Calculate the pH - Chemistry help

The [OH-]= 6 x 10 ^-11 and the H3O is 1.67 x 10^-4

How do I do this?

Also, how do I calculate the OH of a solution having a H3O of 2.85 x 10^-6?

Please explain. Thank you.

If the [H3O+] is 2.85 x 10^-6 then the pH is the -log(2.85 x 10^-6) = 5.55

If the pH is 5.55 then the pOH is 14 - 5.55 = 8.45

The [OH-] is going to be 10^-8.45 = 3.55 x 10^-9

Ok, some basics:

Kw = [H+][OH-] =1.0*10^-14 mol/L

[ H+ is analogous to H3O+]

pH+pOH=14

pH = -log[H+] by log I mean the log with the base of 10
[H+] = 10^(-pH)

I am a little confused by what you are asking for the first part. Are you needing to find out how those numbers come to be? You have more than enough information to calculate pH:

[OH-]=6e-11 mol/L 1e-14 mol/L/6e-11 mol/L= 1.67 e-4 mol/L <----this is to show you how those two numbers can be determined from one another using the Kw....the opposite can be done using your hydronium ion concentration and solving for hydroxide concentration.
The pH for this system is:
pH=-log(1.67e-4 mol/L)=3.78

For the second problem: you want the same formula, just solve for [OH-] concentration and determine pH?
then: 1e-14 mol/L / 2.85e-6 mol/L = 3.51e-9 mol/L
The pH of this system is simply: pH=-log(2.85e-6 mol/L)=5.55