EXPERIMENT 1

AIM: To determine the no. of water molecules of crystallization in Mohr’s

salt provided standard K2Cr2O7 sol (0.1N). Using diphenylamine as

internal indicator.

APPARATUS: Burette, burette stand, pipette, beaker, conical flask, funnel,

measuring cylinder.

CHEMICAL USED: Mohr’s salt, H2SO4, H3PO3, Dipheyl amine, K2Cr2O7

THEORY: This titration is a type of redox reactions. Here, the principle

involved is that ferrous sulphate (FeSO4) present in Mohr’s salt.

[FeSO4(NH4)2 6H2O] is oxidized by potassium dichromate (K2Cr2O7)

to ferric sulphate [Fe2 (SO4)3] in the presence of dil. H2SO4 using

diphenyl amine [(C6H5)2 NH2] as internal indicator. This reaction

involved:

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

[Fe2+ → Fe3+ + e-] x 6

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

Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ +7H2O

At the end point, all the ferrous ions (Fe2+) present in the solution get

completely oxidized to ferric ions (Fe3+) by chromate ions (Cr2O72-).

As soon as a slight excess of potassium dichromate solution is

added, it brings about the oxidation of diphenylamine resulting in the

formation of a blue coloured complex named diphenyl benzidine. At

the end point a sharp change from colourful solution to deep blue

solution is observed.

H oxidation with N N N Cr2O7

2-

(Colourless) (Blue/Violet)

Diphenylamine Diphenyl benzide

OBSERVTION TABLE:

SNo. Vol. of titrate taken (Mohr’s salt)(ml)

Burette reading (initial)(ml)

Burette reading (final)(ml)

Vol. of titrant (K2Cr2O7) used(ml) (Final-Initial)

1 20ml 5.5 12.8 7.3 2 20ml 12.8 20.3 7.5 3 20ml 20.3 28.6 7.9

CALCULATIONS: Applying law of equivalence,

N1V1 = N2V2 N1= Normality at Mohr’s Salt

N2= Normality of K2Cr2O7

V1= Volume of Mohr’s salt

V2= Volume of K2Cr2O7

N1 x 20 = 110

x V2

N1 = 𝑉2200

= 7.6200

= 0.038

Strength of anhydrated Mohr’s Salt (y) = Normality x eq. wt.

= N1 x 284

= 10.792

Strength of hydrated salt = 20g/l (given)

Strength of hydrated salt

Strength of anhydrated salt =

284 + 18𝑥284

20𝑦

= 284+18𝑥284

18x = 20 x 284

𝑦 – 284

18x = 20 x 28410.79

– 284

x = 13.46

RESULT:

The no. of water molecules of crystallization in Mohr’s Salt is 6.73

EXPERIMENT 2

AIM: Determination of iron content in an iron are by titrating it against

standard K2Cr2O7 solution using potassium ferricyanide [K3Fe(CN)6],

H2SO4, FeSO4 as an external indicator.

APPARATUS: Burette, burette stand, conical flask, white glazed tile, beaker, glass

rod, measuring cylinder, funnel.

CHEMICAL USED: K2Cr2O7, [K3Fe(CN)6], H2SO4, FeSO4

THEORY: This estimation is based on the principle that the solution containing

ferrous ammonium sulphate can be quantitatively titrated against

standard K2Cr2O7 soln in the presence of H2SO4 using potassium

ferrocynide as an external indicator.

Oxides of ferrous sulphate present in Mohr’s salt into ferric

sulphate in the presence of dil H2SO4.[K3Fe(CN)6] is used as an

external indicator gives a greenish blue colour due to formation of

ferro-ferricyanide complex.

OBSERVATION TABLE:

S.No. Vol. of titrate taken (Mohr’s salt)(ml)

Burette reading (initial)(ml)

Burette reading (final)(ml)

Vol. of titrant (K2Cr2O7) used(ml) (Final-Initial)

1 20ml 0 10.1 10.1 2 20ml 10.1 19.9 9.8 3 20ml 19.9 30.2 10.3 CALCULATIONS:

Applying law of chemical equivalence,

N1V1 = N2V2

N1 x 20 = 110

x V2

N1 = 𝑉2200

= 10200

= 0.05

Strength of Fe2+ in the soln = N1 x eq.wt.

= 𝑉2200

x 56

= 120

x 56 = 2.8 g/l

% of Fe = 56 200

x V2 x 100𝑥

(x = 20 given)

= 2.8 x 5 = 14 %

Hence % of iron in iron soln is 14%

RESULT:

The % of iron in iron soln is 14%.

Screw Pinch Cock

B

EXPERIMENT 3

0

20

40

60

80

100

120

140

160

0 50 100

Dro

ps

Composition of A in mixture

Graph

Linear (Graph)

The Redwood viscometer was made by Sir Boverton Redwood in about 1880.

EXPERIMENT 7

AIM:

To determine the type & extent of alkalinity of given water sample.

APPARATUS:

Burette, pipette, conical flask, beaker, measuring flask.

CHEMICALS:

Water sample, V/10 HCl, phenolphthalein and methyl orange

indicator.

THEORY:

Alkalinity of water is mainly due to the presence of the following.

(i) Hydroxides only

(ii) Carbonates only

(iii) Bicarbonates only

(iv) Hydroxides and carbonates

(v) Carbonates & bicarbonates

Since OH- & HCO3- ion cannot co-exist because both combine

together to form carbonates.

OH + HCO3- CO3

2- +H2O

The extent of alkanity present in a water sample is determined by

titrating the water sample (titrate) with a standard acid (titrant) using

phenolphthalein indicator and alkalinity if found out in terms of

CaCO3 equivalent by using normality equation. This is called

phenolphthalein alkalinity (P). At this point, complete neutralization

of hydroxide and conversion carbonate to bicarbonate takes place.

OH- + H+ → H2O -----(i)

CO32- + H+ → HCO3

- -----(ii)

HCO3-+ H+ → H2O + CO2

-----(iii)

Now titrate the same alkality soln using methyl orange indicator &

alkalinity is calculated in terms of caco3 equivalents. This alkalinity is

called mothyl orange alkalinity, alkalinity due to diff. ions can be

calculated. The results are summarized in the following table.

Case I P=O NIL NIL M Case II P=1/2M NIL 2 PORM NIL Case III P<1/2M NIL 2 P (M-2P) Case IV P>1/2M (2P-M) 2 (M-P) NIL Case V P=M P=M NIL NIL

INDICATORS: Phenol phtalien and methyl orange.

END POINTS: Pink to colourless (phenolphthalein),

Yellow to red (methyl orange)

OBSERVATIONS:

(a) Reading using phenolphthalein

S.No. Vol. of soln titrate)token in the titration flask(ml)

Burette reading Vol. of the titration

used(V1-V1)

Initial (V1) Final (V1) Final-Initial 1 20 0 7 7 2 20 11.5 18.2 6.7 3 20 22.5 29.0 6.5

Mean=6.73

(b) Reading using methyl orange

S.No. Vol. Soln taken(ml) Initial Final Final-Initial 1 20 7 11.5 4.5 2 20 18.2 22.5 4.3 3 20 29 33.2 4.2

Mean=4.33

CALCULATIONS:

1. Phenolphthalein alkalinity in terms of CaCO3 equivalent

Acc. to Law of Equivalence,

N1V1 = N2V2

(acid) (water)

1/10(V2-V1) = N2 x 20

N2 = 6.73/200 = 0.03365

Strength in terms of CaCO3 eq. = N2 x eq. wt of CaCO3

= 0.03365 x 50 = 1.6825 g/l

P = 1.6825 x 1000 mg/l = 1682.5 mg/l

= 1.6825 x 1000 ppm = 168.25 ppm

2. Methyl orange alkanity in terms of CaCO3 equivalent

Acc. to Law of Equivalence,

N1’ V1’= N2’ V2’

(acid) (water)

1/10(V3-V1) = N2’ x 20

N2’ = 1/10 x (V3-V1)/20

RESULT:

Phenolphthalein alkalinity (P) = 1682.5ppm of CaCO3

Methyl orange alkalinity (M) = 1082.5 ppm of CaCO3

(To find out the alkalinity in terms of individual fans, find out to which

case the values of P & M falls out from table, calculate the amount of

individual ions as below.)

Alkalinity due to OH- ions = 2P-M = 2282.5 ppm

Alkalinity due to CO32- ions = 2(M-P) = 1200 ppm

Alkalinity due to HCO3- ions = Nil

PRECAUTIONS:

1. Phenolphthalein indicator should be added first & then methyl

orange.

2. The vol. of indicator should be same in all observations.

3. Constant shaking of soln mixture should be done.

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