[BHAVESH]

Kw of water is given at various temperatures
Like...
@ 0°C - 1.5 X 10^-15
@ 25°C - 1.0 X 10^-14

So how do I calculate at KW AT any temperature???

Thanks

[mjc123]

Have you heard of the van't Hoff equation?

[Borek]

Either from the thermodynamical data (as mjc123 suggested) or by fitting some function to the experimentally determined values (like these listed here: water ion product).

[BHAVESH]

Thank you mjc123 & Borek.
Matter Cleared.
Thanks again!!!

[BHAVESH]

Hi mjc123,
I applied Van't Hoff Equation but getting difficulties....

[mjc123]

Care to elaborate?
Can you show your working, and where the difficulty comes?
(note: your value at 0°C is different from e.g. what is found here: https://en.wikipedia.org/wiki/Self-ionization_of_water. Maybe that is causing dicrepancies?)

[BHAVESH]

Dear mjc123,
attached my working.

[mjc123]

Where do you get your figures for enthalpy from? The figures of ca. 100 kJ/kg you quote are wildly inaccurate, and the figure you use in calculation of 62.7 kJ/mol is also inaccurate (though not wildly so). Using a value of 57 kJ/mol, I get Kw at 40°C = 3.01 x 10^-14.

[BHAVESH]

There is an article published by Indian Institute of Technologies Bombay  "Thermodynamic properties of Water substance" and I took ΔH from Enthalpy @ 40°C is 167.53 & @ 25°C it is 104.83 so the diff of enthalpy is 62.7 KJ/mol.

Can you please elaborate your calculation?

[BHAVESH]

So, pH at 40 °C at Kw 3.01 X 10^-14 will be 6.76 AND pH with Kw I calculated 4.35718 X 10^-14 will be 6.68. There is a difference of 0.08 which I want to understand.

Moreover, when the temp increases, the difference of pH with given in table and the way I calculated increases.

[BHAVESH]

I have attached pdf of my working. I cannot attach excel sheet here.

[mjc123]

Above, you wrote kJ/kg, not kJ/mol, which was confusing - which is it? In either case, it's the wrong value to use. ΔH in the van't Hoff equation is not the difference between the enthalpy of water at 25 and 40°C. It is the enthalpy change for the reaction whose equilibrium constant is under consideration, i.e. for the reaction
H₂O(l)  H⁺(aq) + OH^-(aq).
57 kJ/mol was a value I used from memory. Using the values of enthalpy of formation from Johnson, "Thermodynamic aspects of inorganic chemistry", I get a value of 55.84 kJ/mol.

[BHAVESH]

Ok. Thank you very much.
So, -55.84 KJ/mol is basically Neutralization Enthalpy of H₂O.

Thank yo very much.