The document describes an experiment to determine the water of crystallization in Mohr's salt, (NH4)2Fe(SO4)2.XH2O, by titrating a solution of the salt with standardized potassium manganate(VII). The experiment involves standardizing the KMnO4 solution using sodium oxalate, then titrating samples of Mohr's salt solution and calculating the iron content. Based on comparing the experimental iron percentage to theoretical percentages for different values of X, the water of crystallization is determined to be 6 moles, giving the formula (NH4)2Fe(SO4)2.6H2O.

1. Estimation of water of
crystallization in Mohr’s salt by
titrating with standardized KMnO4
Dr. Mithil Fal Desai
Shree Mallikarjun and Shri Chetan Manju
Desai College Canacona Goa
(NH4)2Fe(SO4)2.’X’H2O

2. Theory
The aqueous solution of KMnO4 needs to be standardized using sodium
oxalate. The amount of Fe is quantitatively measured in the fixed amount of
Mohr’s salt. The Mohr’s salt molecular formula is (NH4)2Fe(SO4)2.’X’H2O,
thus X can be determined by comparing the percentage of Fe with different X
values.
Standardization of KMnO4
2KMnO4 + 5H2C2O4 + 3H2SO4 → 2MnSO4 + K2SO4 + 10CO2 + 8H2O
Estimation of Fe(II) in Mohr’s salt
2KMnO4 + (NH4)2Fe(SO4)2 + 8H2SO4 → 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 10(NH4)2SO4
+ 8H2O
Reaction

3. Procedure
i) Preparation of Mohr’s salt solution :)Dissolve 0.4 g of Mohr’s salt (note the
exact weight) in ~5 mL distilled water and add 2-3 drops of concentrated
sulphuric acid. Quantitatively transfer this solution in a 100 mL standard flask
and dilute up to the mark.
ii) Standardization of KMnO4 :Rinse the clean burette with a minimum quantity
of KMnO4 and fill it appropriately. Pipette out 10 mL of 0.01 N sodium oxalate
solution in a clean conical flask. Add about 20 mL of 2N H2SO4. Heat the
solution in flask to about 60 °C. Titrate it against KMnO4, until light pink colour
permanently develops in the flask. Repeat the titration to get constant readings.
iii) Estimation of Fe(II) in Mohr’s salt :Rinse the burette with a minimum
quantity of standardized KMnO4 and fill it appropriately. Pipette out 10 mL of
given Mohr’s salt solution in a clean conical flask. Add about 20 mL of 2N
H2SO4. Heat the solution in flask Titrate it against KMnO4, until light pink colour
permanently develops in the flask. Repeat the titration to get constant readings.

4. Observation (Standardisation of KMnO4)
1) Solution in burette: ~0.01N KMnO4
2) Solution in conical flask: 0.01N sodium oxalate + 20 mL of 2N H+
3) Indicator: KMnO4 is self indicator
4) Colour change: colourless to Pink
5) Reaction
MnO4
-
(aq) + C2O4
2-
(aq)+ H+
(aq) → Mn2+
(aq) + CO2 (g) + H2O (l) balance?
Observation table
Burette reading
(B.R.)
Piolet reading I (mL) II (mL) III (mL) Constant
‘A’
Initial
9-10 mL
10.0 20.0 30.1
10.0 mL
Final 20.0 30.1 40.1
Difference 10.0 10.1 10.0

5. Observation (Estimation of Fe)
1) Solution in burette: ~0.01N KMnO4
2) Solution in conical flask: Fe2+ solution + 20 mL of 2N H+
3) Indicator: KMnO4 is self indicator
4) Colour change: colourless to Pink
5) Reaction
MnO4
-
(aq) + Fe2+
(aq)+ H+
(aq) → Mn2+
(aq) + Fe3+
(aq) + H2O (l) balance?
Observation table
Burette reading
(B.R.)
Piolet reading I (mL) II (mL) III (mL) Constant
‘B’
Initial
9-10 mL
10.0 20.0 30.0
10.0 mL
Final 20.0 30.0 40.0
Difference 10.0 10.0 10.0

6. Calculations
1.Normality of KMnO4
N1XV1 (sodium oxalate) = N2 X V2 (KMnO4) V2= B.R.=A= 10 mL
0.01X 10 mL = N2 X 10, → N2= 0.01 N
2. Normality of Fe2+ solution
N1XV1 (Fe2+) = N2 X V2 (KMnO4) V2= B.R.=B= 10 mL
N1 X 10 mL = 0.01 X 10, → N1= 0.01 N
3. Amount of Fe2+ in 1000 mL
=0.01X equivalent weight of Fe (55.85/1)
=0.01 X 55.85
= 0.5585 g/L
4. Amount of Fe2+ in 100 mL (i.e. in 0.4 g of Mohr’s salt)
0.5585 → 1000mL
‘X’ →100 mL
X= 0.05585 g

7. Calculations
Experimental percentage of Fe in 0.4000 g of Mohr’s salt
=
0.05585
0.4000
× 100 = 13.96 %
Molecular formula
(NH4)2Fe(SO4)2.’X’H2O
Molecular
weight
of
(NH4)2Fe(SO4)2
‘A’ in g
Molecular
mass of
XH2O
‘B’ in g
A + B = ‘C’ Atomic
weight of Fe
% of Fe
=(55.85/C)*100
(NH4)2Fe(SO4)2.0H2O
283.93
0*18.02 283.93
55.85 g
19.67
(NH4)2Fe(SO4)2.1H2O 1*18.02 301.95 18.49
(NH4)2Fe(SO4)2.6H2O 6*18.02 392.05 14.24
(NH4)2Fe(SO4)2.7H2O 7*18.02 410.07 13.61
14.24- 13.96= |0.28|= 0.28
13.61-13.96=|-0.35|= 0.35 13.96 ~14.24 ↔ 6 moles H2O

8. Result
i) Experimental percentage of Fe in ~0.4000g of
Mohr’s salt is 13.96% which corresponds to 6-7
moles of water of crystallization.
ii) Water of crystallization in Mohr’s salt is X = 6.
(NH4)2Fe(SO4)2.’6’H2O