Soil Science 100 Exam 2 Answers

Soil Science 100
Second Exam
20 November 1998

Six digit ID: Answers
Value: 100 points

1.     A soil has the following cations on the exchange complex: H⁺ (13 cmolc/kg soil), Na⁺ (7 cmolc/kg soil), Ca²⁺ (26 cmolc/kg soil), K⁺ (12 cmolc/kg soil), and Mg²⁺ (19 cmolc/kg soil).

(a)     What is the total cation exchange capacity of the soil? (3 points)

It is the sum of cmolc for all the cations. 13 + 7 + 26 + 12 + 19 = 77 cmolc/kg.

(b)     How many cmol of Na are on 1 kg of soil? (3 points)

There are 7 cmolc/kg and each cmol of Na contributes 1 cmolc. Therefore, there are 7 cmol of Na on the exchange complex of 1 kg.

Give credit if only the number is written.

(c)     How many cmol of Ca are required to replace all of the Na on the soil exchange complex? (3 points)

The 7 cmolc of Na require 7 cmolc of Ca. Each cmol of Ca can replace 2 cmolc of Na. Therefore, 3.5 cmol of Ca are required to replace all of the Na on 1 kg of soil.

(d)     What mass of Ca is required to replace all of the Na on the soil exchange complex. The atomic weight of Ca is 40. (3 points)

The mass of 3.5 cmol = 3.5 cmol * 40 cg/cmol = 140 cg of Ca. This is equal to 1.40 g of Ca. Give partial credit if g are substituted for cg.

(e)     What experimental method would you use to determine the amount of exchangeable hydrogen on the soil exchange complex? (3 points)

I would use a cation other than hydrogen (such as calcium, Ca) to displace the hydrogen on the exchange complex. I could measure the amount of hydrogen displaced by measuring the pH of the soil solution and comparing it to the soil pH measured in water. Measuring pH without displacing hydrogen is incorrect

2.     The following figure shows the water content (Q) of two soils as a function of water potential (Y).

(a)     Label the diagram with high, low for both water content and water potential. (4 points)

(b)     Label each curve with the likely texture (clay or sandy loam) (2 points)

(c)     Note on the water potential axis the field capacity and permanent wilting points. (2 points)

(d)     Which soil has the higher plant available water holding capacity? (4 points)

(Clay). The AWC is the difference in theta between FC and PWP.

3. The water infiltration rate into soil invariably slows from the initial application of water until the final water entry. Explain why this occurs. (7 points)

The gradient at the wetting front decreases as the soil wets. The gradient is the driving force that drives the process of water entry.

4. Consider the following diagram and answer the questions a through c. The soil in all pots is the same, and all have the same low fertility.

Pot 1 had organic nitrogen fertilizer added and was properly watered.
Pot 2 had the same organic nitrogen fertilizer added plus was overwatered.
Pot 3 had the same organic nitrogen fertilizer plus wheat straw mixed into the soil.
Pot 4 had wheat straw plus twice as much organic nitrogen fertilizer as pot 1.

(a)     Why is the plant in pot 2 smaller than the plant in pot 1?(5 points)

The wet conditions were unsuitable for the aerobic bacteria that mineralize organic N and that convert NH₄ to NO₃ in pot 2. Therefore, the plant was nitrogen deficient compared to the plant in pot 1 and did not grow as well.

(b)     Why is the plant in pot 4 equal in height to the plant in pot 1? (5 points)

The additional N fertilizer overcame the problem of the high C/N ratio caused by the wheat straw. N was no more limiting in pot 4 than in pot 1, thus the plants grew about the same.

(c)     Why is the plant in pot 3 taller than the plant in pot 2 but not as tall as the plant in pot 1? (5 points)

The aerobic conditions were more suitable for N-mineralization in pot 3 compared to pot 2, but the C/N ratio in pot 3 caused immobilization of N and this reduced growth in pot 3 compared to pot 1.

5.     The following data were collected using soil X during a microbial respiration experiment. Answer questions a through e using the data in the figure. Treatment A is the addition of glucose to the soil. Treatment B is the addition of ground oat straw and treatment C is the addition of water. Soil X was used in all three treatments.

(a)     Why was little CO₂ produced from treatment C? (5 points)

The microorganisms were limited by lack of carbon.

(b)     Why was additional CO₂ produced in treatment B compared to treatment C? (5 points)

Carbon was added which stimulated microbial growth but something else was limiting.

(c)     How do you explain the rapid rise and sudden drop in CO₂ production in treatment A? (5 points)

The carbon source is easily utilized as a substrate. Microbial growth is stimulated quickly but then either the organisms deplete the supply of carbon or something else becomes limiting to their growth.

(d)     Why do we measure CO₂ production when we want to know the effects of these additions on microorganisms? (5 points)

It is easily measured and can be directly related to microbial growth.

(e)     Assume you extracted 20 g of soil X with 200 mL of 0.25 M NaOH. You took 5 mL of this extract and diluted it to 50 mL and you determined that the base extractable C was 6 mg C per liter of extract. What is the NaOH extractable C content per g of soil? (5 points)

6 mg/L * .001L/mL = .006 mg/mL

0.006 ug/mL *50 mL/5mL * 200 mL/20g = 0.6 mg C/g soil

6.     Why does nitrogen fertilization with NH₃ acidify soil? (7 points)

The oxidation of NH₃ to NO₃^- produces protons.

7.     Complete the following table.

Nutrient	Macro or Micro (1 point each)	Source in soil (1 point each)
N	Macro	O.M., NO₃, NH₄
P	Macro	O.M., minerals like apatite
Ca	Macro	Exchange, minerals
S	Macro	O.M., gypsum in some soils
Cu	Micro	Exchange, chelates,
Fe	Micro	Oxides, exchange (little)

Several potential correct answers under source. Only one is required.

8. In most soils, deep rooting is more beneficial to a plant’s N nutrition than to its P nutrition. Why is this so? (7 points)

N is much more mobile than P. Deep rooted plants can capture N that leaches below a shallow-rooted plant’s roots. Because P is much less mobile in soil, deep rooting is not an important strategy in plant P-nutritition.