From: Richard Wood (rwoodphd_at_yahoo.com)
Date: Thu Feb 08 2007 - 10:24:29 CST
Ok, I got the volume right, not the number of H3O+ ions. It's been a long time since I did any wet chemistry. And so, I forgot about Avogadro's number. The pH of 6 is 1 x 10^-6 MOLES of H3O+, and not MOLECULES of H3O+. Thus, I would need 6.022 X 10^23 * 1 x 10^-6, or 6.022 x 10^17 H3O+ ions in a liter, or 8.325 x 10^-6 ions in my box. of 512 waters.
I would thus need a box 120122 times this, or a box of 61502884 waters to have a one H3O= ion and a pH of 6.
Richard L. Wood, Ph. D.
Cockeysville, MD 21030
----- Original Message ----
From: Jerome Henin <jhenin_at_cmm.chem.upenn.edu>
To: Richard Wood <rwoodphd_at_yahoo.com>
Sent: Thursday, February 8, 2007 10:58:50 AM
Subject: Re: namd-l: pH in NAMD
On Thursday 08 February 2007 10:36, Richard Wood wrote:
> Hi Mark,
> If, theoretically, your pH was 6, you'd have 1 x 10^-6 H3O+ ions per liter
> of solution. So, you'd need to know what the volume of your simulation
> was, i.e., how many liters (a really small number), meaning you'd have much
> less than 10^-6 H3O+ ions in your simulation!!!
> For example, I have a box of water that is 24A per side, which is a volume
> of about 13824 A^3. One 1 liter is 1000 mL, or 1000 cm^3. Conversely, 1 A
> is 1 x 10^-8 cm, so 1A^3 is 1 x 10^-24 cm^3. My box of water is therefore
> 1.3824 x 10^-20 cm^3. This turns out to be 1.3824 x 10^-23 L. Thus, I
> should have 1.3824 x 10^-29 H3O+ ions in my box to have a pH of 6. I'd
> obviously need to simulate a very large box in order to have the right pH,
> as my 24 A box of waters had 512 waters. I'd need to have a box of 3.7 x
> 10^31 waters in order to have ONE H3O+ for me to simulate.
> (I think I did the chemistry math correctly...)
No you didn't :-) although you do get the qualitative conclusion right. Think
about the order of magnitude again. 10^-29 molecules? Does this sound right?
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