Concentration of water vs. equilibrium

Hi,
I understand from calculation that the molar concentration of Pure water is 55.5 moles/Liter
Then how come in equilibrium reactions when calculation the dissociation constant, we say that the concentration of water is 1 Molar? This seems like a huge difference to me? What is it I need to understand here?
 

DrDu

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Then how come in equilibrium reactions when calculation the dissociation constant, we say that the concentration of water is 1 Molar?
This is not correct.
What enters the equilibrium constant is the ratio of the concentration of a substance c relative to some standard concentration ##c_0##, i.e. ##c/c_0##. For diluted substances this standard concentration is ##c_0=##1 mol/l (molarity) or 1 mol/ kg (molality) or the like (or more precisely the behaviour at infinite dilution extrapolated to a concentration of 1 mol/l). For solvents etc. we use as a standard state the pure substance, i.e. ##c_0=55,5## mol/l for water. In a dilute solution, the concentration ##c## of water is to an excellent extent equal to ##c_0## so that we can set the ratio equal to 1, at least for calculations with chemical precision.
 
This is not correct.
What enters the equilibrium constant is the ratio of the concentration of a substance c relative to some standard concentration ##c_0##, i.e. ##c/c_0##. For diluted substances this standard concentration is ##c_0=##1 mol/l (molarity) or 1 mol/ kg (molality) or the like (or more precisely the behaviour at infinite dilution extrapolated to a concentration of 1 mol/l). For solvents etc. we use as a standard state the pure substance, i.e. ##c_0=55,5## mol/l for water. In a dilute solution, the concentration ##c## of water is to an excellent extent equal to ##c_0## so that we can set the ratio equal to 1, at least for calculations with chemical precision.
Okay, so I might understand what you are saying in 2 possible ways, which of the following two ways is the correct?

So if we are measuring the equilibrium H2O + CO2 ⇔ H(+) + HCO3(-) which is a reaction taking place in the blood of a person, then the concentration of water is only 1Molar because a) we have a diluted solution, where water is the solvent so [H2O] =1M ?
or b) the ratio between the initial concentration of water and final concentration ( at products) is almost the same so the ratio between the two concentrations is 1?
 

DrDu

Science Advisor
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The equilibrium constant is a dimensionless quantity as it only depends on the ratios of concentrations to their respective standard concentrations.
Specifically for your reaction

## K=(c_\mathrm{H^+}/1\mathrm{ mol/l})\cdot (c_\mathrm{HCO_3^-}/1\mathrm{ mol/l})/[(c_\mathrm{H_2O}/ 55,5\mathrm{ mol/l})\cdot (c_\mathrm{CO_2}/1\mathrm{ mol/l})]\approx (c_\mathrm{H^+}/1\mathrm{ mol/l})\cdot (c_\mathrm{HCO_3^-}/1\mathrm{ mol/l})/(c_\mathrm{CO_2}/1\mathrm{ mol/l})
##
 
That's exactly the part I don't understand: so you have 55.5M H2O in the denominator in your equation (which makes the fraction 55 times as small) and then you can say it's approximately the same as "≈" removing the 55.5M H2O? How come we can just do that?
 

DrDu

Science Advisor
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The point is that c is to a very good approximation equal to c_0. So their ratio is 1.
E.g. the concentration of water in a solution of CO2 containing 1 mol/l is still about c=(55,5-1) mol/l=54,5 mol/l, hence it differs very little from c_0
 
PErfect, so the ratio between [H2O]_Start and [H2O]end is 1 :)
Thank you.
 

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