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**To**:**Aquatic-Plants at actwin_com****Subject**:**Re: Aquatic Plants Digest V2 #879****From**:**Eric Jezek <jezek at aries_scs.uiuc.edu>**- Date: Wed, 6 Aug 1997 19:29:34 -0500 (CDT)
- In-Reply-To: <199708061948.PAA17660 at acme_actwin.com>

On Wed, 6 Aug 1997, Roger Miller wrote: > > Actually, for some of us it wouldn't even be much of a pH "spike". Based > just on simple mixing calculations, if your tank starts out with a pH of > 7, you drain off 15% of that and replace it with water with a pH of 10, > then the pH of the resulting mix is 7.07 (gasp!). Unfortunately, your simple mixing calculations are wrong. It looks like you calculated [H+] = 0.85*(1.0e-7) + 0.15*(1.0e-10), take the negative log of this number and indeed 7.07 pops right out. You can do the same for hydroxide ion, [OH-] = 0.85*(1.0e-7) + 0.15*(1e-4) = 1.5085e-5 and -log[OH-] = pOH = 4.82. Now pH = 14 - pOH, so what gives? You have to do an equilibrium calculation by satisfying this equation [H+][OH-] = 1.0e-14 (assuming standard temp) let x be the amount of H+ and OH- ions that react to form water and the equation looks like this [H+ - x][OH- - x] = 1.0e-14 Plug in your values for [H+] and [OH-] that you obtained from above and solve for x. You get two answers but one doesn't make any sense, so you you plug the other in to get [H+] = 6.67e-10 which give pH = 9.18. Of course, both calculations, yours and mine, make a lot of irrelevant assumptions so neither are really applicable to water changes in an aquarium. > Once its mixed, the buffers take over and the pH will likely shift some, > with the final number depending on the buffer capacity in the mixed water. Just be careful, that initial pH spike might be a little higher than you thought.

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