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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: THC on March 16, 2008, 12:28:26 PM

Title: pH for water at different temperatures
Post by: THC 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 pK_w... but why?

Example: The K_w 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?

Title: Re: pH for water at different temperatures
Post by: Borek on March 16, 2008, 01:26:22 PM

Quote from: THC on March 16, 2008, 12:28:26 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

Title: Re: pH for water at different temperatures
Post by: THC on March 16, 2008, 02:38:07 PM

Quote from: Borek on March 16, 2008, 01:26:22 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