Explaining the dissociation constant of water

[Sconesnatcher]

The question is The dissociation constant of water is 10^-14, explain what this means using a chemical equation.

Heres how I'd answer this. The dissociation of water into the ions H3O+ and OH- is in dynamic equilibrium. The concentration of H3O+ and OH- ions present in 1L of water is said to be 10^-7 each and since Kw = [H3O+][OH-] / [H2O].

 

Do I even need to say all that bullshit or would I get away with writing

[H3O+][OH-] / [H2O]

 

[H3O+] = 10^-7

[OH-] = 10^-7

[hermanntrude]

well for a start, you need to write the equation.

 

Also remember that Kw is simply Kc for that particular equation. Also remember that the concentration of water doesn't appear in the equilibrium expression because it's a pure liquid.

 

The perfect answer would include your first sentence (which is very correct), followed by the equation for the equilibrium, with the Kw beside it.

 

You can then say that Kw = [H3O+][OH-] = 1 x 10^-14, and that since the [H3O+] and [OH-] should be equal, you can say that the [H3O+] is equal to 1 x 10^-7 and therefore the pH is 7. You should also add a note that the above is only true at 25°C. Kw, like any equilibrium constant, is dependant on temperature. In fact, the pH AND the pOH of water at higher temperatures are BOTH less than 7, because the autoionization is endothermic.