Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
I read that $\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ in water. Solubility of $\ce{MgCl2}$ is $\pu{543 g/L}$ and that of $\ce{NaCl}$ is $\pu{360 g/L}$ (both at $20^{\circ} \pu{C}$).
I think that $\ce{NaCl}$ should be more soluble due to its higher ionic nature. $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ are more polarizing and have more covalent nature and thus should be less soluble.
I want to know the reason for why the opposite is happening.
There are no answers to such why questions, in case anyone is looking for an "answer". Although it is very tempting to rationalize every laboratory observation, there are no definite answers to questions about why certain chemical phenomena occur. While it may be tempting to come up with explanations that allow us to pass exams or understand concepts on a surface level, the truth is that our current understanding of chemistry is not advanced enough to accurately predict the solubility of compounds in water or other solvents. The same goes for melting and boiling points. Tools such as ChemDraw and SciFinder can provide estimates, but there is a significant margin of error in those solubility numbers and the results may be incorrect.
Let us see the complications, e.g.,
Take anhydrous $\ce{MgCl2}$, add it to water and dissolve, the reaction is exothermic, and solution can be shown to be slightly basic.
Take the hydrated version, $\ce{MgCl2.6H2O}$, and add it to water, the reaction is not exothermic in water, and water's pH does not change. So may even ask, which magnesium chloride is being discussed?
As a result, many chemists rely on macroscopic measurements such as thermodynamic properties but such properties are quiet about microscopic details and "whies". A paper titled The solubilities of some inorganic halides, Trans. Faraday Soc., 1958,54, 34-39, is relevant the OP's question with actual experimental data. The abstract reads,
"
They have the thermodynamic data of Group I and Group II halides in Table 1.
I don't think there is any simple answer.There are some rules of thumb, all of which have exceptions. Most chlorides are soluble. That includes both MgCl2 and CaCl2.
Solubility is the result of the balance between competing interactions: between the ions in the salt and between those ions and the water.
Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.
The longer answer is to do with something called the "enthalpy of solvation".'
you can think it this way that the lattice enthalpy in MgCl2 and CaCl2 will be less and can be easily be overpowered by hydration enthalpy
$\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ because In $\ce{MgCl2}$ and $\ce{CaCl2}$ the ionic charge is +2 hence it can interact with more number of $\ce{H2O}$ molecules, as the number of interactions increases more amount of energy is liberated which is a hydration energy which is sufficient enough to overcome the lattice energy of the compound. Even solubility increases with increase in charge on ion.
