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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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