Calculating [H3O+] and [OH-]

Excuse me, /sci/, I need some help calculating the amount of H30+ and OH- in an acid/base.  How would you fine fellows solve the following: "An aqueous solution of Ba(OH)2 has a [H30+] of 1x10^-11 M.  What is the [OH-]?  What is the molarity of the Ba(OH)2 solution?"

Excuse me, that's supposed to be H3O+, not H30.

God knows I'm rusty at this but here' goes

It gives you [H3O+], meaning it gives you [H+], so pH = -log[H3O+]. The relationship between pH and pOH is 14 = pH + pOH. Subtract pH from 14 to get pOH, then do 10^(-pOH) to get [OH-].

The Last part of your question requires you to notice that Ba(OH)2 has produces 2 moles of OH- for every 1 mole of Ba(OH)2. So [Ba(OH)2] = [OH-]/2

Alright, okay, -log(1e-11) = 11

pH = 11

11 + pOH = 3, pOH = 3

10^(-3) = 1e-3 / 2 = 5e-4 M = [OH-]

So if the [OH-] is 5e-4, then -log(5e-4) should be the pOH which is 3, right? Oh god, it's 3.3.  Have I gone wrong somewhere?

Alright, I think I understand what I'm doing here now.

Moving on, lets say I have to calculate the pH of Ba(OH)2 given 4.5e-4 M Ba(OH)2.  My thought process goes like this:

//Adjust for the two moles of OH
• 4.5e-4 M * 2 =  9.0e-4 M
//Use the formula -log(OH)= pOH
• -log(9.0e-4) = 3.045757491
//Subtract the pOH from 14 to determine pH
• 14 - 3.045757491 = 10.95424251 = pH

Have I done the calculations correctly?  Reworking the problem yields the same answer, and working it backwards yields 4.5e-4 M.