# Solution of Carbonates

Steven Dutch, Professor Emeritus, Natural and Applied Sciences, University of Wisconsin - Green Bay

## Basic Data

(CO2aq)/cO2air = 0.034

Solubility Constant of CaCO3  = (Ca+2)(CO3-2) = 4.8x10-9

(H+)(HCO3-)/(H2CO3) = 4.3 x 10-7

(H+)(CO3-2)/(HCO3-) = 4.7 x 10-11

## Carbon Dioxide and Surface Water

In normal air, CO2 is 300 ppm (before the recent rise in CO2 content) or in other words, exerts 3 x 10-4 atmospheres of partial pressure. Therefore we can expect the dissolved concentration of CO2 in water to be 0.034  x 3 x 10-4 = (10-5)

The term "carbonic acid" conventionally includes both dissolved CO2 and true H2CO3. Actually only 0.4% of the CO2 in solution goes into H2CO3; the rest is simply dissolved CO2. With that understood, we know

(H+)(HCO3-)/(H2CO3) = 4.3 x 10-7

(H+)(HCO3-) = 4.3 x 10-7 (H2CO3) = 4.3 x 10-7 (10-5) = 4.3 x 10-12

Since equal numbers of H and HCO3 ions form, we can write

(H+) = (HCO3-) = sqrt(4.3 x 10-12) = 2.1 x 10-6

Note that in neutral water the concentration of H+ is 10-7, so this ionization results in 21 times as much hydrogen ion as is normally present. Thus we can safely ignore the original hydrogen ion concentration.

Log 2.1 x 10-6 = -5.7, or pH = 5.7

Therefore all water in equilibrium with the air is slightly acidic due to the presence of carbonic acid. This has nothing to do with acid rain.

The next breakdown is easy to calculate. We have

(H+)(CO3=) /(HCO3 -) = 4.7 x 10-11

Since (H+) = (HCO3 -), these two terms cancel out, leaving (CO3-2) = 4.7 x 10-11. Period. So regardless of whatever else is going on, in any solution containing carbonic acid, the CO3-2 concentration is 4.7 x 10-11

### A Good Thing

The fact that H2CO3 only dissociates to a small extent makes it a weak acid, in contrast to nitric, sulfuric, and hydrochloric acids, which ionize completely in solution. Acetic acid in vinegar is another familiar weak acid. Being a weak acid doesn't make an acid harmless - hydrogen cyanide and hydrofluoric acid are also weak acids. One will make you dead and the other will make you wish you were dead.

Imagine what the world would be like if carbonic acid were a strong acid. It would break down completely to carbonate ion and hydrogen ions. We wouldn't be worried about global warming because the oceans would be a perpetual carbon dioxide sink, since carbonic acid would break down as soon as carbon dioxide dissolved.

Let's say we had H2CO3 = 2H+ + CO3= and

(H+)2(CO3=)/(H2CO3) = 1 (A good value for a very strong acid. If anything, it's generous, since it allows for some carbonic acid.)

Since there are two hydrogen ions for every carbonate, (H+) = 2(CO3=)

If we assume we have (H2CO3) = 10-5, the present value, we'd have

(H+)2(H+/2)/(10-5) = 1 or (H+)3 = 2 x 10-5 and (H+) = .027, meaning we'd have a pH of 1.6. We'd have oceans like battery acid. It would be a wonderful place for extremophiles.

## Solution of Limestone on the Surface

We can't just take the solubility constant of calcite, the concentration of CO3, and calculate the concentration of dissolved Ca, because as we liberate CO3 from the limestone, some CO3 will react to form HCO3 and thence carbonic acid. Calcite will continue to react until all three species are in equilibrium. The complete reaction, taking HCO3 into account, is

CaCO3 + H2O + CO2 (or H2CO3) <--> Ca++ + 2HCO3 -

It's not the carbon dioxide that drives everything, but the hydrogens. It takes two carbon dioxides to strip the hydrogens off the water: one from the air and one from the limestone.

We ignore excess CaCO3 and water in the calculation, and that leaves us with the need to determine (Ca++)(HCO3 -)2/(H2CO3)

We have

• (Ca++)(CO3=) = 4.8 x10-9
• (H+)(HCO3 -) /(H2CO3) = 4.3 x 10-7
• (H+)(CO3= )/(HCO3 -) = 4.7 x 10-11

Putting them together to get Ca and HCO3 on top and H2CO3 on the bottom, we get

• [(Ca++)(CO3=)] [(HCO3 -)/((H+)(CO3=  ))] [(H+)(HCO3 -) /(H2CO3)] =
• (Ca++)(HCO3 -)2/(H2CO3) =
• (4.8 x 10-9)(4.3 x 10-7)/(4.7 x 10-11) = 4.4 x 10-5

We can refer to this as the grand equilibrium constant for calcium carbonate. On the surface we can assume that whatever CO2 is consumed will be replaced from the air, so we have

• (Ca++)(HCO3 -)2/(H2CO3) = (Ca++)(HCO3 -)2/(10-5) = 4.4 x 10-5 or
• (Ca++)(HCO3 -)2 = 4.4 x 10-10

Since each Ca ion is accompanied by 2HCO3, obviously (HCO3) = 2(Ca), and we have

• (Ca++)(2Ca)2 = 4(Ca)3 = 4.4 x 10-10
• (Ca)3 = 1.1 x 10-10 or (Ca++) = 4.8 x 10-4

Thus a cubic meter of water (1000 kg = 1000 liters) could dissolve 0.48 moles of calcium (19 grams) or 48 grams of limestone.

In a warm rainy climate with 1 m of precipitation per year, we could dissolve 48 grams of limestone per square meter per year or about 18 cm3. That works out to .0018 cm of material removed from the surface per square meter, or 1.8 cm per 1000 years, or 18 meters per million years. That's actually slow compared to some observed weathering rates.

The answer to the apparent paradox that carbon dioxide helps dissolve calcium carbonate, where most chemical intuition would suggest it should go the other way, is that the carbon dioxide goes into an intermediate ion, the bicarbonate ion. Let's revisit the grand equilibrium constant again:

• (Ca++)(HCO3 -)2/(H2CO3) = 4.4 x 10-5.
• Also (CO2aq)/cO2air = (H2CO3)/cO2air = 0.034, so (H2CO3) = 0.034 CO2air
• Thus (Ca++)(HCO3 -)2/(H2CO3) = (Ca++)(HCO3 -)2/(0.034 CO2air) = 4.4 x 10-5.
• And (Ca++)(HCO3 -)2/(0.034 CO2air) = 4.4 x 10-5 so (Ca++)(HCO3 -)2/(CO2air) = 1.5 x 10-6.
• Finally (Ca++)(HCO3 -)2 = 1.5 x 10-6 (CO2air)

In other words, the calcium in solution is directly proportional to the partial pressure of CO2. If we assume the bicarbonate forms solely from the equilibrium between carbonic acid and calcium carbonate, then the calcium content is proportional to the cube root of the partial pressure. In other words, a big increase in atmospheric carbon dioxide will produce a smaller change in dissolved calcium. At the few hundred parts per million of atmospheric carbon dioxide, a change will result in about one third as much change in the solubility of limestone.

What does this do to the water? We have (Ca++) = 4.8 x 10-4, (HCO3) = 2(Ca) = 9.6 x 10-4 and (H2CO3) = (10-5). Also (H+)(HCO3-)/(H2CO3) = 4.3 x 10-7, and we can plug in numbers to get (H+)(9.6 x 10-4)/(10-5) = 4.3 x 10-7, or (H+) = 4.5 x 10-9. Thus the pH = 8.4, somewhat alkaline.

### What If?

What would happen if we had no atmospheric carbon dioxide? Say we take neutral pH 7 distilled water, seal it in a vessel with a nitrogen atmosphere, then drop in some limestone?

If we naively use the solution constant for calcium carbonate, we will get a wrong answer. Some of the carbonate ion will combine with water to make bicarbonate, allowing more calcium carbonate to dissolve and also leaving OH- left over, making the solution more alkaline. Some of the bicarbonate will react again to make carbonic acid, producing still more OH- and making the solution yet more alkaline. Some of the carbonic acid will finally escape into the gas phase, producing a carbon dioxide vapor pressure.

We have CO3= + H2O = HCO3- + OH- and HCO3- + H2O = H2CO3 + OH-

Since we know that (H+)(CO3= )/(HCO3 -) = 4.7 x 10-11, we can presume that bicarbonate ion will be far more abundant. Now we won't have an atmospheric source of carbon dioxide to make bicarbonate ions. They will have to come solely from the breakdown of the limestone.

CaCO3 + H2O  = Ca++ + HCO3- + OH-.

We can see this is going to make the solution alkaline and thus favor low solubility.

Thus (Ca++) = (HCO3-) = (OH-) = 10-14/(H+)

We can combine

• (Ca++)(CO3=) = 4.8 x10-9 and
• (H+)(CO3= )/(HCO3 -) = 4.7 x 10-11

We want something involving (Ca++), (HCO3-) and (H+). We see an obvious way to do it by dividing one expression by the other to eliminate (CO3=). We obtain (Ca++)(HCO3-)/(H+) = 102. Now  (Ca++) = (HCO3-) =  (OH-) = 10-14/(H+) so we have (Ca++)(Ca++)/(10-14/(Ca++)) = 102 or (Ca++)3 = 10-12. Thus  (Ca++) = (HCO3-) = (OH-) = 10-4, (H+) = 10-10 and pH =10. It will be pretty alkaline because every bicarbonate also results in the creation of a hydroxyl ion. It won't be a terribly concentrated solution because calcium carbonate isn't very soluble in alkaline solutions, because it really takes hydrogen ions to create bicarbonate ions and dissolve limestone.

Running the math to completion, we know (H+)(HCO3 -)/(H2CO3) = 4.3 x 10-7, which we can rearrange to give us (H2CO3) = (H+)(HCO3 -)/4.3 x 10-7 = 10-10*10-4 /4.3 x 10-7 = 2.3 x 10-8. Some of the carbonic acid in turn will exsolve to gaseous carbon dioxide according to (CO2aq)/cO2air = (H2CO3)/cO2air = 0.034. So CO2air = 2.3 x 10-8/0.034 = 6.7 x 10-7 or about 0.7 ppm. This is about 1/200 the present value of atmospheric carbon dioxide.

Also, (Ca++)(CO3=) = 4.8 x10-9, so (CO3=) = 4.8 x10-9/10-4 = 4.8 x10-5. This is surprisingly large, about half the concentration of bicarbonate, but since we never made any specific assumptions about carbonate ion like assuming we could neglect it, we can still be confident of our results. For example, if we check against the dissociation constant of bicarbonate, we have (H+)(CO3= )/(HCO3 -) =(10-10)(4.8 x10-5)/(10-4) = 4.8 x10-11, only about 2% different from the actual constant of 4.7 x 10-11. So the results check out.

## Solution of Limestone in the Subsurface

What happens as water circulates underground in limestone? At a depth of ten meters, the water will be under a pressure of one atmosphere. The water is already charged with carbon dioxide in equilibrium with air at 300 parts per million. But the water can now hold a lot more carbon dioxide. If we were to add carbon dioxide, it wouldn't begin to escape from solution until there was enough to generate one atmosphere of pressure, enough to overcome the confining pressure on the water.

Fortunately, there's an abundant source of carbon dioxide readily at hand - in the limestone. The solution constant for carbon dioxide in equilibrium with air is:

(CO2aq)/cO2air = 0.034

But now the water is free to pick up carbon dioxide until the water is saturated, as if it were in contact with one atmosphere of CO2. Thus we have

(CO2aq)/cO2air = (CO2aq)/1 = (CO2aq) = (H2CO3) = 0.034

We already worked out the grand equilibrium constant for the solution of limestone above, so we can just assume we have .034 molar carbonic acid instead of (10-5). The carbon dioxide comes from solution of the limestone. Now we have:

n the surface we can assume that whatever CO2 is consumed will be replaced from the air, so we have

• (Ca++)(HCO3-)2/(H2CO3) = 4.4 x 10-5
• (Ca++)(HCO3-)2/(.034)
• (Ca++)(HCO3-)2 = .034 x 4.4 x 10-5 or
• (Ca++)(HCO3-)2 = 1.5 x 10-6

Since each Ca combines with 2HCO3, again, obviously (HCO3) = 2(Ca), and we have

• (Ca++)(2Ca)2 = 4(Ca)3 = 1.5 x 10-6
• (Ca)3 = 3.74 x 10-7 or (Ca++) = 0.007

Thus a cubic meter of water (1000 kg = 1000 liters) could dissolve 7 moles of calcium (280 grams) or 700 grams of limestone. This is nearly fifteen times as much as we could dissolve out in the open air. And it gets more effective the deeper we go.

As for the rest of the chemistry, (HCO3-) = 2(Ca) = 0.014, and (H2CO3) = 0.034. We also know that

• (H+)(HCO3 -) /(H2CO3) = 4.3 x 10-7

Plugging into the first equation, we have (H+)(0.014) /(0.034) = 4.3 x 10-7 or (H+) = 10-6. The pH rises a little bit. We also have:

• (H+)(CO3= )/(HCO3 -) = 4.7 x 10-11
• (Ca++)(CO3=) = 4.8 x10-9

Let's see how they agree. We know (Ca++) = 0.007, (HCO3-) = 0.014 and (H+) = 10-6. So we have

• (10-6)(CO3= )/(0.014) = 4.7 x 10-11 or (CO3=) = 6.6 x 10-7.
• (0.007)(CO3=) = 4.8 x10-9 or (CO3=) = 6.8 x 10-7.

The slight disagreement is due to rounding and approximation errors.

### Caves

Now the water oozes out of the rock into a cavern. Ten meters below the surface the extra atmospheric pressure won't be significant but the solution is now about 15 times oversaturated in calcium carbonate. So calcium carbonate precipitates. This has nothing to do with evaporation - relative humidity in caves is often close to 100 per cent - it's solely a matter of oversaturation.

## The Deep Oceans

Calcium in the oceans is about 0.040% by weight, meaning in a liter of sea water there will be 0.4 grams or .001 moles of calcium. So in sea water we have (Ca++) = 0.001, which is about twice the concentration we have in surface water in equilibrium with limestone. Thus, at the surface, sea water is supersaturated with respect to calcium carbonate.

However, calcium carbonate becomes more soluble at low temperatures and also under increasing pressure. At about 4-5 km depth, it becomes soluble in sea water. The depth where calcium carbonate becomes soluble is called the carbonate compensation depth (CCD).

Carbon dioxide is about twice as soluble near 0 C as it is at 25 C. Also, carbon dioxide becomes more soluble in water with pressure. However, we can't assume the deep oceans are saturated in carbon dioxide since their carbon dioxide is transported down from the surface, plus a little created biologically and by the solution of calcium carbonate.

Created 03 April 2006, Last Update