I have several litres of ammonium nitrate solution that I do not know the concentration of, but is approximately 0.6M. I don't have precise enough instruments to determine it via its density. I titrated it with NaOH, since the ammonium ion is a weak acid, but it is too weak and the endpoint is impossible to discern clearly. I then realised of course I could just evaporate the water and weigh the solute, but the question still stuck with me: how could I determine its concentration by chemical means? The only other way I can think of is to measure its conductance and compare this value to existing data.

  • $\begingroup$ Try adding a precise amount to excess standard NaOH, carefully distilling off the ammonia and titrating the remaining NaOH with HCl. $\endgroup$
    – jimchmst
    Nov 30, 2023 at 20:32
  • $\begingroup$ If you have a good balance and a 100 ml volumetric flask you can determine the density of the solution directly. $\endgroup$
    – MaxW
    Dec 2, 2023 at 10:17
  • $\begingroup$ @MaxW, I tried that but the values I got are way off -- less dense than pure water. The balance is only to 2 d.p. so I suspect it is not good enough. $\endgroup$
    – Rafael
    Dec 2, 2023 at 15:49

1 Answer 1


As long as the solution is pure, your idea of making standardized $\ce{NH4NO3}$ solutions and comparing conductance should work.

Caveat: This works as long as conductivity vs. concentration is a monotonic function. For some other substances is not, e.g., for $\ce{H2SO4}$, which, at STP, has a distinct peak in conductance at ~30%, and then drops off, so that there are two points on the curve at which conductivity is the same, but concentration quite different.

This is also the case for $\ce{NH4NO3}$ above 8.0 m/mol kg-1, but, in the region in which you are interested, it should not be an issue.

However, there is another, simpler, rather old alternative: use a hydrometer. The density of $\ce{NH4NO3}$ is a monotonic function. And others have already done the work for you, charting density against concentration.


  • $\begingroup$ Thank you @DrMoishe. I have learned something new about monotonic functions! $\endgroup$
    – Rafael
    Dec 2, 2023 at 16:16

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