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Topic: pH for water at different temperatures  (Read 26195 times)

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Offline THC

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pH for water at different temperatures
« on: March 16, 2008, 12:28:26 PM »
I don't understand how the pH value ("Neutral pH", right) is calculated in this table: http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

I see that the pH is half the pKw... but why?

Example: The Kw at 100 C is 51.3 *10^(-14), so [H+] =  sqrt(51.3 *10^(-14)) = 7.16*10^(-7). The pH value is therefore -log(7.16*10^(-7)) = 6.14 as it says in the table. No problems so far.
But now I want to calculate the pOH value. Since pH + pOH = 14, pOH = 14 - 6,14 = 7.86. But I also know that [H+] = [OH-] = 7.16*10^(-7), so pOH = -log(7.16*10^(-7)) = 6.14?! That can't be right.

What am I missing here?

Offline Borek

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Re: pH for water at different temperatures
« Reply #1 on: March 16, 2008, 01:26:22 PM »
Since pH + pOH = 14

No.

[H+][OH-] = Kw

Taking logs, changing signs:

pH + pOH = pKw

Now, it happens that for 25 deg C pKw = 14, but at 100 deg C pKw = 12.29 - so pH=pOH=12.29/2=6.14
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Offline THC

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Re: pH for water at different temperatures
« Reply #2 on: March 16, 2008, 02:38:07 PM »
[...]
Now, it happens that for 25 deg C pKw = 14, but at 100 deg C pKw = 12.29 - so pH=pOH=12.29/2=6.14

Ah yes! Thanks, that makes sense :D

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