In arid areas, the soils usually contain large amounts of basic reacting salts. These salts contain calcium, magnesium, barium, and other Group 2 metal cations (positive ions). The primary reason these salts are still found in these areas is the lack of rainfall. Soils in southern New Mexico may have pHs in the range of 8.3 and in some sodium-affected soils, the pH can go as high as 10.
A question arises concerning the possible affects of acid rain in the southwest where soils have basic pHs. During each of my courses, I take a little time to review the probability of altering the pH of one of our soils.
Take for instance soils on the mesa areas just east of Las Cruces. A common constituent of these soils is calcium carbonate. The chemical formula is CaCO3. The local name for this substance is caliché. Amounts of CaCO3 in the soil may vary from a few percent up to nearly 30% of the solid mass.
Lets look at the required acidity which would of necessity be added to remove the CaCO3 from a soil containing 1% by weight.
The chemical reaction is:
CaCO3 has a molar mass of 100 g. H2SO4 has a molar mass of 98 g. For this example, it is sufficient to assume that the two reactants have equivalent masses. For every 100 g of CaCO3, we need 98 g of H2SO4 or about 1 g to 1 g.
So how much CaCO3 is there in a field with 1% calcium carbonate? The answer is a whole lot!!
Using the English acre system, it is estimated that an acre furrow slice, that is an area of 1 acre, 6 inches deep, has a mass of 2.2 x 106 pounds. Roughly 2,200,000 lbs is about 1,000,000 kg. One percent of 1,000,000 kg is 10,000 kg. So in the top 6 inches of soil there are 10,000 kg of CaCO3 in a soil with 1% calcium carbonate. Thus to neutralize the calcium carbonate one would need about 10,000 kg of sulfuric acid. A metric ton is 1,000 kg so the amount needed is close to 10 metric tons per acre. This much acid would only neutralize the basicity of the first 6 inches and only for one acre.
If the field is 100 acres then we need 100 times as much acid which is 1000 metric tons.
If we wish to neutralize the calcium carbonate to a depth of 2 feet we need to increase our acid by a factor of 4 so now we need 4,000 metric tons for a 100 acre field to a depth of 2 ft.
What happens if the soil contains 20% CaCO3? One needs 20 times the amount above or 80,000 metric tons for every 100 acres.
Acid rain with a pH of 3 contains 0.001 moles of H+ per liter of rain. We need 2 moles of H+ for every mole of CaCO3. We calculated above that one acre furrow slice would have 10,000 kg of CaCO3. This equals 100,000 moles of CaCO3 in the top 6 inches of a single acre. We need twice this many moles of H+ to neutralize the calcium carbonate. If we have rain with a pH of 3 we are receiving 0.001 moles in each liter of rain so we need
1 L 100,000 moles x ------------- = 100,000,000 L of rain 0.001 molesThe area of 1 acre is 4048.5 m2. 100,000,000 L is 100,000 m3. The depth of rain necessary to supply the acid would therefore be 24.7 m or 973 inches. Recall this is for 6 inches depth and 1% by weight. So for 2 feet of depth and 20% by weight CaCO3 we would need 77,840 inches of acid rain with a pH of 3 to neutralize 20% CaCO3 to a depth of 2 feet.
Think it's going to rain!