Ammonia is a weak base that reacts with water according to this equation:

NH3(aq)  +   H2O(l)     ⇌     NH4+(aq)  +    OH −(aq)

Will Fe(NO3)3 decrease the percent of ammonia that is converted to the ammonium ion in water?

  • $\begingroup$ @Zenix ecause I'm only considering Fe(NO3)3 or NaOH $\endgroup$ Apr 16 '20 at 12:31
  • $\begingroup$ I expect $\ce{Fe^3+ + 3 NH3 + 2 H2O -> FeO(OH) + 3 NH4+}$ $\endgroup$
    – Poutnik
    Apr 16 '20 at 12:54
  • 1
    $\begingroup$ Well, if the substance on the left is called ammonia and on the right ammonium, and if the substance on the left is converted to the substance on the right, then the conclusion is ? $\endgroup$
    – Poutnik
    Apr 16 '20 at 12:57
  • 1
    $\begingroup$ I reversed my downvote in response. And @MaxW nailed it in his comments: adding hydroxide ions forces the equilibrium to the left. $\endgroup$
    – Ed V
    Apr 16 '20 at 19:29
  • 1
    $\begingroup$ (+1) For doing it right! $\endgroup$
    – Ed V
    Apr 16 '20 at 22:34

If ammonia gas is bubbled into pure water, the following reaction takes place:

$$\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH−(aq)}\tag{1}$$

How to decrease the percent of ammonia that is converted to the ammonium ion in water?

Le Chatelier's Principle indicates that adding $\ce{OH-}$ would push the equilibrium to the left.

I lifted the following figure from:

by Airton Kunz and Saqib Mukhtar
Eng. Agríc. vol.36 no.2 Jaboticabal Mar./Apr. 2016

enter image description here

As can be seen in the figure, an increase in pH favors formation of $\ce{NH3}$ and a decrease in the pH favors $\ce{NH4+}$.

Since the OP asked only for the percent of $\ce{NH4+}$ to be reduced, there is another way. Dilute the solution.

Will $\ce{Fe(NO3)3}$ decrease the percent of ammonia that is converted to the ammonium ion in water?

Iron (III) forms an relatively insoluble hydroxide as shown by the following equation:

$$\ce{Fe^{3+} + 3OH- <=>[K_{\mathrm{sp}} = 2.79\times10^{−39}] Fe(OH)3(s)}\tag{2}$$

So again according to Le Chatelier's Principle removing $\ce{OH-}$ from reaction (1) favors the reaction moving to the right. Thus the precipitation of $\ce{Fe(OH)3}$ will increase the formation of $\ce{NH4+}$.

In a comment on the question user Poutnik indicated the reaction $\ce{Fe^{3+} + 3NH3 + 2H2O -> FeO(OH) + 3NH4+}$. I'm sure there is a study somewhere on $\ce{Fe(OH)3 \pu{ vs } FeO(OH)}$ formation, but I didn't find it in 15 minutes of fooling around. I'd expect that strong $\ce{OH-}$ concentrations and higher temperatures (boiling solution) would favor $\ce{FeO(OH)}$.


Short answer lower pH! [EDIT Correction] raise pH.


$\ce{ NH4+ <=> H+ + NH3 (g) }$

In neutral to basic conditions, this reaction undesirably [EDIT Correction] desirably moves to the right liberating ammonia, especially on warming.

Good news, in acidic conditions the NH4+ ion is hard to chemically attack. To quote a supporting source:

Acidic salts, such as AMMONIUM CHLORIDE, are generally soluble in water. The resulting solutions contain moderate concentrations of hydrogen ions and have pH's of less than 7.0. They react as acids to neutralize bases... They usually do not react as either oxidizing agents or reducing agents but such behavior is not impossible.

So, paths to lower pH, like adding CO2 or salts like Ferric nitrate that undergo a complex formation with water liberating H+. Also, Aluminum sulfate is commonly used to adjust pH and results in a helpful precipitation of Al(OH)3 to quote a source:

Freshly precipitated aluminium hydroxide forms gels, which are the basis for the application of aluminium salts as flocculants in water purification.

  • 1
    $\begingroup$ Zenix: Seemingly, as Aluminum sulfate in dilute solutions breakdowns liberating Al(OH)3 and acidic H2SO4 which has also been used to treat ground soil that is too alkaline. The 1st comment relates to the stability of the NH4+ anion (very hard generally to attack, but I have accomplished it in practice with HOCl). $\endgroup$
    – AJKOER
    Apr 16 '20 at 18:00
  • 3
    $\begingroup$ WRONG!!! You have it backwards. Lowering the pH favors production of $\ce{NH4+}$. Raising the pH favors production of $\ce{NH3}$. $$\ce{NH3 +H2O <=>[lower pH][raise pH] NH4+ + OH-}$$ $\endgroup$
    – MaxW
    Apr 16 '20 at 18:07
  • 2
    $\begingroup$ I agree with@MaxW $\endgroup$
    – Zenix
    Apr 16 '20 at 18:09
  • $\begingroup$ Agree, but my use of the word 'So', may suggest otherwise, so slight rewording. Deleted 'So' and added 'undesirably'. $\endgroup$
    – AJKOER
    Apr 16 '20 at 18:33
  • 4
    $\begingroup$ The OP asked for less $\ce{NH4+}$ and more $\ce{NH3}$. Lowering the pH means that you add an acid to the solution which favors $\ce{NH4+}$ production. // The OP should raise the pH. $$\ce{NH4+ <=>[raise pH (add base)][lower pH (add acid)] NH3 + H+}$$ look at reaction this way. $$\ce{NH4+ + OH- <=> NH3 + H2O}$$ $\endgroup$
    – MaxW
    Apr 16 '20 at 19:01

Here is a source: 'Ionization of Ammonia in Seawater: Effects of Temperature, pH, and Salinity' which may be of value. To quote from the abstract:

Previous studies have shown that the toxicity of ammonia to freshwater fishes depends on the ambient concentration of the un-ionized fraction (NH3) and have quantified the percent NH3 in solution as a function of temperature and pH. If NH3 is also toxic to marine fishes, the effect of salinity on the ionization of ammonia must also be considered. The percent NH3 over ranges of temperature, pH, and salinity common in seawater-culture situations is presented here.


In the water, ammonia can be found in its ionized (NH4+) or unionized form (NH3); the sum of the two is the total ammonia. Their proportion in the water depends on pH and temperature (Bower and Bidwell, 1978). Many studies confirm that non-ionized ammonia is chemically more toxic due to its ability to diffuse through cell membranes,...Studies show that at lower salinity levels, higher levels of ammonia nitrogen are measured...The reason for these differences should be at least partially, the solubility of ammonia, lower in water containing salt (Bower and Bidwell, 1978). Consistently, sea water (32-40‰) displays~20% less NH3 than in fresh water with pH 7.5-8.5 and at 25°C (Bower and Bidwell, 1978).

I do not have access, but you may wish to gain access to examine the charts.

So, both increasing temperature and pH, increases free NH3 and seawater decreases free ammonia by roughly 20%. This agrees with a temperature vs. pH table on Page 3 of this available EPA report. To quote a comment relating to salt:

In saline or very hard waters there will be small but measurable decreases in the percent NH3.

Note: NH3 (aq) will react with Fe(NO3)3 consuming ammonia and producing NH4NO3 (aq).


Not the answer you're looking for? Browse other questions tagged or ask your own question.