Chemistry
posted by Julia .
A solution is prepared by dissolving NH3(g) in water according to the following equation. NH3(g) + H2O(l)=NH4^+(aq)+ OH^(aq)
If the pH= 9.60 what is the [OH] in the solution.

pH + pOH = pKw = 14. You know pH, solve for pOH and convert to OH^.
pOH = log(OH^). 
is this right
pH+pOH =14
pOH=149.60=4.4
[OH] =10^pOH
[OH] = 10^4.4 NOW IM STUCK? 
I'm stuck there too Ryan

pH + pOH = 14
9.60 + pOH = 14
pOH = 14  9.60 = 4.40
pOH = 4.4 = log(OH^)
4.40 = log(OH^)
Key in 4.4 on your calculator and hit the 10^{x} button. It should return 3.981E5 which I would round to 3.98E5. 
I get error when I type in 4.4 log and when I put in 4.4 I get 0.6434

I got error also

You aren't following instructions. I did NOT say take the log of 4.4. Doing that gives you an error message. You want the NUMBER whose LOG is 4.4. To get that you DON'T punch in the log of 4.4. You key in 4.4 and hit the 10^{x} button (that's 10^x button). That give you the number whose log is 4.4. Your display should read 3.98E5. You may have seen this notation, which is the same thing, as (OH^) = 10^{4.40}.
(Note that if you punch in 3.98E5 and hit the log key, you will get 4.4).