lab manual sem ii - · pdf filethe titration is repeated to get concordant values. let the...

The Great Chemist

ALFRED NOBEL

CHEMISTRY

LABORATORY-II

WORK SHEET

Titration 1 : Standardization of EDTA

Burette solution : EDTA solution

Pipette solution : Standard hard water

Addition solution : 5 ml of ammonia buffer solution

Indicator : Eriochrome Black -T

End point : Change of colour from wine red to steel blue

Standard hard water Vs EDTA solution

Sl.

No

Volume of

Std. hard

water (mL)

Burette readings (mL ) Volume of

EDTA

(V1 mL )

Concordant

Value Initial Final

1 mL of standard hard water = 1 mg of CaCO3

20 mL of standard hard water contains 20 mg of CaCO3

20 mL of standard hard water consumes --------------- (V1) mL of EDTA

Hence,

------------- (V1) ml of EDTA solution = 20 mg of CaCO3

1 mL of EDTA Solution = (20/V1) mg of CaCO3

= 20 / ……..( V1)

= …………. mg of CaCO3

-2-

Ex.No. Date:

ESTIMATION OF HARDNESS OF WATER SAMPLE

BY EDTA METHOD

Aim

To Estimate the amount of total, permanent and temporary hardness in

the given sample of water provided that Standard hard water and a link

solution of Ethylene Diamine Tetra Acetic acid (EDTA).

Principle

The total hardness of water is estimated by titrating against EDTA using

Eriochrome Black –T indicator.

Structure of disodium salt of EDTA

NaOOCH2C CH2COOH

N-CH2-CH2-N

HOOCH2C CH2COONa

Calcium (Ca2+

) and Magnesium (Mg2+

) ions present in the hard water

form a stable complex ions with EDTA. The complexation of the reaction is

indicated by Eriochrome Black-T indicator. EBT forms an unstable wine-red

colour complex with Ca and Mg ions at pH 9-10.

NH4Cl -NH4OH buffer

Ca2+

/ Mg 2+

+ EBT [Ca2+

/ Mg2+

+EBT]

pH 9-10 less stable

(Wine-red colour complex)

-3-

WORK SHEET

Titration 2 : Estimation of Total hardness


Pipette solution : Sample hard water




Sample Water Vs Standard EDTA solution

Sl.

No

Volume of

Water

Sample(mL)


EDTA

(V2 mL )

Concordant

Value Initial Final

Volume of EDTA consumed in titration 2 = ---------- (V2) mL

Total hardness of sample

water in 1 litre = ( 20 / V1) xV2 x 1000 mg of CaCO3

20

= [20 / …….(V1 ) x …….( V2 ) x 1000 ]

20

= ----------- mg of CaCO3

Total hardness = ---------- mg / L (or) ppm

-4-

When EDTA is added into the hard water, the metal ions form a stable

metal complex with EDTA by leaving the indicator. When all the metal ions

are taken by EDTA from the indicator metal ion complex, the wine red

colour changes into steel blue, which indicates the end point

[Ca2+

/ Mg 2+

+ EBT]+ EDTA [Ca2+

/ Mg 2+

+EDTA] +EBT

More stable steel blue

(colourless complex) colour

Procedure

Titration 1 Standardization of EDTA

A 50 mL burette is washed with distilled water and then rinsed with the

EDTA solution. It is then filled with the same EDTA solution up to zero

level without air bubbles. Initial reading of the burette is noted. 20 mL of

standard hard water solution is pipetted out into clean conical flask.5 mL of

ammonia buffer solution and two drops of EBT indicator is added. The

solution turns into wine- red colour. The solution is titrated against EDTA

solution taken in the burette. The end point is change of colour from wine

red to steel blue. The final burette reading is noted. The titration is repeated

to get concordant values. Let the volume of EDTA be V1 mL

Titration 2 Estimation of Total Hardness

The burette is filled with the same EDTA solution. 20 mL of the given

water sample is pipetted out into a clean conical flask. 5 mL of ammonia

buffer and two drops of EBT indicator is added. The solution is titrated

against standard EDTA solution taken in the burette. The end point is the

change of colour from wine red to steel blue. The final burette reading is

noted. The titration is repeated to get concordant values. Let the volume of

EDTA be V2 mL

-5-

WORK SHEET

Titration 3 : Estimation of Permanent hardness


Pipette solution : Boiled Sample hard water




Boiled Sample water Vs Standard EDTA solution

Sl.

No

Volume of

Boiled Sample

water (mL)


EDTA

(V3 mL )

Concordant

Value Initial Final

Volume of EDTA consumed in titration 3 = ---------- (V3) mL

Permanent hardness of

Sample water in 1 litre = [(20 / V1) x V3 x 1000]

20

= [(20 / .……(V1) x …….(V3 )x 1000]

20

= ---------- mg of CaCO3

Permanent hardness = ---------- mg / L (or) ppm

So, Temporary hardness = Total hardness – Permanent hardness

= -------------(ppm) – -------------(ppm)

= ------------- (ppm)

-6-

Titration 3 Estimation of Permanent Hardness

The burette is filled with same EDTA solution. 20 mL of the given

sample water taken in a 100 mL beaker and boiled until the volume is

reduced to 10 mL. Then it is cooled and filtered using filter paper. To the

filtrate, 2 mL of ammonia buffer and two drops of EBT indicator are added

and titrated against EDTA solution taken in the burette. The end point is the

change of colour from wine red to steel blue. The final Burette reading is

noted. Let the volume of EDTA be V3 mL

Result

(1) Total hardness of the given sample water = ---------------- ppm

(2) Permanent hardness of the given sample water = ---------------- ppm

(3) Temporary hardness of the given sample water = ---------------- ppm

Viva voce questions

1. What is potable water?

2. Why disodium salt of EDTA is chosen for determination of hardness?

3. Why does the colour of the solution change from wine red to steel

blue?

4. Why does hard water not lather with soap?

5. Which is the best method of hardness determination and why?

6. What is buffer solution? How and why the pH of the solution is

maintained at 10?

7. Which salt produces temporary and permanent hardness?

8. Draw the structure of EDTA and EDTA – metal ion complex.

9. What are units of hardness?

10. Draw the structure of EBT

11. What is the relationship between mg/l and ppm?

-7-

WORK SHEET

Standard H2SO4 Vs Water Sample :

Sl.

No

Vol.of

Sample

Water

(mL)

Burette readings (mL)

Volume of

H2SO4

Concordant

value

Initial

Final

At Phenol-

phthalein

end point

(P)

At Methyl

orange

end point

(M)

(P)

mL

(M)

mL

(P)

mL

(M)

mL

-8-

Ex.No. Date:

DETERMINATION OF ALKALINITY

OF A WATER SAMPLE

Aim

To determine the different types of alkalinity of a given sample of

water.

Principle

Alkalinity in water is due to the presence of OH-, CO3

2-, HCO3

- ions.

The alkalinity of a given sample of water can be obtained by neutralizing the

above mentioned ions with standard H2SO4. Titrating given sample of water

at a pH of 8.3 or till the decolourization of phenolphthalein indicator will

indicate complete neutralization of OH- ions and half of CO3

2- ions. Titrating

the same sample of water at pH of 4.4 or till a sharp color change from

yellow to pink on methyl orange indicator, indicates the total alkalinity i.e.

the amount of

OH-, CO3

2- and HCO3

- present in the given sample.

OH- + H

+ H2O

CO32-

+ H+ HCO3

–

HCO3- + H

+ (H2CO3) H2O + CO2

OH- and HCO3

- ions cannot exist in water together because they combine

instantaneously to form CO32-

ions.

OH- + HCO3

- CO3

2- + H2O

NaOH + NaHCO3 Na2CO3 +H2O

It is for this same reason, the three ions OH-, CO3

2- and HCO3

- cannot exist

together.

-9-

WORK SHEET

Calculation:

Phenolphthalein alkalinity (mg/ L) as CaCO3

P ml of H2SO4 X Normality of H2SO4 X 50 X 1000

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

Volume of water sample taken

…… X …… X 50 X 1000

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

……..

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

Calculation:

Methyl orange alkalinity (mg/ L) as CaCO3 (Total alkalinity)

M ml of H2SO4 X Normality of H2SO4 X 50 X 1000

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

Volume of water sample taken

…… X …… X 50 X 1000

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

……..

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

-10-

Procedure

The burette is washed with the distilled water and then rinsed with

H2SO4. It is then filled with H2SO4 upto the zero mark. 20 mL of the given

water sample is pipetted out into a clean conical flask. To this, 2 drops of

phenolphthalein indicator is added and the solution is titrated against H2SO4

taken in the burette. The end point is just disappearance of pink colour. The

corresponding burette reading is noted and it is denoted as phenolphthalein

end point (P).

To the same solution, two drops of methyl orange indicator is added

and the titration is continued. The end point is colour change from pale

yellow to pale pink. This end point indicates complete neutralization of

alkalinities present in water sample and is known as methyl orange end point

(M)

Result

a) Amount of Phenolphthalein alkalinity = ----------- ppm

b) Amount of Methyl orange alkalinity = ----------- ppm

-11-

WORK SHEET

Observation

Volume of

K2Cr2O7

solution

added (mL)

Potential of

the test cell

Ecell (mV)

∆E

∆V

∆E/∆V

Average

volume of

K2Cr2O7 (mL)

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

∆E = difference in two consecutive reading in millivolts

-12-

Ex.No. Date:

POTENTIOMETRIC TITRATION

(ESTIMATION OF FERROUS ION)

Aim

To estimate the amount of ferrous ions present in the whole of the given

solution potentiometrically by titrating it against standard potassium

dichromate solution (0.05N).

Principle

When ferrous ion is titrated against potassium dichromate in acidic

medium, it gets oxidized into ferric ion.

6FeSO4 +K2Cr2O7+ 7H2SO4 K2SO4 + Cr2 (SO4)3+ 3 Fe2 (SO4)3 + 7 H2O

The ferrous- ferric system establishes equilibrium as

Fe2+

Fe3+

+ e-

and the electrode potential is given by the Nernst equation

E= EO – (0.0591/1) log (Fe2+

/ Fe3+

)

The reduction potential value depends on the concentration of ferrous

and ferric ions present. While adding dichromate, ferrous ions are converted

into ferric ions and hence the potential increase gradually and then steeply at

the end point.

A plot of potential Vs. Volume of potassium dichromate is used to

assess the end point. More accurate titre value can be derived from

derivative graph.

-13-

WORK SHEET

Calculation:

Volume of K2Cr2O7 V1 = ------------- (mL) from graph

Strength of K2Cr2O7 N1 = ------------- (N)

Volume of Fe2+

taken V2 = 20 mL

Strength of Fe2+

taken N2 = ------?------ (N)

Strength of Fe2+

N2 = [-------- (V1) x -------- (N1)] /--------- (V2)

Strength of Fe2+

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

Amount of ferrous ion present in the whole of the given solution

= [Normality x Eq.weight of Fe2+

x 100] / 1000

= [----------- (N2) x ----------- x 100] / 1000

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

(Equivalent Weight of Fe2+

= 55.84)

-14-

Procedure

The given ferrous ion solution is made up to 100mL in a standard

flask using distilled water. 20 mL of the made up solution is pipetted out into

a clean beaker and a test tube full of dilute sulphuric acid is added. Platinum

(inert indicator electrode) and calomel (secondary reference electrode)

electrodes are immersed in solution and connected to the potentiometer.

Potassium dichromate solution is added from the burette in 1 mL portions.

The solution is stirred well and after each addition, the emf value is noted.

Near the end point, emf changes rapidly. Therefore, add 0.2 mL increments

near the end point and note the sudden increase in emf. Continue the titration

with 1 mL addition and take 3 to 4 readings.

The potential (E) is plotted against the volume of potassium dichromate

added. The end point of the titration is obtained from the midpoint of

inflection. The accurate end point can be obtained from the peak point by

plotting ∆E/∆V against the average volume of K2Cr2O7 added. The volume

of K2Cr2O7 corresponding to the peak of the curve is the end point of the

titration.

-15-

WORK SHEET

-16-

For better accuracy, the end point obtained from the differential plot is

taken for calculation. From the titre value, the normality of ferrous ion can

be calculated. From this value, the amount of ferrous ion present in the

whole of the given solution is calculated.

Experimental Setup

Note

At the start of the experiment (Before adding K2Cr2O7), theoretically

there should not be any Fe3+

ions in the solution. So, no potential is

developed for the Fe2+

/ Fe3+

equilibrium. However the meter reads some

sensible potential even before the addition of K2Cr2O7 because of the

presence of Fe3+

ions as impurities in the ferrous sample. So, the zero should

not be recorded in a potentiometric titration.

Result

The amount of the ferrous ion present in the

whole of the given solution is ------------ g

-17-

WORK SHEET

Observation

BaCl2 Vs Na2SO4

Sl.

No

Volume of Na2SO4 added

(mL) Observed Conductance (ohm

-1)

-18-

Ex.NO. Date:

CONDUCTOMETRIC TITRATION USING BaCl2 Vs Na2SO4

Aim

To determine the amount of BaCl2 present in one litre of the given

solution by conductometric titration using standard Na2SO4 of -----------N

Principle

Solution of electrolytes conducts electricity due to the presence of ions.

Since specific conductance of a solution is proportional to the concentration

of ions in it, conductance of the solution is measured during titration.

In the precipitation titration, the ions are converted to insoluble

precipitate, which will not contribute in the conductance.

When Na2SO4 is added slowly from the burette to the solution of BaCl2,

BaSO4 gets precipitated while the chloride ions remain unchanged.

[Ba2+

+ 2Cl] + [2Na+ SO4

2-] BaSO4 + 2Na

+ + 2Cl

Unionized

The Ba2+

ions in the solution are replaced by free Na+ ions. Since the

mobility of Na+ ions is less than that of Ba

2+ ions the conductance of the

solution decrease.

After the end point, when all the Ba2+

ions are replaced, further addition

of Na2SO4 increases the conductance. This is due to the increase of Na+ and

SO42-

ions in the solution.

-19-

WORK SHEET

Calculation

V1N1 = V2N2

Volume of Na2SO4 V1 = ---------- mL (from graph)

Strength of Na2SO4 N1 = ---------- N

Volume of BaCl2 V2 = --------- mL

Strength of BaCl2 N2 = -----?----- N

N2 = (V1 X N1) / V2

= [……..( V1) X ……..( N1)] / …….( V2)

Strength of unknown BaCl2 (N2) = --------- N

Total amount of BaCl2 present in

the given solution = [N2 x Equivalent of BaCl2 x 50]/1000

= [………..X ……… X 50]/1000

= ………..g

(Equivalent Weight of BaCl2 = 122.14)

-20-

Procedure

The burette is filled with Standard N/10 Na2SO4 solution upto the zero

level. 50mL of the given BaCl2 solution is pipetted out into a clean 100mL

beaker. The conductivity cell is placed in it and then diluted to 50mL by

adding conductivity water. The two terminals of the cell are connected with

a conductivity bridge.

Now 1ml of Standard N/10 Na2SO4 from the burette is added to the

solution, taken in the beaker, stirred, and then conductivity is measured. This

is continued upto the end point. (The conductivity is going on decreasing

upto to the end point). After the end point, again Standard N/10 Na2SO4 is

gradually added and few more reading are noted.

Thus the conductivity is continuously measured for each addition of

Standard N/10 Na2SO4 and are tabulated. Now the graph is plotted between

the volume of Standard N/10 Na2SO4 and conductivity. From the graph, end

point is noted and hence amount of BaCl2 present in 1 litre is calculated.

Result

The amount of BaCl2 present in 1litre of the given solution ---------g

-21-

WORK SHEET

Titration I : Standardization of EDTA

Burette solution : EDTA

Pipette solution : Standard ZnSO4




Standard ZnSO4 Vs EDTA solution

Sl.

No

Volume of

ZnSO4 (mL)


EDTA

(mL )

Concordant

Value (mL) Initial Final

Calculations : Volume of ZnSO4 V1 = 20 ml

Strength of ZnSO4 N1 = 0.01 N

Volume of EDTA V2 = ……

Strength of EDTA N2 = ……?

= (V1 x N1) / V2

= (…… x …….) / …..

= ……..N

-22-

Ex.No. Date:

ESTIMATION OF COPPER IN BRASS BY EDTA METHOD

Aim To estimate the amount of copper present in the given solution being

supplied with a standard zinc sulphate and EDTA solution

Principle

Brass is an alloy of Cu and Zn. It also contains small amounts of lead,

tin or aluminium. In this method, brass is dissolved in concentrated nitric

acid so as to bring copper into cupric ions. Fast sulphon black- F is added to

this solution to impart a purple color by the formation of a weak complex.

Cu2+

+ FSB-F � [ Cu2+

- FSB –F]

Purple complex

When EDTA is adder to this, FSB- F in the complex is replaced by EDTA as

per the following reaction

[ Cu2+

- FSB –F] + EDTA � [ Cu2+

- EDTA] + FSB –F

Green Color

Procedure

Titration I : Standardisation of EDTA solution

The given EDTA solution was taken in the burette. 20 mL of the

standard zinc sulphate solution was pipetted out into a clean and dry conical

flask. To this added a 5 ml of ammonia buffer and 2 drops of EBT indicator.

The solution was then titrated against the EDTA solution taken in the burette

till the color changes from wine red to steel blue. Titrations were repeated

till a concordant value was obtained.

-23-

WORK SHEET

Titration II : Estimation of Copper

Burette solution : EDTA

Pipette solution : Copper Sulphate


Indicator : FSB -F

End point : Change of colour from Purple to Green

Standard EDTA Vs CuSO4 Solution

Sl.

No

Volume of

CuSO4 (mL)


EDTA

(mL )

Concordant

Value (mL) Initial Final

Calculations : Volume of EDTA V1 = ……..

Strength of EDTA N1 = ……..

Volume of CuSO4 V2 = 20 mL

Strength of CuSO4 N2 = ……? (V1 x N1) / V2

= (…… x …….) / …..

= ……..N

Hence, Weight of copper present in the

whole of the given solution = 63.54x (----------N2) = -----------g

10

The percentage of copper in brass = [Wt.of copper / Wt.of brass] x 100

= [………… / 1] X 100

= -------------%

(Equivalent weight of copper = 63.54)

-24-

Titration II : Estimation of copper in brass

20 mL of the given copper sulphate solution was pipetted out into a

clean and dry conical flask. To this added a 5 mL of ammonia buffer and

two drops of Fast sulphon black - F indicator. The solution was then titrated

against the EDTA solution taken in the burette till the color changes from

purple to green. Titrations were repeated till a concordant value was

obtained.

Result

Amount of copper present in the whole of the given solution = ……

Percentage of copper in the given brass sample = ……%

-25-

lab manual sem ii - · pdf filethe titration is repeated to get concordant values. let the...

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