양이온치환능력

Practice Problems for Cation Exchange Capacity

Agronomy 354 

 

 

화학에서 물질의 농도를 표시할 때 %나 ppm 같은 단위는 적절하지 않은 경우가 있다. 예를 들면 2 분자의 수소와 1 분자의 산소가 결합해서 물이 만들어질 경우 2.018 g의 수소와 15.999 g의 산소가 결합하여 18.017 g의 물이 생긴다고 쓴다. 그러나 2 당량의 수소와 1 당량(equivalent)의 산소가 결합하여 1 분자량의 물이 생긴다고 생각하는 것이 편하다. 그런데 당량이란 단위은 너무 커서 종전에는 밀리당량이란 단위를 썼는데 SI에서는 밀리당량 대신 센치당량을 쓰기로 하고 그것을 cmol+로 표기하기로 정했다. 사실은 mol은 mole 즉 분자량에서 온 것이다.(당량은 분자량을 이온가 또는 이온값 즉 valency로 나눈 것. valency: 이온이 갖는 결합손의 수. 예컨대 Ca+의 경우에는 결합손의 수 즉 이온가는 2, H+의 경우는 1)

종전에는 토양의 치환성 양이온의 농도를 표시할 때 me/100g이라는 단위를 썼는데 SI에서는 100 g 대신에 kg을 쓰고 me 대신에 cmol+를 쓴다. 그러므로  cmol+/kg과 me/100g의 수치는 같다. 즉 종전에 예컨대 종전의 4.5 me/100g을 SI로 표시하면 4.5 cmol+/kg이다. 왜냐하면 cmol+는 me의 10 배이고 kg은 100 g의 10 배이기 때문이다.

1 mg/kg은  1kg에 1mg이 들어 있음을 나타내므로 ppm이라고도 표현 한다. 왜냐하면 1kg은 1,000,000 mg이기 때문이다. 1 mg/kg 또는 ppm을 cmol+/kg로 나태내는 것은 대개 토양 중의 치환성 양이온들에 대해서만 적용된다. 이 경우에 알아야 할 것이 있다. 즉 Ca++, Mg++, K+, Na+들의 1 cmol+이 몇 mg인지를 알아야 한다. Ca++이온의 원자량은 40 이므로 1 cmol은 400 mg이지만 1 cmol+은 200 mg이다. (+ 하나만 따지는 게 1 cmol+) 같은 이치로 Mg++의 원자량은 24로 1 cmol+는 120 mg이고 K+의 원자량은 39.1로 1cmol+는 391 mg이고  Na+의 원자량은 22.9로 1 cmol+는 229 mg이다.
따라서 Ca++가 200 mg/kg이라면 이것은 Ca++ 1 cmol+/kg입니다. 만약 Ca++가 300 mg/kg이라면 Ca++ 는 1.5(300/200) cmol+/kg일 것입니다.

예전에는 치환성 이온의 양을 나타낼 me/100 g(milliequivalent/100 g)이라는 단위를 썼는데 요즘에는 cmol+/kg으로 쓰고 있습니다. (cmol+/kg은 centiequivalent/kg), milliequivalent는 1/1000 equivalent, centiequivalent는 1/100 equivalent인 것이다. (출처 : http://www.soilove.com)

 

 

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