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SOIL ORGANIC CARBON

Carbon occurs in soils in 4 forms of mineral and organic matter:

1)Carbonate mineral forms (chiefly CaCO3 and MgCO3. CaCO3). However, highly active and importantsmall amounts also occur as CO2, HCO3- and CO3- ions of more soluble salts.

2)Highly condensed, nearly elemental organic C (eg., charcoal, graphite, coal etc.).3)Altered and rather resistant organic residues of plants, animals and microorganisms (sometimes called

humus).4)Little altered organic residues of plants, animals, and microorganisms (both living and dead) subject to

rather decomposition in soil.The total carbon of soils obviously includes all the above 4 forms. Total organic C includes the latter 3

forms, the mineral form being eliminated by treatment with dilute acid prior to the organic C determination. For

ordinary soil, no pretreatment is required to eliminate the inorganic fraction of soil C as they occur in relatively

negligible amount. For these soils, total C of soil means approximately the organic C fraction. However, to determineorganic C of calcareous soil the carbonate fraction must first be eliminated before organic carbon determination. In

such case, free carbonate is determined separately and the value is minus from the organic C result.

Soils containing less than 20% organic matter by weight are classified as mineral soil and more than 20%

organic matter is commonly called organic soil. Agricultural soils are essentially mineral soils and fertile mineral soils

on an average contain about 5% organic matter.

Among various methods to determine soil organic carbon, two commonly used methods are described

below.

Walkley & Blacks wet oxidation method

Organic carbon compounds are highly reducing substances. Chromic acid (K 2Cr2O7 + H2SO4) is a suitable

oxidizing agent to oxidize soil organic carbon according to the following reaction. The excess chromic acid left after

the oxidation of organic carbon may be determined volumetrically with standard ferrous solution (reducing agent).

The quantity of substance oxidized is calculated from the amount of chromic acid reduced.

3C + 2K2Cr2O7 + 6H2SO4

4Cr2(SO4)3 + 3CO2 + 8H2O + H2Cr2O7 (unreduced)

K2Cr2O7 + 7H2SO4 + 6FeSO4= K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O

[Overall reaction: K2Cr2O7 + 7H2SO4 + 6FeSO4 K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O]

Reagents:

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1) Potassium dichromate (K2Cr2O7) solution [Powerful oxidizing agent, obtains pure, stable up to its fusion point]

250 ml 1N solution

mw = (39.1 2) + (52 2) + (16 7) = 294.20

ew = mw Change in oxidation number = 294.20 6 = 49.033333 g

1000 ml 1N solution requires 49.033333 g K2Cr2O7.

Therefore, 250 ml 1N solution requires = 49.033333 4 = 12.258333 g K2Cr2O7

If 12.26 g K2Cr2O7 is measured with the help of a two digit balance, the actual strength of the solution can be

corrected by normality factor (f). This is a bright orange solution.

f = 12.26 12.258333 1 N

2) Ferrous sulfate (FeSO4.7H2O) solution [Powerful reducing agent, secondary standard substance (impure)]

250 ml 1N solution

mw = 55.85 + 32.07 + (16 4) + (18 7) = 277.92

ew = mw Change in oxidation number = 277.92 1 = 277.92 g

1000 ml 1N solution requires 277.92 g FeSO4.7H2O

Therefore, 250 ml 1N solution requires = 277.92 4 = 69.48 g FeSO4.7H2O

Take 69.48 g FeSO4.7H2O into 250 ml volumetric flask, add 3.75 ml (15 ml per liter) concentrated H 2SO4

and finally volume 250 ml with dw. Commercially available FeSO4.7H2O contains suspended

impurities. Therefore, filtration is needed to get pure FeSO4.7H2O solution which is bright light green in

colour. Then standardize the solution against standard K2Cr2O7 solution.

3) Diphenylamine indicator solution: Dissolve 0.25 g indicator into 10 ml dw and slowly add 50 mlconcentrated H2SO4 into it.

4) H2SO4(concentrated)

5) H3PO4(85%)

6) NaF(powder)

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Procedure:

1)Take 2 g air dried soil into a 500 ml conical flask.2)Add 10 ml 1N K2Cr2O7 solution and carefully add 10 ml concentrated H2SO4 and mix thoroughly with slight

rotating.3)Allow the mixture to cool for half an hour with occasional slight shaking. (If the colour of the mixture appears

green then add an additional 10 ml of 1N K2Cr2O7 solution. Green colour is an indication that all oxidizingagent added is used up to oxidize organic carbon.)

4)Add 150 ml dw, 10 ml H3PO4 (85%) and 0.2 g NaF in sequence.5)Add 3 ml diphenylamine indicator prior to titration. The mixture appears deep violet in colour.6)Titrate excess K2Cr2O7 solution left in the flask with FeSO4.7H2O solution.7)Run a blank titration (This step is essential to standardize FeSO 4.7H2O solution against standard K2Cr2O7

solution).

Calculation:

Soil organic carbon:

ew of C = 12 4 = 3

1000 ml 1N FeSO4solution 1000 ml 1N C = 3 g C.

1 ml 1N FeSO4 solution = 0.003 g C.

(B-T) ml S N FeSO4 solution = (B-T) S 0.003 g C.

Therefore, W g soil contains [(B-T) S 0.003] W g C.

100 g soil contains [(B-T) S 0.003 100] W g C.

It has been estimated that only about 77% C of the organic compounds in soil is oxidized by normal K2Cr2O7

solution in this method. A factor is used to correct this (100 77 = 1.3). The constant (1.3) is thus

called recovery factor.

So, Organic C in soil (%) = [(B-T) S 0.003 1.3 100] W

Where,

B = Amount of FeSO4 required in blank titration.

T = Amount of FeSO4 required in soil titration.

S = Strength of FeSO4 (from blank titration).

W = Weight of the soil.

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Soil organic matter:

The organic matter of soil is determined by multiplying the content of organic carbon with a factor 1.724 (100

58) on the assumption that, organic matter of average soil contains 58% organic C by weight. This

constant is called Van Bemmelen factor.

Therefore, organic matter in soil (%) = % organic C 1.724

Notes:

Stepwise reaction:

K2Cr2O7 + H2SO4 K2SO4 + H2Cr2O7

3C + 2H2Cr2O7 3CO2 + 2H2O + 2Cr2O3 [Change in oxidation number C(0) + Cr(6+) C(4+) + Cr(3+) ]

Unreduced H2Cr2O7 + FeSO4 Fe2(SO4)3[Change in oxidation numberFe(2+) Fe(3+)]

NaF: No specific reaction of NaF is known so far. However, it is expected to sharp the colour of the indicator.

H3PO4: To avoid the effect of the brown colour [due to the presence of Fe2(SO4)3] on the colour of the indicatorH3PO4 is used, where colourless FePO4 is formed that has no effect on the colour change of the

indicator.

Fe2(SO4)3 + 2 H3PO4 2FePO4 + 3H2SO4

(brown) (colourless)

Tyrins method

This method is suitable to determine organic C even at lower concentration compared with themethod described by Walkley and Black. However, under diluted conditions Tyrines method does notfunction well.

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Reagents:

1) K2Cr2O7 solution(0.4 N, mw = 294.19, ew = 49.03): Dissolve 9.806 g K2Cr2O7 in 200 ml dw. Then add 250ml (1:1 ratio) concentrated H2SO4 slowly in a porcelin basin. Volume the solution 500 ml with dw and after

cooling, preserve it in an amber bottle.

2) Mohrs salt [(NH4)2 SO4. FeSO4.6H2O] solution (0.1 N, mw = 392.14, ew = 392.14):Dissolve39.214 g Mohrs salt in 900 ml dw and add 20 ml concentrated H2SO4. Volume the solution 1000 ml with

dw. Filter the solution, if necessary. Use freshly prepared solution.

3) Phenylanthranylic acid [C13H11O2N] indicator solution (mw = 213): Dissolve 0.2 g Na2CO3 in100 ml dw. Add 0.2 g phenylanthranylic acid with small amount of 0.2% Na2CO3 solution in a morter, grind

and then add the rest of the Na2CO3 solution.

Procedure:

1. Take about 0.5 g air dried soil (or dried extract) into a 100 ml conical flask.2. 10 ml 0.4 N K2Cr2O7 is added followed by the addition of some quartz sand to reduce over frothing.3. Place a funnel into the neck of the conical flask and heat the mixture at boiling point for exactly 5 minutes on

hot plate.4. Wash the funnel and the neck of the flask with minimum amount of dw (not more than 10 ml). At this point,

the solution should appear orange yellow or greenish brown.

5. Add three drops of phenylanthranylic acid indicator and titrate against 0.1 N Mohrs salt solution.6. Run a blank titration to standardize Mohrs salt solution.

Calculation:

Equivalent weight of C = 12.01 4 3

1000 ml 1 N Mohrs salt 3 g C

1 ml 1 N Mohrs salt 0.003 g C.

Organic C in soil (%) = [(B-T) S 0.003 1.17 100] W

Where,

C = Organic C as % of air-dried soil.

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B = Amount of Mohrs salt required in blank titration.

T = Amount of Mohrs salt required in sample titration.

S = Strength of Mohrs salt.

W = Weight of the soil.

1.17 = A recovery factor as no Ag2SO4 is used in the determination.

% Humus (OM) = %C X 1.724 (As humus contain 58% C)

SOIL ORGANIC MATTER FRACTIONATION

Organic matter fractionation can be carried out on different basis. The fractionation schemedescribed in the following section is based on their solubility in acid and alkali.

Soil sample:

Separate rootlets from air-dried soil and pass through 1 mesh sieve (0.25 mm).

Extraction:

Soil sample taken for extraction depends upon the total OM content of the soil as shown in the following

table.

% OM of soil Soil taken for extraction (g)

7 10

4-7 20-30

0.5-3.0 50

< 0.5 Not applicable

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The extractant is 0.1 M sodium pyrophosphate (Na4P2O7.10 H20) and 0.1 N NaOH mixture, which has a pH

of 13.

1. 20 g air-dried soil is taken into a 500 ml conical flask.2. 100 ml freshly prepared extractant is added and mix thoroughly by shaking. The content is left overnight (for

16-18 hours) after sealing the conical flask with rubber bung.

3. Filter the content using WN-42 filter paper. Extract may be separated from soil through centrifuging.4. Soil residue on the filter paper is discarded.

The extract is analyzed to determine organic carbon of the following components by Tyrins method,which is more sensitive than Walkley & Blacks method.

Total organic C in the extract. Humic acid C in the extract. Fulvic acid C in the extract and Carbon in the soil residue.

Total organic carbon in the extract:

1. 10 ml extract (if extracts are of light colored) is taken into a 100 ml conical flask.2. Add concentrated H2SO4 drop by drop until a slight cloudiness appears.3. Boil the content in water bath until the content are evaporated to dryness.4. The content of the conical flask is diluted with 5-10 ml dw and titrated with 0.1 N Mohrs salt (Tyrins

method).

Express the results in both A) % of soil weight, and B) % of total organic C in the original soil.

Humic acid C in the extract:

1. 50 ml extract (in case of moderate to low humus content) is pipetted into a beaker.2. Humic acid gel is coagulated by adding concentrated H 2SO4 drop by drop, stirring the solution with a glass

rod. No excess of acid should be added. [Cloudiness appears at pH 2-3. Approximately 0.2 to 0.5 mlconcentrated H2SO4 is required.]

3. Heat the beaker (< 800 C) for 30 minutes with continuous stirring with glass rod and left the content at roomtemperature overnight to complete the precipitation of the humic gel.

4. Next morning the extract is filtered. The filter paper is first moistened with 0.05 N H 2SO4 and then the acidsolution is filtered followed by the filtration of humic gel.

5. The precipitate (black) on the filter paper is washed several times with cold 0.05 N H 2SO4 until the filtrate iscolorless (as fulvic acid is yellow colored the colorless filtrate indicate the complete wash of fulvic acid fromthe mixture).

6. The filtrate is discarded.7. The funnel with the filter paper is placed into the neck of a 50 ml volumetric flask.8. The precipitate is dissolved in warm 0.05 N NaOH solution. [Small amount of NaOH solution is first added to

the beaker in which the humic acid precipitated and the solution is then transferred to the filter. The filter

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paper is washed with 0.05 N NaOH until the gel of humic acid is completely dissolved. The filter papershould become colorless. Care should be taken not to add NaOH excess of the volume.]

9. The Na-humate solution is cooled to room temperature and volume with dw.10. 5 to 20 ml of the solution is taken to determine the organic C content by Tyrins method.Express the results in both A) % of soil weight, and B) % of total organic C in the original soil.

Fulvic acid carbon in the extract:

Fulvic acid C = Total organic carbon in the extract - Humic acid C in the extract.

Express the results in both A) % of soil weight, and B) % of total organic C in the original soil.

Carbon in the soil residue:

Soil residual C = Organic C in the original soil - Total organic carbon in the extract.

Express the results in both A) % of soil weight, and B) % of total organic C in the original soil.

Characterization of humic substances

There are various parameters to characterize the nature of humic substances.Among these E4:E6 ratio andcoagulation threshold are outlined below.

The magnitude of the E4:E6 is related to the degree of condensation of the aromatic C network of the humic

substances.

Low E4:E6A low value of E4:E6 (3 to 5) is usually observed for humic acids and other related compounds that hasrelatively high degree of condensation of aromatic constituents. Humic acids with high molecular weight (

30,000) have lower E4:E6 values (4.32 to 4.45) than humic acids with lower molecular weights ( 15,000). The

lower molecular weight fractions of humic acids exhibit E4:E6 values of 5.47 to 5.49. Such molecules are relatively

large molecular size or high molecular weight that has high C content, but is relatively low in O, -COOH groups

and total acidity.

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High E4:E6Conversely, a high E4:E6 reflects a low degree of aromatic condensation and infers the presence ofrelatively large proportion of aliphatic structures. On the other hand, fulvic acid has an extremely low optical

density. A high colour ratio (E4:E6), 7 to 8 or higher is usually observed for fulvic acids or humic fractions with

relatively low molecular weight. Such molecules are smaller in molecular weight that contain less C but more O, -COOH groups and total acidity than the larger molecules.

The absorbance of the humate solution is read both at either 400 and 600 nm (Tan) or 465 and 665 nm

(Kononova). E4:E6 ratio obtained by using 400 and 600 nm produces lower value than that obtained with 465 and 665

nm. Moreover, absorbance becomes unstable using 400 nm wavelength in some cases. Therefore, it is

recommended to use 465 and 665 nm wavelength to determine E4:E6 ratio.

Procedure:

1.The C content of the Na-humate solution is expressed in g/L.2.The C content of the solution is then adjusted to 0.136 g/L either by dilution or condensation. During this also

adjust the pH of the solution in between 7.2 to 9.8.3.Then the absorbance of the solution is read at 465 and 665 nm.4.Then calculate the E4:E6 ratio by using the following formula.

Calculation:

E4:E6 ratio = (Absorbance at 465 nm) (Absorbance at 665 nm)

Coagulation threshold is another important parameter to characterize humic substances. Most of the

functional groups of alkaline humic solutions are negatively charged. These charges play a vital role to keep the

molecules soluble due to repulsive forces. If the charges are neutralized by strong electrolytes then these molecules

loose their solubility and hence, become precipitated. Molecules with more negatively charged functional groups

require higher amount of electrolyte to precipitate the molecules.

Reagents:

Calcium chloride solution(60 me): Dissolve 3.33 g anhydrous CaCl2 in 500 ml solution with dw.

Procedure:

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1. Take 0.25, 0.50, 0.75 and 1.00 ml 60 me CaCl2 solution into four separate screw cap test tubes.2. Then add 0.75, 0.50, 0.25 and 0 ml dw into the test tubes in sequence. (The volume of the electrolyte

solution and dw together make 1 ml in each test tube.)3. Add 5 ml humate solution (containing 0.136 g/L C) into each test tube so the final volume in each tube

becomes 6 ml. The solutions are delivered in the order: CaCl2, dw and humate solution.4. Cap the test tubes and mix the solutions by inverting the tubes 2 to 3 times.5. After 4 hours observe the tubes and note both the time and corresponding amount of electrolyte (in me) for

complete coagulation (The solution above the precipitate becomes transparent and clear.)

Observation:

Appearance of coagulation threshold after 4 fours for Kalma, Chandina and Melandaha soil series are

shown in the following figures.

The coagulation threshold as observed from the above figures is 12 me for Kalma and 18 me for Chandina.

Coagulation threshold for Melandaha soil is above 24 me.