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In light of this question and its answers, I hope to get some insights into the $\mathrm{pH}$ of $\ce{LaCl3}$ in water and its $\mathrm{pH}$ relationship to $\ce{La(OH)3}$ precipitant's $\mathrm{p}K_\mathrm{a}$?

What can be done in order to prevent $\ce{LaCl3}$ from forming $\ce{La(OH)3}$ and maintaining the $\mathrm{pH}$ $\pu{24^\circ C} - \pu{25^\circ C}$?

When creating salt solutions of $\ce{AlCl3}$, $\ce{LaCl3}$ and $\ce{FeCl3}$; $\ce{AlCl3}$ seem to have a pretty low $\mathrm{pH}$ in water while $\ce{LaCl3}$ rapidly precipitate into $\ce{La(OH)3}$. With the rapid formation of $\ce{La(OH)3}$, $\ce{LaCl3}$ seems to show extremely low solubility in water.

Does $\ce{LaCl3}$ also acts as a Lewis acid?

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    $\begingroup$ $\ce{LaCl3}$ doesn't hydrolyze to $\ce{La(OH)3}$, you need to really make solution alkaline for that. Even if you boil the solution for a long time, you'll likely only end up with hydroxochloride $\ce{La(OH)2Cl}$. In distilled water lanthane (III) should be fine. The aquacation participates in equilibrium $$\ce{[La(H2O)9]^3+ + H2O<=> [La(H2O)8(OH)]^2+ + H3O+}$$ which can be easily shifted to the left by acidifying the solution ($\mathrm{p}K_\mathrm{a}$ of hydroxoaquacation is approx. 11). And yes, it's a Lewis acid. Also, what is "percipient" in this context? Do you mean "precipitate"? $\endgroup$
    – andselisk
    Commented Nov 13, 2019 at 14:16
  • $\begingroup$ @andselisk great reply, well "precipitate", yes. I am using LaCl3·7H2O. That's why I suspect La(OH)3 is the formed precipitate. What happens is, I am trying to create a LaCl3 in pH 7-7.4. It is not only impossible (due to the starting pH of the solution being low pH, there's no way or not make sense to use HCl), but solubility is so low, a precipitate is formed rapidly. Are you saying the precipitate is [La(H2O)8(OH)]2+ ? $\endgroup$
    – bonCodigo
    Commented Nov 13, 2019 at 22:47
  • $\begingroup$ Formation of $\ce{La(H2O)8(OH)]^2+}$ is just a first step in olation process which I used for illustration, it's not a precipitate. I'm wondering how you get pH 7.0–7.4. By dissolving the salt in pure distilled water the pH should be well below 7. Do you add a base? If so, then yes, you do get $\ce{La(OH)3}$ all right. And what do you mean by "I am trying to create a LaCl3 in pH 7-7.4"? JIC, $\ce{LaCl3}$ crystallizes as heptahydrate from aqueous solutions, and if you try to dehydrate it by heating up, you end up with oxosalt $\ce{LaOCl}.$ $\endgroup$
    – andselisk
    Commented Nov 14, 2019 at 5:25
  • $\begingroup$ @andselisk, for all my salts, I want to maintain a pH 7-7.4 range. Of course in this case for LaCl3.7H2O, it's impossible to reach even above pH 5. No adding of base. In distilled water the (aq) of LaCl3 is well below 7, yes. My question and perplexity is that, why there's a white precipitate, that too rapidly forming, when dissolving the salt in water at temperature (24'c-27'c)? LaCl3 might be crystallising as heptahydrate from aqueous solutions. Is there a reference that can be used as evidence for this reaction of forming crystals or/and, what occasion LaCl3 in water may form La(OH)3? $\endgroup$
    – bonCodigo
    Commented Nov 16, 2019 at 22:03
  • $\begingroup$ @andselisk is it possible to have a chat with you in the chemistry blackboard Periodic table? $\endgroup$
    – bonCodigo
    Commented Nov 23, 2019 at 6:30

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