Midquarter Exam #1
28 October 1998

Value: 100 points


Please answer all questions in the space provided. Write short, clear answers. Answer with a single word or phrase where appropriate. Show all work for full credit on calculations.

(8 points)
1.     List two soil properties that clearly distinguish A from B horizons and list two that distinguish B from C horizons.

A from B; A horizons have smaller structural units, usually grannular and fine blocky while B horizons have coarser blocky, prismatic and columnar. A horizons are typically darker than B horizons. B horizons are typically browner or redder than A horizons. A horizons have more organic matter than B horizons. Texture is not a good distinquishing feature among soils.

B from C horizons; B horizons are better structured than C. B horizons have accumulated weathering products and may have more clay, iron oxides, carbonate and/or silica than C horizons. B horizons are browner or redder than C horizons.


(9 points)
2.     In what soil horizon would you expect to find the following structure types?

fine granular A horizon
coarse subangular blocky B horizon
medium prismatic B horizon


(9 points)
3.     What are three soil science meanings of the term "clay"?

Clay is a particle size e.g. <2 mm

Clay is a texture class name

Clay is a name for a specific group of alumino-silicate minerals


(3 points)
4.(a)     What is isomorphous substitution in a clay mineral?

It is the process that substitutes one atom for another in the structure of clay minerals at the time the clay minerals form. This leads to net negative charge in the mineral. Examples are the substitution of Al3+ for Si4+ in the tetrahedra and Mg2+, Fe2+, Fe3+, and others for Al3+ in octahedra within clay minerals.

(6 points)
(b)     Give two examples of why isomorphous substitution is important in agriculture and/or environmental protection.

The importance of isomorphous substitution is that it produces a net negative charge in clays. This charge must be balanced by positive charged ions from solution. This is important to retention of plant nutrients (soil fertility) and the retention of charged polutants. Any examples along this line are ok.


(10 points)
5.     Explain why some 2:1 clay minerals have high shrink-swell capacity while 1:1 clay minerals do not.

The 2:1 clays that shrink and swell have an interlayer that can be penetrated by water and hydrated ions. Two layers of oxygen atoms face each other across the interlayer space. They are slightly negative and repel each other, making it possible for the clays to shrink and swell. The 1:1 clays are built differently. In these, a layer of hydroxyl molecules faces a layer of oxygen atoms across the interlayer space. The hydrogen atoms from the hydroxyl and the oxygen atoms in the opposite sheet hydrogen bond, resulting in strong interaction and lack of shrinking and swelling.


(5 points)
6.     What was the effect on water flow rate of raising the Mariotte bottle, relative to your soil column in your column experiment? Explain the effect based on your knowledge of Darcy’s law.

As you raise the Mariotte bottle, the pressure head on the inflow side of the column increases and the water flows out faster. Total head, which is the gravity head plus pressure head in this system is higher when the Mariotte bottle is raised. According to Darcy’s law, the rate of flow is a function of the cross sectional area, the hydraulic conductivity, and the potential gradient. By increasing the gradient (the difference in head at the inflow and outflow ends of the column) we increase the flow rate.


(9 points)
7.       How does the rate of convective flow of gas change as the following soil properties change?

(a)     Pore size distribution gets skewed towards smaller pore sizes.
         Slower

(b)     Soil becomes wet.
         Slower

(c)     The pressure gradient increases.
         Faster


(20 points)
8.     Use the data in Table 1 to make calculations and answer the following questions about soil sample 1. Be sure to include proper units for full credit.

Table 1. Data for soil sample 1.

Total sample
Air-dry weight		 2 mm sample
Air-dry weight		 2 mm sample
Oven-dry weight		 Total sample
volume		 Particle density
(g) (g) (g) (cm3) (g/cm3)
325 290 275 190 2.63

(a)     Calculate the coarse fragment percentage.

Coarse fragments are everything that does not pass the 2 mm sieve. The data show that there is 325 - 290 = 35 g of coarse fragments in a total weight of 325 g or 35/325 *100 = 10.77% or round up to 11%

(b)     Calculate the bulk density.

Bulk density is the mass of oven dry solids divided by the volume. 275 g/190 cm3 = 1.45 g/cm3

(c)     Calculate the porosity.

Porosity is 1 minus the bulk density divided by the particle density. 1- 1.45/2.63 = .45

(d)     Calculate the gravimetric water content in the air-dry soil sample.

GMC = weight of water/weight of oven dry sample = 290 -275 g water /275 g soil = 15/275 = 0.0545 g/g or 5.45%


(12 points)
9.     You measured particle size distribution in the laboratory based on Stoke's law. Answer the following questions about the analysis.

(a)     Why was sodium hexametaphosphate (metaphosphate) added to the soil suspension? Also, why was it added to the blank?

It was added to the sample to disperse particles so that particles would settle individually rather than as aggregates. The material was added to the blank to account for its affect on the density of the solution. [The hydrometer measures the density of the suspension which includes the density of the fluid and the suspended solids.]

(b)     Why were readings taken at 40 seconds and 2 hours?

Particles settle as a function of their diameter. The 40 second reading was taken because by that time all of the sand sized particles had settled beyond the depth where the hydrometer measured. After 2 hours, all of the silt had also settled beyond the depth where the hydrometer measured the suspension density.

(c)     Would you overestimate or underestimate the sand content if you took the 40 second reading at 38 seconds?

You would underestimate the sand content. The first reading measures silt plus clay in suspension. If you overestimate the silt plus clay content, you must underestimate the sand content.

(d)     Why did you use corrected hydrometer readings for your calculations?

You needed to account for temperature and hydometer calibration.


10.     The following data are from soil horizons from soils illustrated in your text. Answer the questions following the data table based on your knowledge of soils and the data in the table.


Soil Horizon Soil CEC Clay Content Organic Carbon
  (cmolc/kg) (%) >(%)
1 6.3 19.7 0.14
2 48.7 59.4 0.84
3 6.0 2.0 1.20


(3 points)
(a)     Based on these data, do you expect the clay types to be the same for the three horizons? Explain your answer.

NO. The ratio of clay content to CEC is quite different for these three soils, indicating differences in clay mineralogy.

(3 points)
(b)     What kind of clay is horizon 1 most likely to contain? Explain your answer.

The dominant clay type is Kaolinite. It has about 320 cmolc/kg clay if you assign all of the CEC to the clay and none to the organic matter. The second soil has about 820 cmolc/kg clay and thus much have a more "active" clay than the first soil. [This explanation is more detailed than most students will give].

(3 points)
(c)     Which soil horizon is likely to have the highest pH dependent exchange capacity? Explain your answer.

Horizon three because it has the highest organic carbon content. Organic matter typically has a higher pH dependent charge than the clay minerals.