[romanenko]

Calculate the pH of a 1.2 multiplied by 10-7 M solution of NaOH in water.

Im stumped   

[istrydummy]

I think the forum rules are that you need to show some idea of what you already know, and the people here will be able to correct your mistakes and/or point you in right direction.

[romanenko]

Oh alright.
Well the reaction goes to completion because its a strong base, so the OH- concentration will also be 1.2 x 10-7 M. But you also need to account for the self dissociation of water, which will make it 2.2e-7.

Then, -log(2.2e-7) = pOH = 6.6575

14-6.6575 = 7.3424, which should be pH.

Anyone know what i did wrong?

[Borek]

You can't just add these two concentrations. Presence of OH^- from the base shifts water autodissociation.

http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base