Exercise 9: Determination of Carbon Dioxide Evolution in Soils

Carbon dioxide evolution from soil is one of the best simple indicators for total microbial activity. Activity requires energy. Many soil microbes obtain energy by oxidizing carbon from organic matter. The more organic matter they decompose, the more carbon dioxide they produce. The microbes include actinomycetes, archaea, bacteria, fungi, protozoa, and others. About three-fourths of the carbon in the decomposing organic matter is given off as carbon dioxide. The remainder becomes part of new microbial tissue.

Microbial activity in soil drops to a low level when some condition is unfavorable. The unfavorable condition might be a shortage of readily decomposable organic matter, a nutrient deficiency, cold temperatures, dry soil, etc. The activity speeds up when all of the essential conditions are favorable. An upsurge of microbial activity is likely to occur when fresh organic materials such as plant residues are added to the soil.

The rate of an initial upsurge of microbial activity is likely to be limited by the number of microorganisms present. But, microorganisms multiply rapidly and soon some other factor becomes limiting. Many residues such as small grain straws and corn stalks are low in nitrogen. Microorganisms need nitrogen to live, so the supply of nitrogen may become a limiting factor. Strong competition often arises when plants are growing and microbes are multiplying in a soil with an inadequate nitrogen supply. Microorganisms usually “win” the battle because they are much more uniformly distributed in the soil than plant roots and get to the nitrogen first. As a result, the plants are certain to suffer from nitrogen deficiency in such circumstances. The solution, of course, is to add nitrogen fertilizer when low nitrogen residues such as straw are incorporated into the soil, or wait on planting until after the microorganisms have decomposed the residue. Eventually, the microorganisms themselves are decomposed and released their nitrogen.

The effects of nitrate release (Exercise 12) and carbon dioxide evolution are closely related, both being microbial processes. Soil samples often are incubated with and without the addition of plant residues to compare the effects of different residues on microbial activity. Various types of residues may be used by different members of the class to find the effects of both low-nitrogen and high-nitrogen residues.

Procedure

1. Weigh 40 g (+0.02 g) of the pre-dried and 2-mm sieved soil into each plastic beaker.
2. To one jar only, add 0.10 ± 0.01 g (0.5% by wt or 10,000 ppm) of ground organic material (straw, cornstalks, leaves, or other material of interest). This rate of residue addition equals 5 tons/acre. The second jar serves as a control with no residue added. Mix the contents of the jar. Take note of the C and N contents of the added residue as provided by the instructor.
3. Place the plastic beakers into the glass jars.
4. Label the jars with and without the residue. Label – Don’t guess.
5. Add 9 mL of DI water on the top of the fiber filter slowly. Be careful NOT to pour water quickly. 9 mL of water added to 30 mL soil of varying bulk-densities is equivalent to 44% to 54% water filled pore space.
6. Holding the edge of the plastic, place the Solvita paddle (probe) next to the plastic beaker in the glass jar.
7. Put the paddle with gel facing out next to the clear side of the jar.
8. Screw the lid on the jar tightly
9. Store jars at room temperature (25 °C) for 24 hours.
10. After 24 hours (+ 30 minutes), remove the paddles from the glass jar.
11. Insert the paddle (gel up) into the DCR.
12. Press the Low CO₂ button and then press the Read button.
13. The DCR reports the color chart number on the first line and the ppm CO₂ -C on the second line. Record both.
14. Use the appropriate interpretation tables from the Solvita booklet for test results.
15. Calculate the amount (mg) of CO₂ evolved per 40 g of soil.

mg/kg CO₂ = mg/L CO₂ ÷ 0.735

16. Calculate the amount in mg of C evolved per 40 g of soil.

The molecular mass of CO₂ is 44, and the atomic mass of C is 12; thus, CO₂ is 12/44 or 27.3% C.

18. Calculate the percentage decomposition of the original organic material added.

Most natural organics contain approximately 40% C. Without a specific analysis, this is a good value to use. The blank soil also evolved C as its organic matter decomposed. Thus, the difference between the soil with amendment and the blank soil is the amount of C coming from the residue.

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 What is the interpretation of potential biological activity of this soil (Table 6)?

Notes or comments (may continue on back or use separate sheet):