Practice Problems for Cation Exchange Capacity
Agronomy 354

1. What is the valence of Ca and what is its milliequivalent weight if the atomic weight is 40?

Calcium often occurs as a divalent, positively charged cation (Ca2+). It appears in Group II of a Periodic Table and readily looses two electrons from its outer orbital, giving it a divalent (2) valence. The meq wt can be determined by dividing the atomic weight by the valence.

Therefore, the meq wt is 40/2 or 20. Be sure to use the appropriate units; that is, 20 mg/meq.

2. If a soil has 20 meq of CEC/100 g, how many milligrams of Ca2+ will this equal? How many grams?

CEC determines the amount of negative charge of 100 g of soil. The CEC often is occupied by a variety of cations, such as H+, Ca2+, K+, Mg2+, etc., but may be occupied by only one cation.

The meq wt of Ca2+ is 20 mg/meq (see above). Therefore, a soil able to hold 20 meq of Ca2+ would hold 400 mg of Ca2+

20 meq CEC/100 g soil X 20 mg/meq = 400 mg Ca2+/100 g soil or 0.4 g of Ca2+

3.  If a soil contains 5% organic matter and 10% kaolinite clay, what is a reasonable estimate of its CEC per 100 g?

Reasonable estimates of the CEC of pure organic matter is 200 meq/100 g and for kaolinite it is only 8 meq/100 g. (NOTE: you should know the appropriate values for common soil colloids.)

Therefore the CEC contribution from these two components is:

Organic matter: 200 meq/100 g X 0.05 = 10 meq

Kaolinite: 8 meq/100 g X 0.10 = 0.8 meq

10 meq from OM and 0.8 meq from kaolinite = 10.8 meq/100 of soil

4.  A soil has a CEC of 24 meq/100 g.  How many grams of Na+ will it take to saturate the CEC?  (At wt Na+ = 23)

This problem is calculated similarlly to problem #2.

The meq wt of Na+ = 23/1 or 23 mg/meq

24 meq CEC/100 g soil X 23 mg/meq = 552 mg of Na+/100 g soil

5. If a soil can hold 50 mg of Mg2+ per 100 g, how many mg of K+ could it hold?

First, we must determine how many meq of charge the soil needs to hold 50 mg of Mg2+, then we must calculate the weight of K+ to equal this amount of charge.

At wt of Mg2+ = 24; Mg2+ is a divalent cation; therefore, 24/2 = 12 mg/meq

50 mg of Mg/12 mg/meq = 4.2 meq of negative charge occupied by Mg2+

At wt of K+ = 39; K+ is a monovalent cation; therefore, 39/1 = 39 mg/meq

4.2 meq of charge X 39 mg/meq = 163.9 mg of K+ (this really says it takes 163.9 mg of K+ to occupy the same amount of negative exchange sites in soil as occupied by 50 mg of Mg2+)

6. A quantity of 100 g of soil has a CEC of 5 meq. How may negatively charged sites does this soil have?

Avogadro's number is 6 X 1023 (this will react with or replace one gram of hydrogen (H+). H+ has an atomic weight of one and it has one charge per atom). Thus, if each H+ occupies one negatively charged site, it would take the same number (6 X 1023) negative sites to hold one gram of hydrogen, which equals one equivalent of charge. CEC of soil is expressed as milliequivalents of charge and not equivalents of charge. Therefore, one meq of charge equals 6 X 1023/1000 or 6 X 1020 charges. (NOTE: remember when dividing, one subtrates exponents and when multipling one adds exponents.)

5 meq X (6 X 1020) = 30 X 1020 or 3 X 1021 negative charges

7. A soil was determined to have 12.5 meq/100 g of CEC. How many cmole/kg would this equal?

The scientific community often expresses CEC of a soil as cmol/kg. This is centimoles (cmol) of charge per kilogram of soil. Many soil testing laboratories (and your textbook), however, express CEC as meq/100 g. This should not cause confusion since: 1 meq/100 g = 1 cmol/kg. (NOTE: 'milli' charge multiplied by 10 equals 'centi' charge and 100 g multiplied by 10 equals kg; thus, the proportions remain the same.)

Thus 12.5 meq/100 g = 12.5 cmol/kg

8. Convert the following concentrations into parts per million (ppm):
At wts:  Ca2+ = 40, H+ = 1, Mg2+ = 24, K+ = 39

10 meq Ca/100 g soil
Recall that parts per million equals mg/kg; mg per 100 g soil X 10 = mg per kg
40/2 = 20 mg/meq
10 meq x 20 mg/meq = 200 mg/100 g soil x 10 = 2000 ppm

1 meq H/100 g soil
1/l = 1 mg/meq
1 meq x 1 mg/meq = 1 mg/100 g soil x 10 = 10 ppm

2 meq Mg/100 g soil
24/2 = 12 mg/meq
2 meq x 12 mg/meq = 24 mg/100 g soil x 10 = 240 ppm

2 meq K/100 g soil
39/1 = 39 mg/meq
2 meq x 39 mg/meq = 78 mg/100 g soil x 10 = 780 ppm

9. Convert 3 cmol/kg soil Ca2+ into parts per million (ppm).

This is a two step problem. One must first convert 3 cmol/kg Ca to mg Ca/kg, or 3 meq/100 g to mg Ca/100 g. The second step is to convert to ppm.

Step 1. The atomic weight of Ca2+ is 40. Therefore, the meq wt is 40/2, or 20 mg/meq

20 mg/meq X 3 meq/100 g soil = 60 mg Ca/100 g soil

Step 2. Since ppm is parts per million, it is the weight of Ca proportional to 1 million weights of soil. Let's convert 100 g of soil to 100,000 mg of soil. Now we have 60 mg of Ca per 100,000 mg soil. If we multiply the demoninator by 10, we get 1 million. We must also multiply the numerator by 10 to keep the same proportion. Thus:

60 mg Ca/100,000 mg soil X 10/10 = 600 parts Ca/1,000,000 parts soil, or 600 ppm.