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I'm currently taking chemistry 12. On our test we were asked, when given a $0.1~\mathrm{M}$ solution of certain compounds whether the resulting solution when added to water would be acidic, basic, or neutral.

At first, I thought the solution would be neutral, because we learnt in class that alkali earth metals and chloride are spectator ions and therefore don't interact with water and pH.

However, I then realized that $\ce{Mg(OH)2}$ has low solubility in water, and I assume that adding $\ce{MgCl2}$, would dissociate and then precipitate $\ce{Mg(OH)2}$ from the solution, therefore the resulting solution would be acidic. My final answer was acidic.

The teacher marked my answer wrong, but when questioned could only say that $\ce{Mg}$ and $\ce{Cl}$ are both spectators, but was unable to give direct contradictory proof against my answer.

I looked elsewhere online and Wikipedia states that anhydrous $\ce{MgCl2}$ is a Lewis acid, despite being a weak one. But anhydrous seems to mean without water, so I currently have no idea which answer is correct.

I would like an explanation to why it is acidic or neutral and why the contradicting argument (both spectators vs low solubility of $\ce{Mg(OH)2}$ ) is invalid or flawed.

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  • $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I edited your post to include MathJax for chemical formulae. You can learn more about that in the help center, this FAQ post or this meta-post. $\endgroup$ – Jan Dec 4 '15 at 22:17
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    $\begingroup$ What kind of proof would you accept? $\ce{MgCl2}$ added to pure water would not precipitate $\ce{Mg(OH)2}$, that much I can attest from firsthand experience. (This does not really prove anything, but whatever.) $\endgroup$ – Ivan Neretin Dec 4 '15 at 22:40
  • $\begingroup$ @IvanNeretin Well I guess I'd like an explanation as to why either argument is invalid besides just experimental proof. But can you give me details about your experience? $\endgroup$ – user2804925 Dec 4 '15 at 22:43
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Long story short, the ultimate cause of non-neutral pH in the solutions of some salts is hydrolysis. If a salt is made from a weak acid and strong base, then the acid anion would get partially protonated by water, like $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$, and that $\ce{OH-}$ would be responsible for the overall alkaline pH of the resulting solution. If we have a salt of a strong acid and weak base, it is the other way around.

Which acids or bases are strong and which are weak is determined by looking up their $pK_a$ or $pK_b$ in the reference books. I don't want to get deeper into this because running far ahead of your syllabus might do more harm than good. Just remember that neither $\ce{Mg^2+}$ nor $\ce{Cl-}$ would hydrolyze significantly, so the solution would remain neutral.

The precipitation is not all that important, as hydrolysis of most common salts rarely goes all the way up to metal hydroxide. For example, $\ce{AlCl3}$ would make the solution pretty acidic, but still produce no precipitate. Also, consider $\ce{NH4+}$ which is prone to hydrolysis despite having soluble hydroxide. (One might argue this does not quite qualify as hydrolysis, but anyway.) If you want an example when the precipitation actually does occur, think of $\ce{TiCl4}$.

As for the spectator ions, that's a lame argument indeed. There is no such thing as "spectator ion in general". True, certain ions are spectators in certain reactions, but there is no ion that would always remain a spectator, no matter what the situation is.

As for the Lewis acids, that's a different concept altogether, to the point that you shouldn't automatically think of "Lewis acids" as acids (though many of them are).

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  • $\begingroup$ Okay, I know about hydrolysis, but that still doesn't seem to answer my question about MgCl2, whether or not the pH of the solution will be changed? $\endgroup$ – user2804925 Dec 4 '15 at 23:42
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    $\begingroup$ @user2804925 I’ll admit it’s not well written, but the answer to that question is hidden in Ivan’s answer. $\endgroup$ – Jan Dec 4 '15 at 23:47
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    $\begingroup$ @Jan okay I found the answer I accidentally skipped over it the first time. Still wondering why my argument would be invalid though. From albapa's answer, it seems that the concentration isn't high enough for precipitation to occur, but Ivan's answer doesn't seem to take the fact that Mg(OH)2 precipitates into account at all. $\endgroup$ – user2804925 Dec 4 '15 at 23:51
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As a physics teacher turned chemistry teacher, I cannot be 100% sure, but I think the answer is getting at the complex formed with water.

$$\ce{MgCl2 -> Mg^2+ + 2Cl-}$$

$\ce{Cl-}$ is pretty unreactive with its high electronegativity and completes outer shell of electrons (hence why $\ce{HCl}$ for example gives almost 100% dissoscation and does not reform).

Well, for the $\ce{Mg^2+}$ on the other hand, water molecules can now use their lone pair of electrons on the oxygen atom to form a dative bond with the metal ion (this seems to happen all the time!) forming $\ce{[Mg(H2O)6]^2+}$.

If it accepts a lone pair, it is a Lewis acid (perhaps this is what your wikipedia research showed).

ANYWAY, now the acidic bit... So you have an $\ce{Mg^2+}$ ion, surrounded by 6 water molecules. Due to the high charge density around the ion (I guess this part of my thinking means that the likes of strontium, further down the period are less acidic?), it can actually help pull the electrons in the $\ce{O-H}$ covalent bonds in the water molecule (which are already being pulled away from the H atoms due to the high electronegativity on oxygen). If the electron pair is pulled enough, it will be beyond the electrostatic force of the proton in the hydrogen atom and hence the proton will dissociate into the solution leaving behind $\ce{OH-}$ (no longer water). This $\ce{H+}$ (no longer $\ce{H}$, because it loses its electron in its shared electron pair) goes into the solution, thus increasing the $\ce{H+}$ concentration, thus decreasing the pH.

SHORT ANSWER:

  • $\ce{Mg^2+}$ forms a complex with water - $\ce{[Mg(H2O)6]^2+}$
  • The high (positive) charge density on the $\ce{Mg}$ ion hydrolyses the surround water ligands
  • This releases $\ce{H+}$ (protons) into the solution, decreasing the pH slightly.
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protected by orthocresol Jul 13 '17 at 14:32

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