Question

Please Answer:

CALCULATION QUESTION 1

(A) Calculate the number of cmol of Al3+ cations needed to replace 10 cmolc of Ca2+ by cation exchange in a soil.

(B) How many grams of Al3+ does this correspond to? (you’ll need the atomic weight of Al, which is 27 g/mol, to calculate this)

CALCULATION QUESTION 2

A soil sample (500 g) contains the following amounts of exchangeable cations: Ca2+= 9 cmolc , Mg2+= 3 cmolc , K+= 1 cmolc , Al3+= 3 cmolc .

(A) What is the CEC of this soil (in cmolc/kg)?

(B) What is the aluminum saturation of this soil?

(C) What is the amount (in cmol) of exchangeable Ca2+ present in 1 kg of this soil?

CALCULATION QUESTION 3

A 100 gram sample of a soil has been determined to contain exchangeable cations in the following amounts: Ca2+= 90 mg; Mg2+= 35 mg; K+= 28 mg; Al3+= 60 mg.

(A) What is the CEC of this soil (in cmolc/kg)?

(B) What is the aluminum saturation of this soil?

(Note that 1 mg=1 milligram = 1/1000 g = 0.001 g) (You will need the atomic weights of the cations, which are as follows: Ca = 40.1; Mg = 24.3; K = 39.1; Al = 27.0)

Answer 1

First problem has been solved here. Please post other questions separately

cmol = centi moles = 10-2 x moles

Also, note that Moles of charge are the same as moles of anything else.

That is One mole of charge is the charge (positive or negative) of one mole of an ion with either a +1 or -1 charge.

Example: An ion such as Ca2+ has 2 moles of charge for each mole of calcium atoms because it has two positive charges. And aluminum has 3 moles of charge per mole of Al.

one mole of charge is denoted as (molc) and one centi mole of charge is denoted as cmolc.

One mole of charge of any ion will always be equivalent to 1 mole of charge of any other ion.

Now, 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); so 20 is molc of Ca2+ cation.

Ion	Mass (g/mol)	Mass of molc (g)	Mass of cmolc (g)
Ca2+	40	(40/2)= 20	0.20
Al3+	27	(27/3) =9	0.09

10 cmolc of Ca2+ will be replaced by 10 cmolc of Al3+ ; as It always takes 1 molc (or cmolc) to replace 1 molc (or cmolc) regardless of the ion carrying the charge.

Since 1 cmolc of Al3+ has a mass of 0.09 g. We need 10 *(0.09) = 0.9 g of Al3+

And 0.9 g of Al3+ is (0.9/27) = 0.0333 mol = 3.33 cmol

So, A) number of cmol of Al3+ cations required = 3.33 cmol

And, B) this correspond to 0.9 g of Al3+

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