# Quick question about equilibrium of a reaction

1. May 13, 2008

### mjolnir80

1. The problem statement, all variables and given/known data
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?

3. 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

2. May 13, 2008

### eli64

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

3. May 13, 2008

### mjolnir80

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

4. May 13, 2008

### 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?

Last edited: May 13, 2008
5. May 14, 2008

### Staff: 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?

6. May 14, 2008

### 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.

water is a reactant here.

7. May 14, 2008

### chemisttree

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)

Last edited: May 14, 2008