# H2O in Equilibria

1. Apr 5, 2013

If we have an acid-base aqueous equilibrium system and decide to treat the concentration of H2O as a variable, does it then become impossible to solve for all the concentrations? Or can we write an equation to help us?

e.g. HA is added to water.

Mass Balance: CHA=[HA]+[A-]
Charge Balance: [H3O+]=[A-]+[OH-]
Equilibria: Ka=[A-]*[H3O+]/([HA]*[H2O])
Kw=[H3O+]*[OH-]/([H2O]2)

I'm not sure the auto-dissociation of water is called Kw still - certainly the constant I have written will not value 10-14 at 298.15 K and 1 bar pressure - but I still used the same equilibrium constant name.

Our variables are: [H3O+], [OH-], [H2O], [A-], and [HA]. 5 variables, 4 equations - any way to do this?

2. Apr 5, 2013

### DrDu

What you call mass balance is in reality a balance of the number of moles of radicals A.
To include the change in the number of moles of water
you have to write down also a balance for the number of moles of oxygen atoms.

The equation for Kw *assuming that it is sufficiently dilute to behave like an ideal solution) includes the molar fraction $x_{H2O}=n(H2O)/\sum_i n(i)$ where i counts all components in the solution,
In a dilute solution, it differs very little from one, whence it is often ignored. If you want the deviations from 1, you should also figure out what kind of concentration the brakets [...] really stand for (e.g. molarity, molality etc) and consider also the activity coefficients.

3. Apr 5, 2013

So it becomes possible to the system if I add the equation CH2O=[H3O+]+[H2O]+[OH-]. The value of CH2O is well known to be 55.51 M roughly. What we lack is the value for the constant that I have called Kw.

By the way why don't I have yet another equation specifying the mass balance of my H atoms? e.g. CHA+2*CH2O=[HA]+3*[H3O+]+2*[H2O]+[OH-]

4. Apr 6, 2013

### DrDu

The balance of amount of substance of H is simply not an independent equation, so it doesn't give you more information.
About the precise meaning of K_w you should consult a book on chemical thermodynamics, e.g. Klotz Rosenberg, Chemical thermodynamics. The point is that you use different standard states for the solvent and the solutes.
I.e. the concentration of the solvent is not number of moles per volume but number of moles per total number of moles.

5. Apr 6, 2013

Thanks.

What do you mean? Why is it not independent?

6. Apr 6, 2013

### DrDu

Because the equation you wrote down for H is expressible in terms of the equations for O, A and charge.

7. Apr 6, 2013

Ah I see, thanks. And just to check, the reason we can't write an O mass balance if we are treating [H2O] as constant is because we have no term for [H2O] then (which is the majority of our O atoms).

Back to the original problem: once I've written my O balance starting from H2O, you're saying I need to redefine [H2O] because Concentration=Moles/Volume is not sufficient here. So is concentration of H2O a unitless fraction then (according to the definition you just gave)?

What data exactly do we need to calculate it?

Last edited: Apr 6, 2013
8. Apr 6, 2013

### DrDu

I think the definition of K_w is $(\gamma_{H^+}c_{H^+}/1 mol/)\cdot( \gamma_{OH^-}c_{OH^-}/1 mol/l)/\gamma_{H_2O}^2x_{H_2O}^2$ where
$x_{H_2O}=c_{H_2O}/(c_{H_2O}+c_{H^+}+\ldots)$ and the gammas are activity coefficients which approach 1 in the limit of infinite dilution. For diluted ionic solutions they can be calculated using e.g. Debye Hueckel theory.
The precise definition of K_w may vary from source to source, e.g. it is quite common to use molalities instead of molarities, that's why you will have to go into experimental papers where this constant was measured or into specialized books like Klotz and Rosenberg.
If you are assuming ideal behaviour (i.e. assuming all gammas to be 1), you would have to express the molar fraction x of H2O in terms of the other concentrations. You can probably replace x by its first order Taylor expansion in the other concentrations.

9. Apr 7, 2013

### Darwin123

As a first order approximation, maybe you can use the density of liquid water at different concentrations of solute. From the density of water, one can calculate the "molarity" of the water in an aqueous fluid. The formula for equilibrium constant contains terms that include the molarity of the different components. The molarity of the solutes is going to change a great deal. The molarity of a liquid solvent won't change that much.

Liquid water is effectively incompressible. It is slightly compressible, but it has a very large bulk modulus. Even if the chemical reaction generates or disassociates water molecules, the density of the water molecules won't change very much.

The density of the liquid changes slightly with the concentration of solutes, but not much. Liquid water has a concentration of water molecules that changes very little. The solutes in an aqueous solution act effectively like a gas.

The situation would be different if the water is in vapor form. Water vapor is as compressible as any other gas. At atmospheric pressure and typical temperatures, water vapor is close to an ideal gas.