I am a high school student and I am a little confused in solubility product.

My confusion is that when we talk of solubility of salts we say that suppose if any salt have solubility $s$ it means that whatever concentration of it we add on it will dissolve it will be dissolved by only $s$ concentration into the solution and other will precipitate.

This dissolved form will be present in ionized from in the solution, as degree of dissociation of salt is 100% so it will not be present as $\ce{AB(aq)}$ it will automatically dissociate.

But my question is that when we talk of solubility why do we treat weak bases like $\ce{Mg(OH)2}$ as sparingly soluble salts?

I.e they can be present in molecular form in aqueous solution without dissociating as well, as their degree of dissociation is not 100%. But still, we treat them exactly like salts and apply all those things which I have discussed above.

  • 2
    $\begingroup$ Hm, try to read your question after a weak or a month - to simulate it is not your own question. Then ask yourself, if its formulation still seems OK for you. Like proper usage of diacritics, choice of the sentence lengths, paragraph separation of ideas/topics/questions. Even if you understand each word and the topic, it is difficult to get the idea unless you are highly focused. As a guidance, the recommended average sentence length of scientific texts is 20-25 words, with the advice to avoid sentences longer than 35 words. $\endgroup$
    – Poutnik
    Feb 1, 2021 at 15:46

1 Answer 1


Let assume the equlibrium chain

$$\ce{Mg(OH)2(s) <=>[K_\mathrm{sp}]Mg(OH)2 <=>[K_\mathrm{b1}] Mg(OH)+ + OH- <=>[K_\mathrm{b2}] Mg^2+ + 2 OH-}$$

.. and evaluate the related equilibrium equations:


$$K_\mathrm{b1} = \frac {\ce{[Mg(OH)+][OH-]}}{\ce{[Mg(OH)2]}}=\frac {\ce{[Mg(OH)+][OH-]}}{K_\mathrm{sp}}$$

$$K_\mathrm{b2} = \frac {\ce{[Mg^2+][OH-]}}{\ce{[Mg(OH)+]}}=\frac {\ce{[Mg^2+]{[OH-]}^2}}{K_\mathrm{b1} \cdot K_\mathrm{sp}}$$

$$K_\mathrm{b2} \cdot K_\mathrm{b1} \cdot K_\mathrm{sp} =K_\mathrm{sp}^{*} = \ce{[Mg^2+]{[OH-]}^2}$$

You can see that effectively, for practical purposes, $\ce{Mg(OH)2}$ acts like a salt, with minor glitches of occurence of $\ce{Mg(OH)2(aq)}$ + $\ce{Mg(OH)+(aq)}$


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