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 CaCO_{3}. The local
name for this substance is caliché.
Amounts of CaCO_{3} 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 CaCO_{3} from a soil containing 1% by
weight.

The chemical reaction is:

CaCO_{3} has a molar mass of 100 g.
H_{2}SO_{4} 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 CaCO_{3}, we
need 98 g of H_{2}SO_{4} or about 1 g to 1 g.

So how much CaCO_{3} 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 10^{6} 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
CaCO_{3} 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% CaCO_{3}? 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 CaCO_{3}. We calculated above that one acre
furrow slice would have 10,000 kg of CaCO_{3}. This
equals 100,000 moles of CaCO_{3} 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 m

Think it's going to rain!