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    fu(acid) =1 / 1+antilog(ph-pka)
    how do I do an antilog?
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    (Original post by asaaal)
    fu(acid) =1 / 1+antilog(ph-pka)
    how do I do an antilog?
    Antilog is just inverse log.

    \ anti\log_a x = a^x
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    (Original post by joostan)
    Antilog is just inverse log.

    \ anti\log_a x = a^x

    so what do i type into my calculator exactly?
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    (Original post by asaaal)
    so what do i type into my calculator exactly?
    Combine the pKa and the pH into one logarithm.

    \ log_a x + log_a y = log_a xy

    Then use the antilog.

    \ a^{log_a xy} = xy
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    (Original post by joostan)
    Combine the pKa and the pH into one logarithm.

    \ log_a x + log_a y = log_a xy

    Then use the antilog.

    \ a^{log_a xy} = xy
    I'm not following - a) under what circumstances would you want to "combine" the pH and pKa, b) they are both -log10 of something so (I have no idea what kind of calculation you're trying to do here by the way) if you're for some reason adding them you'll get -log10([H3O+])-log10(Ka)=-log10([H3O+]*Ka). No need for an "antilog", whatever that is - if it's just the same as raising something to the power, then why call it something different?
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    Oh OP you might be looking for 10pKa-pH which is a useful way of determining ratios of [HA]/[A-] (or [H2A]/[HA-], [H7A2-]/[H6A3-], whatever acid/conjugate base pair you like) in solution. I don't know why you'd call it an antilog or whatever, it's just 10 ^ (pKa-pH).
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    (Original post by Big-Daddy)
    Oh OP you might be looking for 10pKa-pH which is a useful way of determining ratios of [HA]/[A-] (or [H2A]/[HA-], [H7A2-]/[H6A3-], whatever acid/conjugate base pair you like) in solution. I don't know why you'd call it an antilog or whatever, it's just 10 ^ (pKa-pH).
    Ther term antilog has been around for a long time to describe the operation which is the reverse (or inverse) of a logarithm.

    A logarithm is the number to which 10 (if that's the base) must be raised to make the target value.

    Hence an antilog is the value obtained by 10 raised to the power of its log.
 
 
 
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