Question
Asked 19th Sep, 2013

What is the best way to test the presence of Nitric acid (HNO3) in solution?

Are there any indicators that can be used to check the presence of nitric acid in liquid?

Most recent answer

31st Jul, 2019
Rajeev Jain
Central Forensic Science Laboratory Chandigarh
Diphenylamine may be used as a wet chemical test for the presence of the nitrate ion. In this test, a solution of diphenylamine and ammonium chloride in sulfuric acid is used. In the presence of nitrates, diphenylamine is oxidized, giving a blue coloration.

All Answers (5)

19th Sep, 2013
Andrzej Szymanski
Poznan University of Technology
Okky
I agree that the ringed method (the method of brown ring) is fast and effective for determining the presence of nitrate(V) ions in the solution. However, this is not definitive method, and its positive result can not be authoritatively regarded as evidence of the presence of nitric(V) acid in a solution. A very similar effect also give NO2-ions, and their presence is not necessarily related to the presence of nitric(V) acid.
23rd Sep, 2013
Jörg Acker
Brandenburg University of Technology Cottbus - Senftenberg
The tests described here are suitible to determine the presence of nitrate ion. If you with to determine this, then it is OK. However, if you want to know if you have nitric acid present, i.e. the undissociated acid, HNO3, then you shoud use Raman spectroscopy. You can find the bands at 1304 cm-1, 960 cm-1, 690 cm-1 and 645 cm-1. Refs: N. Minogue, E. Riordan, J.R. Sodeau, J. Phys. Chem. A 107 (2003) 4436, M. Steinert, J. Acker, M. Krause, S. Oswald, K. Wetzig, J. Phys. Chem. B 110 (2006) 11377.
1 Recommendation
31st Jul, 2019
Rajeev Jain
Central Forensic Science Laboratory Chandigarh
Diphenylamine may be used as a wet chemical test for the presence of the nitrate ion. In this test, a solution of diphenylamine and ammonium chloride in sulfuric acid is used. In the presence of nitrates, diphenylamine is oxidized, giving a blue coloration.

Similar questions and discussions

How do I calculate the gas concentration in PPM?
Question
6 answers
  • Kaleem UllahKaleem Ullah
I have asked two times the following question:
I want to calculate the gas concentration, in ppm, in the closed chamber of volume 1025 cm3. I have a cylinder of mixture of NH3 and air with the 10 and 90% ratio respectively. we are flowing this gas mixture to the chamber at flow rate of 100 sccm. Now how can i calculate the NH3 concentration in ppm inside the closed chamber ?
But my seniors and teachers answer it in a different manner I don’t know to which answer I believe, I am bit confused… the answers I got in response of my questions are as:
Hi Kaleem,
the concentration in ppm does not depend on flow rate or volume as long as the chamber has been evacuated before you introduce the mixture of NH3/air. The ppm measure simply says how many µg/g (weight) or molecules per million molecules (molar ratio) of NH3 there are in the mix.
Assuming that your 1:10 mix is calculated in weight, the concentration is 100,000 µg/g or ~170000 molecules/1,000,000 molecules of mix. The latter, higher molecular ratio arises from the fact that the molar mass of NH3 is lower than the average for air, i.e. per unit weight there are more molecules of NH3 than of the gas mix that makes up air.
If your gas mix of 1:10 is in molecules/molecules (partial pressures), the concentration of NH3 is ~59,000 µg/g or 100,000 molecules/1,000,000 molecules of mix.
The 2nd way of calculations of the question (We have a closed chamber having volume 1000 cm3. 1 sccm of pure oxygen was then incorporated in chamber from flow controller. Air is already present in chamber in which oxygen is 20.95%. How we can calculate the concentration of oxygen in ppm in that chamber?):
a) If we suppose that ppm are cm^3/m^3, we can calculate oxygen concentration as:
V(O2)/V(total) = (209.5+x)cm^3/10^-3m^3.
b) If ppm = mg/l:
m(O2)/V(total) = (209.5+x)*32/(Vm*1).
Here x= is volume of oxygen added, x(cm^3)=1*t ((cm^3/min)*min), t(min) is time during which oxygen was added, 32 is molar mass of O2, Vm is molar volume, 1 = 1litre, volume of your chamber.
The 3rd way of calculations :
xA = mA / mtotal
Parts Per Million (ppm) = xA*106
i.e mole fraction of component (XA *106 )
There are other answers even I know that they are not right….… Now tell me the right way which is most accurate to calculate the gas concentration in PPM......Thanx

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