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Name: Bill
Status: educator
Grade: 9-12
Location: CA
Date: April 2008

I have checked the archives and found one related entry: Kris N asked about the cobaltous chloride equilibrium on 18 March, 2003 [1672 Cobalt Chloride], [ ] but his question was only partially answered by Vince Calder since the question was not entirely clear.

My question is about the same equilibria: 1 mole of hexahydrated cobalt (II) ion reacts with 2 moles of chloride ion to produce in a reversible process 1 mole of tetrahydrated cobalt (II) chloride and 2 moles of water [Co(H2O)6]2+(aq) + 2Cl-(aq) <==> Co(H2O)4Cl2(aq) + 2H2O(l) Adding a source of chloride ion (e.g. hydrochloric acid) drives the equilibrium forward indicated by the resulting blue-colored solution.

This is predicted by Le Chatelier's Principle and expected since the increased concentration of chloride ion increases the rate of the forward reaction. Subsequently adding water to the acidified equilibrium appears to drive the system in reverse since the color reverts back to the pink associated with the hydrated cobalt ion. I require an explanation why this happens. Le Chatelier's Principle might predict the shift to the left as a response to the addition of water, a 'reactant' for the 'reverse' reaction. However, this is a heterogeneous equilibrium, the concentration of the water is not altered by the addition of more. The reverse reaction should not increase its rate due to the addition of water. Certainly adding water lowers the concentration of all the aqueous species, but I have assumed that this affects the forward and reverse reaction equally and there should be no shift in equilibrium. Is it a matter that there are more moles of reactants (for the reaction as written) and the effect of dilution has a greater impact on the left side slowing down the forward reaction more that the reverse? Or am I missing the point altogether and it is a matter of a simple dilution altering the color intensity (in this case somehow changing it from blue through purple to pink)?

Note that the system has been observed at constant temperature. That is to say there has been no temperature stress applies, although since adding water drives the system in reverse (the exothermic direction) there is a small temperature rise as a *result* of the shift.

You need to keep track of which concentrations are constant and which are not.

When water is added to the solution, the concentrations of all reacting species are lowered, EXCEPT WATER. So the water concentration stays approximately the same, all the rest are lowered, and the tetrahydrated species replaces chloride with water, allowing the chloride to dissociate.

Richard Barrans, Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming


You are correct in your statement that the concentration of water is not affected by adding more water. As such, pure liquids are not included in the expression of Keq because such systems do not change in concentration over the course of the reaction and do not, in of themselves, affect rates.

You are also correct in stating that adding more water to the system results in the further dilution of all aqueous reactants and products. Thus, both the forward and reverse reaction rates are slowed down. However -as you have thought- the rates are not slowed down *equally*. The forward reaction rate is slowed down more since it is dependent on essentially two reactants (one of which is raised to the second power) while the reverse rate is a function of only one reactant. Consequently there is a net production of more reactants (as written).

Or if you want to look at it from the perspective of Keq, the numerator has [Co(H2O)4Cl2] while the denominator has: [Co(H2O)6]2+]*[Cl-]^2. Reducing all the molarities by the amount of water added would lessen the numerical values of the denominator more. To require equilibria (set the equation to equal to Keq), there would need to be a shift toward the production of more reactants as written.

Greg (Roberto Gregorius)

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