Chemistry - Why doesn't calcium carbonate dissolve in water even though it is an ionic compound?

Solution 1:

As someone said here, this:

The teacher stated that the ionic compounds dissolve in water except some carbonates.

Is indeed an oversimplification. First of all, the distinction between an "ionic compound" to other compounds isn't too defined. What your teacher probably said, or didn't say but wanted to, is that some ionic compounds easily dissolve in water. Salt (halite - NaCl) is the best example.

Calcium carbonate, in nature, also commonly dissolves. It's just not as immediate as dissolution of the more soluble ionic compounds. You are probably familiar with this phenomenon:

This forms because calcium carbonate dissolves. The rock is limestone, which is usually composed of pure calcium carbonate. Acidic water greatly enhances the solubility of calcium carbonate, and it doesn't even need to be highly acidic. Rain or river water that come into contact with the atmosphere absorb the $\ce{CO2}$ as $$\ce{H2O + CO2 <=> H2CO3},$$ which then facilitates calcium carbonate dissolution with $$\ce{CaCO3 + H2CO3 <=> Ca^2+ + 2HCO3-}.$$

Solution 2:

For an ionic substance to dissolve in water, there are two competing factors that determine the enthalpy of solution $\Delta{H}_\mathrm{sol}$ which is the enthalpy (energy) change when a solute is dissolved in a solvent:

1. The lattice energy (LE), the energy of formation of the crystal between infinitely separated ions. As LE is proportional to the charges of its ions, for calcium carbonate, its LE would be roughly 4× the LE of sodium chloride. $$\ce{CaCO3 (s) -> Ca^2+ (g) + CO3^2- (g)}$$ This value is always positive, as energy is required to separate the ions.
2. The hydration energy of gaseous ions, which is the enthalpy change when gaseous ions dissolve in sufficient water to give an infinitely dilute solution. $$\ce{Ca^2+ (g) + CO3^2- (g) -> Ca^2+ (aq) + CO3^2- (aq)}$$ These values are always negative, as energy is released upon hydration of ions.

As the LE of calcium carbonate is so large, a great amount of free energy would be required to break the strongly attracted ions apart, and this energy has to come from the enthalpy of hydration. In the case of calcium carbonate, the enthalpy of hydration is not large enough to overcome the large lattice energy, hence it exists as a solid.

Do note that solubility of ionic compounds are not absolute. In fact, solubility in water is often given as a value in g/L of water.

However, generally, we define solubility as:

1. Soluble substances can form a 0.1 M solution at 25 °C.
2. Insoluble substances cannot form a 0.1 M solution at 25 °C.

---

Solution 3:

Teacher stated that the ionic compounds dissolve in water except some carbonates.

That is, at best, an oversimplification. Other ionic compounds such as silver sulfide are sparingly soluble in water. Note that this isn't a carbonate.

And sodium hydrogen carbonate - $\ce{NaHCO3}$ - is soluble in water.

Sodium carbonate ($\ce{Na2CO3}$) is also soluble in water.

I want a clear elaboration of this phenomenon can somebody help me??

Solubility depends on:

1) The strength of the intermolecular attractions within the solute and the strength of the solvent-solvent interactions. How strongly is the calcium ion and the carbonate ion bound together? You might think that the force of attraction between a +2 and a -2 ion is strong, but remember, covalent bonds are generally stronger than ionic interactions. How strong are the solvent-solvent molecule interactions - which must be broken before something can be solvated?

2) Solvation. How well can the solvent lower the energy of the dissociated solute ions? How well can calcium ion and carbonate ion be solvated (by water in this case?) How strong are the solvent-solute ion interactions? What about the entropy of solvation? (Solvation shells are highly ordered).

If you wish to predict the solubility of some compound, try using hard soft acid/base theory.

---

Solution 4:

Solubility of salts is not a prediction that I like to make. It all comes down to $\Delta{G}_{sol} < 0$. That condition is fulfilled when $\Delta{H}_{sol} < 0$ and $T \Delta{S}_{sol} > 0$ or one factor outweighs the other.

There's two problems. Lattice enthalpies are large, as are solvation enthalpies, and the common observation is that $\Delta{H}_{sol}$ in most cases is small. We are calculating differences of large numbers, and the one fact we can confidently assert is that the difference is impossible to predict and generally meaningless.

Same thing for $\Delta{S}_{sol}$. Lattice entropies are small, but entropies of solution are lower than one would think thanks to the short-range order the solvated cation imposes on the solvent. You are gaining translational entropy from dissolving the lattice and losing translational entropy from the solvent through the inner coordination shell. You are calculating differences of numbers of equal magnitude again, and just by staring at reagents it's impossible to tell which way the pendulum is going to swing and if the result is positive or negative.