Re: [APD] Re: pH at which CaCO3 dissolves

```Thanks, Roger.

I think I know why I couldn't remember the value. ;-)

So I'll write this all down again. Then next time it comes
up, try to remember where I wrote it down.

Scott H.

--- roger <roger at spinn_net> wrote:
>
> Augustine wrote:
>
> >At what pH does CaCO3 dissolve? pH=8.8
>
> Scott and Augustine, it isn't that simple.  Augustine,
> 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.
>
>
> Roger Miller
>
>
> 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
> degrees C)
>
> K=10^(-3.44)
>
> 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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