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chemistry

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give the [OH] of a vinegar solution that has a pH of 3.2. i got 6.3 *10^4. is that right?

  • chemistry - ,

    I don't get your number. I think you did this.
    pH = -log(H^+)
    3.2 = -log(H^+)
    -3.2 = log(H^+)
    (H^+) = 5.86 x 10^-4
    The problem asks for OH, so
    (H^+)(OH^-) = Kw.
    Kw = 1 x 10^-14
    You know (H^+). Calculate (OH^-)

    A far easier way to do it is this.
    pH = 3.2; therefore, pOH = 14 - 3.2 = 10.8.
    Then pOH = 10.8 = -log)OH^-)
    -10.8 = log(OH^-)
    (OH^-) = ??

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