# Quick question about equilibrium of a reaction

mjolnir80

## 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 can't really change the concentration of the water so nothing would happen?

edit: sorry abot the bad reaction formula

eli64
this is an equilibrium, what happens when one side of the equilibrium is disturbed? what chemical principle is this related to?

mjolnir80
its related to le chatelires principle but I am not sure whether the water has any effect at all because we can't really change the concentration of the water

eli64
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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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?

eli64
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.

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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