Soil
Science 1002002 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 (ovendry weight)
sample of soil by substituting ammonium (NH_{4}^{+}) on the
exchange complex. The following
weights of cations were removed from the sample: Na^{+} (1.51 g) K^{+}
(1.56 g), Ca^{2+} (4.40 g), Mg^{2+} (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 Mg^{2+} Mg^{2+} 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% Ca^{2+}, 20% Mg^{2+},
5% Na^{+} and 15% K^{+}.
How many grams of each of these cations are on the exchange complex of
1 kg of this ovendry 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 CaSO_{4}. What is the minimum
amount of CaSO_{4} needed to remove the Na^{+}? Assume the
bulk density of the A horizon is 1.2 Mg/m^{3} 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 CaSO_{4}
is 136g. Assume the CaSO_{4}
is 100% pure. Here is one solution to the problem. The mass of soil to be treated is
100m*100m*.25m*1.2 Mg/m^{3} =3000 Mg A Mg is 1000 kg.
The soil to be treated has 35000 cmolc/Mg or 105x10^{6} 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 CaSO_{4} is not
pure calcium so we need to add more CaSO_{4}. 40g Ca/136g CaSO_{4}
= 3150000/X g CaSO_{4} = 10,710,000 g/ha or 10,710kg/ha or 10.7 Mg/ha of CaSO_{4}.
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 Ca^{2+}
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 CaCO_{3} to get that amount of Ca
because Ca is 40% of the molecular weight of CaCO_{3}. You therefore need 2.5*0.15 g of CaCO_{3}/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 CaCO_{3} per hectare and you are treating 5 hectares, so
you need 952*5=4,762.5 or 4763
Mg of CaCO_{3}_{ }to replace the H^{+} on the
exchange complex in this area. 