Soil Science 100-2002 ID____KEY

Homework 4 Lab Section_____________

Due Friday October 25

Value 20 points

You must show all calculations for full credit. Each problem is worth five points.

1. Cations are removed from the soil exchange complex of a 1000g (oven-dry weight) sample of soil by substituting ammonium (NH₄⁺) on the exchange complex. The following weights of cations were removed from the sample: Na⁺ (1.51 g) K⁺ (1.56 g), Ca²⁺ (4.40 g), Mg²⁺ (0.72 g), and H⁺ (0.05 g). What is the cation exchange capacity of the soil?

Ion	Atomic mass (g)	charge	Mass of 1 cmolc (g/cmolc)	Mass on exchange (g/kg)	cmolc/kg
Na^+	23	1	0.23	1.51	6.6
K^+	39	1	0.39	1.56	4.0
Ca^2+	40	2	0.20	4.40	22.0
Mg^2+	24	2	0.12	0.72	6.0
H^+	1	2	0.01	0.05	5.0
				Total CEC	43.6

Example Calculation using Na⁺

Na⁺ 23g/mole charge or 0.23g/cmolc

0.23 g/cmolc = 1.51g/kg/x cmolc =6.56 cmolc/kg soil

Example using Mg²⁺

Mg²⁺ 24g/mole but 12g/molc because Mg is divalent therefore 0.12g/cmolc

0.12g/cmolc = 0.72g/kg/X cmolc = 6cmolc/kg

2. A soil sample has a CEC of 37 cmol_c/kg. The exchange complex consists of 60% Ca²⁺, 20% Mg²⁺, 5% Na⁺ and 15% K⁺. How many grams of each of these cations are on the exchange complex of 1 kg of this oven-dry soil?

Ca: 37 cmolc/kg *0.60 = 22.2 cmolc/kg *0.20g/cmolc =4.44g/kg

Mg: 37 cmolc/kg * 0.20 =7.4 cmolc/kg *0.12g/cmolc =0.89g/kg

Na: 37cmolc/kg * 0.05 =1.85 cmolc/kg * 0.23g/cmolc =0.42g/kg

K: 37 cmolc/kg * 0.15 =5.55 cmolc/kg* 0.39g/cmolc =2.16g/kg

3. Assume you have a 1 hectare field and 15% of the cation exchange capacity of the 25 cm thick A horizon is occupied by sodium (Na⁺). You want to remove all of the Na⁺, using CaSO₄. What is the minimum amount of CaSO₄ needed to remove the Na⁺? Assume the bulk density of the A horizon is 1.2 Mg/m³ and the cation exchange capacity is 35 cmol_c/kg of soil. Some helpful numbers: One hectare is 100 meters by 100 meters, the molecular weight of CaSO₄ is 136g. Assume the CaSO₄ is 100% pure.

Here is one solution to the problem.

The mass of soil to be treated is 100m*100m*.25m*1.2 Mg/m³ =3000 Mg

A Mg is 1000 kg. The soil to be treated has 35000 cmolc/Mg or 105x10⁶ cmolc

15% of this charge is Na or 15,750,000 cmolc that must be replaced with Ca

We need to supply a minimum of 15,750,000 cmolc of Calcium and 1 cmolc Ca has a mass of 0.20g

15,750,000 cmolc * 0.20g/cmolc =3,150,000 g of Calcium or 3,150 Kg or 3.15Mg Calcium/hectare BUT CaSO₄ is not pure calcium so we need to add more CaSO₄.

40g Ca/136g CaSO₄ = 3150000/X g CaSO₄ = 10,710,000 g/ha or 10,710kg/ha or 10.7 Mg/ha of CaSO₄.

4. You have been called upon to recommend how much crushed limestone (CaCO₃) to add to the A horizon of a soil in a 5 ha field to reduce the exchangeable acidity by 10%. The 20 cm thick A horizon has an exchange capacity of 30 cmol_c/kg of soil. The exchangeable acidity is 25% of the exchange capacity. The bulk density of the A horizon is 1.27 Mg/m³. What is the minimum amount of limestone you recommend to add? What assumptions have you made about the limestone?

Assume the limestone is pure and 100% soluble. These calculations are the same as the previous problem.

CEC is 30 cmolc/kg and exchangeable acidity is 25% or 7.5 cmolc/kg. You want to replace 10% of that with calcium so you need to provide 0.75 cmolc/kg of Ca²⁺ to the five hectares.

You know that 0.20g Ca provides 1 cmolc, thus you require 0.75*0.20 = 0.15 g Ca/kg soil. You need 2.5x as much CaCO₃ to get that amount of Ca because Ca is 40% of the molecular weight of CaCO₃. You therefore need 2.5*0.15 g of CaCO₃/kg of soil = 0.375 g/kg.

One hectare (100m x 100m), 0,20 m thick with a bulk density of 1.27 Mg/m3 weighs 2,540 Mg or 2,540,000 kg.

You require 0.375 g/kg 2,540.000 kg =952,500 kg or 952 Mg of CaCO₃ per hectare and you are treating 5 hectares, so you need 952*5=4,762.5 or 4763 Mg of CaCO₃ to replace the H⁺ on the exchange complex in this area.