# College Chemistry

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Calculate both [H30] and [OH-] for a solution that is:
pH=5.50

ph=7.00

• College Chemistry -

pH = -log(H^+)
Plug and chug for (H^+).
Then Kw = (H^+)(OH^-). Substitute (H^+) and solve for (OH^-).

• College Chemistry -

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 10x 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.

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