In normal displacement reactions, reactivity plays a large role and sometimes the reaction doesn't even happen. So is there anything limiting double displacement reactions? For example $\ce{2KI + Pb(NO3)2 -> 2KNO3 + PbI2}$. If reactivity plays a role in this reaction, why does it occur? Isn't the halogen ion (iodide) more reactive than $\ce{NO3-}$?

(I'm a high school chem student so this may come off as a newbie question, sorry)

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    $\begingroup$ That name (double displacement) is nothing more than a name. At university level it is not used to the best of my knowledge. It is always the overall balance of reactivity or in your case solubility that matters. $\endgroup$
    – Jan
    Commented Nov 30, 2015 at 13:10
  • $\begingroup$ @Jan could you elaborate on why potassium ion prefers the nitrate ion, if that is the case? $\endgroup$
    – Airdish
    Commented Nov 30, 2015 at 13:13
  • $\begingroup$ There is no $\ce{NO3^2-}$, only $\ce{NO3^-}$. Also, there is no such thing as reactivity in general. As for potassium ion, it does not prefer nitrate. Potassium ion does not really "know" that a reaction takes place at all. $\endgroup$ Commented Nov 30, 2015 at 13:14
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    $\begingroup$ @S.Mo It is not potassium ‘preferring’ nitrate — in fact there are no ‘preferences’ (i.e. choices) involved at all. Rather, $\ce{PbI2}$ has a very low solubility, thus iodide and lead(II) in solution will meet each other and precipitate. Potassium and nitrate are merely spectators. $\endgroup$
    – Jan
    Commented Nov 30, 2015 at 13:16
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    $\begingroup$ @S.Mo generally speaking, the alkali metal ions (Li+, Na+, K+, etc.) and ammonium as well as nitrate and perchlorate will remain in solution. chem.sc.edu/faculty/morgan/resources/solubility $\endgroup$
    – DavePhD
    Commented Nov 30, 2015 at 19:25

2 Answers 2


The simple answer is whether the reaction leads to the formation of a solid (precipitate) or a gas otherwise there is really no reaction

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    $\begingroup$ Generally speaking, that's correct, but it would be great if you could elaborate a little bit or perhaps give some examples. $\endgroup$ Commented Nov 30, 2015 at 20:15

It's often helpful to use state symbols when dealing with solutions, especially in this case. Let's say you mix $\ce{KI}$ with $\ce{NaNO_3}$. Because the $\ce{KI}$ and $\ce{NaNO_3}$ are soluble and fully dissolved (or assumed to be) in the solutions, before mixing, we write them as $\ce{KI}$ $\ce{(aq)}$ and $\ce{NaNO_3}$ $\ce{(aq)}$. The aq stands for aqueous. After you pour them together, the ions in both solutions mix into the larger volume. However, they're still fully dissolved, so no reaction happens. Before, you had $\ce{K^+}$ $\ce{(aq)}$ and $\ce{I^-}$ $\ce{(aq)}$ with $\ce{Na^+}$ $\ce{(aq)}$ and $\ce{NO_3^-}$ $\ce{(aq)}$, and that's exactly what you have afterward. ($\ce{KI (aq)}$ means the same as $\ce{K^+}$ $\ce{(aq)}$ + $\ce{I^-}$ $\ce{(aq)}$, and the same with sodium nitrate).

However, with $\ce{KI}$ $\ce{(aq)}$ and $\ce{PbNO_3}$ $\ce{(aq)}$, they are both initially fully dissolved, but after pouring them together, the $\ce{Pb^{2+}}$$\ce{(aq)}$ ions react with the $\ce{I^-}$ $\ce{(aq)}$ ions and form a yellow solid ($\ce{PbI_2}$) that IS NOT SOLUBLE in water. It comes out of solution as a precipitate. Thus, we write it as $\ce{PbI_2}$ (s). Something actually happened: you had ions that mixed to form a solid.

The equation you have is better written this way:

$\ce{KI (aq) + Pb(NO_3)_2 (aq) -> KNO_3 (aq) + PbI_2 (s)}$. Notice the solid. If a reaction produces only aqueous products, then nothing happens beyond mixing. If it produces anything that's non-aqeuous (like in this case), THEN there's a chemical reaction.

  • $\begingroup$ Even better is to cancel the spectator ions, i.e., potassium ion and nitrate ion. $\endgroup$
    – Ed V
    Commented May 25, 2020 at 13:18

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