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I've often seen the hydroxides of Calcium, Strontium and Barium described as "strong bases that are only slightly soluble," or similar. But strong bases dissociate completely in solution, so I would guess this means that such hydroxides dissolve only a small amount, but the dissolved portion dissociates completely.

But the problem with this interpretation is that these hydroxides are ionic compounds, and my understanding is that "dissolve" is essentially equivalent to "dissociate" for ionic compounds, i.e. ionic compounds dissolve by dissociating.

If there's no distinction between dissolving and dissociating for ionic compounds, what does the description "strong base that is only slightly soluble" mean? How does it differ from the term "weak base?"

If there is a distinction, and it is possible for ionic compounds to dissolve without dissociating, how does such dissolution work? I have a model for the distinction between dissolution and dissociation of covalent compounds: first the compound dissolves into separate molecules, then each molecule dissociates into two or more ions. If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve? Formula units? Small clusters of ions? Something else?

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  • $\begingroup$ And you guess well. What IS dissolved IS fully dissociated. Stop here and you see your puzzle is solved :) (being fully dissociated does not require that more material, then, has to dissolve) $\endgroup$ – Alchimista Jul 28 '17 at 20:09
  • $\begingroup$ @Alchimista If an ionic compound can dissolve without dissociating, how do we model the dissolution process? Into what sort of particles does the compound dissolve? $\endgroup$ – Chad Jul 28 '17 at 20:57
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    $\begingroup$ No one says that! I say that IF it get dissolved, THEN (implies) it is dissociated. If it would be more, it will dissociate. But there is not more! That when it in liquid phase it is dussociated does not imply that it has to be in liquid phase. You are mixing two equilibria here. Between solid hydroxide and its ions in solutions (Kps) AND and the one in acid-base reaction. They are two different things. Of course when your hydroxide does not dissolve anymore, if you add acid you force it to do it (Le Chatellier). This is what links the two equilibria. Look also at answer by a-cyclohexane-mol.. $\endgroup$ – Alchimista Jul 28 '17 at 21:12
  • $\begingroup$ You know what, read about coordination chemistry then aquocomplexes and polycentric clusters and you'll probably get what happens with cations in solutions. $\endgroup$ – Mithoron Aug 1 '17 at 23:48
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Solubility is different from dissolution, which is itself different from dissociation.

Solubility tells you the extent to which a substance dissolves. Dissolution is the kinetic process of solvation, when solvent molecules interact with solute molecules or ions. Dissociation is, from IUPAC,

  1. The separation of a molecular entity into two or more molecular entities (or any similar separation within a polyatomic molecular entity). Examples include unimolecular heterolysis and homolysis, and the separation of the constituents of an ion pair into free ions.
  2. The separation of the constituents of any aggregate of molecular entities.

Dissociation is the mode of solvation for ionic compounds.


I think your confusion stems from associating solubility and dissolution improperly. ChemGuide directly answers your question here:

Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation.


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    $\begingroup$ Ok. But what does dissolution of consist of, if not dissociation? In the case of sucrose, I have a good model for understanding dissolution and dissociation as separate processes. Dissolution is the process of breaking down into separate sucrose molecules, whereas dissociation involves the further breakdown of sucrose molecules into ions. If Ca(OH)₂ can dissolve without dissociating, what are the particles into which it dissolves? Formula units? Small clusters of ions, larger than formula units but still microscopic? Something else? $\endgroup$ – Chad Jul 28 '17 at 20:45
  • $\begingroup$ *If Ca(OH)2 can dissolve without dissociating : It CANNOT. That is the simple point, and you more or less got it, in the first part of your question $\endgroup$ – Alchimista Jul 28 '17 at 21:16
  • $\begingroup$ @Chad, that's a fair question. I think my answer is quite misleading as-is, and I'm quite uncertain as to how well $\ce{Ca(OH)2 (aq)}$ holds up under scrutiny, but let me edit my answer and see if it resolves your concerns. $\endgroup$ – a-cyclohexane-molecule Jul 28 '17 at 21:24
  • $\begingroup$ @Alchimista What about a compound like Fe(OH)₂, which is commonly described as a weak base, not a "strong base that is only slightly soluble." Can Fe(OH)₂ dissolve without dissociating? If so, how does that process work? If not, what is the distinction between a weak base and "strong base that is only slightly soluble?" $\endgroup$ – Chad Jul 28 '17 at 21:25
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    $\begingroup$ It seems a quite general statement: OH- which is in both Kb and Kps. Then we likely run into formal hydroxides with covalent bonds. Let us try solve clearly your original question first. To me the answer is that by a cyclohexane mol. Perhaps tomorrow I ll post an answer, at the moment I am getting crazy with the small box in the app. Just let me know if shall I refer to the original question although edited or you came up with a new one or a new generalisation. See you asap $\endgroup$ – Alchimista Jul 28 '17 at 23:37
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Qualitative descriptors such as "slightly soluble" usually refer to the maximum amount of a substance that can be dissolved in water. The use of the word "slightly" indicates that, for example, less $\ce{Ca(OH)2}$ can dissolve in a given volume of water than a comparable molecule like $\ce{NaOH}$ in the same amount of water.

If you were to continually add small amounts of $\ce{Ca(OH)2}$ to 1L of water (numbers are just picked at random) at some point no more $\ce{Ca(OH)2}$ would dissolve. All the $\ce{Ca(OH)2}$ that dissolved at that point DID completely dissociate, but after the solution is saturated, more $\ce{Ca(OH)2}$ will not continue to dissolve. After this point, the excess $\ce{Ca(OH)2}$ that you added would just sink to the bottom of the beaker and stay there.

Note: My description relates to what is observable in the macro view. The microscopic view shows that an equilibrium between the solid and ions will be established once the saturation limit is reached. The equilibrium will not cause the average concentration to exceed the saturation limit.

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Actually, when I learned my strong bases in high school my teacher told me that those "slightly soluble" Ca(OH)2 etc were not strong bases. But I think the idea is even though they are only slightly soluble, they are still more soluble than other even weaker bases such as say Fe(OH)3.

All solids dissociate completely until a saturation point; they do not dissolve proportionally like aqueous acids since the concentrations of the solids (concentration is not amount!) do not change. So, what you call a "strong" base is based on how strong you have to get.

List of Ksp (dissociation constant; generally greater means more dissociation):

Aluminum hydroxide Al(OH)3 1.3×10–33

Barium hydroxide Ba(OH)2 5×10–3

Cadmium hydroxide Cd(OH)2 2.5×10–14

Calcium hydroxide Ca(OH)2 5.5×10–6

Chromium(II) hydroxide Cr(OH)2 2×10–16

Chromium(III) hydroxide Cr(OH)3 6.3×10–31

Cobalt(II) hydroxide Co(OH)2 1.6×10–15

Copper(II) hydroxide Cu(OH)2 2.2×10–20

Iron(II) hydroxide Fe(OH)2 8.0×10–16

You can find more at: http://webcache.googleusercontent.com/search?q=cache:http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

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  • $\begingroup$ @Chad. I should have commented below, not here. Sorry. Where is the problem with the equilibria AB(solid) double arrow AB(solv) double arrow A(solv) + B(solv)? In case of full dissociation you end always with higher solubility, as is indeed the case. Yes you have a neutral entity involved, the smaller one or as cluster, within their solvation sphere. This should be experimentally verified. Anyway you are dealing with the Kps value. $\endgroup$ – Alchimista Aug 1 '17 at 18:26
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I have to admit that jumping between the various parts of the question as well as the editing/comments almost confused me, or at least gave me some discomfort. So I try to clarify the situation although I am fond of reading some of the previous answers, in particular that given by a_cyclohexane_molecule. To this end, I go through your text. Please note that capital is not shouting but an attention driver.

1 - "But strong bases dissociate completely in solution, so I would guess this means that such hydroxides dissolve only a small amount, but the dissolved portion dissociates completely."

Yes, this is the meaning and what happens.

2 - ".... problem with this interpretation is that these hydroxides are ionic compounds, and my understanding is that "dissolve" is essentially equivalent to "dissociate" for ionic compounds, i.e. ionic compounds dissolve by dissociating.

Not completely right. The ACTUAL amount of THESE hydroxides - NaOH and Ca(OH)2 - that goes in solution (solubility) is fully dissociated. It is again point 1), that you inadvertently generalise to all hydroxides, whatever their base strength is.

Think of Me(OH)n in water. Me ions coexist with Me(OH)n-1, Me(OH)n-2 etc. positively charged ions (according to the stoichiometry). In which sense the dissolved hydroxide is fully dissociated? It is not, even if you postulate that what goes in solution undergoes at least one complete first dissociation process, i.e. no neutral entities can dive in solution (I think this is matter for a separated question. We are in the case "to which extent a bond is ionic?", at the end. Or even as to ask what is the biggest neutral agglomerate of ionic compounds that can be filtered out of a solution. It is not a coincidence that most hydroxides are tricky to handle :) Unfortunately real things rarely fit in absolute definition).

3 - " if there's no distinction between dissolving and dissociating for ionic compounds, what does the description "strong base that is only slightly soluble" mean? How does it differ from the term "weak base?"

Here is where some definitively confusion comes in. First question in the paragraph. We answered already. There is a distinction, as above. 100% dissociation does NOT imply that the same amount is in solution. Microscopically, let us take for granted that each whatever small portion of the compound that gets dissolved does dissociate. It does ONLY IF gets dissolved, AND IF is not like the case Me(OH)n above.

Second question in the paragraph. Let me stay in the realm of Arrhenius: a strong base definitively gives you a solution with a high pH. Differently, a very weak base does not generate a considerable amount of OH-, their final concentration it is about the natural one in water (from Kw). (You get see this in B-L or L theories, too, and you need those theories to rationalise the behaviour of ammonia, for instance).

3 - " If there is a distinction, and it is possible for ionic compounds to dissolve without dissociating, how does such dissolution work? I have a model for the distinction between dissolution and dissociation of covalent compounds: first the compound dissolves into separate molecules, then each molecule dissociates into two or more ions. If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve? Formula units? Small clusters of ions? Something else?"

I think that answer 2 applies here, too. The point is that if you stick to hydroxides (except alkaline and earth-alkaline), metal and amphoteric hydroxides are weak bases because

  • are so sparingly soluble in water that by a practical, Arrhenius based viewpoint, do not gives considerable amount of OH-. This will be true even assuming their complete dissociation!

  • a finer analysis of their equilibria in water does reveal the coexistence of different degree of dissociation, so they can be classified as weak indipendently of their limited or negligible solubility.

Although the OP did not have troubles in cases in which he could easily identify the covalent molecule that goes in solution without (or better, partially) dissociate, let me conclude by pointing to the fact that a weak base it is not necessarily sparingly soluble (as for a weak acid it is not necessary sparingly soluble), as this will not leave room for ammonia (acetic acid) etc. Ergo we still need a definition of weak bases independently of their solubility.

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  • $\begingroup$ I'm still trying to sort out the question "If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve?" I get that Me(OH)n can exist in equilibrium with Me(OH)n-1 positively charged ions. Me(OH)n dissolve into individual particles of Me(OH)n before dissociating into Me(OH)n-1 and hydroxide. If the metal hydroxides (other than alkaline and alkaline earth) can have "different degrees of dissociation... independently of their limited or negligible solubility," then do they first dissolve into some particle before dissociating? $\endgroup$ – Chad Aug 1 '17 at 15:17
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There are a couple of things you need to carefully think about. You claim that Ca(OH)2 is an ionic compound. Yet I doubt you believe it exists as separate ions of Ca++, O=, and H+. If that is your belief, you are wrong. If it isn't your belief, then you probably recognize that the anion OH- is covalent. Allow me to give you a counter example: NH4OH exists in aqueous solution mostly as NH3 molecules. [perhaps this isn't the best counter-example, since ammonium hydroxide isn't a stable isolable chemical compound; perhaps I should have used tetramethyl ammonium hydroxide, (CH3)4OH instead]. One (counter-)example disproves the assertion. Solubility of "ionic" compounds does not necessarily require dissociation. Since water is such an excellent solvent for screening electric charges, water solutions of ionic compounds are often, perhaps even usually, dissociated. But water is only one of (nearly) countless solvents. (Granted, liquid water comprises over 70% of the surface of this planet, so it's arguably the most important solvent as far as you and I are concerned.) Here's another example: acetic acid. It isn't a strong acid (pKa ~4.76) and neither is it ionic (but, obviously, it does ionize in (aqueous) solution. Are there weak ionic bases which are highly soluble in water? Hmmm. If we agree that anything which contains the hydroxyl radical can not be purely ionic, then I'd have to do a bit of searching to see if I could find a binary (purely ionic) compound which was itself soluble and yet not strongly dissociated in aqueous solution. Hmmm. pKb of LiOH is ~0 meaning LiOH exists in aq. solution as a molecule (but not purely ionic, but certainly purely inorganic). I guess another way to answer your question is to agree that for purely ionic compounds (not that common, as it turns out) dissociating and dissolving in aqueous solution is common, but not necessary. But in general ionic shouldn't be confused with inorganic, dissolved shouldn't be confused with dissociated (nor ionized) and solvent shouldn't be confused with water.

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  • $\begingroup$ My understanding is that Ca(OH)₂ is an ionic compound composed of one Ca²⁺ ion and two OH⁻ ions. Each OH⁻ is itself a polyatomic ion held together by covalent bonds. Are you saying that Ca(OH)₂ can dissolve without dissociating into Ca²⁺ and OH⁻? If so, what are the particles into which it dissolves? $\endgroup$ – Chad Jul 28 '17 at 19:10

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