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    Hi,
    I'm trying to calculate pH on a Casio fx-85 and I'm not getting the right answer. I'm doing pKa = -log(10)*(2.3x10(-1)) (for example)
    and I'm getting -0.23

    What am I doing wrong
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    http://en.wikipedia.org/wiki/Acid_dissociation_constant
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    Is the answer 0.638...?
    You aren't putting the 10 in are you? Just do -log(conc. H+)
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    That asterisk implies you're putting a multiplication sign in there? That's where you're going wrong.
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    (Original post by smoozth)
    Hi,
    I'm trying to calculate pH on a Casio fx-85 and I'm not getting the right answer. I'm doing pKa = -log(10)*(2.3x10(-1)) (for example)
    and I'm getting -0.23

    What am I doing wrong
    I really have absolutely no idea why you're putting that sum in, it makes very little sense! What you might be doing wrong is that the log button on a casio calculator automatically makes it log to the base 10, so the 10 is unnecessary, so long as you just use the button with 'log' written on it, NOT the one with log and two strange bars diagonally right and up. So if you were trying to get pKa from a Ka, then it's just -log(2.3x10^-1) to use your numbers, I'm not sure why you're multiplying. If you're trying to get pH from a pKa....

    pKa= -log(Ka)

    To find the pH from a pKa, you need to do 10^-pKa, which gives you the Ka.

    Then, bearing in mind that Ka=[H+]x[A-] all over [HA], and that [H+] = [A-] under most circumstances, we get Ka=([H+]^2) / [HA] , where [HA] is the intial conc. of the acid.

    Just rearrange this to get [H+]^2 = Ka x [HA]

    Then, find the square root of that to give you [H+] and then put it into the pH=-log[H+] to get the pH.
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    (Original post by JK471993)
    Is the answer 0.638...?
    You aren't putting the 10 in are you? Just do -log(conc. H+)
    That gives pH, they're trying to find pKa! Which is -log(Ka). I think she's doing log to the base 10 of 10 and then multiplying it by the Ka...
 
 
 
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