soil science mid term 2 study guide
soil science mid term 2 study guide SOIL 205
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This 11 page Class Notes was uploaded by Julia Notetaker on Sunday February 21, 2016. The Class Notes belongs to SOIL 205 at Oregon State University taught by Meg Mobley in Winter 2016. Since its upload, it has received 60 views. For similar materials see soil science in Agriculture and Forestry at Oregon State University.
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Date Created: 02/21/16
SOIL 205 Midterm 2 Study Guide ANSWERS 1. List three things that affect the amount of water stored in soil, and EXPLAIN how each affects water storage. (That is, don’t just say “organic matter,” say HOW OM affects water storage. And now that I used that one, you have to come up with three OTHERS.) (Chapter 5) Pore size: Larger pores drain faster, smaller pores hold water against the force of gravity. Soils wither smaller pores hold more water. Total Pore volume: Soils with larger pore volume can potentially store more water. Depth to bedrock or a restrictive layer Timing and Intensity of Precipitation: If rainfall is heavy and comes over short periods of time may supply water faster than the soil can absorb it. Type of Vegetation: Vegetation and its surface residue help reduce the intensity of rain hitting the soil, thus reduce runoff and promote water infiltration. Soil Texture: Particle sizes affect pore sizes. Clays hold more water than sand. 2. What is “field capacity” of a soil? How is it determined? (Chapter 5) Field capacity can be defined as the proportion of water remaining in soil twothree days after saturation and after free drainage has stopped. Field capacity can also be considered as the point when soils are holding the maximum amount of water available for plants. In terms of structure properties field capacity is the point when soil is consider to be at the plastic limit. Above field capacity the soil acts plastic puttylike that easily turns to mud and below field capacity acts as a crumbly semisolid. 3. Soils rich in what type of clay would provide the best site on which to build a house? Kaolinite: No shrinking or swelling. Smectites would be the worst because it has the greatest shrinkswelling capacity. 1 4. The shrinkingswelling tendency of some clay minerals is due primarily to: a) the presence of anions attracted by the negative charges on the internal surfaces b) the movement of water molecules in and out of the interlayers of the crystals c) varying thickness of the film of water covering the external surfaces of the particles d) expansion in the width of the interlayers sue to movement of large ions such as K + e) the high Mg content in the octahedral layers 5. HARD. Of the formulae below that represent different clay minerals, which represents the LEAST isomorphous substitution? a) Si O2 a4d (OH) Al O 2 2 2 b) Si O2 a4d (OH) AlMg2 2 c) SiAlO an4 (OH) Al O 2 2 2 d) SiAlO an4 (OH) AlMgO2 2 6. Which of the clays we covered has a CEC that is highly pHdependent? WHY?? Kaolinite: There is no permanent charge but there are edge charges. Edge charges are dependent on pH (see question 8). 7. Explain the difference between cation exchange, anion exchange, and cation exchange capacity? What is anion exchange capacity? Cation exchange is the swapping of a cation on a charged surface (e.g., clay, organic matter, humus) with another cation that is in soil solution. Cation exchange capacity is the sum total of exchangeable cations that a soil is capable of adsorbing at a particular pH. It is expressed in terms of of positive charge adsorbed per unit mass [cmol charge/kg soil]. CEC can also be called baseexchange capacity or the total exchange capacity. Anion exchange is the swapping of a anion on a charged surface (e.g., clay, organic matter) with another anion that is in soil solution. Anion exchange capacity is the sum total of exchangeable anions that a soil is capable of adsorbing at a particular pH. It is expressed in terms of charge adsorbed per unit mass [cmol charge/kg soil]. 8. Briefly explain the relationship between CEC and pH. In most soils, the CEC increases as pH increases. In acidic environments (low pH) 1:1 clays and humus (edge charge only) are protonated, thus the all the charge is satisfied. Permanent charges on 2:1 clays hold exchangeable ions regardless of pH. When you increase pH (less acidity), negative charges on 1:1 clays, humus and some oxides increase because of dissociation of hydroxides , thus increasing the amount of sites available for cation exchange. Most of the time CEC is measured in the lab at pH 7 or slightly alkaline because at these pHs the CEC reflects most of those pH dependent charges as well as permanent charges. 2 9. What is the relationship between charge location and CEC? Charge due to isomorphous substitution (permanent charge) is internal to the clay structure, and provides CEC regardless of pH. Charge on clay or colloid surfaces is variable depending on pH (see above), and so provides variable CEC depending on pH. 10. You are given a silty clay loam soil that contains 27% clay and 4% humus. You get data back from the geosciences lab that indicates 18% of the clay portion is vermiculite (CEC of 120 cmolc/kg) and 82% is chlorite (CEC of 30 cmcl /kg). The CEC of humus is 200 cmol ckg. Calculate the CEC of the soil. SHOW YOUR WORK. CEC = (% O.M. x CEC ) +OM% clay x CEC ) clay CEC= (0.04 x 200) + (0.0486 x 120)+ (0.2214 x 30) CEC= 8 + 21.6 + 24.6 CEC= 20.47 cmcl /kg soil 11. Listed below are a number of common soil management practices. For each one, tell what effect the practice has on 1) texture; 2) bulk density; 3) particle density; 4) % pore space; 5) other effects you want to note? a) Working peat moss and manure into your garden soil b) Twenty years’ use of a hiking trail in the forest of the Three Sisters Wilderness area c) Removing the clay soil from your back yard and replacing it with sandy loam d) Trying to plow a soil when it is too wet e) Changing the tires on your tractor to a wider size (compare this to the effect of using narrow tires, pulling the same load) f) Adding mulch to the soil surface Practice Texture Db Dp %PS Others a) no NC + More water change holding capacity (NC) b) NC + NC Runoff c) sandier + NC Lower water holding capacity, better drainage d) NC + NC Destroys structure e) NC NC + Less negative impact 3 f) NC in time (not NC + in time (not Prevents immediately) immediately) evaporation 12. In many soil profiles, the subsoil is high in clay, but is also quite permeable to percolating water. Why? A wellaggregated soil will have a relatively high infiltration capacity, regardless of texture. 13. If two soil cores were taken for the purpose of measuring bulk density, how would the results differ if one core got bumped around and the other core from the same soil was carefully transported? If the bump caused loss of soil, then the mass would be lower than it “should be,” and if you measured the volume of the corer and assumed it was full (which it wasn’t), then the calculated bulk density (dry mass/ total volume) would be lower than if you hadn’t lost soil. That is, you would UNDERestimate the bulk density. 14. Calculate bulk density, the volume of water held at saturation (hint %PS = θv at saturation), and gravimetric water content. Identify which case is an Andisol, which is a typical clay loam, and which is not possible. Case 1: wet mass = 1153.3 g water mass = 346.0 g dry mass = ? (1153.3 – 346 = 807.3 g) diameter = 10.2 cm height = 7.6 cm D b= 1.3, %PS = 51%, clay loam Ө m= (wet – dry)/dry = water mass/dry mass = 346/807.3 = 42.8% Case 2: wet mass = 2420.6 g water mass = 558.9 g dry mass = ? diameter = 7.6 cm height = 15.2 cm Db = 2.7, %PS = 1.8%, impossible Ө m(if it were possible)= 558.9/1861.7 = 30% Case 3: wet mass = 1003.6 g water mass = 420.1 g 4 dry mass = ? diameter = 10.2 cm height = 10.2 cm Db = 0.7, %PS = 73.6%, Andisol; note how high the porosity is!!! Ө m= water/dry = 420.1/583.5 = 72% 15. HARD. As part of a cleanup of an agricultural field that was contaminated with a load of sewage sludge containing heavy metals you will be paying $20/m to haul the soil to a 4 2 treatment facility. The field in 50 ha in size (1 ha = 10 m ) and the contamination has been tilled to a depth of 0.1 m. The bulk density of the soil is 1.4 Mg/m . The particle 3 density is 2.6 Mg/m . The soil texture is classified as sandy clay loam. How much does it cost to haul the soil? 50 ha * 10 m /ha * 0.1 m depth = 50,000 m of soil that needs to be cleaned up. 3 3 3 The cost is $20 per m , so: 50,000m * $20/m = $1,000,000 5 16. Which soil is the BRIGHTEST? a. 10YR4/3 b. 2.5Y6/2 c. 7.5YR2/1 d. 5R4/6 e. 10YR7/5 17. Which soil is the LIGHTEST? a. 10YR4/3 b. 2.5Y6/2 c. 7.5YR2/1 d. 5R4/6 e. 10YR7/5 18. Which soil is most likely found at the bottom of a hill derived from shale parent material in an area with LOTS of rainfall? a. 10YR4/3 b. 2.5Y6/2 c. 7.5YR2/1 d. 5R4/6 e. 10YR7/5 19. Check the 3 characteristics that are most likely to decrease as aggregation increases. _____total porosity _____particle size _____waterholding capacity __x___bulk density _____water infiltration _____particle density 6 __x___unavailable water __x___number of small pores 20. Choose increase/increases or decrease/decreases for each of the blanks below. As clay content in a soil decreases, the bulk density of the soil _increases_, the porosity of the soil _decreases_, and the water holding capacity of the soil _decreases_, but the availability of water for plant uptake may _increase_. 21. Describe capillarity in soils. What does it depend upon? What does it result in? WHY does it occur? (What is it, anyway???) Capillarity is the attraction of water to soil particles, due to forces of adhesion and cohesion. Capillarity holds some water in soil after the gravitational water has drained out, and this is water that is available to plants. It depends on the pore sizes: smaller pore sizes hold more capillary water in soils. 22. Two soil samples, A & B, at different soil moisture levels are placed in contact with each other. Water will more likely move from soil A to soil B if their water potentials (expressed in kPa) are: a. A= 50, B= 30 b. A= 20, B= +5 c. A= 60, B= 0 d. A= 30, B= 60 e. A= 150, B= 40 23. Soil water potential (matric potential, ψ) is a measure of the amount of force / suction / tension that must be exerted by a plant in order to extract water from a soil. 24. The volumetric water content of a sandy loam soil is 29%, the bulk density is 1.4 g/cm , 3 and the wet mass is 32.2 g. What is the dry mass of the soil? volumetric = gravimetric * D b 0.29 / 1.4 = gravimetric = 0.21 7 gravimetric = (wetdry)/dry 0.21 = (32.2x)/x 0.21x = 32.2 – x (now add 1x to both sides) 1.21x = 32.2, so x = 26.6 g 25. What is the percentage pore space filled with air when ψ = 0? ___0__% 26. You dry a soil sample and find the mass is 64.7 g. The estimated volume of the sample is 22.8 cm . 3 a. The bulk density of the sample is: 2.8 g/cm b. Is this reasonable? Yes / No (circle one) c. If yes, what is the likely soil texture? If no, why not? The soil bulk density can’t be higher than the particle density. This is more like the density of a dense rock. 27. An oven dry soil sample weighs 66.2 g. How much water must be added to get it to 26.4% moisture content on a volumetric basis? The bulk density of the soil is 1.2 g/cm . 3 Gravimetric*bulk density= 0.264 OR Gravimetric*1.2 = 0.264 gravimetric = 0.22 (wetdry)/dry = 0.22 (wet66.2)/66.2 = 0.22 wet weight = 80.8 g wetdry = water weight 80.8 – 66.2 = 14.6 g 14.6 ml water must be added 28. Given the following set of data for a soil sample answer the following questions. Mass of sample moist from the field = 48.2 g; Mass of soil sample oven dry = 37.9g 3 Volume of the soil sample = 29.3 cm ; a. What is the gravimetric water content (θ )? m θ m= (wet –dry)/dry θ m= (48.2 g – 37.9 g)/37.9 g θ = 27.2% (or 0.272 g water/g dry soil) m b. What is the volumetric water content (θ )? v 8 θ v= θ m* D b θ v= 27.2 * (37.9/29.3) θ v= 35.2% c. The % pore space (porosity) in this sample is: %PS = [1 – (D /D )] * 100 b p %PS = [1 – ((37.9/29.3)/2.65)] * 100 %PS = 51.2% 29. Is water labeled “C” available to plants? Why or Why not? A C E C is the available water (between field capacity and wilting point), held in the soil by capillary forces 30. Describe the water in the area “A”, and in the area “E” above. A is the gravitational water, which runs out of a soil within a few days of a rain or irrigation event. E is the hygroscopic water, which exists as very thin films and in very tiny pores, this water is held at too great a water potential (very large negative value) for plants to be able to take it up. 31. A soil is said to be at field capacity just after all the gravitational water has drained from the soil. Water potential (suction) is 10 cbar at field capacity. Many crop plants are 9 most productive when soil moisture levels are kept at levels of at least 60 90% of field capacity. Using the graph below, answer these questions. (cbar) -105 -104 -103 b -100 a -10 -1 0 10 20 30 40 50 60 (%) a. If you are farming soil a, what is the minimum range of soil volumetric water content you would aim for to keep your crops most productive? For soil a, field capacity (10 cbar) is at at ~38%. 60% of 38% is 0.6*38% or ~ 23% 90% of 38% is 0.9*38% or ~34% You would want to keep your volumetric moisture content up at at least 23 to 34%. b. Which soil (a or b) has a higher bulk density? Soil a has higher bulk density. You can tell because it’s maximum water content is lower. c. HARD. What is the bulk density of soil a? Max volumetric water content for soil a is ~45%, which means the %PS for soil a is ~45%. %PS = [1 – (D bDp)] * 100 45% = [1 – (X/2.65)] * 100 3 X= 1.46 g/cm d. HARD. If your field started completely dry, and your irrigation system delivers 1 cm water per hour, how long would you have to irrigate soil a for to bring the top 30 cm of your field to a moisture content that is 60% of field capacity? 10 θv = ~38% at field capacity for soil a 60% of 38% is 23% 23% is the volumetric water content you are aiming for, and you are starting at 0% (dry soil) 30 cm * 23% = ~6.8 cm water needed at 1 cm/ hour irrigation rate, you need to irrigate for ~6.8 hours e. The infiltration rate of soil a at saturation, which is a function of its K satue, or intrinsic permeability, is 6 cm/hr. Will your irrigation system (described in the previous question) lead to surface runoff? Why or why not? It is unlikely that there will be any surface runoff (unless your soil has an impermeable layer in the top 30 cm) because the irrigation rate is much lower than the slowest infiltration rate for the soil, which occurs at saturation. f. Which soil (a or b) holds more water at wilting point? (Wilting point is at a suction of 1500 cbar). Is this water available to plants? Soil b holds more than twice as much water at wilting point, compared to soil a, but this water is unavailable to plants (held at too great suction or tension). g. Which soil (a or b) has a finer texture? Soil b has a finer texture. Soils that hold more water at wilting point have finer textures, because this is the water held in tiny pores and on thin films on mineral surfaces. Finer textured soils have higher surface area, and more tiny pores. 11
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