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[APD] Re: pH at which CaCO3 dissolves
>At what pH does CaCO3 dissolve? pH=8.8
Scott and Augustine, it isn't that simple. Augustine, your first step assumes
that the molarity of calcium and carbonate are equal, which is rarely the
case. I'm not sure what you are doing in the second step. If CaCO3 dissolved
at pH 8.8 then it would dissolve in sea water (pH ~8.4). The reefs, shells
and banks would not exist.
I came up with
pH < 3.44 - log([Ca++][HCO3-])
would dissolve CaCO3. Expressing the calcium concentration as ppm hardness
and the bicarbonate concentration as ppm alkalinity the equation is
pH < 13.44-log(alkalinity*hardness).
If the water is soft then CaCO3 will dissolve under most conditions. For
instance, if alkalinity is 50 and calcium hardness is 30 the calcium carbonate
will dissolve at a pH less than 10.3. If the water is hard then CaCO3 won't
dissolve unless the pH is a little lower; if alkalinity is 300 and calcium
hardness is 250 then calcium carbonate won't dissolve unless the pH is less
than 8.6. My assumptions aren't particularly good in a solution as
concentrated as sea water, but with seawater calcium hardness (1000 ppm) and
alkalinity (117 ppm) calcium carbonate would dissolve at a pH less than 8.37.
There's a big difference between dissolving, and dissolving at a rate that
makes anyone happy. It's the rate of dissolution and not whether CaCO3 will
dissolve that really makes the difference.
Details are below.
Equilibrium between CaCO3 and it's parts can be expressed as:
Ca2+ + HCO3- <-> H+ + CaCO3
So the pH at equilibrium varies with the concentrations of both calcium and
bicarbonate. Assuming that the activity of CaCO3 is 1 and that other
activities are equal to molarity produces
K = [H+]/([Ca+2][HCO3-])
Thermodynamic quantities from my old CRC Handbook gives me (at 1 atm and 25
Recalling that pH=-log([H+}) you can get my second equation above.
The second formula comes from
molarity Ca++ = calcium hardness in ppm divided by 100000.
molarity HCO3- = alkalinity in ppm divided by 100000
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