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
[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 10x button. It should return 3.981E-5 which I would round to 3.98E-5.
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 10x button (that's 10^x button). That give you the number whose log is -4.4. Your display should read 3.98E-5. You may have seen this notation, which is the same thing, as (OH^-) = 10-4.40.
(Note that if you punch in 3.98E-5 and hit the log key, you will get -4.4).