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My son asked this over the weekend and I didn't have a good answer: how come with polyprotic acids we assume that the release of each $\ce{H+}$ is progressively harder, but with bases like $\ce{Ca(OH)2}$ we write $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?

What I know about, say, sulfuric acid, seems to suggest that when we put calcium hydroxide in water we should get $\ce{CaOH+}$ and a single hydroxide, and it should be much harder to release that second hydroxide. What am I missing?

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The $\ce{H+}$ is bonded covalently to the rest of the molecule. So, when you take off the first $\ce{H+}$, the molecule fundamentally changes. It becomes another molecule. The second $\ce{H+}$ is being taken away from a different species and so it is governed by a different equilibrium.

In $\ce{Ca(OH)2}$, $\ce{Ca^2+}$ and $\ce{OH-}$ are held together via ionic bonds. When you dissolve that in water, all the ionic bonds are going to break at the same time. You cannot break some of the ionic bonds without also breaking the rest.

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This is not a general rule. Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base.

In the case of $\ce{Ca(OH)2}$ you don’t have a molecule like $\ce{HO-Ca-OH}$ or something like you can assume for $\ce{HO-SO2-OH}$ (sulfuric acid). Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in $2~\mathrm{mol}$ of $\ce{OH-}$ per mole of $\ce{Ca(OH)2}$ dissolved.

As a different example take sodium oxide $\ce{Na2O}$. If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2 NaOH}$ first before a second set of protons will generate water. This is much the same mechanism as for diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)

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  • $\begingroup$ It would be nice to tell about aquo complexes also @ortho $\endgroup$ – Mithoron Apr 18 '16 at 20:46
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    $\begingroup$ @Mithoron I don’t think that that belongs here …? $\endgroup$ – Jan Apr 19 '16 at 11:46
  • $\begingroup$ I think he meant it as a cc to me, since we basically wrote the same thing within 1 minute of each other (mine being a bit less expanded) :D @Mith, what about aquo complexes? $\endgroup$ – orthocresol Apr 19 '16 at 18:13
  • $\begingroup$ @orthocresol I was actually saying that aquacomplexes don't necessarily belong here ;) $\endgroup$ – Jan Apr 19 '16 at 18:15
  • $\begingroup$ In water there are equilibria of protonation involving complexes of Ca. Maybe I'll make an answer with reasoning involving it if you don't want ;) $\endgroup$ – Mithoron Apr 19 '16 at 18:58
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If I would conclude from Jan's statement

Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in 2 mol of $\ce{OH-}$ ion per mole of $\ce{Ca(OH)2}$ dissolved. [$\ldots$]

then it would be not that difficult to give the following two reactions to answer the original question.

1. $$\ce{Ca(OH)2 (aq) <=> Ca(OH)+ (aq) + OH- (aq)}$$ 2. $$\ce{Ca(OH)+ (aq) <=> Ca^2+ (aq) + OH- (aq)}$$

Water is already in an ionised-equilibrium state, so there may be availability of protons and hydroxides in the (universal) solvent

Why not try to rearrange the above equation by taking the help of protons contributed from water, let's do it.

  1. Rearranging 1st equation with the help of, protons and hydroxides, on either sides of equation; we get: $$\ce{Ca(OH)2 (aq) + H+ (aq) <=> Ca(OH)+ + H2O}$$

  2. Again, rearranging 2nd equation with the help of, protons and hydroxides, on either sides of equation; we get: $$\ce{Ca(OH)+ (aq) + H+ (aq) <=> Ca^2+ + H2O}$$

It would now be easy for us to conclude why the reaction must proceed harder progressively; because as soon as we see the modified the form of reaction 1&2: we notice the repulsive interaction between $\ce{Ca(OH)+}$ ions and $\ce{H+}$ ions. So one may conclude with this pointed statement:

In the case of polyprotic acids we assume that the release of $\ce{H+}$ ions is progressively harder.

said by Asker, has justified why must not "strong bases like Ca(OH)2", "directly form $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?", further added by Asker.

Jan says

This is not a general rule.

Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base. In the case of Ca(OH)2 you don’t have a molecule like HO−Ca−OH or something like you can assume for HO−SO2−OH (sulfuric acid).

Rather, you have an ionic compound composed of two hydroxide ions per a calcium ion. Once we dissolve this, all the ions are released at the same time resulting in 2 mol of hydroxide per mole of $\ce{Ca(OH)2}$ dissolved. As a different example take sodium oxide $\ce{Na2O}$.

If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2NaOH}$ first before a second set of protons will generate water. This is pretty much the same thing that happens to diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)*

Jan statement is much correct and easily explains my asserted warning at last sentence under the conclusion.

Conclusion:

So let me again repeat the whole equation but with a slight change, this time for $\ce{H2SO4}$, a strong acid:

  1. Rearranging 1st equation with the help of, protons and hydroxides, on either sides of equation; we get: $$\ce{HO−SO2−OH (aq) + OH- (aq) <=> O−SO2−OH- + H2O}$$
  2. Again, Rearranging 2nd equation with the help of, protons and hydroxides, on either sides of equation; we get: $$\ce{O−SO2−OH- (aq) + H+ (aq) <=> O−SO2−O^2- + H2O}$$

Note that ${\rm p}K_{\rm a}$ and ${\rm p}K_{\rm a}$ are like constants whose value does not change until either temperature or some other thermodynamic variable changes (do not confuse with Kinetic Variables, acid dissociation constants have nothing to do with it) are changed (https://en.wikipedia.org/wiki/Acid_dissociation_constant)

TL;DR;

  • Firstly, determine the Acid Dissociation Constant constant of the given base, acid or mixed salt.

  • Secondly, determine the Thermodynamic conditions and values of its variables to help you go into detail of the reaction (not necessary, just knowing that it is constant or not, is enough)

  • Thirdly, write the whole dissociation reaction (only, because those that does not dissociate or whose dissociation constant is too small is generally ignored, i.e., Weak acids and Weak Bases and their mixed salts, can be ignored).

  • Finally, interpret the interactions between the filtered out dissociation-reactions, then cross out those that are against any of the 1st three rules. You may also filter out on the basis of what are you actually trying to obtain from these reactions. Just keep in mind, that the fundamental rules of Ionic Equilibrium in the context of whole Chemistry. must not be violated.

    You will be having only the most important and relevant reactions. Just a few steps.

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  • $\begingroup$ There is no reason to assume statements 1 and 2 have any physical significance. $\endgroup$ – Lighthart Apr 19 '16 at 6:08
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    $\begingroup$ I dunno, let's face it, it's bad. Putting the formatting and grammar aside, essentially everything here is irrelevant. Half of it is just quoting and claiming that the quoted sections prove some unknown "point". The other half is a disordered collection of assertions about pKa values, which are entirely irrelevant to the question to begin with. As Lighthart says, the first two equations have absolutely zero physical significance. $\endgroup$ – orthocresol Apr 19 '16 at 19:29

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