# Concentration of water vs. equilibrium

1. Mar 21, 2013

### christian0710

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?

2. Mar 21, 2013

### DrDu

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.

3. Mar 21, 2013

### christian0710

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?

4. Mar 21, 2013

### DrDu

The equilibrium constant is a dimensionless quantity as it only depends on the ratios of concentrations to their respective standard concentrations.

$K=(c_\mathrm{H^+}/1\mathrm{ mol/l})\cdot (c_\mathrm{HCO_3^-}/1\mathrm{ mol/l})/[(c_\mathrm{H_2O}/ 55,5\mathrm{ mol/l})\cdot (c_\mathrm{CO_2}/1\mathrm{ mol/l})]\approx (c_\mathrm{H^+}/1\mathrm{ mol/l})\cdot (c_\mathrm{HCO_3^-}/1\mathrm{ mol/l})/(c_\mathrm{CO_2}/1\mathrm{ mol/l})$

5. Mar 21, 2013

### DrDu

6. Mar 21, 2013

### christian0710

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?

7. Mar 21, 2013

### DrDu

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

8. Mar 21, 2013

### christian0710

PErfect, so the ratio between [H2O]_Start and [H2O]end is 1 :)
Thank you.