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You are not adding any hydrogen ions, so why does the pH of the solution decrease? Thank you in advance.

EDIT: I will add the context of my asking this question to clarify some details. This is not a homework question. I am preparing for an upcoming examination in Chemistry, and I found this statement in the mark scheme of a practice exam I was doing, and yet I don't understand why it is true.

I have learnt pH as simply being the negative log of the concentration of H+ ions. I fail to see how adding the above salt changes the pH, especially since HCl is a strong acid (hence it completely dissociates in solution).

Thought process: Is the pH change a result of Aluminium forming Aluminium Oxide?

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    $\begingroup$ This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$
    – M.A.R.
    Commented Apr 29, 2017 at 18:46
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    $\begingroup$ @M.A.R. Hi, thanks for your comment. Please see my edit, marked "EDIT" $\endgroup$
    – K. T.
    Commented Apr 29, 2017 at 18:54
  • $\begingroup$ chemistry.stackexchange.com/questions/41675/… $\endgroup$
    – Mithoron
    Commented Apr 29, 2017 at 19:00
  • $\begingroup$ Your thought process shouldn't be another question! $\endgroup$
    – Pritt says Reinstate Monica
    Commented Apr 30, 2017 at 2:18
  • $\begingroup$ @PrittBalagopal Well, answering either question would in the process answer the other, i.e. they are linked. And as it is, my thought process is only questions, no answers! $\endgroup$
    – K. T.
    Commented Apr 30, 2017 at 5:48

3 Answers 3

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A more clarified reaction in addition to Berry Holmes' answer:

$$\ce{AlCl3 + 4H2O <=> [Al(OH)4]- + 3Cl- + 4H+}$$

The reason this happens is the Lewis-Acidity of $\ce{Al^3+}$ as well as the higher affinity for $\ce{OH-}$ ions when compared with $\ce{Cl-}$.

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  • $\begingroup$ Equilibrium of dissociation of complex won't get that far here, remember it's acidic solution Daniel's answer gets this right. $\endgroup$
    – Mithoron
    Commented Apr 29, 2017 at 22:30
  • $\begingroup$ Hi tillyboy, thanks for your answer! Right now, I have 2 answers and I am not sure if they conflict. Could you thus clarify the following thought: Does Aluminium directly form the Al(OH)4 complex or does it first form Al(H2O)4 which then releases the H+ to form Al(H2O)3(OH) and so on until 4 protons are released? Thank you very much again. $\endgroup$
    – K. T.
    Commented Apr 30, 2017 at 5:52
  • $\begingroup$ @Mithoron: Well, that is the general equation of the complete reaction. Of course that reaction consists of many individual steps with their own equilibrium constant. You could determine through thermodynamical calculations where the equilibria of the individual constituent reactions are, although I am way to lazy to conduct said calculations :P (Actually after re-reading the question two times, where is it said, that the solution is acidic?) $\endgroup$
    – tifrel
    Commented Apr 30, 2017 at 6:50
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    $\begingroup$ @K.T. I am not quite sure about the mechanism here, but my best guess is actually that it is a mixture of both: The hydroxyl ions from dissociated $\ce{H2O}$ molecules attach to the aluminium, shifting the equilibrium of auto-dissociation, while other water molecules attach to the aluminium, gradually throwing away their protons. Note however that the equation above is a equation that sums up a lot of reactions but doesn't give any information about the equilibrium state. In the end you will be left with a solution of $\ce{[Al(H2O)4]^{3+}}$, $\ce{[Al(H2O)3OH]^{2+}}$ $\endgroup$
    – tifrel
    Commented Apr 30, 2017 at 6:59
  • $\begingroup$ $\ce{[Al(H2O)2(OH)2]^+}$, $\ce{[Al(H2O)(OH)3]}$ and $\ce{[Al(OH)4]-}$, as well as your $\ce{Cl-}$, $\ce{H+}$ ions. $\endgroup$
    – tifrel
    Commented Apr 30, 2017 at 7:02
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When you add $\ce{AlCl3}$ to water a reaction takes place:

$$ \ce{AlCl3 + H2O <=> Al(OH)3 + HCl} $$

But wait, now I've an acid, i.e. hydrochloric acid ($\ce{HCl}$) and probably a base ($\ce{Al(OH)3}$). Close enough, but $\ce{Al(OH)3}$ is actually amphoteric in nature, which means it can behave both as an acid and a base.

Think of it as a tennis match, on one side you've a strong player (Novak Djokovic, Roger Federer or Rafael Nadal – you name it) and on the other side you've a teen who has barely known tennis for a month. For sure, the strong player is going to win. But which of the two is the strong player in our case?

If you're guessing it to be $\ce{HCl}$, I'll give you points. I think $\ce{Al(OH)3}$ is double minded, it doesn't know for sure if it's going to act as an acid or a base. $\ce{HCl}$ on the other hand knows that it is a strong acid. $\ce{HCl}$ wins the game, making the solution acidic and so, the pH decreases.

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    $\begingroup$ Hi Berry, thank you for your very fun to read and pedagogical answer! In fact, my thought process was similar (although I did not state this in the post): Adding AlCl3 to water forms HCl, which is an acid. However, since it is a strong acid, it is completely dissociated in solution, so I don't understand why the presence of Cl makes a difference to the pH? In other words, before adding AlCl3, there were a certain number of H+ ions, and after there are still these many ions left. Why does the pH change? $\endgroup$
    – K. T.
    Commented Apr 29, 2017 at 19:01
  • $\begingroup$ @berryholmes Indeed great answer, any journal reference to this behaviour of AlCl3? What's your opinion about LaCl3 in water (aq) and its pH? Can you comment on the pH of LaCl3 in water and pKa of La(OH)3? Because LaCl3 (aq) rapidly forms La(OH)3 percipient. $\endgroup$
    – bonCodigo
    Commented Nov 13, 2019 at 7:56
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First of all, $\ce{Al^3+}$ will separate itself from $\ce{Cl^-}$ in water, due to the nature of this salt (don't know why, the other answers have included this I believe).

Because in water, the $\ce{Al^3+}$ ion surrounds itself with $\ce{H2O}$ molecules, with the partially positively charged H-atoms turned towards the $\ce{Al^3+}$.

With enough electric force pulling all of the $\ce{H2O}$ atoms in there will be so many H2O molecules surrounding the $\ce{Al^3+}$, all being pulled in by the positive charge, that the molecules are too close together, and eventually one of the $\ce{H^+}$ ions is whipped out of this "$\ce{H2O}$ circle".

So $\ce{[Al(H2O)6]^3+}$ will become $\ce{[Al(H2O)5(OH)]^2+}$ + $\ce{H^+}$

Not all $\ce{[Al(H2O)6]^3+}$ 's will perform this action, and therefore it, like all metal aquo complexes, is a weak acid. For example, it works with $\ce{Fe^3+}$ as well.

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  • $\begingroup$ Hi, thank you for your answer. However, I have some queries: 1) How does Al forming a complex with H2O ligands alter the pH? 2) I don't understand what you mean by "metals are all weak acids". Thanks for your answer, my knowledge is roughly secondary school level as well. $\endgroup$
    – K. T.
    Commented Apr 29, 2017 at 18:57
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    $\begingroup$ @K.T. 1) because there are so many H2O molecules surrounding the Al3+, all being pulled in by the positive charge, that the molecules are too close together, and eventually one of the H+ ions is whipped out (will edit this bit in) 2) all positive metal-ions are weak acids, it's just a fact, and it's an explanation to the "do sometimes" bit before the brackets. $\endgroup$
    – Daniel Glas
    Commented Apr 29, 2017 at 19:00
  • $\begingroup$ @airhuff ty for the edit, haven't got a clue how to do subscript and superscript. $\endgroup$
    – Daniel Glas
    Commented Apr 29, 2017 at 19:06
  • $\begingroup$ No problem. See this for a discussion/tutorial on how to use MathJax formatting for chemical and mathematical equations and such. It's a pretty small learning curve ;) $\endgroup$
    – airhuff
    Commented Apr 29, 2017 at 19:08
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    $\begingroup$ "don't know why, the other answers have included this I believe" Do you answer based on other answers to the question? See this for why ions split: en.m.wikipedia.org/wiki/Dissociation_(chemistry) $\endgroup$
    – Pritt says Reinstate Monica
    Commented Apr 30, 2017 at 2:22

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