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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
    Commented Dec 4, 2015 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.) $\endgroup$ Commented Dec 4, 2015 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? $\endgroup$ Commented Dec 4, 2015 at 22:43

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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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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? $\endgroup$ Commented Dec 4, 2015 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
    Commented Dec 4, 2015 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. $\endgroup$ Commented Dec 4, 2015 at 23:51
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Technically, the magnesium ion will act as a weak acid and thus magnesium chloride solutions are going to be at least slightly acidic. However, the acidity will be less than that with most transition metal chlorides, and at least in dilute solution it's a bit of a borderline case. Calculations below indicate that the pH is estimated as 6.24, so if we regard pH between 6 and 8 as essentially neutral, then 0.1 M magnesium chloride solution would be in that range.

A compilation by IUPAC gives several values for the $\mathrm{p}K_\mathrm{a}$ of magnesium ion in water, these mostly clustering around 11.5. The measurement used based on the hydrolysis reaction which is predominantly the first stage of proton dissociation:

$\ce{Mg^{2+} + 2H2O <=> MgOH^+ + H3O^+}$

The second stage of proton dissociation is essentially inaccessible by normal techniques in aqueous solution because it equilibrates only at pH values where the solubility of magnesium hydroxide has become very low.

With the first stage of dissociation we may compute for 0.1 M magnesium chloride solution:

$[\ce{Mg^{2+}}]+[\ce{MgOH^+}]=0.1$

$2[\ce{Mg^{2+}}]+[\ce{MgOH^+}]+[\ce{H^+}]-[\ce{OH^-}]=0.2$

$[\ce{MgOH^+}][\ce{H^+}]/[\ce{Mg^{2+}}]=10^{-11.5}=3.2×10^{-12}$

$[\ce{H^+}][\ce{OH^-}]=10^{-14}$

The first equation is the magnesium ion balance, the second gives electroneutrality with 0.2 molar chloride ion, and the last two equations represent the equilibrium realtions. From the fourth equation we immediately have

$[\ce{OH^-}]=10^{-14}/[\ce{H^+}].$

Putting this into the first two equations and taking appropriate linear combinations gives

$[\ce{Mg^{2+}}]=0.1-[\ce{H^+}]+10^{-14}/[\ce{H^+}]$

$[\ce{MgOH^+}]=[\ce{H^+}]-10^{-14}/[\ce{H^+}]$

Substituting these into the third equation and simplifying gives

$a^3+3.2×10^{-5}a^2-33a-320=0$

where $a=10^7[\ce{H^+}]$ is introduced to enable better numerical calculation. This gives a root

$[\ce{H^+}]=10^{-7}a\approx5.7×10^{-7} M,$

from which the pH is $\approx6.24$.

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  • $\begingroup$ :When substitute $a=10^7[\ce{H^+}]$ in the following equation : $[\ce{H^+}]=10^{-7}a= 10^{-7}10^7[\ce{H^+}] \approx5.7×10^{-7} M,$? help me to clarify it $\endgroup$ Commented Mar 21, 2022 at 23:19
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The pH of water is defined as neutral, because it contains H$^+$ and OH$^-$ ions in equal abundance, 10$^{-7}$M. We would probably call a pH of 6.9 or 7.1 neutral. Maybe even 6.5 and 7.5 are essentially neutral.

The questions become "How wide is your neutral pH range?" and "What is the actual pH of 0.1M MgCl2?" Neither is answered; however, it is possible to make a case for either of two answers to the OP's question.

In the first case, neutral is a wide range, say from 6 to 8. This neutrality is neither very acidic nor very alkaline, perhaps a pH that registers neutral with pH paper. In this case, MgCl2 is probably neutral, because we consider Mg(OH)2 to be a strong base and HCl is considered to be a strong acid. "Spectator ions", both of them. Very approximate language, but descriptive enough for many, if not most purposes. So, with this approach, MgCl2 is neutral.

But then someone might suggest using a pH meter with very high accuracy, MgCl2 of very high purity, and distilled water freed of CO2 by boiling or repeated distillations, and setting the neutral pH at 7.00, because you want the observed pH to fall on one side or the other.

Then we discover a report that Mg(OH)2 can be considered to be a weak base (Ref 1), so we would expect a pH somewhat below 7.00, and the answer would be "0.1M MgCl2 is acidic." (But not very acidic.) An elegant way of describing the situation is that Mg(OH)2 is a strong electrolyte but a weak base (Ref 2 and 3).

None of this so far has directly addressed the issue of low solubility, which seems to be relevant, somehow. We know that Mg$^{2+}$ has a great affinity for OH$^-$ ions, because they attract each other in great numbers, precipitate, and resist the solubilizing action of water. So in solution, along with 5 water molecules, a Mg$^{2+}$ ion probably attracts one of the few (10$^{-7}$M) OH$^-$ ions tightly enough to lower the pH a bit, increase the concentration of H$^+$ by a little.

So, the correct answer is neutral if you measure pH with paper (wide neutral range), or slightly acidic if you measure with a sensitive pH meter (sharp neutrality), because Mg(OH)2 can be considered either a strong base or a weak base.

Ref 1. https://www.numerade.com/questions/why-can-magnesium-hydroxide-be-described-as-a-strong-base-even-though-it-is-only-slightly-soluble-in/

Ref 2. https://www.toppr.com/ask/question/a-solution-of-mgcl2-in-water-has-ph/

Ref 3. https://www.sciencedirect.com/topics/chemistry/magnesium-hydroxide

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A very quick answer to this question is that $\ce{MgCl2}$ is the salt formed from $\ce{Mg(OH)2}$ and $\ce{HCl}$. Both of these are strong bases and acids respectively, so when their salt is formed, it is made up of a very weak conjugate acid ($\ce{Mg^2+}$) and a very weak conjugate base ($\ce{Cl^-}$). Since these are both very weak conjugate acids and bases, the salt is neutral and you should see a pH close to 7 (maybe slightly below 7 due to dissolved acidic $\ce{CO2}$, a small amount of hydrolysis, and whatnot).

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