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So, lead iodide is insoluble. I see conflicting answers online. Some say it's a weak electrolyte because it is insoluble, others say it is a strong electrolyte because it is an ionic compound and any small amount that dissolves does so in the form of ions.

Which one is the correct answer?

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    $\begingroup$ Solubility and being strong/weak electrolyte are 2 independent properties. $\endgroup$
    – Poutnik
    Sep 18, 2022 at 16:28
  • $\begingroup$ Thank you for your answer Poutnik. So, is it a strong or weak electrolyte then? $\endgroup$
    – blazerunner
    Sep 18, 2022 at 16:54
  • $\begingroup$ en.wikipedia.org/wiki/Strong_electrolyte $\endgroup$
    – Poutnik
    Sep 18, 2022 at 17:08
  • $\begingroup$ From that Wikipedia article: Strong acids, strong bases and soluble ionic salts that are not weak acids or weak bases are strong electrolytes. It says soluble ionic salts..lead iodide is insoluble. I did my research, I just asked because I couldn't find a definitive answer! $\endgroup$
    – blazerunner
    Sep 18, 2022 at 17:15
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    $\begingroup$ Saying European cars are good does not say that American cars are not good. // Insoluble salts are not usually considered as electrolytes at all, but if they are, they usually are strong electrolytes. In fact, I cannot remember a case that would not be. It would mean both ions would hydrolyze to neutral weak acid and base. $\endgroup$
    – Poutnik
    Sep 18, 2022 at 18:35

1 Answer 1

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Lead(II) iodide is sparingly soluble in water at room temperature $(\pu{0.76 g/L}$ at $\pu{20 °C})$ and a bit more soluble in hot water $(\pu{4.1 g/L}$ at $\pu{100 °C}).$ In solution, all of the lead iodide is dissociated into ions. At high concentrations of iodide, the lead ions form two complexes, $\ce{PbI3-}$ and $\ce{PbI4^2-}$ [1]:

The increase of solubility of lead iodide caused by the presence of iodide ion in concentration greater than 0.1 molal may be explained by the formation of the complex ions $\ce{PbI3-}$ and $\ce{PbI4^=}.$

In any case, all of the dissolved lead iodide is present as ions (there is no neutral $\ce{PbI2(aq)}$ species), so it is a strong electrolyte. This is different from weak electrolytes such as, say, acetic acid, where there is a neutral species $\ce{CH3COOH(aq)}$ in solution which dissociates into the charged $\ce{CH3COO-(aq)}$ and $\ce{H+(aq)}$.

As an aside, the difference in color (colorless in solution, yellow as a solid) and the temperature-dependence of the solubility leads to a beautiful effect that has been called toxic golden rain, see e.g. video YouTube — Golden Rain - Growing crystals of lead iodide. Chemical reaction.

Reference

  1. Lanford, O. E.; Kiehl, S. J. The Solubility of Lead Iodide in Solutions of Potassium Iodide-Complex Lead Iodide Ions. J. Am. Chem. Soc. 1941, 63 (3), 667–669. DOI: 10.1021/ja01848a010.
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    $\begingroup$ A devil advocate may object how there can be $\ce{Pb^2+(aq)}$, $\ce{[PbI3]-(aq)}$ and $\ce{[PbI4]^2-(aq)}$ without $\ce{[PbI2](aq)}$. $\endgroup$
    – Poutnik
    Sep 19, 2022 at 10:06
  • $\begingroup$ @Poutnik You may ponder the opposite: Why do the complex ions not precipitate with a counter ion? The cite paper mentions double salts such as $\ce{RI.PbI2.H2O}$, and methylammonium lead iodide is an example. $\endgroup$
    – Karsten ♦
    Sep 19, 2022 at 10:36

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