO(OH )2

MnO(OH )2 + H 2 SO4 → MnSO4 + 2H 2O +[O]

2KI + H 2 SO4 +[O] → K2 SO4 + H 2O + I2

2Na2 S2O3 + I 2 → Na2 S4O6 + 2NaI

Starch + I2 → Blue colored complex.

The Nitrites present in water interfere with the titration as these can also liberate I2 from KI.

2 HNO2 + H2SO4 + 2 KI → 2 NO + K2SO4 + 2 H2O + I2.

PROCEDURE:

TITRATION –

1. STANDARDIZATION OF SODIUM THIOSULPHATE

a. The burette is washed and rinsed with sodium thiosulphate solution.

b. Then the burette is filled with given sodium thiosulphate solution.



LABORATORY

MANUAL




PAGE:


nd

c. 20ml of 0.01N k2Cr2O7 solution is pipette out into a clean conical flask.

d. To this 5ml of sulphuric acid and 15 ml of 5% KI are added, this is titrated against sodium

thiosulphate solution.

e. When the solutions become straw yellow color, starch indicator is added and then the titration

is continued.

f. The end is disappearance of blue color and appearance of light yellow color, the titration is

repeated to get the concordant value.

2. ESTIMATION OF DISSOLVED OXYGEN

a. 100-150ml of water sample is taken in flask, 2ml of MgSo4 and 2ml of alkali-iodide are

added. The stopper is replaced and the flask is inverted and shacked several times for the

rough mixing of reagents.

b. The flask is left aside for some times. When half of the precipitate is Settle down, the stopper

is removed and 2ml of conc. H2SO4 is added.

c. The stopper is replacer and kept inverted for several times for complete dissolution of the

precipitate.

d. 100ml of this solution is pipette out and titrated against standard sodium thiosulphate

solution.

e. When the solution becomes light yellow starch indicator is added .The titration is repeated

until the blue color is disappears. From the titrant value the strength of dissolved oxygen is

calculated and hence the amount of dissolved oxygen in the water sample is calculated.

RESULT:

Amount of dissolved oxygen in tap water is =...........................mgs/lit.

Step 1- OBSERVATION TABLE: STANDARDIZATION OF SODIUM THIOSULPHATE

TITRATION – 1

POTASSIUM DICHROMATE VS SODIUM THIOSULPHATE

S.No

Volume of K2Cr2O7 (in

mL)

Burette reading Volume of

sodium

thiosulphate (in

mL)

Initial reading Final reading

1

2

4

Calculation:

Volume of potassium dichromate v1 = 20ml

Strength of potassium dichromate N1 = 0.01N

Volume of sodium thiosulphate V2 =................ml

Strength of sodium thiosulphate N2=......

𝑽𝟏𝑵𝟏 = 𝑽𝟐𝑵𝟐

𝑵𝟐 =𝑽𝟏𝑵𝟏

𝑽𝟐

𝑵𝟐 =𝟐𝟎×𝟎.𝟎𝟏

𝑽𝟐

Strength of sodium thiosulphate

=.................

=...................

Step 2 - Estimation of dissolved oxygen

Titration – 2 Water sample Vs sodium thiosulphate

S,No

Volume of the

solution taken in

titration flask

Burette reading Volume of titrant

used (in mL) Initial reading Final reading

1 100

2 100

3 100

Calculation:

Volume of sodium thiosulphate V1 = ...................... ml

Strength of sodium thiosulphate N1 =..................... N

Volume of water sample V2 = 100ml

Strength of dissolved water sample N2 = -----------------?

𝑽𝟏𝑵𝟏 = 𝑽𝟐𝑵𝟐

𝑵𝟐 =𝑽𝟏𝑵𝟏

𝑽𝟐= − − − − −𝒎𝒍 × − − − −

𝑵

𝟏𝟎𝟎

Amount of dissolved oxygen = normality × eq.wt.of o2×1000mgs

In one L of tap water = − − − − −𝑁 × 𝑔 × 1000

RESULT: Dissolve oxygen in the sol. is____gm/l.

PRECAUTIONS:

1. Accurate reading should be taken.

2. The indicator to be added should be adjusted in a manner to give accurate end point.

3. be careful while sucking KI sol. as it is highly poisonous.

VIVA QUESTIONS:

1. Why MnSO4 Is used in above determination?

2. What is the significance of above determination?

3. What is the end point if we use methyl orange as indicator in titration?

AIM: To determine the percentage of available chlorine in a given sample of bleaching powder.

(iodometrically).

REQUIREMENTS:

1. Chemical or reagents: given sample of bleaching powder, KI, dil. Acetic acid (5%), N/10

sodium thiosulphate solution. Distilled water.

2. Indicator: freshly prepared starch solution.

3. Apparatus: burette, pipette, measuring flask, conical flask, funnel, pastle, and mortar.

PRINCIPLE:

Commercially available bleaching powder is a mixture of Calcium hydrochloride (CaOCl2), the active

constituent of bleaching powder is hypochlorite (CaOCl2), which produces hypochlorous acid (HOCl),

responsible for disinfectant properties.

The amount of chlorine liberated by the action of acetic acid on hypochlorites is termed as the available

chlorine.

It is expressed as the percentage by weight of bleaching powder. In principle, the determination of the

available chlorine is done by treating the hypochlorite solution with an excess of a solution of KI. The

liberated chlorine reacts with KI solution giving free iodine. This free iodine is then titrated with the

standard sodium thiosulphate solution.

CaOCl2 + 2CH3COOH → (CH3COO) 2Ca + H2O + Cl2

(Available chlorine)

2KI + Cl2 → 2KCL + I2

2Na2S2O3 + I2 → 2NaI + Na2S4O6

(Sodium tetrathionate)

Na2S2O3 = Cl = I

1 gm equiv 1 gm equiv 1 gm equiv



LABORATORY

MANUAL




PAGE:


nd

PROCEDURE:

1. Weight accurately 3gm of sample of bleaching powder in a clean and pre-weighing tube.

2. Transfer the sample in a motor and make a paste with little distilled water.

3. Transfer the mixture to 250 ml measuring flask.

4. Rinse well the motor with distilled water.

5. Fill the flask with distilled water up to the mark and make until a homogeneous suspension is

obtained.

6. Pipette on 25ml of this solution into 250ml conical flask, add 10 ml of KI and 10 ml of dilute

acetic acid to this solution.

7. Titrated the liberated iodine immediately with standard N/10 freshly prepared starch solution. A

blue color immediately appears

8. Continue titration till the blue color disappears, note down the reading of burette and repeat the

titration until at least two concordant reading are obtained.

OBSERVATIONS:

Weight of the bleaching powder sample taken =------

Weight of weighing tube =-----------

Weight of weighing tube + bleaching powder = --------------gm

Weight of bleaching powder =-------------------

Normality of hypo solution (Na2S2O3) = N/10

OBSERVATION TABLE:

S.No

Volume of CaOCl2

solution (mL)

V1

Reading of burette in mL

Volume of sodium thiosulphate

solution used in mL

𝑽 − 𝑽𝟎 = 𝑽𝟐

Initial

reading V0

Final

reading V

1 25

2 25

3 25

CALCULATION:

𝑵𝟏𝑽𝟏 = 𝑵𝟐𝑽𝟐

(Bleaching powder) (Sodium thiosulphate)

Normality of available chlorine,

𝑵𝟏 =𝑵𝟐𝑽𝟐

𝑽𝟏

𝑵𝟏 =𝑵

𝟏𝟎×

𝑽𝟐

𝟐𝟓

Amount of chlorine per litre of solution = 𝑁1 ×35.5

250𝑚𝑙

% of available chlorine =𝑁1×35.5×250×100

1000 × 𝑦−𝑥

RESULT:

The % of available chlorine present in given sample of bleaching powder is ----------------------------%

PRECAUTIONS:

1. All the glass apparatus should be washed first with chromic acid and then thoroughly with

distilled water.

2. Sample of bleaching powder must not be kept open as it absorbs moisture from the atmosphere.

3. Rinse the burette with standard N/10 sodium thiosulphate solution.

4. The reagent must be prepared freshly.

5. The starch solution and acetic acid must be added in equal amount in each experiment.

6. Titration must be started immediately after the addition of acid because chlorine is liberated

instantly.

AIM: To prepare Iodoform in the laboratory from acetone.

REAGENTS REQUIRED: Beaker, stirrer.

APPARATUS: Acetone, NaOH, 5% Iodine solution.

PRINCIPLE:

CH3COCH3 + 4 NaOH + 3I2 CHI3 + CH3COONa + 3 H2O

(Acetone) (Iodoform)

PROCEDURE:

To a mixture of 10 ml acetone and a solution of 6 g NaOH in 40 ml water, add 5% of iodine solution with

constant stirring until saturation where a yellow ppt. of iodoform separate immediately. Allow the mixture

to stand at room temperature for 15 minutes. Then filter, wash with cold water and recrystallize from

methylate spirit where yellow glistening scales are obtained.

RESULT:

Weight of the filter paper = ------------gm.

Weight of filter paper + precipitate = --------------gm

Weight of precipitate = ----------------gm

The amount of Prepare Iodoform is ----------gm.



LABORATORY

MANUAL




PAGE:


nd

PRECAUTION:

1. The amount of chemicals must be taken accurately.

2. The beaker must be clean and dry before starting the experiment.

The precipitation of Iodoform should be complete.

VIVA QUESTIONS:

1. Chemical formula of iodoform?

2. Chemical reaction between acetone and iodine?

AIM: To prepare sodium cobaltinitrite, Na3 {Co (NO2)6}.

APPARATUS: beaker, conical flask, stirrer, funnel.

REAGENTS: sodium nitrite, 15gm cobalt nitrate, glacial acetic acid, 22 mL of alcohol.

THEORY:

Cobalt nitrate and sodium nitrite solutions are mixed together in cold and the mixture acidified with

glacial acetic acid solution. Cobalt nitrate is oxidized by HNO3 to cobaltic nitrate which at ones forms a

complex ion with nitrite ions.

PROCEDURE:

Take 12 gm of sodium nitrite and dissolve it in 15ml of warm water, cool to 50 C, Now add 5 gm of

cobalt nitrate, stirring until the salt is completely dissolved. Acetic acid (2.5 ml glacial acetic acid + 2.5

ml of water) is gradually added to this solution with constant stirring. Filter it to remove any impurities

(potassium cobaltinitrite). Add 20ml of alcohol in small precipitate to settle. Collect the yellow precipitate

on a filter plate and wash it with small quantities of alcohol until the washing become colorless. Spread

the yellow crystalline precipitate on filter paper and allow it to dry in air.

RESULT:

Weight of the filter paper = ------------gm.

Weight of filter paper + precipitate = --------------gm

Weight of precipitate = ----------------gm

The amount of Prepare sodium cobaltinitrite is ----------gm.



LABORATORY

MANUAL




PAGE:


nd

PRECAUTION:

1. The amount of chemicals must be taken accurately.

2. The beaker must be clean and dry before starting the experiment.

The precipitation of sodium cobaltinitrite should be complete.

VIVA QUESTIONS:

1. Chemical formula of sodium cobaltinitrite?

2. Chemical formula of glacial acetic acid?

3. Use of nitric acid?

AIM: To prepare phenol formaldehyde resin. (Bakelite)

APPARATUS REQUIRED:

Glass rod, beaker, funnel, measuring cylinder, roper and filter paper.

CHEMICAL REAGENTS: Glacial acetic acid, 40% formaldehyde solution, Phenol, conc. H2SO4.

PRINCIPLE:

Phenol formaldehyde resins (PFs) are condensation polymers and are obtained by condensing phenol with

formaldehyde in the presence of an acidic or alkaline catalyst. They were first prepared by Backeland, an

American Chemist who gave them the name as Bakelite. These are thermosetting polymers.

Thermosets: The polymers which on heating change irreversibly into hard rigid and infusible materials

are called thermosetting polymers. These polymers are usually prepared by heating relatively low

molecular mass, semi fluid polymers, which becomes infusible and form an insoluble hard mass on

heating. The hardening on heating is due to the formation of extensive cross-linking between different

polymeric chains.

Some important examples of thermosetting polymers are

Urea - Formaldehyde resin and Melamine - Formaldehyde resins.

Properties:

1. Phenol- formaldehyde resins having low degree of polymerization are soft. They possess

excellent adhesive properties and are usually used as bonding glue for laminated wooden planks

and in varnishes and lacquers.

2. Phenol- formaldehyde resins having high degree of polymerization are hard, rigid, scratch

resistant and infusible. They are resistant to non-oxidizing acids, salts and many organic solvents.

They can withstand very high temperatures. They act as excellent electrical insulators also.

Uses:

1. They are used for making molded articles such as radio and TV parts, combs, fountain pen

barrels, phonograph records etc.

2. They are used for making decorative laminates, wall coverings etc.



LABORATORY

MANUAL




PAGE:


nd

3. They are used for making electrical goods such as switches, plugs etc.

4. They are used for impregnating fabrics wood and paper.

5. They are used as bonding glue for laminated wooden planks and in varnishes and lacquers.

6. Sulphonated phenol-formaldehyde resins are use as ion-exchange resins.

Preparation:

PFs are prepared by reaction of phenol with formaldehyde in the presence of acidic or basic catalyst. The

process may be carried out as follows A mixture of phenol and formaldehyde are allowed to react in the

presence of a catalyst. The process involves formation of methylene bridges in ortho, Para or both ortho

and Para positions. Results first in the formation of linear polymer and then in to cross-linked polymer

called phenol-formaldehyde resin or Bakelite.

PROCEDURE:

1. Place 5ml of glacial acetic acid and 2.5ml of 40% formaldehyde solution in a 500ml beaker and add

2 grams of phenol.

2. Add few ml of conc. Sulphuric acid into the mixture carefully. Within 5 min. a large mass of plastic

is formed.

3. The residue obtained is washed several times with distilled water, and filtered product is dried and

yield is calculated.

RESULT: The weight of the phenol formaldehyde resin is ___ g.

PRECAUTIONS: 1.The reaction is sometimes vigorous and it is better to be a few feet away from

the beaker while adding the H2SO4 and until the reaction is complete.

2. The experiment should be preferably carried out in fume cupboard.

VIVA QUESTIONS:

1. Phenol formaldehyde is also called as?

2. What do you understand by resin?

3. Give main uses of the phenol formaldehyde resin.

4. What type of co-polymer is phenol formaldehyde resin?

5. Briefly describe the properties of Phenolic resins.

6. Write chemical equations for preparation of phenolic resins.

Safety Instructions

1. Phenol: Most phenols are harmful if inhaled, ingested or absorbed through skin. They

cause severe irritation or damage to skin and eyes. Some phenols are suspected carcinogen, should

not inhale its dust or vapor, wear gloves and avoid contact.

2. H2SO4: It is poisonous and corrosive. Contact or inhalation can cause severe damage to the eyes,

skin and respiratory tract. Wear gloves and dispense under a hood, avoid contact and do not breathe

the gas.

AIM: separation of metal ions by paper chromatography.

APPARATUS: Boiling tubes or chromatographic jar, Measuring cylinder. Pipette, Spotting capillaries,

test tubes, whatsman No. 1 Filter paper.

REAGENT: Potassium chromate, Lead nitrate, Silver nitrate, Mercurous nitrate, Nitric acid.

PRINCIPLE:

In paper chromatography of cations, the principles of partition, adsorption and ion exchange may be

exploited, out of these the most important is partition the involves the distribution of a solute between a

mobile liquid phase and a gel (a kind of water cellulose complex) as the stationary phase. The different

components of the sample are distributed across the paper depending on their partition coefficients.

Attempts have been to employ paper chromatography for the systematic qualitative analysis of metal

cations. However, it is not possible to separate all the cations simultaneously and therefore; the separation

of a group of cations can be handled. Most frequent of these is the preliminary separation into the current

analytical groups, each of which is then subjected to a separate chromatographic analysis.

PROCEDURE:

Proceed according to the following steps:

1. Preparation of solutions-

Prepare 1 cm3 aqueous solutions of –

a). lead nitrate b). Silver nitrate

c). mercurous nitrate

by dissolving about l g crystals in a small test tube. Add few drops of HN03 to prevent

hydrolysis. For the preparation of mixture of cations, add few drops of each cation solutions in a

test tube. Developer: Distilled water.

2. Cut what man No. 1 filter paper strips of the required size: 15 cm x 2 cm to fit usual boiling tube

or 15 cm x 3 cm for chromatographic jar.

3. On each strip draw a line at about 1 cm of the one and put a dot in the centre of line. This and will

be the bottom of the strip and development will take place from this end.

4. Apply the solutions of pb+2

, Ag+1

and Hg+2

separately on 3 strips with the help of a fine capillary.

Use a fresh capillary for each solution..

5. On the 4th paper strip apply the mixture of the three cations.



LABORATORY

MANUAL




PAGE:


nd

6. Apply the unknown solution (e,g. containing any one or two on the 5th paper strip, Place 5 dry

boiling tubes vertically in a stand.

7. Add distilled water, with the help of a pipette, to each of the boiling tubes so that the height of

the developer (distilled water) in each of the boiling tubes is less than 1 cm. The sides of the

oiling tube must be dry as far as possible.

8. Suspend the spotted and dried paper strips in the respective boiling tubes containing distilled

water with the upper end and the lower end touching the developer. And see that this is done

gently and the strip is vertical. The spot should always be above the developer level.

9. Allow the developer to rise along the paper and wait till the developer (solvent front) reaches near

the upper and of the paper.

10. Remove the paper from the boiling tube and mark the solvent front with the help of a pencil.

11. Get the dried to evaporate the developer.

12. Take potassium chromate solution in a Petri dish (or a watch glass) and dip the dried paper in the

detector.

13. Encircle the colored zones with pencil and mark the centre of the zone.

14. Calculate the Rf,

15. Compare the Rf values of individual cations with that of their Rf values in known mixture and in

unknown.

16. Identify the cations present in the unknown on the basis of Rf values.

OBSERVATION:

Observe the colored spots of different cations.

pb+2

will appear as yellow

Ag+1

as orange-red and

Hg+2

as orange zones.

Measure the distance of the centre of each solute zone from the point of application call this distance

ds.

Measure the distance between the solvent front and the starting line and call this distance as dm.

Calculate the Rf value of each solute by the relation:

Rf = Distance travelled by the centre of solute zone = ds

Distance travelled by the solvent front dm

Observation Table:

Sample ds dm Rf = ds/dm

Ag+1

Pb+2

Hg+2

mixture

Unknown sample

RESULT: Metal ions present in the unknown sample are:

1. ------------------------

2. -------------------------

VIVA Questions:

1. What is the basic principle of chromatography?

2. What do you mean by stationary phase and mobile phase?

3. What is absorption and adsorption?

4. What is retention factor?

AIM: Determination of heat of neutralization of hydrochloric acid and sodium hydroxide.

APPARATUS: Weighing bottle, three 250 mL standard flasks, 250 mL beakers, thermos flask, rubber

cork (3” dia), glass stirrer, two 0.1oC thermometers, stop watch / stop clock, burette, 10 mL pipette, 100

mL conical flask, tripod stand, wire gauze.

REAGENTS: Conc. HCl, NaOH, oxalic acid, phenolphthalein indicator.

THEORY:

The enthalpy of neutralization is defined as the change in enthalpy when one mole of H+ (aq) ions from an

acid is neutralized by one mole of OH(aq) ions from a strong base in their dilute solutions at a constant

temperature and under 1 bar pressure. The neutralization reaction between a strong acid HA and a strong

base BOH occurs as:

HA (aq) + BOH (aq) BA (aq) + H2O(l)

It is observed that the enthalpy of neutralization is the same in almost all cases of neutralization reaction

between a strong acid and a strong base and is found to be – 57.2 kJ mol-1 . The reason for this can be

understood as follows. All strong acids (HA), and bases (BOH) are completely ionized in their aqueous

solutions as:

HA (aq) H + (aq) + A(aq)

BOH (aq) B + (aq) + OH(aq)

The neutralization reaction can be rewritten as:

H + (aq) + A(aq) + B + (aq) + OH(aq) B + (aq) + A(aq) + H2O(l)

Since B+

(aq) and A(aq) ions are common on both the sides of this equation, they can be cancelled to obtain

the equation.

H +

(aq) + OH(aq) H2O(l) nH o = –57.2 kJ mol

-1

Since this reaction is common to all neutralization reactions between strong acids and strong bases, their

enthalpy of neutralization is also the same.



LABORATORY

MANUAL




PAGE:


nd

‘When dilute solutions of strong acids are neutralized with dilute solutions of strong bases at room

temperature, when one mole of a strong mono-base is neutralized by one mole of a strong

monoprotonic acid in a sufficiently dilute solution, a constant amount of heat is evolved, termed as

the heat of neutralization’.

This experiment is performed in the following two steps:

1. Determination of heat capacity of calorimeter.

2. Determination of enthalpy of neutralization of HCl and NaOH .

A. Determination of heat capacity of calorimeter

The amount of heat absorbed or evolved during a thermo chemical process (chemical or physical) is

carried out by measuring the change in temperature. This method is called calorimetry and the

apparatus used is called calorimeter.

Calorimetric Calculations:

Calculations are based upon the first law of thermodynamics according to which the total energy of a

system and its surroundings in conserved. In an adiabatic calorimeter the heat exchange with the

surroundings is zero. Hence, the heat lost or gained inside the calorimeter must balance each other.

(i) Heat lost or gained by water

It is calculated with the help of calorimetric formula

qwater= m x s xt

Where,

m = mass of water taken in the calorimetric vessel. Since density of water is very nearly 1g mL-1 at

room temperature, the volume of water (or solution) in milliliter is numerically equal to its mass in

grams. The slight inaccuracy in omitting the presence of solutes is negligibly small.

s = Specific heat capacity of water. It is the amount of heat required to raise the temperature of unit mass

(1g) of water by 1oC. In joule (SI unit) the specific heat capacity of water is 4.18 J

oC

-1 g

-1 and in calorie

it is 1 cal oC -1 g

-1.

t = Change in temperature = tfinal – tinitial

(ii) Heat lost or gained by calorimeter

The above formula cannot be used for the calorimeter used here (glass beaker along with the

stirrer and the thermometer). It is because the glass is a thermal insulator and heat is not

uniformly gained or lost by the entire vessel. It is almost only that part of the beaker which is

actually in contact with the reacting system which absorbs or loses heat. Mass of this portion

cannot be determined. For this reason instead of using mass and specific heat capacity, use is

made of heat capacity of calorimeter, C in the calorimetric formula i.e.

qcalorimeter = C x t

Where,

qcalorimeter is the heat lost or gained by calorimeter, C is its heat capacity, which is the amount of

heat required to raise the temperature by 1oC. t is the change in temperature.

PROCEDURE:

1. Take a clean and dry 250 mL glass beaker and place it in a thermos flask on a 3” diameter rubber

cork which is kept at the bottom of the flask.

2. Measure 100 mL distilled water into it with the help of a burette or a 50 mL pipette.

3. Cover of the thermos flask with a lid with two holes in it through which a glass stirrer and a 0.1oC

thermometer are fixed.

4. Take a 250 mL glass beaker, rinse it with distilled water and then measure 100 mL distilled water

into it.

5. Keep this beaker on a tripod stand over wire gauze. Fix a 0.1oC thermometer in it and note down

the temperatures of the water.

6. Heat the water till its temperature rises by 8o -10oC. Now remove the beaker from the tripod

stand and keep it on the laboratory table.

7. Start a stop clock and record the temperatures of the cold water in the thermos flask and the hot

water kept out side with constant stirring, alternately every 15 seconds. Continue recording the

temperature till 15 seconds before the time of mixing (i.e. up to 4 min 45 s, if the time of mixing

has been fixed as 5 min).

8. Remove the thermometer from the beaker containing hot water. Hold this beaker in your hand

near the mouth of thermos flask.

9. At about 3-4 seconds before the time of mixing (5 min), remove the lid of the thermos flask partly

and start pouring the hot water gently into the beaker kept in the thermos flask. When completely

poured, quickly replace the lid and start stirring the water in the thermos flask gently but

continuously.

10. After 30 seconds from the time of mixing (i.e. from 5 min 30 s) start recording the temperature of

mixed water at regular intervals of 30 seconds. Take 8-10 readings and record them in the table.

11. Temperature versus time graph for determination of heat capacity of calorimeter giving a vertical

line.

OBSERVATIONS:

1. Masses of cold and hot water:

Mass of cold water taken = 100 g

Mass of hot water taken = 100 g

2. Temperature - Time Data:

Time (in

min)

Cold

water

Temperature Time (in

min)

Temperature of

Mixture / oC Recorded Corrected

CALCULATIONS:

Density of water = 1 g mL-1

Mass of cold water = 100 g

Temperature of cold water at the time of mixing (from graph) = tc=----

Mass of hot water = 100 g

Temperature of hot water at the time of mixing (from graph) = th =----

Temperature of mixture at the time of mixing (from graph) = tm =------

Specific heat capacity of water = 4.18 J 0C

-1 g

-1

Let the heat capacity of calorimeter be = C J 0C

-1

Heat lost by hot water= 𝑚 × 𝑠 × ∆𝑡

= 100𝑔 × 4.18 𝐽°𝐶−1𝑔−1 × 𝑡𝑏 − 𝑡𝑚 𝐽

Heat gained by cold water = 𝑚 × 𝑠 × ∆𝑡

= 100𝑔 × 4.18 𝐽°𝐶−1𝑔−1 × 𝑡𝑚 − 𝑡𝑐 𝐽

Heat gained by calorimeter = 𝑐 × ∆𝑡

= 𝑐 × 𝑡𝑚 − 𝑡𝑐 𝐽

Heat gained by cold water and calorimeter = Heat lost by hot water

100 × 4.18 × 𝑡𝑚 − 𝑡𝑐 + 𝐶 × 𝑡𝑚 − 𝑡𝑐 = 100 × 4.18 × 𝑡𝑏 − 𝑡𝑚

100 × 4.18 + 𝐶 𝑡𝑚 − 𝑡𝑐 = 100 × 4.18 × 𝑡𝑏 − 𝑡𝑚 𝐶

100 × 4.18 𝑡𝑚 − 𝑡𝑐

𝑡𝑚 − 𝑡𝑐 − 100 × 4.18 𝐽𝐶−1

Calculate C using data from 2 sets as C1 and C2

Average heat capacity = 𝐶2+𝐶1

2𝐽°𝐶−1

RESULT:

The average heat capacity of calorimeter for 200 mL of water (100 mL cold water +100 mL hot water)

was found to be …. J 0C

-1.

B. Determination of the Enthalpy of Neutralization of Hydrochloric Acid with Sodium Hydroxide.

It can be done by the following procedure:

1. Take the same calorimeter setup (250 mL beaker, thermos flask, thermometer and stirrer whose

heat capacity was determined in experiment No.A

2. Measure 100 mL 1 M HCl solution in the 250 mL beaker used as calorimeter and keep it inside

the thermos flask.

3. Fix a glass stirrer and 0.10C thermometer in it with the help of the lid of the thermos flask.

4. Rinse a 250 mL beaker with 1 M NaOH solution and then measure 100 mL of this solution in it.

Fix the second 0.10C thermometer in it.

5. Start the stop clock and record the temperatures of HCl and NaOH solutions alternately every 15

seconds till the last reading is taken 15 seconds before the time of mixing.

6. Remove the thermometer from NaOH solution, open the lid of thermos flask partly and pour it

into the beaker containing HCl solution such that one half of mixing would occur by the time of

mixing.

7. Replace the lid as soon as the pouring is complete and start stirring the mixture solution in the

thermos flask gently but continuously.

8. Starting from 30 seconds after the time of mixing, record the temperature after every 30 seconds

for next 4-5 minutes.

9. Plot graphs of temperatures of HCl, NaOH and their mixture solutions versus time. Draw a

vertical line at the time of mixing and determine the temperatures of the acid (ta), base (tb) and

mixture tm at the time of mixing from the graph.

10. Repeat the experiment (steps 10-17) once more time to obtain second value of enthalpy of

neutralization using the same solutions

OBSERVATIONS AND CALCULATIONS:

In the neutralization reaction the heat is liberated which raises the temperature of the calorimeter and the

resulting solution.

Total heat liberated, q = - (heat gained by calorimeter + heat gained by water)

Heat capacity of calorimeter = C J oC

-1 (as determined in part I)

Mass of water in the mixture of acid and base solutions = 200 g

(Mass of 200 mL water 200 g)

Specific heat capacity of water = 4.18 J oC

-1 g

-1

Rise in temperature = 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙

Where,

𝑡𝑓𝑖𝑛𝑎 𝑙 = 𝑡𝑚

the temperature of mixture at the time of mixing

𝑡𝑓𝑖𝑛𝑎𝑙 =𝑡𝑏 +𝑡𝑎

2 , the average initial temperature at the time of mixing.

∆𝑡 = 𝑡𝑚 −𝑡𝑎 +𝑡𝑏

2

Heat gained by calorimeter = 𝑐 × ∆𝑡

Heat gained by water = m x s x Heat gained by water = 𝑚 × 𝑠 × ∆𝑡

= 200𝑔 × 4.18𝑗°𝑐−1𝑔−1 × ∆𝑡

Total heat gained by water and calorimeter= 200 × 4.18 + 𝐶 ∆𝑡

Heat evolved during neutralization = q = (200 x 4.18 + C) t

Enthalpy of neutralization is the heat evolved when 1 mole of H+ (aq) ions from the acid is

neutralized by 1 mole of OH (aq) ions from the base. This can be obtained by calculating the heat

evolved, q, when one litre volume of 1 M HCl is neutralized by 1 litre volume of 1 M NaOH.

∆𝑛𝐻° =𝑞 × 1000

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑁 𝑎𝑐𝑖𝑑 𝑜𝑟 𝑏𝑎𝑠𝑒 𝑢𝑠𝑒𝑑=

𝑞 × 1000

100= 10𝑞

Calculate n Ho for both the sets and calculate the average value.

RESULT

The enthalpy of neutralization of HCl and NaOH was found to be ------------------… kJ mol-1

Viva Questions:

1. Define Hess's Law.

2. Define heat capacity and specific heat.

3. Define endothermic and exothermic reactions in terms of the sign of H.

AIM: To determine the amount of chloride present in 100 ml of the given water sample, being

supplied with standard solution of sodium chloride of strength …………N and an approximately

N/20 solution of silver nitrate.

APPARATUS: Pipette, Burette, Conical flask

REAGENT: NaCl solution, AgNO3 solution, 2% K2CrO4 indicator.

PRINCIPLE:

Generally water contains chloride ions (Cl–) in the form of NaCl, KCl, CaCl2 and MgCl2.

The concentration of chloride ion in water, more than 250 ppm, is not desirable for drinking purpose. The

total chloride ion can be determined by argentometric method (Mohr’s Method).

In this method Cl– ion solution is directly titrated against AgNO3 using potassium chromate (K2CrO4) as

an indicator.

AgNO3 + NaCl AgCl + NaNO3

(In water) (White precipitate)

At the end point, when all the Cl– ions are removed. The yellow colour of chromate changes into reddish

brown due to the following reaction.

2AgNO3 + K2CrO4 Ag2CrO4 + 2KNO3

(Yellow) (Reddish brown precipitate)



LABORATORY

MANUAL




PAGE:


nd

PROCEDURE:

Step I

TITRATION – I

Standardization of Silver nitrate

The burette is washed well with distilled water and rinsed with the small amount of AgNO3 solution. It is

then filled with the same solution up to the zero mark without any air bubbles. The pipette is washed well

with distilled water and rinsed with the small amount of standard NaCl solution. 20 ml of this solution is

pipette out into a clean conical flask. 1ml of 2% K2CrO4 indicator is added and titrated against AgNO3

solution taken in the burette. The end point is the change of color from yellow to reddish brown. The

titration is repeated for concordant values.

Step II

TITRATION – II

Determination of chloride ions

20 ml of the given water sample is pipette out into a clean conical flask and 1ml of 2% K2CrO4 indicator

is added. It is then titrated against standardized AgNO3 solution taken in the burette. The end point is the

change of color from yellow to reddish brown. The titration is repeated for concordant values.

Step I: STANDARDISATION OF SILVER NITRATE

TITRATION – I

Standard NaCl vs. AgNO3

S.No Volume of

NaCl ml

Burette Readings Concordant

volume of

AgNO3 ml

Indicator

Initial ml Final ml

1 20 0

K2CrO4

2 20 0

CALCULATION:

Volume of Sodium Chloride V1 = 20 ml

Strength of Sodium Chloride N1 = ……………N

Volume of Silver Nitrate V2 = …………..ml

Strength of Silver Nitrate N2 -------------?

According to the law of volumetric analysis

𝑽𝟏𝑵𝟏 = 𝑽𝟐𝑵𝟐

𝑵𝟐 =𝑽𝟏𝑵𝟏

𝑽𝟐

𝑵𝟐=

𝟐𝟎×−−−𝑵−−−

∴ Strength of Silver Nitrate = ……………….N.

Step II: DETERMINATION OF CHLORIDE ION

TITRATION – II

Water sample vs. Std. AgNO3

S.No

Volume of

water sample

ml

Burette Readings Concordant

volume of

AgNO3 ml

Indicator

Initial ml Final ml

1 20 0

K2CrO4 2 20 0

Calculation of Normality of the water sample (Chloride ion)

Volume of water sample V1 = 20 ml

Strength of water sample N1 =?

Volume of Silver Nitrate V2 = …………..ml

Strength of Silver Nitrate N2 =……………N

According to the law of volumetric analysis

𝑵𝟏𝑽𝟏 = 𝑵𝟐𝑽𝟐

𝑵𝟏 =𝑽𝟐𝑵𝟐

𝑽𝟏

𝑁2 =−−−−−−−−−−−−−−×−−−−−−−−−−−−−𝑁

20

Calculation of amount of the chloride

Amount of the chloride ion present in = Equivalent weight of chloride ion X

1 litre of the water sample Strength of the chloride ion

= 35.46 x ……………..N

= ………………gms

Amount of the chloride ion present in100 ml of the given water sample

= ………………gms x 100/1000

= ………………gms

RESULT:

1. Amount of chloride ion present in the whole of the given water sample

= ………….gms/mgs/ppm

2. Amount of chloride ion present in the 100 ml of the given water sample

= ………....gms/mgs/ppm

Viva Questions:

1. What do you mean by hard water?

2. How hardness cause?

3. Write chemical formula of mohr’s salt?

Date post:	11-Apr-2018
Category:	Documents
Upload:	dinhnhi
View:	217 times
Download:	2 times

Lab Manual - dbit.ac.in of alkalinity in the given water sample. 2. Determination of temporary and...

Documents