https://www.slideshare.net/mithilfaldesai/to-estimate-the-amount-of-nitrite-present-in-the-given-nano2-solution-by-titrating-vs-ceric-ammmonium-sulphate-ceric-sulphate https://youtu.be/RO1ydGn9S48

1. Dr. Mithil Fal Desai
Shree Mallikarjun and Shri Chetan Manju Desai College
Canacona Goa
To estimate the amount of nitrite present
in the given sodium nitrite solution by
titrating v/s Ceric ammonium sulphate.
1

2. Theory: In the determination of nitrite, the ceric sulphate solution is
standardised against ferrous ammonium sulphate dissolved in air-free
dilute sulphuric acid. The effect of atmospheric oxidation during the
titration is negligible. The known volume of unknown concentration of
nitrite is oxidized with an excess of ceric ammonium sulphate and the
excess of Ce4+ is determined by titrating it against standardized ferrous
ammonium sulphate solution using a ferroin indicator
Reactions:
Ce4+ + Fe2+  Ce3+ + Fe3+
(NO2)- + 2Ce4+ + H2O  (NO3)- + 2Ce3++ 2H+
[Ph3Fe]3+
(blue) + e-  [Ph3Fe]2+
(red)
2

3. Observation (Titration 2 Estimation of nitrite.)
1) Solution in burette: 0.5N Fe2+
2) Solution in conical flask: 25 mL 0.05N Ce4+ + 10 mL (NO2)- + 2 drops of ferroin indicator
3) Indicator: Ferroin
4) Colour change: blue to red
5) Reaction
Ce4+ + Fe2+  Ce3+ + Fe3+
(NO2)- + 2Ce4+ + H2O  (NO3)- + 2Ce3++ 2H+
Observation table
Burette reading Piolet reading I (mL) II (mL) III
(mL)
Constant ‘A’
Initial
14.0-15.0 mL
15.0 30.0 45.0
15.0 mL
Final 30.0 45.0 60.0
Difference 15.0 15.0 15.0

4. 4
Calculation
A) Standardisation of Ce4+ solution
N1XV1 (Ce4+) = N2 X V2 (Fe2+)
N1 X 10 mL = 0.05 X B.R. of standardisation
For now consider it is 0.05N.

5. 5
Calculation
B) Estimation of nitrite
i) Out of 25.0 mL of 0.05N ceric salt, a =(25- Y mL) of 0.05N ceric salt solution reacted with 10 mL of XN
nitrite solution. Y mL of this excess 0.05 N ceric solution reacted with 15 mL 0.05N ferrous solution.
N1V1(excess ceric solution) = N2V2 (ferrous solution)
0.05N X Y mL =0.05N X 15.0 mL
Y = 15 mL
Volume of 0.05N ceric solution reacted with 10 mL of XN nitrite solution is a = 25.0-15.0 mL= 10 mL
ii) Therefore the strength of nitrite solution
N1V1( nitrite solution) = N2V2 (ceric solution)
XN X 10 mL = 0.05 X 10 mL
XN = 0.05 N
iii) Amount of NaNO2 in 1000 mL = N X eq. wt.