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There was this exercise which asked to find the OH- concentration given the H+ concentration, i firstly calculated the pH as -log[H+], did 14-pH to find the pOH and then did 10^(-pOH) to find the concentration of OH-.

My textbook used the fact that Kw (water equilibrium constant) is essentially Kw = [OH-] * [H+] so it did [OH-] = Kw/[H+]

The fact is...isn't Kw temperature dependent? My textbook says that it is

25 celsius --> Kw --> 1*10^-14

whereas

60 celsius --> 9.61*10^-14

So my question is...why is my procedure temperature independent while the book one temperature dependent? My procedure would always give the same concentration of OH whereas changing the Kw the textbook would also get other ones.

Thank you.

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    $\begingroup$ Where does your formula $pOH=14-pH$ come from in the first place ? ;-) $\endgroup$
    – Hippalectryon
    Commented Jun 30, 2016 at 16:55
  • $\begingroup$ chem.purdue.edu/gchelp/howtosolveit/Equilibrium/… My textbook also states that $\endgroup$
    – Nur Bedeir
    Commented Jun 30, 2016 at 17:09
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    $\begingroup$ Click the "show me why" section of that link you provided :) $\endgroup$
    – Hippalectryon
    Commented Jun 30, 2016 at 17:55
  • $\begingroup$ Ok, even my method is temperature dependent :P Thank you really much for letting me know :) $\endgroup$
    – Nur Bedeir
    Commented Jun 30, 2016 at 20:31
  • $\begingroup$ Exactly, both methods are equivalent in the end. $\endgroup$
    – Hippalectryon
    Commented Jun 30, 2016 at 20:43

1 Answer 1

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It is because you assumed the number 14 as a constant while it is actually $-\log(K_\mathrm w)$. It is $-\log(1 \times 10^{-14}) = 14$ at 25 degrees Celsius but changed to $-\log(9.61 \times10^{-14}) = 13.02$ at 60 degrees Celsius.

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