A2 Chem: 0.3M of NaOH. calculate the pH when Kw = 1x10^14 at 298K

use pOH = -log [OH-] then pH + pOH = 14

eh?

surely its pH = -log(OH), so pH = 0.522

just like it is for acids?

nope:

potential of hydrogen pH = -log [H+]

potential of hydroxide pOH = -log [OH-]

what!?

wait hang on.

If there is 1.4 M of H2SO4, what is its pH? -0.442 right?

yeah (well -0.447)

Ok, so how is the method I used to get that answer different to the method I use to get the answer to the question in the opening post?

0.3M of NaOH.

Calculate the pH when Kw = 1x10^14 at 298K.


I dont get how to do this.

I know that Kw = [H+] x [OH-]

But since I only know the conc. of the NaOH, I dont understand how to use that identity above to work out the pH of NaOH.

Or am I being stupid and confusing myself over information that has no bearing on the answer?


Thanks

You want to find the pH, and you're given the concentration of NaOH (a strong base so can assume [NaOH] = [OH-]). You cannot find the pH directly from this and so you are given that K_w = [OH-][H+] = 10^-14.

So, firstly find the pOH then use pH + pOH = 14 to find the pH.

If pH = -log [OH] then > pH = 0.55 is a bit low for a strong base

The underlines pH and the boldend pH are different arnt they?