# Question a98a1

Sep 23, 2015

Yes, it is.

#### Explanation:

Anhydrous cobalt(II) chloride, ${\text{CoCl}}_{2}$, is actually hygroscopic, which means that it will readily attract water from its surroundings, usually moisture from the air, to form the hexahydrate $\text{CoCl"""_2 * 6"H"_2"O}$.

The hexahydrate is deliquescent, which means that it will continue to absorb water from it surroundings and eventually dissolve in the absorbed water.

So, when you heat the the hexahydrate, you essentially drive off the water of hydration and keep the anhydrous cobalt(II) chloride, ${\text{CoCl}}_{2}$.

Upon heating, the hexahydrate, which is light purple in color, will turn into the dihydrate form, $\text{CoCl"""_2 * 2"H"_2"O}$, which is violet in color, and eventually into the anhydrous form, which is blue in color.

The anhydrous salt is blue and the hexahydrate is light purple.

This is the dehydration reaction. The hydration reaction will take place in the exact opposite direction.

The blue anhydrous salt will form the dihydrate first, then finally the hyxahydrate.

Depending on how much water you add, you can form an aqueous solution of the hexahydrate. This solution will contain hydrated ${\text{Co}}^{2 +}$ ions in the form ["Co"("H"_2"O")_6]^(2+)#, and it will be light purple in color as well.

So, to answer your question, the dehydration and hydration of cobalt(II) chloride are reversible because by adding water to the anhydrous form you can reform the hexahydrate.

You drive off the water of hydration upon heating to get the anhydrous salt, then add water to the anhydrous salt to reform the hexahydrate.