# Quick question about equilibrium of a reaction

## Homework Statement

we have a solution of
Cocl$$^{}2-_{}4$$(alcohol) + 6H$$_{}2$$O(l) <---> Co(H$$_{}2+$$O)$$_{}6^{}2+$$(aq) + 4Cl-(aq)
at equilibrium
what would happen if we were to add more H2O?

## The Attempt at a Solution

we cant really change the concentration of the water so nothing would happen?

edit: sorry abot the bad reaction formula

this is an equilibrium, what happens when one side of the equilibrium is disturbed? what chemical principle is this related to?

its related to le chatelires principle but im not sure whether the water has any effect at all because we cant really change the concentration of the water

any reactant or solvent in a chemical reaction has a concentration, even water. So what will the equilibrium do if more water is added?

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Borek
Mentor
It is hard to tell from your reaction, but isn't it done in alcohol, in which case water concentration can be easily controlled?

likely the CoCl2 4- was prepared with an alcohol content in its solution to maintain this species and not revert to the water product.

## Homework Statement

we have a solution of
Cocl$$^{}2-_{}4$$(alcohol) + 6H$$_{}2$$O(l) <---> Co(H$$_{}2+$$O)$$_{}6^{}2+$$(aq) + 4Cl-(aq)
at equilibrium
what would happen if we were to add more H2O?
water is a reactant here.

chemisttree
Homework Helper
Gold Member

## Homework Statement

we have a solution of
CoCl2-4(alcohol) + 6H2O(l) <---> Co(H2O)6+2(aq) + 4Cl-(aq)
at equilibrium
what would happen if we were to add more H2O?
Here is a cleaned up version of your equation. Hit 'quote' to see how I did it.

I think it should be:

CoCl4-2(alcoholic) + 6H2O(l) <---> Co(H2O)6+2(aq) + 4Cl-(alcoholic)

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