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Re: Aquatic Plants Digest V2 #879
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.