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### Topic: Equilibrium constant, Kc, and water  (Read 18981 times)

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#### cliverlong

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##### Equilibrium constant, Kc, and water
« on: April 14, 2008, 05:43:31 PM »
Hi,

I'm finding it difficult to explain when the concentration of water does and does not need to be included in determination of equilibrium constant, Kc.

The texts I have say ignore concentration of water when it acts as a solvent. Now, sorry this is a naive question, which reactions are those where water is acting as a solvent?

If I take the solution of sodium chloride or ethanoic acid then I can say that solution will produce a disassociation of H20 liquid into hydroxide OH- and hydronium H3O+ ions. So in this case if I "see" H2O on one side and the ionic species on the other side of the equation, then this process is solution and I ignore concentration of water when determining Kc

Similarly, if I consider the solution of ethanol by water, is this also (partially) an "ionic solution process"? So again concentration of water is ignored in determining Kc ??

Whereas, if I consider the reaction of a primary alcohol with an aldehyde to produce an equilibrium product of ester and water, the water does not exist either as a liquid or ionised on one "side" of the reaction. The water is created by the esterification process. Hence, this is not solution and in this case I include the concentration of water in the determination of Kc

Thanks

Clive

#### Kyle1990

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##### Re: Equilibrium constant, Kc, and water
« Reply #1 on: April 14, 2008, 06:45:15 PM »
water is omitted from the equilibrium expression because it is a pure liquid. You can't increase the concentration of a pure liquid or solid, (they are pure) so they are omitted from the expression.
thus for the equation:
H2O--->H+ +OH-

the equilibrium expression is written as
K=[H+][OH-]

for a solution of sodium chloride and acetic acid dissolved in water:
NaCl + CH3COOH----> Na+  +Cl- +  H+   + CH3COO-
the species merely dissociate in the water, they do not necessarily react with it. Thus,

K= [Na+][Cl-][H+][CH3COO-]/[CH3COOH]

note NaCl is not included in the expression; it is a solid
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#### Borek

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##### Re: Equilibrium constant, Kc, and water
« Reply #2 on: April 14, 2008, 07:02:21 PM »
You can't increase the concentration of a pure liquid or solid, (they are pure) so they are omitted from the expression.

What about decreasing them? Is it impossible as well? If it is possible to decrease - why we are omitting water concentrations?

Water concentration is usually assumed to be constant, so it is ignored, or - more precisely - moved into equilibrium constant. It works for diluted solutions, it doesn't work for more concentrated ones. However, in the case of concentrated solutions other effects start to play important role, so calculations are already difficult.

Quote
thus for the equation:
H2O--->H+ +OH-

the equilibrium expression is written as
K=[H+][OH-]

Spmetimes as K=[H+][OH-]/[H2O], that's why pKw is sometimes listed as 14 and sometimes as 15.7.

Quote
for a solution of sodium chloride and acetic acid dissolved in water:
NaCl + CH3COOH----> Na+  +Cl- +  H+   + CH3COO-

You can't combine two separate reactions into one.
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