11
$\begingroup$

I know that, solubility of alkaline earth metal hydroxides increases down the group and solubility of alkaline earth metal sulfates decreases down the group.

I wish to understand the reason for these trends.

$\endgroup$
4
$\begingroup$

A few weeks ago I would have given the hard soft acid base theory (HSAB) explanation.

I have heard this "trend" explained in this way: Hydroxide is hard and sulfate is soft. So, the harder ions are, the more soluble with sulfate and less with hydroxide.

About a week ago I was playing around with some DH, lattice energy, and Ksp data and I found that barium seems to go against the "trend".

I found these posts that show that many of the "trends" we are taught in chem class aren't as trendy as we are told.

Look at this: http://www.chemguide.co.uk/inorganic/group2/solubility.html

Is this a homework question or a curiosity question? If the former, the "trend" isn't that trendy and the teacher should read the links above. If the latter, do you know that barium salts tend to go against the "trend"?

$\endgroup$
  • $\begingroup$ This answer could be improved by adding more information from the hyperlink. If the link breaks, then this answer isn't helpful. $\endgroup$ – Swedish Architect Mar 18 at 18:51
18
$\begingroup$

Okay, for this scenario, the two concepts that play a key role are the lattice enthalpy and the hydration enthalpy. Lattice enthalpy is the energy released when one mole of a compound is formed from its constituent gaseous ions under standard conditions. Lattice enthalpy is dependent on the length of the ionic bond between the cation and the anion in the compound. If we look at Coulomb's law, we find that the attractive force between two bodies with different charges is inversely proportional to the square of the distance apart between their centres of mass, i.e. $F\propto 1/r^{2}$. A shorter bond would require more heat energy to be broken. Therefore, the shorter the bond, the higher the lattice enthalpy.

Hydration enthalpy is the energy released when one mole of gaseous ions is completely hydrated by water molecules under standard conditions. If two ions of different sizes were surrounded by an equal number of water molecules each, the smaller ion would be more hydrated compared to the larger one due to the size differential. Therefore, the smaller the ion, the higher the hydration enthalpy.

Let's deal with $\ce{OH-}$ first. While progressing down Group 2, the size of the cation increases since the number of shells increases to accommodate to the extra electrons. Due to this, the distance between the centres of mass of the cation and the anion increases and by definition, the lattice enthalpy decreases. The hydration enthalpy also decreases since the size of the cation increases. However, due to the square factor, the lattice enthalpy decreases faster than the hydration enthalpy. This is why the solubility of Group 2 hydroxides increases while progressing down the group.

Now let's look at $\ce{SO4^2-}$. The same thing applies to the cation while progressing down the group. However, the $\ce{SO4^2-}$ ion is shaped in such a way that the change in the distance between the centres of mass of the ions is far less significant compared to that in the case of Group 2 compounds containing $\ce{OH-}$ ions. Therefore, the lattice enthalpy decreases much slower when progressing down the group compared to the case of Group 2 $\ce{OH-}$ compounds. Again, the hydration enthalpy decreases the same way as it does in the case of Group 2 cations bonded to OH⁻ ions. Since the hydration enthalpy decreases faster than the lattice enthalpy in the case of Group 2 sulphates, the solubility of Group 2 sulphates decreases while progressing down the group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.