Observation (Standardisation of KMnO
4)
1)Solution in burette: ~0.01N KMnO
4
2)Solution in conical flask: 0.01N sodium oxalate + 20 mL of 2N H
+
3) Indicator: KMnO
4 is self indicator
4) Colour change: colourless to Pink
5) Reaction
MnO
4
-
(aq)+ C
2O
4
2-
(aq)+ H
+
(aq)→Mn
2+
(aq)+ CO
2(g)+ H
2O
(l)
balance?
Observation table
Burette reading
(B.R.)
Piolet readingI (mL)II (mL)III (mL)Constant
‘A’
Initial
9-10 mL
10.020.030.1
10.0 mLFinal 20.030.1 40.1
Difference 10.010.1 10.0
Observation (Estimation of Fe)
1)Solution in burette: ~0.01N KMnO
4
2)Solution in conical flask: Fe
2+
solution + 20 mL of 2N H
+
3) Indicator: KMnO
4 is self indicator
4) Colour change: colourless to Pink
5) Reaction
MnO
4
-
(aq)+ Fe
2+
(aq)+ H
+
(aq)→Mn
2+
(aq)+ Fe
3+
(aq)+ H
2O
(l)
balance?
Observation table
Burette reading
(B.R.)
Piolet readingI (mL)II (mL)III (mL)Constant
‘B’
Initial
9-10 mL
10.020.030.0
10.0 mLFinal 20.030.0 40.0
Difference 10.010.0 10.0
Calculations
1.Normality of KMnO
4
N1XV1 (sodium oxalate) = N2 X V2 (KMnO
4) V2= B.R.=A= 10 mL
0.01X 10 mL = N2 X 10, →N2= 0.01 N
2. Normality of Fe
2+
solution
N1XV1 (Fe
2+
) = N2 X V2 (KMnO
4) V2= B.R.=B=10 mL
N1 X 10 mL = 0.01 X 10, →N1= 0.01 N
3. Amount of Fe
2+
in 1000 mL
=0.01X equivalent weight of Fe (55.85/1)
=0.01 X 55.85
= 0.5585 g/L
4. Amount of Fe
2+
in 100 mL (i.e. in 0.4 g of Mohr’s salt)
0.5585 →1000mL
‘X’ →100 mL
X= 0.05585 g
Calculations
Experimental percentage of Fe in 0.4000 g of Mohr’s salt
=
0.05585
0.4000
×100=13.96 %
Molecular formula
(NH
4)
2Fe(SO
4)
2.’X’H
2O
Molecular
weight
of
(NH
4)
2Fe(SO
4)
2
‘A’ in g
Molecular
mass of
XH
2O
‘B’ in g
A + B = ‘C’Atomic
weight of Fe
% of Fe
=(55.85/C)*100
(NH
4)
2Fe(SO
4)
2.0H
2O
283.93
0*18.02283.93
55.85 g
19.67
(NH
4)
2Fe(SO
4)
2.1H
2O 1*18.02 301.95 18.49
(NH
4)
2Fe(SO
4)
2.6H
2O 6*18.02392.05 14.24
(NH
4)
2Fe(SO
4)
2.7H
2O 7*18.02410.07 13.61
14.24-13.96=|0.28|=0.28
13.61-13.96=|-0.35|=0.35
13.96~14.24 ↔6 moles H
2O