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22 October, 20:11

What is the [oh-] of a solution if its ph = 8.78? answer in units of m?

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  1. 22 October, 20:36
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    PH = - log ([H+])

    [H+] = 10^ (-pH)

    [H+] = 10^ (-8.78) = 1.65*10^-9

    [H+][OH-] = Kw

    Kw = 1.0*10^-14 at 25 degrees celsius.

    [OH-] = Kw/[H+] = (1.0*10^-14) / (1.65*10^-9) = 6.06*10^-6

    The concentration of OH - ions is 6.1*10^-6 M.
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