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The Great Chemist ALFRED NOBEL CHEMISTRY LABORATORY - I

CHEMISTRY LABORATORY - I - chettinadtech.ac.inchettinadtech.ac.in/storage/14-11-03/14-11-03-14-43-30-senthil... · Viva questions 1. Name the sources of ... What is the significance

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The Great Chemist

ALFRED NOBEL

CHEMISTRY

LABORATORY - I

-1-

WORK SHEET

Titration 1 : Standardization of AgNO3

Standard Sodium chloride Vs AgNO3

Calculation

Volume of Sodium chloride V1 = ------- mL

Strength of Sodium chloride N1 = -------N

Volume of AgNO3 V2 = …… mL

Strength of AgNO3 N2 = …?….N

V1 N1 = V2 N2

N2 = V1 N1 / V2

= [------- (V1) X ------ (N1)] / ------(V2)

Strength of AgNO3 N2 = ----------- N

-2-

Sl.No

Vol.of

Sodium

chloride

V1 (mL)

Burette reading (mL) Volume of

AgNO3

V2 (mL)

Concordant

value Initial Final

Ex.No. Date:

DETERMINATION OF CHLORIDE CONTENT OF WATER

SAMPLE BY ARGENTOMETRIC METHOD

Aim

To determine the chloride content of water sample by argentometric

method (Mohr’s method).

Principle

Chloride is present in water usually as NaCl, MgCl2 and CaCl2.

Although chlorides are not harmful as such, their concentrations beyond 250

ppm impart a peculiar taste of the water, thus rendering the water

unacceptable for drinking purposes.

Further, existence of unusually high concentration of chloride in a

water sample indicates pollution from domestic sewage or industrial waste

water. Salts like MgCl2may undergo hydrolysis under the high pressure and

temperature prevailing in the boiler, generating hydrochloric acid which

causes corrosion in boiler parts. Chlorides in the form of MgCl2 and CaCl2

cause various disadvantages in household and industrial use of water.

By argentometric method chloride ions in water sample, which is

neutral or slightly alkaline can be determined by titrating it against standard

silver nitrate solution using potassium chromate as an indicator. The pH

should be between 7 and 8.

At higher pH Ag+

+ OH- Ag OH

and at lower pH K2CrO4 indicator is converted to K2Cr2O7

2 K2CrO4 + 2 HCl 2 KCl + K2Cr2O7 +H2O

-3-

WORK SHEET

Titration 2 :

Standard AgNO3 Vs Distilled water

Titration 3:

Standard AgNO3 Vs given Water Sample

-4-

Sl.No

Vol.of

Distilled

Water (mL)

Burette reading (mL) Volume of

AgNO3

V1 (mL)

Concordant

value Initial Final

Sl.No

Vol.of

Sample

Water (mL)

Burette reading (mL) Volume of

AgNO3

V2 (mL)

Concordant

value Initial Final

As KHCrO4 is weak in nature, concentration of CrO42-

decreases and

therefore, higher concentration of Ag+ is needed for the solubility product of

Ag2CrO4 to be exceeded.

Ag+ reacts with Cl

- and CrO4

2-, when AgNO3 solution is released from

the burette sample solution which has Cl-

Ag+ + Cl

- AgCl (K sp = 3X10

-10)

White ppt

2Ag+ + CrO4

2-, Ag2 CrO4 (K sp = 5X10

-12)

Red ppt

Red colour formed because of formation of silver chromate, disappears

as the solution contains high concentration of Cl-

Ag2 CrO4 +2Cl- 2AgCl +CrO4

2-

As the concentration Cl-ion decreases, the red colour disappears more

slowly and when all the chloride is precipitated, a faint reddish or pinkish

tinge persisting the white precipitate even after brisk shaking.

Procedure :

Titration 1 : Standardization of AgNO3

Transfer 20 mL of standard NaCl in a conical flask and add about 3-4

drops of potassium chromate indicator. Then slowly add standard solution of

AgNO3 solution from the burette and the value of titrant is noted as end

point, when yellow colour changes to red colour. The titration is repeated

until a concordant volume V2 is obtained.

-5-

WORK SHEET

Calculation :

Volume of AgNO3 consumed by given sample V2 =……….. mL

Volume of AgNO3 consumed by dis.water V1 =……….. mL

Net volume (V2 – V1) = ………. mL (V2 )

Volume of given sample V1 = 50mL

Strength of given sample N1 = …?…….N

Volume of Net AgNO3 consumed V2 = ………...mL

Strength of AgNO3 consumed N2 = …………N

V1 N1 = V2 N2

N1 = V2 N2 / V1

= [------- (V2) X ------ (N2)] / ------(V1)

N1 = ---------- N

The amount of chloride in the given sample = …………(N1) X 35.5 X 1000

= ------------ ppm

Precautions

1. The whole apparatus must be washed with distilled water.

2. The volume of the indicator should be same in all titrations.

3. The reaction mixture should be briskly shaken during the titration.

-6-

Titration 2 : Distilled water Vs AgNO3

Transfer 50 mL of standard distilled water in a conical flask and add

about 3-4 drops of potassium chromate indicator. Then slowly add standard

solution of AgNO3 solution from the burette and the value of titrant is noted

as end point, when yellow colour changes to red colour. The titration is

repeated until a concordant volume V1 is obtained.

Titration 3: Estimation of chloride present in the given sample.

Transfer 50 ml of given water sample in a conical flask and add about

3-4 drops of potassium chromate indicator. Then slowly add standard

solution of AgNO3 solution from the burette and the value of titrant is noted

as end point, when yellow colour changes to red colour. The titration is

repeated until a concordant volume V2 is obtained.

Result

The amount of Chloride present in the given water sample

= -------------- ppm.

Viva questions

1. Name the sources of chlorides in water

2. Why does AgNO3 combine first with chloride ions in the water and not

with K2CrO4?

3. What is the action of chloride on pathogens?

4. How is the pH adjusted when the alkalinity is not being measured?

5. What is the effect of temperature on the determination of chloride content

in water?

6. What is the principle of Mohr’s method?

-7-

WORK SHEET

Observation

Sl. No Volume of NaOH added

(mL) Observed Conductance (ohm

-1)

-8-

Ex.No. Date:

CONDUCTOMETRIC TITRATION

(STRONG ACID Vs STRONG BASE)

Aim

To find out the strength of acid (HCl) (approximately N/10), by

titrating it against sodium hydroxide solution conductometrically.

Principle

When a strong acid like HCl is titrated against a strong base like NaOH,

the neutralization reaction takes place.

HCl +NaOH NaCl +H2O

This neutralization reaction can be followed by conductometric method.

During the course of titration, the conductivity decreases due to the

replacement of H+

ions by Na+ions. After the neutralization, the

conductivity increases steeply due to the excessive presence of OH- ions.

Procedure

0.5 N NaOH solution is taken in the burette. 50 mL of the given HCl is

taken in a 100 mL beaker. Conductivity cell containing two platinum

electrodes, is placed in HCl. The cell is connected to the conductivity meter.

Now add 1mL, 2mL, 3mL and 4mL of NaOH from the burette and stir the

contents of the beaker thoroughly. Note the conductance readings after each

addition.

-9-

WORK SHEET

A graph is plotted with observed conductivity against volume of NaOH

added. The shape of the curve will be as shown in the figure.

Calculation :

V1N1 = V2N2

Volume of NaOH V1 = ---------- mL (from graph)

Strength of NaOH N1 = ---------- N

Volume of HCl V2 = --------- mL

Strength of HCl N2 = -----?----- N

N2 = (V1 X N1) / V2

= [……..( V1) X ……..( N1)] / …….( V2)

Strength of unknown HCl (N2) = ------------- N

Amount of HCl present in the whole of the given solution

= [---------------- (N2) x ------------- (Eq.wt) x 50] / 1000

= ------------------- g

(Equivalent Weight of HCl = 36.5)

-10-

Result

(i) The strength of the given acid = ----------------- N

(ii) Amount of HCl present in

the whole of the given solution = ------------g

-11-

WORK SHEET

Observation

Volume of

NaOH

solution

added (mL)

pH

∆pH

∆V

∆pH/∆V

Average

volume of

NaOH (mL)

Where, ∆V = difference in two consecutive volumes of titrant added in mL

∆pH = difference in two consecutive reading

-12-

EX.NO. Date:

DETERMINATION OF STRENGTH OF

HYDROCHLORIC ACID BY pH TITRATION

Aim

To find out the strength of given hydrochloric acid solution by titrating

it against sodium hydroxide (0.1N) using pH meter.

Principle

When an alkali is added to an acid solution, the pH of the solution

increases slowly, but at vicinity of the end point, the rate of change of pH of

the solution is very rapid. From the sharp break in the curve, we can find out

the end point, from which the strength of HCl can be calculated.

Procedure

First standardize the pH meter using different buffers of known pH,

then wash the glass electrode and reference electrode with distilled water

and then with the acid solution. Take 20 mL of HCl solution in a 250 mL

beaker. Add sufficient distilled water (100mL) so that the glass electrode as

well as the reference electrode is completely dipped.

Note the pH of the pure acid solution. Now add 1 mL of 0.1 N NaOH

from the burette in the beaker. Stir the contents well. Note the pH of the

solution. Now keep on adding NaOH solution from the burette and the note

the pH of the solution, up to 9- 10 mL of the NaOH. Near the end point add

very small amount of sodium hydroxide, because change in pH will be very

much appreciable when the acid is neutralized, further addition of such a

small amount as 0.01 mL raises the pH about 9 to 10.

-13-

WORK SHEET

Graph

Calculation: Volume of NaOH required for complete neutralization of HCl = --------- mL

Volume of NaOH V1 = ------------- (mL) from graph

Strength of NaOH N1 = ------------ N

Volume of HCl V2 = ------------- (mL)

Strength of HCl N2 = ------?------- N

Therefore Strength of HCl N2 = [-------- (V1) x -------- (N1)] / ---------- (V2)

= ------------N

Amount = Normality x Equivalent Weight (HCl)

Therefore Amount of HCl present in the whole of the given solution

= [---------------- (N2) x ------------- (Eq.wt) x 20] / 1000

= ------------------- g

(Equivalent Weight of HCl = 36.5)

-14-

Plot a graph between pH and Volume of NaOH added. From this graph,

determine the volume of NaOH required for the complete neutralization of

HCl.

Result

(i) Strength of the given hydrochloric acid solution = ---------- N

(ii) The amount of hydrochloric acid present in the whole

of the given solution = --------- g

Viva questions

1. What do you mean by pH?

2. What is the effect of temperature on pH?

3. What are the composition of glass electrode and calomel

electrode?

4. What is the effect of dilution on pH of an acidic solution?

5. What is the pOH of pure water at 25o C?

6. What chemicals would you use to make a buffer of pH 10?

7. What is a combined electrode?

8. Why is hydrogen not generally used in pH measurements?

9. What is the significance of pH titration?

10. What are the industrial applications of pH metric measurements?

11. What is the pH value of lime juice, blood and de-ionized water?

12. What is function of calomel electrode?

-15-

WORK SHEET

Observation

Sl.No Volume of NaOH added (mL) Observed Conductance (ohm-1

)

-16-

Ex.No. Date:

CONDUCTOMETRIC TITRATION OF

MIXTURE OF ACIDS

Aim

To find out the strength of mixture of acids (hydrochloric acid and

acetic acid) of approximate strength N/10 each, by titrating it against sodium

hydroxide solution conductometrically.

Principle

The type of titration is just a combination of two separate titrations,

viz. HCl against NaOH and CH3COOH against NaOH. By adding to the

mixture, the conductivity of sodium ion decreases due to the replacement of

H+

ions forms the strong acid. The conductivity of the solution increases as

the weak acid is converted into salt and finally rises more steeply as excess

of alkali is added.

A graph is plotted with conductivity against volume of alkali added. It is

observed that there is a rounding off at both the end points. Usually,

extrapolation of the straight lines of the three branches would lead to a

definite location of the end points. It must be noted that the first end point

will be that of hydrochloric acid (strong) while the second will be that of

acetic acid (weak).

-17-

WORK SHEET

Calculation :

For HCl: (Strength of unknown HCl) V1N1 = V2N2

Volume of NaOH V1 = --------- mL ( from graph )

Strength of NaOH N1 = ---------- N

Volume of HCl V2 = ---------- mL

Strength of HCl N2 = [--------( V1 ) X--------(N1)] / --------(V2)

Strength of unknown HCl N2 = --------- N

Amount of HCl present in the whole of the given solution

= [---------------- (N2) x ------------- (Eq.wt) x 50] / 1000

= ------------------- g

(Equivalent Weight of HCl = 36.5)

For CH3COOH: (Strength of unknown CH3COOH)

Volume of NaOH V3 = -------- mL (from graph)

Strength of NaOH N3 = ---------- N

Volume of CH3COOH V4 = --------- mL

Strength of CH3COOH N4 = [--------(V3) X-------- (N3)] / --------(V4)

Strength of unknown CH3COOH N4 = --------- N

Amount of CH3COOH present in the whole of the given solution

= [---------------- (N4) x ------------- (Eq.wt) x 50] / 1000

= ------------------- g

(Equivalent Weight of CH3COOH = 60.05)

-18-

Procedure

Prepare an exact N/10 Solution of sodium hydroxide. Rinse and fill the

burette with the alkali solution. Now take 10 mL of the mixture of HCl and

CH3COOH in a 200 mL beaker. Add about 100 mL of distilled water.

Immerse the cell in this solution and determine the conductance of the

solution. (The first end point will lie near about 5 mL, while the second will

be near about 10 mL of sodium hydroxide solution).

Now add 1mL, 2mL, 3mL and 4mL of NaOH from the burette and stir

the contents of the beaker thoroughly. Note the conductance readings after

each addition. Now, add NaOH at an increment of 0.2 mL up to 6 mL and

then again between 9 mL and 11 mL.

After adding 11 mL of NaOH solution, add 1mL in each addition till you

have added 18 mL, note the conductance of the solution after each addition

thoroughly stir the contents of the beaker.

Result

The strength of each acid in the given mixture is

a. (i) Strength of hydrochloric acid = ---------------- N

(ii) Amount of HCl present in the whole of the given

solution = ---------------g

b. (i) Strength of acetic acid = ---------------- N

(ii) Amount of CH3COOH present in the whole of the given

solution = ---------------g

-19-

WORK SHEET

Observation

Sl.No.

Vol.of

Stock

Solution

Vol.of

HNO3

(mL )

Vol.of

Potassium

Thiocyanate

(mL)

Vol.of

Distilled

water (mL)

Total

Volume

(mL)

Sl.No. Concentration in ppm Absorbance at 480 nm

-20-

Ex.No. Date:

ESTIMATION OF FERRIC ION BY SPECTROMETRY

Aim

To estimate the amount of ferric ion present in the given sample using

spectrophotometer.

Principle

When a monochromatic light passes through a homogeneous coloured

solution, a portion of incident light is reflected, a portion is absorbed and the

remaining is transmitted.

Io = Ir + Ia + It

Where Io, Ir, Ia, It are the intensities of the incident, reflected,

absorbed and transmitted light, respectively, Ir is usually eliminated and

hence, Io = Ia + It

The mathematical statement of beer- Lambert’s law is given by

Log Io / I = ε Ct

A = ε Ct

Where, log Io /I = A

Io = intensity of incident light

It = intensity of transmitted light

C = Concentration of the solution in moles /L

t = thickness of cell in cm

ε = molar absorption coefficient

A = absorbance or optical of solution density

-21-

WORK SHEET

Graph

-22-

When a ray of monochromatic light passes through an absorbing

medium, its intensity decreases exponentially as the concentration of

absorbing substance and the length or thickness increase independently.

Keeping the path length constant (t=1 cm), the variation is with reference to

only concentration, C.

Fe3+

ion do not give any colour. However, it develops a red colour when

it reacts with potassium thiocyanate solution.

Fe3+

+ 6 KSCN → [Fe (SCN) 6]3-

+ 6K+

Further, this colour is in the blue region (λ- 480 nm). Spectrophotometer

has a wide range of adaptability that allows selection of monochromatic light

of any wavelength in the visible spectrum.

The light source is an ordinary light bulb and monochromatic light is

obtained by using either a prism or a diffraction grating. The monochromatic

light is then passed through the filter and is directed through a cell

containing the sample. The light that penetrates hits the photoelectric cell

and the output of this can be seen in the display.

-23-

WORK SHEET

-24-

Estimation of Ferric ion in the given sample:

The given sample solution is taken in 100mL SMF and 3 mL of strong

HNO3 (1:1 V) and 5mL of potassium thiocyanate solution are added and

made up to the mark using distilled water. The absorbance of the test

solution can be calculated by performing the same procedure as standard.

Result

The amount of iron present in the given test sample = ------------ ppm

-25-