# Chemistry - acids, bases, equilibrium

1. Feb 25, 2008

1. The problem statement, all variables and given/known data
Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.49e-13 at some temperature, compute the pH of a neutral aqueous solution at that temperature.

2. Relevant equations
Kw = [H+][OH-]
pH = -log[H+]

3. The attempt at a solution
3.49e-13 = [H+]*10e-7
[H+] = 3.49e-6
pH = 5.457

2. Feb 26, 2008

### mit

remember Kw is the ionproduct of water at a given temperature. So {H+}*{OH-}= 3.49e-13 next {H+}*{OH-} can be seen as {H+}e2 so what to do next to gain {H+} from Kw?