Lab 6: Measuring CEC

Use soils from last week to – > extract exchangeable cations > measure basic cations by AA spectroscopy > measure acid cations by titration > calculation CEC, % BS for your soil > convert from meq/100g to lbs/acre

Procedure:

1. Weigh 3.0 g soil into 50 mL centrifuge tube 2. Pump in 35 mL NaCl extracting solution 3. Shake for 20 minutes to displace exchangeable cations from colloid surface into solution…

Ion Exchange by Mass Action

Na Na Na Na Na Na Na

Added Na soln …

Mg +2 K + Ca +2 Al +3

Displaced cations

Measurement of exchangeable cations:

1. Centrifuge, filter to separate clear solution.

2. Fill glass vial about ½ full of solution for AA analysis.

3. Measure remaining solution using grad cylinder—record this number, this will be your “volume titrated”. Pour back into Erlenmeyer flask.

Titration of exchangeable acids:

1. To Erlenmeyer flask: add stir bar and 3-4 drops phenolpthalein indicator , turn on stirrer to slow speed.

2. Add NaOH from burette SLOWLY (drop by drop) until solution stays pink for 20-30 seconds. Record volume base used from burette.

At endpoint: pH is near neutral, and AMT BASE ADDED = AMT ACID PRESENT N B x V B = N A x V A

Titration calculations: N

B

x V

B

= N

A

x V

A

N

B: conc of base: 0.005 N (eq/L = meq/mL)

V

B: volume of base (mL, burette)

V

A: volume of acid (“mLs titrated”)

N

A: conc of acid: UNKNOWN… meq acid/mL

Conc (solution)



Conc (soil)

We used 35 mL of NaCl solution to extract 3 g soil to get cations into solution….

Meq acid x 35 mL soln mL soln x 100 = meq acid 3 g soil 100 100 g soil Same for basic cations by AA: Meq Ca x 0.035 L soln L soln x 100 = meq Ca 3 g soil 100 100 g soil

CEC of soil =  cations (in meq/100 g): Ca + Mg + K + acidity (H+Al) Mg and K: written on board (meq/L)…..

% base saturation (BS) =  bases (Ca+Mg+K)/CEC Convert meq/100 g into lbs/acre-furrow slice….

CEC calculations: meq/100 g to lbs/acre Assume a soil has 2 meq K + /100 g: how many lbs/acre? (“acre” = acre-furrow slice= 2x10 6 lbs) Note: Atomic mass of K=39 g/mole, and 1mol=1equiv (K +1 ) So mol X 1 mol 1 equiv X 1000 mg 1 g X 1 equiv 1000 meq 2 meq K x 39 mg K 1 meq x 1000 g 1 kg = = 39 mg meq 780 mg K kg soil Now: “mg/kg” is PPM = parts per million… and LBs/AFS = lbs/2x10 6 lbs, which is “parts per 2 million”…. SO: 780 ppm K x 2 = 1560 lbs K/acre