Aqueous vs. Solid

I've been having a conceptual struggle with these two states for a while now, and I'd like to get it cleared up.

A compound AB dissolves in water.

$$\ce{AB(s) + H2O <-> A+(aq) + B- (aq) + H2O(l)}$$

Now, we know that in the case of salts, there can be strong and weak electrolytes based on the degree of dissociation. For example, NaCl is a strong electrolyte because when dissolved in water it dissociates completely. However, in theory, there should be some salts that do not dissociate completely and thus are weak electrolytes, for example, I think, NaCH3CHOO, despite its high solubility.

$$\ce{NaCH3COO(s) + H2O(l) <-> NaCH3COO(aq) + Na+(aq) + CH3COO- (aq)}$$

Here's an example that confuses me. Assuming that CaO dissociates completely in water:

$$\ce{CaO(s) + H2O(l) <-> Ca^{2+}(aq) +O^{2-} (aq) + H2O(l)}$$

O(2-)(aq) + H2O(l) --> (irreversibly, I suppose because of the general formula for the combination of anhydride metal oxides with water) 2OH-(aq)

Which then finally yields:

$$\ce{CaO(s) + H2O(l) <-> Ca^{2+}(aq) + 2OH- (aq)}$$

Then, here's the tricky part:

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

Now, I don't know whether this is true. Calcium hydroxide, according to solubility rules, is insoluble in water, but as we know, all compounds are at least a little bit soluble. The KSP may be very low, but it dissolves a bit. My confusion is in how we know if a precipitate forms or not without being given any molar concentration of Ca2+ or the KSP of Ca(OH)2. The latter is necessary to calculate QSP, and this is necessary to determine whether a precipitate actually forms. Without being given any other information in a school setting, is it correct to assume that all salts are strong electrolytes and that all low-KSP compounds form a precipitate in solution?

Thanks!

3 Answers

$\ce{CaO}$ (lime) begins by reacting with water to $\ce{Ca(OH)2}$ (slaked lime) with a very large exotherm

http://www.ktf-split.hr/periodni/en/abc/kpt.html
$\ce{Ca(OH)2, K_{sp}} = 5.02×10^{-6}$

If you are not concerned about pH (via ion hydrolysis) and pH buffering, soluble salts are strongly dissociated. As you say, depending on concentration, the answer can go either way. If you are reacting it further (lime water as a $\ce{CO2}$ detector), precipitation of much less soluble $\ce{CaCO3}$ makes the intermediate step unimportant (perhaps aside from kinetics). Add more $\ce{CO2}$ to obtain soluble $\ce{Ca(HCO3)2}$.

Up to my knowledge the reaction mechanism of CaO and $H_{2}O$ is not well understood yet, however seems that for low density CaO, $H_{2}O$ is absorbed over the CaO. The mechanism should be something like this:

$$\ce{CaO + 2H2O -> CaO\cdot2H2O ->Ca(OH)_2 }$$

Your reaction is wrong. Reference here. The reaction is very favorite energetically so you don't have the equilibrium you mentioned: when $Ca(OH)_{2}$ is formed you will not have CaO back until you provide sufficient heat and temperature.

The second reaction is right: $$\ce{ Ca(OH)2(s) <=> Ca^{2+}(aq) + 2OH- (aq)}$$

The solubility of $Ca(OH)_{2}$ is 0.173 g/100 mL (20 °C). This cause the alkalinity of the solution.

How we know if a precipitate forms or not without being given any molar concentration of Ca2+ or the KSP of Ca(OH)2?

You can't know it: if you have less than 0.173 g/100 mL of calcium hydroxide you will not have any precipitate.

Without being given any other information in a school setting, is it correct to assume that all salts are strong electrolytes?

No, some salts are absolutely not good electrolytes $CaCO_{3}$ is an example. You could say that most of the halogen salts are strong electrolytes.

Is it correct to assume that all low-KSP compounds form a precipitate in solution?

No, because if the solution is not saturated you will not have a precipitate, however is correct say that probably a low-KSP compound will form a precipitate in solution.

I think what's confusing you is how you know if an insoluble precipitate will form and the simple answer would be if an insoluble precipitate can form them it will form, and this is almost a direct consequence of le Chatelier's principle.

Consider the reaction,

$\ce{MgSO4(aq) + Na2CO3(aq) -> MgCO3(s) + Na2SO4(aq)}$

What's actually happening is:

$\ce{MgSO4(s) -> Mg^2+(aq) + SO4^{2-}(aq)}$

$\ce{Na2CO3(s) -> 2Na+(aq) + CO3^{2-}(aq)}$

$\ce{Mg^2+ (aq) + CO3^{2-}(aq) -> MgCO3(s)}$

The Sodium and Sulphate ions in solution are spectators and do not 'react', they don't really do anything at all but you have a solution equivalent to that which you would have formed from the dissolution of Sodium Sulphate in water.

Sorry for the poor format, I haven't figured out how to the proper chemical font yet and also you don't have to include water in your dissolution reactions, the involvement of water is implicit since your state symbols are changing