I have one gallon of commercial, surfactant free, ammonium hydroxide solution that I would like to use to prepare a concentrated solution.

I have two options as I see it, and I am still getting into the chemistry side (amateur chemistry), but am pushing ahead and want to do this even without the understanding. I am hoping the solutions will provide adequate explanation as to what is being done, and why, which will help me learn to solve problems like this in the future. Chemistry newbie, here.

Solubility at room temperature is to be assumed.

I see that I have two options.

1) Prepare a receiving flask with cold distilled water (likely in an ice bath) and then prepare a distillation flask with the %10 ammonium hydroxide solution. Slightly heat the distillation flask to the point where the ammonia gas begins to off-gas, which is funneled into the receiving flask and dissolve.

2) Prepare a receiving flask and distillation flask both with an amount of the 10% ammonium hydroxide solution. As above, gently heat the distillation flask to off-gas the ammonia into the receiving flask where it should dissolve, and increase the concentration beyond 10%.

Since I would like a 1 liter solution containing 28% NH3(aq), what volume of cold distilled water should be placed in the receiving flask, which I ask because I assume the solvation of NH3 into water will cause the volume to increase. If I placed 1 liter of water into the flask initially, then this would result in more than 1 liter of solution. This also depends on the amount of ammonia that needs to be dissolved to reach my desired concentration, which varies from option #1 to #2.

So what I want to know is,

Q1) How to determine the volume of a solution given a specific concentration of NH3(aq). This will allow me to account for the increase in volume when adding ammonia to an existing solution, in order to reach a desired concentration.

Q2) How do I know when I have reach %28 NH3(aq) concentration?

Q3) A procedure leading from 10% to 28%, and any advice on more technically correct approaches to this problem. Is it a problem that should be considering molarity, etc.?

Q4) Any quick background info that I can look up to help me understand this problem would be welcome!

Bottom line: Prepare a 1 liter solution of 28% NH3(aq) from 1 gallon of 10% NH3(aq), and please provide an explanation.

Thank you for any help. I am a software programmer, and have delved into he world of chemistry as a hobby, and am getting confused by terms like molarity, normality, and both of those versus concentration, etc. It's maddening to say the least.

  • $\begingroup$ Why are you using archaic "NH4OH"? It's Simply NH3 water solution. $\endgroup$
    – Mithoron
    Jan 15, 2015 at 13:43
  • $\begingroup$ I am learning chemistry here. Thank you for the input. I see it referenced as NH4OH, but I think that is what it exists as in an equilibrium in the solution. And archaic? Not sure how that is archaic, as I would think "liquor ammonia" would be the archaic term. Just answer my question, please. I try to be as accurate as possible in my question/description. $\endgroup$ Jan 15, 2015 at 14:35
  • $\begingroup$ NH3 is a weak base so degree of dissotiation is small - writing NH4OH implies total dissotiation, but it was written like that in old books. $\endgroup$
    – Mithoron
    Jan 15, 2015 at 15:55
  • $\begingroup$ Interesting. I knew NH3 was a weak base, and only partially ionized in solution, but could you point me to a place where I can read about these implications? I didn't know that NH4OH actually meant anything other than a molecule formation. I now see where your comment was coming from, because I wasn't sure how to take it exactly. $\endgroup$ Jan 15, 2015 at 16:02
  • $\begingroup$ NH4OH isn't archaic at all. Aqueous ammonia is sold as ammonium hydroxide: Sigma-Aldrich $\endgroup$
    – Kyle
    Jan 16, 2015 at 2:58

1 Answer 1


Disclaimer: ammonia is extremely hazardous including flammable in some vapor concentrations.

How do I know when I have reach %28 NH4OH concentration?

You can measure the specific gravity and compare to the table on page 2 of this document:

Next, it is important to understand that if you heat/boil/distill the 10% solution the vapor coming off is not 100% ammonia, it is a mixture of ammonia and water, but with increased ammonia concentration.

To predict the percent ammonia in the vapor phase, you need to look at the dew point line in a table like Fig. 1.2 here, but that one is for 16 bars or Fig. 2 of Ganesh et al. Journal of Mechanical Engineering Research Vol. 3. (1), pp. 25-39 that shows many pressures including 1 bar.

As the original solution is heated, the system progresses toward the upper left in Fig. 1.2/Fig.2, the liquid phase along the bubble point line and the vapor phase along the dew point line. The concentration of ammonia in both phases progressively decreases. Eventually, the concentration of ammonia in the vapor phase drops below the concentration of ammonia in the original solution.

From Fig. 2 of Ganesh et al. it can be seen that at 1 bar starting from 10 weight% ammonia liquid phase, the initial vapor phase concentration would be close to 28% ammonia.

This shows that even if you added no water to the condensate, there is no way you can get 1 liter of 28% ammonia from 1 gallon of 10% ammonia in one simple distillation step.

Only through fractional distillation could you get that much 28% ammonia from that little 10% ammonia.

Also, practically speaking, it will be difficult to avoid lossing ammonia vapors to the environment and safety needs to be considered.

  • $\begingroup$ I mean a 28% NH3(aq) solution, where I believe by weight, 28% of its weight consists of dissolved NH3(g). I realize there are NH4+ and OH- ions in solution, but I think I was trying to calculate the amount of actual NH4OH in solution, rather than the dissolved NH3(g), which is partly where my confusion comes from. $\endgroup$ Jan 15, 2015 at 14:42
  • $\begingroup$ I got the idea of gently heating a NH3(aq) solution below the boiling point of water to produce NH3(g) here: flinnsci.com/media/620542/91231A.pdf $\endgroup$ Jan 15, 2015 at 14:45
  • $\begingroup$ My questions are more with how do I predict how much water I will need to arrive at 1 liter of 28% NH3(aq) as I gas it with NH3(g), because of the volume expansion. You answered my question about determining the NH3(aq) concentration, but I was also thinking about weighing it and determining completion based on weight gained. $\endgroup$ Jan 15, 2015 at 14:47
  • $\begingroup$ How do you calculate the volume of a liquid when a gas has become dissolved within it, as in the case here? 100g of water has a volume of approx. 100ml, but what about when I dissolve 100g of NH3(g) into it. What effect on volume does that have? $\endgroup$ Jan 15, 2015 at 14:53
  • $\begingroup$ I just checked out the MSDS, and I must point out it is for Anhydrous Ammonia, not gaseous wet ammonia. I think, but might be wrong, but there is a stark difference. Yes, both dangerous in their own ways, but one won't kill you just by being around it. $\endgroup$ Jan 15, 2015 at 15:09

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