I think my main question/concern really relates to the formation of a product that is listed as "reacting with water" after the combination of two aqueous solutions.

For example, both sodium iodide (NaI) and iron(III) chloride (FeCl3) form aqueous solutions. When combined, the expected products from double replacement would be sodium chloride (NaCl) and iron(III) iodide (FeI3). Equation something like this (Where the ? indicates the root of my question):

$$\ce{3 NaI (aq) + FeCl3 (aq) -> 3 NaCl (aq) + FeI3 (?) }$$

Now, FeI3 is listed as "reacting with water" for its solubility. Does this mean that FeI3 is, in fact, not produced, but rather begins decomposing into iron(II) iodide (FeI2) and elemental iodine (I2)? Or do we get some form of iron compound with an iodine oxyanion like Fe(IO3)3 and hydrogen gas (H2)?

And in either case, would this reaction be said to take place or not? I'm not really sure what my end goal is here, I just don't understand how an ionic compound that might be formed from aqueous solutions will behave in "reacting with water" considering its surrounded by water from the two aqueous solutions.

Thanks in advance for the help!

  • 3
    $\begingroup$ $$\ce{2Fe^3+(aq) + 3I-(aq)-> 2Fe^2+(aq) + I3-(aq)}$$ $\ce{I2}$ is not very soluble in water unless it combines with an $\ce{I-}$ to form $\ce{I3-(aq)}$. The reaction is not with water but rather in aqueous solution. $\endgroup$
    – Karsten
    Commented Jan 19, 2023 at 15:35
  • 1
    $\begingroup$ Related: chemistry.stackexchange.com/q/51610/17175. Note that FeI3 is known only in nonpolar environments or as a soft-base complex. Forming a salt by double displacement in water solution does not offer either stabilizing condition. $\endgroup$ Commented Jan 19, 2023 at 15:49
  • $\begingroup$ @OscarLanzi So then it would be simplest to identify this specific situation, in terms of double displacement for these solutions, as being a situation where "no reaction" takes place? $\endgroup$ Commented Jan 19, 2023 at 16:34
  • 2
    $\begingroup$ I would say it is a reaction, but not double displacement. Instead, with sodium and chloride as spectators, we are looking at a redox reactions between iron(III) and iodide. $\endgroup$ Commented Jan 19, 2023 at 17:29
  • $\begingroup$ Addressing the general topic of the title itself: There is a wide range of reactivity, since "so reactive the solution just does not make sense" to "reaction does not allow to reach dissolution equilibrium". $\endgroup$
    – Poutnik
    Commented Jan 20, 2023 at 9:54

1 Answer 1


It seems that your textbook or teacher is highlighting the fact that some products formed in water-based reactions can decompose into other substances. It is important to note that many salts are not stable in water. For example, anhydrous aluminum chloride will immediatly react with water and iron (III) chloride, although soluble in water, will begin to hydrolyze and form solid, brown-colored hydrated iron oxides within hours at neutral pH.

In the specific reaction mentioned in the question, it's important to keep in mind that it is likely to be a pretty complex reaction with multiple side products. For example, when the two solutions are mixed, free iodine will be released due to the mild oxidizing properties of iron (III) chloride. This free iodine can potentially react with NaI to form a triiodide ion. Additionally, the iron (III) chloride can hydrolyze in water to form other insoluble side products. Overall, it can be difficult to predict the outcome of this reaction at neutral pH. Furthermore, if ferrous iodide is formed, it will quickly react with oxygen in the air. This complexity is a characteristic of many chemical reactions in the real world. This is why chemistry is still an experimental science. Nobody should ask this question in the exam.


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