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Re: pH calculation please - really :)
- To: Aquatic-Plants at actwin_com
- Subject: Re: pH calculation please - really :)
- From: Paul Sears <psears at nrn1_NRCan.gc.ca>
- Date: Fri, 5 Jul 2002 08:33:29 -0400 (EDT)
- In-reply-to: <200207041948.g64Jm1229115 at acme_actwin.com> from "Aquatic Plants Digest" at Jul 04, 2002 03:48:01 PM
> PH Calculation Please
> (I believe) that if I mix 1.5 gallons of 0 dkh RO water with 1 gallon
> dkh tap water the resulting KH will be 6 dkh. Now, if I mix (the same
> as above) 1.5 gallons of 6.0 pH RO water with 1 gallon of 7.6 pH tap
> what will the resulting pH be, and why?
I missed the original, so I'm replying to the reply.
Assuming that the tap water has some sort of buffer system in it, and that
the RO water has effectively nothing in it, the pH of the mixture will be
7.6. This is because the ratio of acid to its anion will not have changed.
Both concentrations will have been reduced by the same factor (1/2.5).
Buffers set the pH in this way:
HA (acid) <-> H+ + A- (anion)
[H+][A-]/[HA] = constant
or [H+] = constant*[HA]/[A-]
Change [HA] and [A-] by the same factor - nothing happens to the pH.
[x] is the concentration of x.
It becomes a little more interesting if the buffer system is CO2/HCO3-:
The immediate effect of the dilution is nil, as before, but
if the CO2 concentration in the water was in equilibrium with the atmosphere
before dilution, it won't be after, and the CO2 concentration will rise
and the pH will fall, by log(2.5) if things go back to where they were.
log(2.5) is about 0.4.
Of course, tap water isn't usually in equilibrium with the atmospheric CO2.
Paul Sears Ottawa, Canada