Posted by Gayla on .
Calculate both [H30] and [OH] for a solution that is:
pH=5.50
ph=7.00

College Chemistry 
DrBob222,
pH = log(H^+)
Plug and chug for (H^+).
Then Kw = (H^+)(OH^). Substitute (H^+) and solve for (OH^). 
College Chemistry 
DrBob222,
pH = 7.00
pH = log(H3O^+)
7.00 = log(H3O^+)
7.00 = log(H3O^+)
Now you take the antilog of both sides of the equation. Do you know how to do that on your calculator. Punch in 7.00, change the sign to 7.00, then punch the 10^{x} button on your calculator. The answer should come up as 1 x 10^7. So 1 x 10^7 M = (H3O^+). Now to calculate (OH^).
(H3O^+)(OH^) = Kw. I'm sure you have had in class that Kw is the ion product for water and that equals 1 x 10^14 at room T. So
(H3O^+)(OH^) = 1 x 10^14
Substitute 1 x 10^7 for (H3O^+) and solve for (OH^). You should get 1 x 10^7 M.
Repost above if you have trouble with any of the others but show your work.