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I need help calculating [H3O] from pH and the pOH scale
using my calculator. I have a TI-30XIIS

  • chemistry - ,

    Given pH = 4.67, pH = ?
    pH = -log(H^+)
    4.67 = -log(H^+)
    -4.67 = log(H^+)
    (H^+) = 2.29E-5

    Here is how you do it on the calculator.
    Punch in 4.67. Hit the change sign to it becomes -. (If you don't have a change sign, punch in -4.67 at the start.)
    Now locate and hit the 10x key. That should return 2.29086E-5 which can be rounded.
    For pOH, the easy way, if you have pH, is
    pH + pOH = pKw = 14
    So 14=pH = pOH.

    If you have OH^- already, then
    pOH = -log(OH^-) and it works just like the pH does. By the way, if you need pKa, it works just like pH and pOH, pKa = -log Ka. pKb = -log Kb.

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