Answers to cation exchange capacity practice problems 1. One mole is the atomic mass or molecular weight per liter of each cation or anion. Use the periodic table on the front cover of the text to find the mass of each element. Sum the masses of the constituent elements in each molecule. Note that mass values have been rounded to the nearest whole number. |
Ion |
Mass of 1 mole (g) |
Mass of 1 centimole (g) |
Mass of 1 millimole (mg) |
H^{+} |
^{1} |
^{0.01} |
^{1} |
Ca^{2+} |
40 |
0.40 |
40 |
Na^{+} |
23 |
0.23 |
23 |
K^{+} |
39^{} |
0.39^{} |
39^{} |
SO_{4}^{2-} |
96 |
0.96 |
96 |
NO_{3}^{-} |
62 |
0.62 |
62 |
HCO_{3}^{-} |
61 |
0.61 |
61 |
2. It is the atomic mass or molecular weight
divided by the charge. For example,
the mass of one mole of calcium ions is 40 g. The charge is plus 2.
The mass of calcium ions required to supply one mole of charge of
calcium ions is the mass divided by the charge, (40/2=20). Both mass of 1 mole of the ion or molecule (determined in question 1) and 1 mole of charge and 1 centimole of charge are shown in this table. |
Ion |
Mass (g) |
Mass of molc (g) |
Mass of cmolc (g) |
H^{+} |
^{1} |
^{1} |
^{0.01} |
Ca^{2+} |
40 |
20 |
0.20 |
Na^{+} |
23 |
23 |
0.23 |
K^{+} |
39^{} |
39 |
0.39 |
SO_{4}^{2-} |
96 |
48 |
0.48 |
NO_{3}^{-} |
62 |
62 |
0.62 |
HCO_{3}^{-} |
61 |
61 |
0.61 |
3. You need to
replace the potassium which is positively charged with positively charged
ions. Each mole of CaCl_{2}
has two mol_{c} (+) supplied by the calcium (Ca^{2+}). You need to replace 4 mol_{c} K^{+},
thus you need two moles of CaCl_{2} to supply the 4 moles of charge.
One mole of CaCl_{2} weighs 111 g.
You need 2* 111 or 222 grams of CaCl_{2} to supply 4 moles of
charge to replace the potassium. 4. 12 mol_{c}
of K+. It always takes 1 mol_{c} to replace 1 mol_{c}
regardless of the ion carrying the charge. 5. In problem 4,
it was determined that 12 mol_{c} K^{+} are required to
replace 12 mol_{c} Ca^{2+}.
From the answer to problem 2, you know that 1 mol_{c} K^{+}
has a mass of 39 g. You need 12 * 39
= 468 g of K^{+}. 6. You require 12
mol_{c} Ca^{2+} to replace 12 mol_{c} K^{+}. BUT, only 6 moles of Ca^{2+} are
required because each mole has 2 mol_{c}. From the answer to problem 2, you know that a mole of Ca^{2+}
is 40 g and that 1 mol_{c} has a mass of 20g. You can solve this problem either by
multiplying 40g/mole *6 moles or 20 g/molc * 12 molc. In either case, the answer is 240 g. Keeping the
units attached to figures will help you to work through these problems. 7. In problem 6
you determined that you need 6 moles of Ca^{2+}. Each mole of CaCO_{3} contains 1
mole of calcium. You need 6 moles of
CaCO_{3} to obtain the required 6 moles of Ca^{2+}. Each mole of CaCO_{3} has a mass
of 100 g. You need 6 moles * 100
g/mole = 600 g. 8. If you assume
that you removed 100% of the ions on the exchange complex of the soil, then
you sum the cmol_{c} for the ions removed from the soil. The answer is 16 cmol_{c} Kg^{-1}. 9. To determine
the mass, you multiply the number of centimoles of charge by the mass per centimole
of charge. Note the units here are
centimoles not moles. From the answer
to problem 2 you have the mass/cmolc.
The mass is the product of the numbers in the second and third
columns. |
Ion |
cmol_{c}
Kg^{-1} on exchange |
mass/cmol^{c
}(g) |
mass (g/kg
soil) |
H^{+} |
6 |
0.01 |
0.06 |
Ca^{2+} |
6 |
0.20 |
1.20 |
Mg^{2+} |
2 |
0.24 |
0.48 |
K^{+} |
1 |
0.39 |
0.39 |
Na^{+} |
1 |
0.23 |
0.23 |
10. First, you
determine the cmolc of each cation on the exchange complex by multiplying the
total CEC by the percentage for that ion.
For hydrogen it is 0.05 * 30 cmol_{c}/kg. You then convert the cmolc/kg for each ion
to mass of ion per kg by multiplying by the mass of 1 cmolc. For hydrogen this is 0.01g/cmol_{c}*1.5
cmol_{c}/kg = 0.015 g/kg. The calculations and answers are shown in
the table below. |
Ion |
Percent of ion on exchange |
cmol_{c} of ion on exchange |
cmolc/kg soil of each ion |
mass (g/kg) of each ion |
H^{+} |
5 |
0.05
* 30 |
1.5 |
0.01*1.5
= 0.015 |
Ca^{2+} |
50 |
0.50
* 30 |
15.0 |
0.2*15
= 3 |
Mg^{2+} |
20 |
0.20
* 30 |
6.0 |
0.12*6
= 0.72 |
K^{+} |
23 |
0.23
* 30 |
6.9 |
0.39*6.9
= 2.691 |
Na^{+} |
2 |
0.02
* 30 |
0.6 |
0.23*0.6
= 0.138 |
11. First find out
how much sodium is on the exchange complex. If 12% of the CEC is occupied by
Na^{+}, then 0.12 * 25 cmol_{c}/kg = 3 cmol_{c}/kg_{
} is Na^{+}. The problem is to replace one-half
of the sodium with calcium from gypsum (CaSO_{4}). You need to replace 0.5 * 3 cmol_{c}/kg
= 1.5 cmol_{c}/kg Na^{+} with the same number of cmol_{c}
of calcium = 1.5 cmol_{c}/kg Ca^{2+}. You need to add 1.5
cmol_{c} calcium in the form of gypsum. Each
cmol of gypsum carries 1 cmol of calcium and 2 cmol_{c} calcium. The mass of 1 cmol gypsum= 1.36 g CaSO_{4}
(0.40 g (Ca) plus 0.96 g (SO_{4})).
The weight of 1 cmolc of Ca^{2+} from the gypsum is 1.36 g
CaSO4/cmol gypsum divided by 2 cmol_{c}/cmol gypsum. You need to add 0.68 g gypsum/cmolc
replaced. You are required to replace
1.5 cmolc. The weight to get 1.5 cmol_{c}.
is 0.68 g/cmol_{c}*1.5 cmolc/kg soil = 1.02 g. You
must add 1.02g of CaSO_{4} to a kg of soil to replace 50% of the
sodium if the efficiency is 100%. |