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How is it possible that some insoluble compounds in water, e.g. $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ are considered strong bases?

Are they even strong bases? How can I tell if a compound is a strong base or not?

Is it possible to explain with equations and/or formulas what happens when these compounds are dissolved into water?

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  • $\begingroup$ $\ce{Ca(OH)2}$ is not insoluble in water --> 0.173 g/100 mL (20 °C). I also do not quite understand your question. $\endgroup$ – Jori Jan 26 '15 at 12:49
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    $\begingroup$ @Jori s\he asks if there exists a general way of predicting if a compound will be a strong base in water. At least that's what I got from it. $\endgroup$ – M.A.R. Jan 26 '15 at 13:17
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The typical definition of strong base that you will see at an introductory level is that there is complete dissociation to hydroxide ions, for example that NaOH completely dissociates into Na+ and OH-.

This is a terrible definition because reactions never go to completion. The definition contradicts the concept of chemical equilibrium.

In reality:

$\frac{[Na+(aq)][OH-(aq)]}{[NaOH(aq)]} = K_B = ~4M$ (from table discussed below)

So even dissolved aqueous NaOH is not all dissociated. It can exist as ion pairs or molecules in solution without being dissociated.

$K_B$ is quantification of how strong or weak the base is. Strong base could be defined as $K_B$ exceeding a certain value. Let us say $K_B>1$ and therefore $pK_B<0$

So the strength of the base only refers to the equilibrium between the undissoicated dissolved species and the dissociated ions. It is independent of how much or how little solid dissolves. Some hydroxides, like NaOH and KOH have high solubilities and some have low solubilities, but $K_B$ is a different equilibrium.

For quantitative data on the $K_B$ values of hydroxides, see: DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION Look up the corresponding metal ion alphabetically, and a $pK_B$ value is given.

However, keep in mind that for each of $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ there will be two $K_B$ values, since there are two hydroxides that can dissociate.

The $pK_B$ values for the second hydroxide dissociating are Mg 2.6, Ca 1.4, Sr 0.8, Ba 0.6 becoming stronger bases as you go down the column in the periodic table, increasing radius stablizing the charge of the ion. The first $pK_B$ would be even lower, but I don't see any quantitative values. It would be so low that it would be difficult to measure. So, yes these are strong bases.

As far as how to tell, other than experimentally, how strong a base of a hydroxide is, there is a good lecture The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution by Pilkington. As you go from left to right across the periodic table (excluding the noble gases), the hydroxides progress from being strong bases to being acids.

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  • $\begingroup$ Can you please explain how, on moving from left to right across the periodic table, "the hydroxides progress from being strong bases to strong acids"? $\endgroup$ – user33789 Oct 3 '16 at 6:20
  • $\begingroup$ @KaumudiHarikumar NaOH (strong base), Mg(OH)2 (base), B(OH)3 (weak acid), CO(OH)2 (weak acid), N(O)2(OH) (Ka =21), S(O)2(OH)2 (strong acid), Cl(O)3(OH) (even stronger acid) $\endgroup$ – DavePhD Oct 3 '16 at 12:26
  • $\begingroup$ OK, but you haven't explained the reason behind this trend... $\endgroup$ – user33789 Oct 3 '16 at 12:59
  • $\begingroup$ @KaumudiHarikumar The bonds between oxygen and the X element become increasingly covalent as you go to the right on the periodic table. In water, Na+ and OH- can easily dissolve as separate ions, but a C-OH bond won't break just by adding a compound to water. $\endgroup$ – DavePhD Oct 3 '16 at 13:16
  • $\begingroup$ Ohh, but it was my understanding that ionic bonds are generally stronger than covalent bonds :/ Secondly, why do the bonds become increasingly covalent? Fajaan's rules? $\endgroup$ – user33789 Oct 3 '16 at 13:18

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