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EXPERIMENT 1 AIM: To determine the no. of water molecules of crystallization in Mohr’s salt provided standard K 2Cr2O7 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, K 2Cr2O7 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 6H2 O] is oxidized by potassium dichromate (K 2Cr2O 7) 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
-------------------------------------------------------------Cr2O 72- + 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
Cr2 O 7 2(Colourless)
(Blue/Violet)
Diphenylamine
Diphenyl benzide
OBSERVTION TABLE: SNo.
Vol. of Burette titrate taken reading (Mohr’s (initial)(ml) salt)(ml)
Burette reading (final)(ml)
1 2 3
20ml 20ml 20ml
12.8 20.3 28.6
5.5 12.8 20.3
Vol. of titrant (K 2Cr2 O7 ) used(ml) (Final-Initial) 7.3 7.5 7.9
CALCULATIONS: Applying law of equivalence, N1V 1 = N2V 2
N1= Normality at Mohr’s Salt N2= Normality of K 2Cr2O 7 V1= Volume of Mohr’s salt V2= Volume of K 2Cr2O 7
N1 x 20 = N1 =
1
x V2
10 𝑉2 7.6
200
=
200
= 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)
284 + 18𝑥 Strength of hydrated salt = 284 Strength of anhydrated salt 20 284+18𝑥 = 𝑦 284 18x =
20 x 284 – 284 𝑦
20 x 284 18x = – 284 10.79
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 K 2Cr2 O7 solution using potassium ferricyanide [K 3Fe(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: K 2Cr2 O7 , [K 3Fe(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 K 2Cr2 O7 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.[K 3Fe(CN)6] is used as an external indicator gives a greenish blue colour due to formation of ferro-ferricyanide complex.
OBSERVATION TABLE: S.No.
Burette Vol. of titrate taken reading (initial)(ml) (Mohr’s salt)(ml)
Burette reading (final)(ml)
1 2 3
20ml 20ml 20ml
10.1 19.9 30.2
0 10.1 19.9
CALCULATIONS: Applying law of chemical equivalence, N1V 1 = N2V 2
N1 x 20 =
N1 =
𝑉2
200
1
10
=
x V2 10
200
= 0.05
Strength of Fe2+ in the soln = N1 x eq.wt. =
=
𝑉2
200 1
20
x 56
x 56
= 2.8 g/l
Vol. of titrant (K 2Cr2 O7 ) used(ml) (Final-Initial) 10.1 9.8 10.3
% of Fe
=
56
200
x V2 x
100 𝑥
= 2.8 x 5 = 14 %
Hence % of iron in iron soln is 14%
RESULT: The % of iron in iron soln is 14%.
(x = 20 given)
EXPERIMENT 3
Screw Pinch Cock
B
160 140 120
Drops
100 80
Graph
60
Linear (Graph)
40 20 0 0
50
100
Composition of A in mixture
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-
CO32- +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 Case II Case III Case IV Case V
P=O P=1/2M P<1/2M P>1/2M P=M
NIL NIL NIL (2P-M) P=M
NIL 2 PORM 2P 2 (M-P) NIL
INDICATORS: Phenol phtalien and methyl orange. END POINTS: Pink to colourless (phenolphthalein), Yellow to red (methyl orange)
M NIL (M-2P) NIL NIL
OBSERVATIONS: (a)
Reading using phenolphthalein
S.No.
1 2 3 (b)
Vol. of soln titrate)token in the titration flask(ml)
Burette reading
Initial (V1) 0 11.5 22.5
20 20 20
Final (V1) 7 18.2 29.0
Vol. of the titration used(V1-V1) Final-Initial 7 6.7 6.5 Mean=6.73
Reading using methyl orange
S.No. 1 2 3
Vol. Soln taken(ml) 20 20 20
Initial 7 18.2 29
Final 11.5 22.5 33.2
Final-Initial 4.5 4.3 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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