Answers to cation exchange capacity practice problems

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
Ca²⁺	40	0.40	40
Na⁺	23	0.23	23
K⁺	39	0.39	39
SO₄^2-	96	0.96	96
NO₃^-	62	0.62	62
HCO₃^-	61	0.61	61

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
Ca²⁺	40	20	0.20
Na⁺	23	23	0.23
K⁺	39	39	0.39
SO₄^2-	96	48	0.48
NO₃^-	62	62	0.62
HCO₃^-	61	61	0.61

You need to replace the potassium which is positively charged with positively charged ions. Each mole of CaCl₂ has two mol_c (+) supplied by the calcium (Ca²⁺). You need to replace 4 mol_c K⁺, thus you need two moles of CaCl₂ to supply the 4 moles of charge. One mole of CaCl₂ weighs 111 g. You need 2* 111 or 222 grams of CaCl₂ to supply 4 moles of charge to replace the potassium.

12 mol_c of K+. It always takes 1 mol_c to replace 1 mol_c regardless of the ion carrying the charge.

In problem 4, it was determined that 12 mol_c K⁺ are required to replace 12 mol_c Ca²⁺. 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⁺.

You require 12 mol_c Ca²⁺ to replace 12 mol_c K⁺. BUT, only 6 moles of Ca²⁺ are required because each mole has 2 mol_c. From the answer to problem 2, you know that a mole of Ca²⁺ 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.

In problem 6 you determined that you need 6 moles of Ca²⁺. Each mole of CaCO₃ contains 1 mole of calcium. You need 6 moles of CaCO₃ to obtain the required 6 moles of Ca²⁺. Each mole of CaCO₃ has a mass of 100 g. You need 6 moles * 100 g/mole = 600 g.

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.

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
Ca²⁺	6	0.20	1.20
Mg²⁺	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	0.011.5 = 0.015*
Ca²⁺	50	0.50 * 30	15.0	0.2*15 = 3
Mg²⁺	20	0.20 * 30	6.0	0.126 = 0.72*
K⁺	23	0.23 * 30	6.9	0.396.9 = 2.691*
Na⁺	2	0.02 * 30	0.6	0.230.6 = 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 (CaSO₄). 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 Ca²⁺. 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 CaSO₄ (0.40 g (Ca) plus 0.96 g (SO₄)). The weight of 1 cmolc of Ca²⁺ 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 CaSO₄ to a kg of soil to replace 50% of the sodium if the efficiency is 100%.