General Soils

 

Quiz 3 Study questions

 

 

 1.  Name and describe four types of soil colloids.

1)  layer silicate clays- dominant inorganic colloids in most soils.  They have a layer-like, crystalline structure and net negative charge.

2)  iron and aluminum oxide clays- Common in highly weathered tropical soils.  They have pH-dependent charge.  At high pH, these particles carry a small negative charge.  In very acid soils some Fe, and Al oxides have a net positive charge and attract negatively charged anions (anion exchange).

3)  Allophane- These are colloidal silicates whose crystalline structure is not sufficient to be detected by X-ray diffraction.  They are most commonly found in volcanic ash soils.  They have pH-dependent charge like the iron and aluminum oxides.

4)  Organic soil colloids- convoluted chains of carbon bonded to H, O, and N.  They have high negative charge (much higher than the layer silicate clays) at neutral pH values.  The charge associated with organic colloids is pH dependent.  Under very acid conditions, they have a very low negative charge (can be lower than that of silicate clays).

2.  Draw a 1:1 and 2:1 type mineral.  List the important characteristics of each.

            1:1 (kaolintie)               2:1 (mica or montmorillonite)

 

 

 

 

 

 

 

 

 

 

 


Kaolinite important characteristics:

1)1 Si tetrahedral sheet: 1 Al octahedral sheet

2)

 H bonding between layers

3)

 Rigid structure (low shrink/swell)

4)

 NO ISOMORPHIC SUBSTITUTION All charge is surface charge from broken bonds. 

This gives Kaolinite a relatively low negative charge

 

Mica important characteristics:

1)  2 Si tetrahedral sheets : 1 Al octahedral sheet

 

2)  Semi-rigid (non-expanding)

 

3)  Isomorphic substitution in tetrahedral layers (leads to internal charge)

4) Surface charge is from broken bonds.

 

5)  Potassium is found between the layers.

 

 

 

Montmorillonite important characteristics:

 

1)2 Si tetrahedral sheets: 1 Al octahedral sheet

2)Non-rigid, expanding mineral

3)Isomorphic substitution in octahedral layer (leads to internal charge)

4)  Surface charge is from broken bonds.

5)Ca, Mg, and/or H2O may be found in the interlayer.

 

 

 

3.  Explain isomorphic substitution and how it influences the properties of a mineral.

Isomorphic substitution is the substitution of one cation for another in the mineral structure.  This occurs when the mineral is forming.  Typical substitutions include Al for Si in the tetrahedral sheet and Mg for Al in the octahedral sheet. The substitution of a lower charged ion for a higher charge ion leaves unsatisfied negative charge.   Isomorphic substitution is a major source of negative, permanent charge in soils.

 

4.  Explain the differences between mica and montmorillonite.

Both mica and montmorillonite are 2:1 minerals.  Mica has substitution in the tetrahedral sheet and montmorillonite has substitution in the octahedral sheet.  Due to the arrangement of oxygens in the tetrahedral sheet of mica, K can become fixed in the interlayer, forming a relatively rigid structure.  Due to the fact that the internal charge of montmorillonite is in the octahedral layers, cations in the interlayer are not held as tightly and the mineral is free to expand as water enters the interlayer space.

 

5.  Fill in the following table.

 

Colloid

Charge (cmol/kg)

Kaolinite

3-15

Smectite (montmorillonite)

80-120

Mica

15-40

Organic matter

200 (at pH = 7)

 

 

6.  What are the two types of charge on soil colloids?

Charge can be either permanent or pH-dependent.  Permanent charge comes from isomorphic substitution.  pH-dependent charge comes from hydroxyls and other such functional groups on the surfaces of the colloidal particles (mineral and organic) that are releasing or accepting H+ ions.

 

7.  The following concentrations of ions were removed from the exchange sites of a soil:

Ca = 3 cmol+/kg

Mg = 2 cmol+/kg

Na = 1.8 cmol+/kg

K = 1 cmol+/kg

Al = 1.5 cmol+/kg

H = 1 cmol+/kg

 

a.  What is the cation exchange capacity of this soil?

= Ca + Mg + Na + K + Al + H = 10.3 cmol+/kg

b.  What is the percent base saturation?

Base saturation = sum of bases/CEC

= (7.8/10.3) * 100 = 76%

c.  Is this soil a sodic soil?

ESP = Na/CEC

= 1.8/10.3 * 100 = 17% Based on the fact that the ESP is greater than 15%, this would be a sodic soil

 

8.  What is the difference between active and exchangeable acidity?

Active acidity is the H+ concentration in solution (this is what is commonly measured)

Exchangeable acidity equals the Al and H on the exchange sites.  They are available to move into the soil solution thereby decreasing the pH.  The exchangeable acidity pool is much larger than the active acidity and needs to be considered when making lime requirements.

 

9.  What is the relationship between base saturation and pH?

As pH increases, there is less H and Al on the exchange sites and the percent base saturation increases.

 

10.  What is meant by soil “buffering”?

Buffering is the ability of a soil to resist changes in pH.  Basic or acidic cations in soil solution can replace cations on the exchange sites.  In the reaction below 2 moles of H+ are added to the soil solution.  The H+ ions exchange with the Ca on the clay and there is no net change in pH.

 

 

 

 

 


11.  How many kg of pure CaCO3 are required to raise the pH of a ha-15cm of soil (DB = 1.33 Mg/m3) from pH 5 (%BS = 40) to pH 7 (%BS=90).  The total CEC of the soil is 20 cmol+/kg.

 

90 – 40 = 50% change

0.5 (20cmol+/kg) = 10 cmol+/kg need to be replaced

10 cmol+/kg (1mol+/100 cmol+)(1 mol CaCO3/2mol+ CaCO3)(100g CaCO3/mol CaCO3) = 5 g  CaCO3/kg soil

 

Kg of soil in 1ha-15 cm (you can do this calculation or simply remember that 1ha-15cm weighs about 2 x 106 kg.

1ha = 10000m2

10000m2 * 0.15 m = 1500 m3

1500m3 (1.3Mg/m3)(1000kg/Mg)= 1950000 kg soil

1950000 kg soil (5g CaCO3/kg soil)(1kg CaCO3/1000g) = 9750 kg CaCO3/ha-15cm

 

12. How many kg of pure CaCO3 are required to completely replace the exchangeable acidity of a ha-15cm of soil with base saturation of 50% and a CEC of 25 cmol+/kg?

 

% Base saturation = 50%, this means that 50% of the exchange sites must be filled with cations.  If this is true than the other 50% must be filled with acid cations (H and Al).  So the exchangeable acidity = 100% - 50% = 50%.  If the exchangeable acidity is 50% and the soil has a CEC of 25 cmol+/kg, then:

0.5* 25 = 12.5 cmol+/kg need to be replaced by Ca+2

 

12.5 cmol+/kg (1mol+/100 cmol+)(1 mol CaCO3/2mol+ Ca)(100g CaCO3/mol CaCO3) = 6.25 g  CaCO3/kg soil

 

there are about 2 x 106 kg soil/ha-15cm (see question 11)

 

6.25 g CaCO3/kg soil (2 x 106 kg soil/ha-15cm) = 1.25 x 107g or 12500 kg CaCO3/ha-15cm

 

13.  Explain “natural” acidifying processes and “human induced” acidifying processes.

Natural processes include the decomposition of organic material (such as the decomposition of leaf litter in forests).  The decomposition process releases strong organic and inorganic acids.  Weathering is a natural process that decreases the pH.  As minerals weather, base cations are released and then possibly removed from the soil profile through leaching.  Arid and semi-arid regions tend to have higher pH than humid, wet regions since there is not enough water to leach salts and base cations from the profile.

 

Human induced:

1) use of chemical fertilizers.  Nitrogen fertilizers produce acidity when they are oxidized.  In the US, 20 mil. Metric tons of limestone would be required to neutralize the acidity produced by N fertilizers every year.

 

2)  Acid rain.  N and S released from cars and manufacturing processes combine with O and water in the atmosphere to form strong acids (nitric and sulfuric acid) that comes back to earth in rainfall.

 

3)  Disposal of acid-forming wastes.  Application of sewage sludge for example, which decomposes and forms strong acids.

 

4)  irrigation can increase the pH if water with a high salt content is applied and the soil is not fully drained.  As the irrigation water evaporates, the salts that it was carrying are left behind and accumulate in the soil.

 

5)  Drainage of some costal wetlands

-problem when soils contain FeS2, FeS and S (formed from microbial reduction of sulfates).  When oxidized, these compounds from sulfuric acid.

 

 

 

 

14.  What are two differences between the exchangeable sodium percentage and the sodium adsorption ratio?

 

The ESP (ESP = Na/CEC * 100) is calculated based on the exchangeable Na relative to the total CEC.  The SAR takes into account the Ca and Mg (which can help alleviate the physical problems that arise when Na content is high) and uses concentrations measured in soil solution.  The SAR is useful in that it can be calculated for irrigation waters as well as for soil solutions.  

 

15.  Why are sodic soils problematic?  How would you reclaim sodic soils?

 

Sodic soils are problematic because of both fertility and physical problems.  Fertility problems include toxicity of Na and deficiency of micronutrients.  Physical problems occur due to dispersion of clays caused by Na.  Dispersion results in the blocking of water-conducting pores and decreased infiltration.

 

Reclamation of sodic soils is extremely difficult.  One approach is to add gypsum (CaSO4).  The Ca will replace the Na on the exchange sites.  Na combines with SO4 to form a leachable compound.

 

 

16.  The addition of a liming materials helps to:

            1)  increase the pH of a soil

            2)  decrease Al toxicity

            3)  increase micronutrient, Ca, and Mg availability