Prac

1

Exp no: 01

Date: 09.07.06

Name of the Experiment: Determination of total nitrogen present in soil sample by modified Kjeldahl method. Most of the nitrogen in soils is in organic form. Relatively small amounts ordinarily occur in ammonium and nitrates, the available forms. Two general types of analytical procedure are used – the Kjeldahl conversion of nitrogen to ( NH 4 ) 2 SO 4 , which is essentially a wet oxidation procedure and Dumas method (conversion of nitrogen gas), which is fundamentally a dry oxidation or combustion technique. The former is now a widely used and accepted method. The Kjeldahl method includes both organic and ammonium form, and with modifications includes the nitrate form. Principle: The Kjedahl procedure generally employed for determination of total N involved two steps: 1. Digestion of sample to convert the N to Ammonium sulphate. 2. Determination of ammonium in the digest. The digestion is usually performed by heat on the sample with H 2 SO 4 containing substance which oxidize the organic matter in the sample and concomitant conversion of the organic nitrogen to ammonium sulphate , the substance generally favored being salts such as K 2 SO 4 or Na 2 SO 4 which increases the temperature of digestion and catalysts such as Hg , Cu or Se , which increase the rate of oxidation of organic matter by H 2 SO 4 . Now a days Hg is not being used, because it is volatile in low temperature and hazardous to health. Reaction In digestion: The temperature of the digestion is very critical. If the temperature is too low (below 360 o C), the release is slow or incomplete, and if too high (over 410o C), some loss of NH3 from the mixture results. The reactions are given below: 0

0

− 410 C Soil-N + H 2 SO4 360    →( NH 4 ) 2 SO4

The amount of N thus converted to ammonium sulfate may be determined by distilling it with alkali and collecting the distilled ammonia in a standard acid (2% boric acid) and collected ammonium is titrated against N 10 H 2 SO4 followed by finding out the strength of the acid. The following reactions occur:

2

on ( NH 4 ) 2 SO4 + NaOH Distillati   → NH 3 + H 2 O + Na 2 SO4

NH 3 + H 3 BO3

    → NH 4 .H 2 BO3 .

NH 4 H 2 BO3 + H 2 SO4    → ( NH 4 ) 2 SO4 + H 3 BO3

Material: Sample: Soil sample (A) Apparatus : 1. Kjeldahl digestion unit 2. Kjeldahl flask 3. Distillation apparatus 4. Measuring cylinder 5. Pipette 6. Micro burette 7. Conical flask (B) Reagent : 1. H 2 SO 4 , concentrated, 36N 2. Digestion mixture: In a mortar 100gm of anhydrous Na2SO4, 6.5 gm of anhydrous CuSO4 3. 40% NaOH 4. Mixed indicator solution 5. 2% Boric acid mixed indicator solution 6. H 2 SO 4 of 0.1 N Procedure: 1. 2gm of finely powdered soil was taken in a Kjedahl flask and 10 ml of distilled water was added to it and left for sometimes. 2. Then 20ml of conc. H 2 SO 4 was added. 3. The flask was gently heated over a low flame in a digestion chamber for about 10 minute and then after off the burner followed by cooled. 4. After that 5 gm of digestion mixture was added, on the burner and temperature was raised. The heat was confined within the portion of the flask that contained liquid. 5. The liquid became colorless and another 1 hour heat was continued. 6. The digest was cooled and when cold, it was diluted with distilled water and made up to volume in a 100 ml volumetric flask. 7. Then 10 ml of digest was distilled with 10 ml of 40% NaOH solution using Kjeldahl’s distillation apparatus.

3 8. About 10-15 ml of distillate (NH3 gas) was collected in a 125 ml conical flask containing 10 ml of 2% boric acid with mixed indicator. 9. The conical flask was removed and the delivery outlet of the distillation apparatus was washed with distilled water. 10. The distillate was then titrated against the standard H 2 SO 4 until the solution color became pink. 11. A blank experiment was also done simultaneously using the entire chemical except soil.

Data:

No. of Observati on Sample Blank

1. 1

Initial burette reading (ml)

Final burette reading (ml)

Difference (ml)

9.6 15.2

12.6 15.5

3.0 0.3

Mean (ml) 3.0 0.3

Calculation: 1000 ml of 1N H2SO4 or 0.01 N HCl = 14 g of nitrogen 14 = 0.014 g of nitrogen 1 ml of 1N H2SO4 or 0.01 N HCl = 1000 ( T − B ) × f × 0.014 × 100 ml Volume × 100 Of N in soil = Amount of soil ( g ) × amount of extract used ( ml ) Where, T = Amount in ml of 0.01N H2SO4 or 0.01 N HCl required in titration of the experiment with soil. B = Amount in ml of 0.01N H2SO4 or 0.01 N HCl required in titration of the blank experiment. f = Normality factor of the acid. N in soil =

( 3.0 − 0.3) × 0.014 × 100 × 100

= 2.7 × 7 = = 18.9 %

2 × 10

Result: The total soil nitrogen of the soil sample was 18.9%.

4

Exp No: 2

Date: 10.07.06

Name of the Experiment: Determination of available P present in plant sample by Vanadomolybdophosphoric Yellow Color method in nitric acid system. Principle: The total quantity of P in most soils is relatively small ranges from 0.03 to 03%. Most soil P determination has two distinct phases: First, the preparation of solution containing the soil P, and Second, the quantitative determination of the P in this solution. The choice of colorimetric method depends on the concentration of P in solution and particular acid system involved in the analytical procedure. The complexes are thought to be formed by coordinating of molybdate ions, with P as the central coordinating atom, the oxygen of the molybdate radicals being substituted for that of PO4: H 3 PO4 + 12 H 2 MoO4  → H 3 P( Mo3O10 ) 4 + 12 H 2 O The complex, before reduction, gives a yellow hue to their water solution. With high P concentration, a yellow precipitate is formed. In solution of low enough concentration to be suitable for determination by reduction to form the blue color, the yellow color is so faint that it is not noticed. At about 100 times this concentration, the yellow color is suitable for spectrophotometer as for molybdosilicic acid color. The vanadomolybdo phosphoric yellow color is greatly intensified by the vanadium component. The extent nature of the yellow chromogen of the vanadomolybdophosphoric system is not known, but the color is attributed to substitution of oxyvanadium and oxymolydenum radicals for the O of PO4to give a heteropoly compound that is chromogic. The vanadomolybdophosphoric yellow color method has the advantages of extreme simplicity, lower sensitivity, stability of color, freedom from interferences with a wide range of ionic species in concentration up to 1000 ppm, and adaptability to HNO3, HCl, and H2SO4 or HClO4 system. Apparatus: 1. Pipettes 2. volumetric flask 3. Spectrophotometer (400 to 490 nm) Reagent:

1. Standard P solution which was made by KH2PO4.

5

2. Standard phosphorus solution of 0, 1, 2,5,10 ppm was made from primary stock solution and secondary solution. 3. The following special combined HNO3- Van date molybdate reagent : Solution A: dissolve 25g of ammonium molybdate in 400ml of water at 50oC, cool and filter if necessary. Solution B: Dissolve 1.25 g of anhydrous ammonium metavanadate in 400ml of hot water. Cool the solution and then add 250ml of HNO3 and again cool it room temperature. Solution-A and solution-B was freshly mixed before work. 400ml of solution-A and 600ml of solution-B was mixed in the ratio of 1:1.5. Procedure: 1. 10ml aliquot was taken in a volumetric flask of 50ml. 2. 10 ml of the vanadomolybdate reagent was added. The color developed rapidly but usually read after 10 min to assure full strength. 3. Reading of the sample was taken in a spectrophotometer using 400-490 nm wavelengths and was compared the reading with standard curve. 4. A standard P curve was prepared using known solution of P and the above reagent. Data: ppm 0 1 2 5 10

Reading for standard solution: Absorbance 0.32 0.135 0.242 0.632 1.234

Absorbance (concentrated) 0 0.103 0.21 0.6 0.9

Reading for sample Absorbance (concentrated) 0.144 Calculation: ppm from the Curve × Volume of extract × 100 Amount of soil × Amount of extract 2.1 × 50 = × 100 0.5 × 10 = 2100 ppm = 0.21%

ppm P =

Result: The total soil P of the soil sample was 0.21%.

ppm from graph 2.1

6 Exp no: 3 Date: 11.07.06 Name of the Experiment: Determination of total sulphur by turbid metric method. Principle: By digestion bound from of sulfur is converted to water soluble SO42- . The concentration of SO42- in the solution is turbid metrically determined by spectrometer. If barium solution is added into the solution, a water insoluble precipitate BaSO 4 is formed which is whitish in color. Ba2++SO42-    → BaSO4 ↓ The amount of precipitate from is directly proportional to the sulfate content of the solution. For analytical purpose, it is necessary to maintain the solution suspended with insoluble BaSO4 so that, it appears turbid and hinders the light to pass through it (spectrphotometric method principle). The suitable surfactant for this purpose is a chemical, commercially known as twin -80. The effectiveness of this surfactant is only 20-40 minutes. Materials: (a) Sample: Extract prepared by perchloric-nitric acid digestion. (b) Apparatus: 1. 2. 3. 4. 5. 6.

Measuring flask Electric balance Beaker Volumetric flask Pipette Spectrophotometer, etc.

(c) Reagent: 1. Standard sulfur solution of Na2SO4. Standard sulfur solution of 0,5,10,15 ppm 2. HCl acid ( 6N) 3. Barium chloride – twin 80 solution: 13.3 gm BaCl2 and 132 ml Twin was made to volume into 1000ml. Procedure: 1. 2. 3. 4.

10 ml of aliquot was taken in a 25 ml volumetric flask with which 3 ml of BaCl2Twen 80 solution was added. Distilled water was added to the solution and the volume was made up to the mark. It was then left for at least half an hour for the formation of turbidity of BaSO4. Standard solution of 0, 5, 10, 15 ppm of sulfur was prepared from primary and secondary stock solution by using Na2SO4 salt.

7 5. 6. 7. 8.

Then 10 ml of standard solution from 0, 5, 10, 15 ppm was taken into four 2.5 ml volumetric flask. Therefore, 3 ml of BaCl2 –Twen80 solution was added into every volumetric flask was made to volume up to the mark. The reading was then taken for the standard solution and the standard curve was made where absorption were taken against the concentration in ppm. Then the reading for the sample solution was taken. Thus reading was put in the standard curve and sulfur content was determined from the standard curve for sulfur.

Data: ppm 0 3 5 7 10

Reading for standard solution: Absorbance 0.012 0.080 0.154 0.185 0.235

Absorbance (concentrated) 0 0.068 0.142 0.173 0. 223

Reading for sample Absorbance (concentrated) 0.17

Calculation: ppm from the Curve × 100 × 50 Amount of soil × amount of extract 6.8 × 50 × 100 = 0.5 × 10 = 6800 ppm = 0.68 %

ppm S =

Result: The total soil S of the soil sample was 0.68 %.

ppm from graph 6.8

8 Exp. No:4 Date:12.07.06 Name of the Experiment: Determination of total Potassium by flame photometry method. Principle: Total potassium in soils commonly ranges from 0.5 to 2.5%. Total potassium in soil is determined by a wet digestion of the sample with nitric and perchloric acid followed by flame photometry. It is a device that soaks out the sample through pipes. Up to burning on the flame K+ gives characteristics color. In this zone other colors also produce if it contains other element. A monochromatic filters adjoining to the flame select only the chromate of K+ and allows it to pass through it. Each penetrating wave makes a pulse when they strike portable cell. This pulse no. is recorded by the machine and is displayed as digital signal. The amount of pulse is directly related to the K+ content of the sample. Material: (a) Sample: Extract prepared by perchloric – nitric acid digestion. (b) Apparatus: 1. Flame photometer with filters of 766 to 769 nm. 2. Volumetric flask (c) Reagent: Standard Potassium solution of KH2PO4 of 5, 10, 20, 30 ppm. Procedure: 1. The extract was 10 times diluted by adding distilled water with 5ml digest and made up to 50 ml volume in 50 ml volumetric flask. 2. The reading for potassium in a flame photometer from the diluted extract and standard solution. 3. A standard potassium curve was prepared by using known solution of potassium. 4. Total potassium was determined from the standard potassium curve. Data: ppm 0 5 10 15 Sample

Reading for standard solution: Intensity 0 3 5.9 8.4 3.9

Intensity (concentrated) 0 3 5.9 8.4 3.9

Reading for sample Intensity (concentrated) 3.9

ppm from graph 6.5

9

Calculation: ppm from the Curve × 100 × 50 Amount of soil × amount of extract 6.5 × 50 × 100 = 0.5 × 10 = 6500 ppm = 0.65 %

ppm K =

Result: The total soil K of the soil sample was 0.65%.

10 Exp. No: 5 Date:15.07.06 Name of the Experiment: Determination of cation exchange capacity of the supplied soil sample. Principle: Cation exchange capacity (CEC) is defined as the sum of the exchangeable cations retained by soil. It is expressed in millequivalent per 100gm or in cmol of cations per kg of soil. Cation exchange in soil is a reversible chemical reaction and corresponds to the negative charge of the soil. The principal factors which determine CEC are amount and types of clay present, organic matter content, and soil pH. The negative charge and hence the CEC, increase as pH rises. The capacity of the soil to adsorb exchangeable cation cannot therefore be determined unless a standard pH its measurement is agreed on (Bache, 1979). Cation exchange capacity is most commonly determined as the quantity of cations absorbed from salt solution buffered at pH -7.0 with NH4OAC or at pH 8.2 with BaCl2-triethanol ammine. The predominant cations involved in exchange are H, Ca, Mg, K, Na and NH4+ and such cations which can replace or be replaced by other in this manner are termed ‘exchangeable’. Element such as zinc, Mn and Cu also occur in exchangeable forms but in small amounts. When soil sample is leached by NH4OAC salt solution to replace all the exchangeable cations from colloid surface; NH4+ takes this place. Leaching accelerate the exchange process. To wash away the NH4+ ions rather than those retained by electrostatic force on colloid surface, the soil is leached with alcohol for 4-5 times. At this stage soil retain only exchangeable NH4+ against negative charges. Now this retained NH4+ ion is replaced by other suitable cation (e.g. K+) and the leached solution containing NH4+ is collected. By alkali distillation method the amount of NH4+ is determined. On the hand , the first leached contains all exchangeable cations .Such as Na+, K+, Ca+, Mg2+, H+ present on colloid surface of the soil are collected for determining those species. Material: (a) Sample: Supplied soil sample (b) Apparatus: 1. 2. 3. 4. 5. 6. 7. (c)

Conical flask Filter paper (whatman no.40) Distillation apparatus Micro burette Measuring cylinder pipette Volumetric flask

Reagent: 1. 2. 3.

Ammonium acetate, 77.08 gm of glacial ammonium acetate was taken 1N in a 1000ml volumetric flask. Then pH was adjusted at 7.0 and finally made volume up to the mark. Ethyl alcohol , 95% NaCl, 1N: 58.5 gm of NaCl was taken in a 1000 ml volumetric flask. About 800 ml of water was added in the volumetric flask and salt was dissolved. Finally it was made volume up to the mark.

11 4. 5. 6. 7.

H2SO4 , 0.01 N NaOH, 10% 2% Boric acid Mixed indicator solution

Procedure: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

10 gm of soil was taken in a 250 ml volumetric flask and 1000ml of 1N neutral ammonium acetate solution was added. Then the content was shaken for an hour. Therefore, the content of the conical flask was filtered through whatman filter paper no. 40. The filtrate was collected in a bottle for determination of exchangeable cations. The soil material on the filter paper was leached out five times with Ethyl alcohol by taking 10 ml each time. The filtrate was discarded. There after the soil material on the filter paper was leached with 1N NaCl solution in a volumetric flask of 100 ml capacity and was made volume up to the mark. 10 ml of leached was taken from volumetric flask and was distilled with 10 ml of 10% NaOH solution using micro-Kjeldaghl’s distillation apparatus. The distilled ammonia was collected in a coined flask containing 10 ml of 2% boric acid with mixed indicator. The conical flask was removed and the delivery outlet of the distillation apparatus was washed out with distilled water. The ammonium borate solution contained in against standard H2SO4 solution until the color changed from blue to pink. A blank experiment was run by same process except the leached.

Data: Strength of H2SO4 = 0.01 Weight of soil = 10gm

No. of observation Sample Blank

1 2 1

Data for burette reading (with soil) Final Initial burette burette Difference reading reading 0 0.6 0.6 5.9 6.5 0.6 15.2 15.5 0.3

Mean 0.6 0.3

12

Calculation: 1000ml 1N H 2 SO4 ≡ 18 gm NH 4 + ≡ 14 gm N 14 × 0.3 ∴0.3 ml 1N H 2 SO4 = gm N 1000 14 × 0.3 × 0.01 ∴0.3 ml 0.00973 N H 2 SO4 ≡ gm N 1000 = 0.000042 gm N 10ml leachate contains 0.0000408 gm N 0.000042 × 100 ∴100 ml leachate contain = gm N 10 = 0.00042 gm N 10 gm soil contain 0.000408 gm N 0.00042 × 100 ∴100 gm soil contain = gm N 10 = 0.0042 gm N ∴ ( % ) of N in soil = 0.0042 gm = 0.0042 ×1000 mg = 4.2mg N So, CEC in meq / 100 gm soil =

4 .2 = 0.3 meq 100 gm soil 14

Result: The cation exchange capacity of the soil sample is 0.3 meq/100gm soil.

13 Exp. No:6 Date: 16.07.06 Name of the Experiment: Determination of ammonium acetate extractable sodium content by flame photometry method. Principle: The ammonium acetate does not interfere in flame photometry method. So ammonium acetate extractable sodium can be determined directly from ammonium acetate extract by flame photometer. Materials: (a) Sample: Ammonium acetate extract (b) Apparatus: Flame photometer (c) Reagent: Standard solutions of sodium of 0, 5, 10, 15, 20, 25, 30 ppm. Standard solution was made by using Na2SO4 salt. At first 500ppm in 1000ml was made as primary stock solution by taking 1.543 gm Na2SO4 in 1000 ml volumetric flask. Then secondary stock solution was made from primary stock solution. And finally 5, 10, 15, 20, 25, 30 ppm was made from by secondary stock solution by serial dilution. Procedure: 1. The reading for Na was taken from flame photometer for both the extract and standard solution. Besides blank reading was taken. 2. A standard curve of sodium prepared by using known concentration of sodium. 3. The amount of sodium was determined from the standard curve. Data: Reading standard solution ppm 0 5 10 15 20 Sample

Intensity 0 3.2 5.9 8.2 10 3.7 Reading for sample:

Intensity 3.7

ppm from graph 6.0

14 Calculation: Conc. from graph × 100 10 6.0 × 100 = 10

µg Na g −1 soil =

= 0.6 µg Na g −1 soil = 0.6 × 100 µg Na 100 gm soil = 0.06 mg Na 100 gm soil 0.06 meq Na = 100 gm soil 23 = 0.0023 meq Na 100 gm soil

Result:

The amount of NH4OAc extractable sodium in sample is 0.0023

meq Na

100 gm soil .

15 Exp. No: 7 Date: 15.07.06 Name of the Experiment: Determination of ammonium acetate extractable potassium content by flame photometry method. Principle: The ammonium acetate does not interfere in flame photometry method. So ammonium acetate extractable potassium can be determined directly from ammonium acetate extract by flame photometer. Materials: (a) Sample: Ammonium acetate extract (b) Apparatus: Flame photometer (c) Reagent: Standard solutions of sodium of 0, 5, 10, 15, 20, 25, 30 ppm. Standard solution was made by using K2SO4 salt. At first 500ppm in 1000ml was made as primary stock solution by taking 1.543 gm K2SO4 in 1000 ml volumetric flask. Then secondary stock solution was made from primary stock solution. And finally 5, 10, 15, 20, 25, 30 ppm was made from by secondary stock solution by serial dilution. Procedure: 1. The reading for K was taken from flame photometer for both the extract and standard solution. Besides blank reading was taken. 2. A standard curve of sodium prepared by using known concentration of K. 3. The amount of K was determined from the standard curve. Data: Reading standard solution ppm 0 5 10 15 20 Sample Intensity 6.5

Intensity 0 3 5.9 8.4 10 6.5 Reading for blank: 0 (intensity) Reading for sample: ppm from graph 11.25

16 Calculation:

µg K g −1 soil =

Conc. from graph × 100 10

= 11.25×100 10 = 112.5 µg Kg¯1 = 11.25 meqk/100 gm 39.1 = 0.287 meqk/100gm soil

Result: The amount of NH4OAc extractable K in sample is 0.287 meqK/100gm soil.

17

Exp. No: 8 Date: 17.07.06 Name of the Experiment: Determination of exchangeable (extractable) Ca by EDTA titration method. Principle: The classical routine method of determining Ca is by complexometric titration using EDTA which forms a slightly ionized, colorless, stable complex with Ca. The classical indicator used for Ca titration is ammonium turtarate, commonly known as murexide. Murexide at a pH of H changes from pink to light blue at the end point of titration. The murexide indicator is very susceptible to oxidation and for this reason is added as a powder and not in solution. The pink color complex formed has an ionization constant greater than the EDTA-Ca complex. Materials: (a) Sample: Ammonium acetate extract (b) Apparatus: 1. 2. 3. 4. 5.

Micro pipette Small porcelains volumetric flask Pipette Measuring cylinder

(c) Reagent: 1. NaOH 2.5N, 10gm of NaOH was taken in a 100 ml volumetric flask and was made volume up to the mark with distilled water. 2. Standard EDTA solution, 0.02M. The molecular weight of EDTA is 372.254. Therefore, 1.8613 gm of EDTA was taken in 250ml volumetric flask; dissolved and was made volume up to the mark. The solution was 0.02 M EDTA in 250ml. 3. Standard calcium solution , .02M 4. Murexide (ammonium turtarate ) indicator Procedure: 1. 5ml of extract was taken in a small porcelain cup and 5 ml of 2.5 N NaOH was added . 2. 0.02gm of murexide indicator was added which give the solution light pink color. 3. It was titrated against EDTA solution. At the end point color changed from pink to light blue.

18

Data: Strength of EDTA = 0.02 M

No. of observation Sample

1 2

Data for burette reading (with soil) Final Initial burette burette Difference reading reading 0.2 0.5 0.3 0.7 1.2 0.5

Mean 0.4

Calculation: % Ca in soil =

0.02 × ml of EDTA used × normality of EDTA × 100 ml volume × 100 ml of extract used × wt. of soil = 0.02×0.4×1×100 ×100 2.5×5 = 6.4%

19

Exp. No: 9 Date: 18.07.06 Name of the Experiment: Determination of lime requirement of supplied soil. Principle: The lime requirement is the amount of lime that must be applied to acid change soil pH from its present value to any desired value. The materials commonly used for liming soils are Ca and /or Mg oxide, hydroxide, carbonates and silicates. The accompanying anion must lower H+ activity and Al3+ in the soil solution. Liming reaction begin with the neutralization of H+ in the soil solution by either OH-or HCO3- originating from the liming materials . For example CaCO3behave as follows: CaCO3 + H2O → Ca2+ +HCO-3+OHThe rate of the reaction is directly related to the rate at the OH- ions are removed from solution by H+ ions to made H2O. Soil pH is the most important criteria determining the lime requirement of soil. Other properties Such as texture, structure, percentage of organic matter amount and type of clay affect lime requirement, CaCO3 is the most common agricultural lime but its incubation in the lab is too much time consuming. To obtain relatively quick determination in laboratory we use Ca (OH)2 for incubation experiment and convert the amount in CaCO 3. Buffering capacity of the soil is the main considerable factors here. The more the buffering capacity, the more the lime required to bring desired change. Material: (a) Soil sample Acid sulphate soil (pH(b) Apparatus: 1. 2. 3. 4. 5.

)

Beaker ,50ml pH –meter Stirring rod Electric balance Para film

(c) Reagent: 1. Ca(OH)2 2. Standard buffer solution of pH 4 and 7. 3. Distilled water Procedure:

20

1. 20gm of soil was taken in each of 7 beakers of 50ml. 2. Then 0, 0.04, 0.08, 0.12, 0.16, 0.20, 0.24 gm of Ca(OH) 2 was added to the beaker one after one. 3. Soil and Ca(OH)2 was mixed by stirring rod. Therefore, 6 ml of water was added to each beaker and the suspension was mixed thoroughly with glass rod. 4. Then the beaker was well covered with Para film and left untouched for 2 months. 5. After 2 months, the pH value were measured the help of the pH meter and a curve was drawn with the obtained values against the amount of Ca(OH)2 used. Data: pH reading for the sample after 2 months No. of observation Amount of Ca(OH)2 In gm 1 0.0 2 0.04 3 0.08 4 0.12 5 0.16 6 0.20 7 0.24

pH 3.39 5.96 6.05 6.62 7.73 7.83 8.59

Calculation: We have raised the pH from 4.0 to6.8. From graph, to increase the soil pH from 4.0 to 6.8, 0.12 gm Ca (OH)2 was needed per 20 gm soil. 20 gm soil require 0.12 gm Ca ( OH ) 2 0.12 ×1000 20 = 6.0 gm Ca ( OH ) 2 kg soil

∴1000 gm , , , , =

1kg soil require 6.0 gm Ca ( OH ) 2 ∴ 2 ×10 6 Kg soil requires = 6.0 × 2 ×10 6 gm Ca ( OH ) 2 = 12.0 ×10 6 gm Ca ( OH ) 2 / hac = 12.0 × 103 Kg Ca ( OH ) 2 hac = 12.0 tons Ca ( OH ) 2 hac soil

Now, to get the calcium carbonate equivalent for Ca (OH)2 , CaCO3 and Ca(OH)2 should be equal in term of acid neutralizing ability. Ca (OH)2+2HCl=CaCl2+2H2O CaCO3 +2 HCl = CaCl2+CO2 + H2O

21 Ca (OH)2 and CaCO3 both can neutralize 2 molecule of HCl acid. Molecule wt. of Ca(OH)2 =74 Molecule wt. of CaCO3 =100 74 gm of Ca (OH)2 = 100 gm of CaCO3 ∴ 74 ton of Ca (OH)2 = 100 ton of CaCO3 100 × 12 ∴ 12 ton of Ca (OH)2 = = 16.21 tons of CaCO3 74 Result: To raise the pH from 4.0 to 6.8 of the supplied acid soil sample, the required amount of Ca (OH)2 is 16.21 tons of calcium carbonate per hectors.

22

Exp. No: 10

Date: 19.07.06

Name of the Experiment: Determination of total organic carbon in soil by Tyurin’s method. Principle: Tyurin’s method (1931) is a modification of volumetric determination of soil organic carbon by oxidation with K2Cr2O7 in strongly acid solution, leading to the formation of CO2 according to the following reaction: 2 K 2 Cr2 O7 + 8 H 2 SO4    → 2 K 2 SO4 + 2 Cr2 ( SO4 ) 3 + 8 H 2 O + 3O2 3 O2 + 3C    → 3CO2 The amount of oxygen consumed during the oxidation of organic carbon is calculated from the difference between the amount of dichromate taken and the amount remaining after oxidation, this is determined by titration with a standard solution of Mohr’s salt. Materials: (a) Sample: soil sample (b) Apparatus: 1. 2. 3. 4. 5.

Beaker Measuring flask Burette, pipette, funnel Hot plate Conical flask

(c) Reagent: 1. 0.4 N K2Cr2O7 solution 2. 0.1 N Mohr’s salt solution, [( NH 4 ) 2 SO4 .FeSO4 .6 H 2O] 3. Indicator : Phenylanthranilic acid Procedure: 1. 2. 3. 4.

0.2gm of soil was taken in a dry 100ml conical flask. 10 ml of 0.4 N K2Cr2O7 solutions was added to it. Then a 4 cm diameter funnel was placed and the flask was placed on a hot plate. The mixture was heated to boiling for 5 minutes.

23 5. The flask was then removed from the plate and the inner and outer surfaces of the funnel and the neck and walls of the flask were carefully washed with about 10 ml of distilled water. 6. The color of the liquid after the oxidation should be orange –yellow. A greenish color of the liquid indicates insufficient oxidizing agent and the analysis must be repeated to take small amounts of soil. 7. The unutilized K2Cr2O7 solution in the oxidation was titrated against 0.1 N Mohr’s salt solution using phenylanthranilic acid as the indicator. 8. At the end point of the titration, the color was changed from brown to green. 9. A blank experiment was run at the same time using all the reagents except soil. Data:

Sample Blank

Initial burette reading (ml) 0 15.5 0

Final

Difference

Mean

15.5 30.0 18.7

15.5 14.5 18.7

15.0 18.7

Using the formula, S1×V1 = S2×V2 Here, S1= Strength of K2Cr2O7 solution = 1N V1= Volume of K2Cr2O7 solution =10ml S2 = Strength of FeSO4 solution V2 = Volume of FeSO4 solution = 18.7 ml So, S 2

S ×V 1 × 10 = 1 1= = 0.547 N V2 18.7

Calculation: % Organic C =

( B − T ) × 0.39 × f W

Where, B= Amount in ml of N FeSO4 solution in blank experiment T= Amount in ml of N FeSO4 solution in soil sample experiment F = Strength of FeSO4 solution W= weight of soil

24 So , % Organic C =

(18.7 − 15.0) × 0.39 × 0.547

2 3.7 × 0.39 × 0.547 = 2 0.789 = 2 = 0.394 %

∴ % Organic matter content of soil = % Organic C ×1.724 = 0.394 ×1.724 = 0.679 % Result: The organic matter content of soil is 0.679 %.

25 Exp. No: 11 Date: 22. 07.06 Name of the Experiment: Determination of total organic carbon in soil. Principle: The principle that organic C –compound are highly reducing substances is employed in determination of organic C, which reduces solution of oxidizing agent when digested with it. This method is known as wet oxidation method. Chromic acid is suitable agent for determining soil organic C. K 2 Cr2 O7 + 3C + 6 H 2 SO4 → 4Cr2 ( SO4 ) 3 + 3CO2 + 8 H 2 O + H 2 Cr2 O7 K 2 Cr2 O7 + FeSO4 + 7 H 2 SO4  → K 2 SO4 + Cr2 ( SO4 ) 3 + 3 Fe2 ( SO4 ) 3 The excess acid left after the oxidation may be determined volumetrically with standard ferrous sulfate solution and the quantity of substances oxidized is calculated from the amount of chromic acid reduced. Materials: (a) Sample: soil sample (b) Apparatus: 1. 2. 3. 4.

Beaker Measuring flask Burette, pipette, funnel Conical flask

(c) Reagent: 1. 2. 3. 4. 5.

1 N K2Cr2O7 solution Conc. H 2 SO4 , 96% Phosphoric acid NaF Ferrous sulfate solution

Procedure: 1. 2. 3. 4. 5. 6. 7. 8.

2gm of soil was taken in a dry 500ml conical flask. 10 ml of K2Cr2O7 solution was added to it. Then add 10 ml Conc. H 2 SO4 , 96% The mixture was cooled to with shaking for half an hour. Then add 150 ml distilled water, 10 ml phosphoric acid and 0.2 gm NaF. Add 3 ml of diphenylamine indicator. The color will be violet. Cool the flask again. Titrate the excess of chromic acid left in the flask with the help of ferrous sulfate solution from burette. 9. At the end point the color of solution will change to deep bottle green.

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Data: Initial burette reading (ml) 0 15.5 0

Sample Blank

Final burette reading (ml) 15.5 30.0 18.7

Using the formula, S1 × V1 = S 2 × V2 Here, S1 = Strength of K2Cr2O7 solution =1 N V1= Volume of K2Cr2O7 solution = 10 ml S2 = Strength of FeSO4 solution V2= Volume of FeSO4 solution = 18.7 ml So, S 2

S ×V 1 × 10 = 1 1= = 0.547 N V2 18.7

Calculation: % Organic C =

( B − T ) × 0.39 × f W

Where, B= Amount in ml of N FeSO4 solution in blank experiment T= Amount in ml of N FeSO4 solution in soil sample experiment F = Strength of FeSO4 solution W= weight of soil

So , % Organic C =

(18.7 − 15.0) × 0.39 × 0.547

2 3.7 × 0.39 × 0.547 = 2 0.789 = 2 = 0.394 %

∴ % Organic matter content of soil = % Organic C ×1.724 = 0.394 ×1.724 = 0.679 % Result: The organic matter content of soil is 0.679 %.

Difference 15.5 14.5 18.7

Mean 15.0 18.7

27 Exp. No: 12

Date: 23.07.06

Name of the Experiment: Determination of Free Carbonate present in soil. Principle: The carbonate can be determined by decomposing them with and the amount of CO2 evolved may be determined volumetrically. The CO3- may be reacted upon by a known amount of standard solution and the acid required to decompose the CO3- may be determined volumetrically. CaCO3+2HCl=CO2. H2O + CaCl2 The unreacted HCl is titrated against 1 N NaOH. Material: (a) Soil sample (b) Apparatus: 1. 2. 3. 4. 5. 6.

Beaker , Stirring rod Electric balance Burette, pipette, funnel

(c) Reagent: 1. 1 N NaOH solution 2. 1 N HCl 3. Bromothymol blue indicator solution Procedure: 1. 2. 3. 4. 5. 6. 7. 8. Data:

Weight 10 gm soil in 400ml beaker Add 100 ml of 1 N HCl and cover the Beaker with a clock glass Stir with a glass rod for 1 hour Filter the suspension through filter paper Add 6 to 8 drop of bromothymol blue indicator. Titrate the solution against 1 N NaOH. At the end point the color of the solution will change to yellowish green.

Soil sample blank

Initial burette reading (ml) 0 0

Final burette reading (ml) 9.0 10.6

difference 9.0 10.6

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mol. wt. of CaCO3 = 100 g Equivalent wt. of CaCO3 = 50 g. 1000 ml of 1N HCl ≡ 50 g o CaCO3 ∴ 1 ml ml of 1N HCl ≡

50 = 0.05 g o CaCO3 1000

Calculation: % of CaCO3 present in soil =

( B − T ) × 0.05 × f × 100 × 100 10 g soil × 10 ml extract

Where, B = Blank titration reading T = Soil sample reading f = Strength of NaOH = 1 N ∴ % of CaCO3 present in soil =

( B − T ) × 0.05 × f × 100 × 100

10 g soil × 10 ml extract (10.6 − 9.0) × 0.05 × 1 × 100 × 100 = 10 × 10 1.6 × 0.05 × 1 × 100 × 100 = 10 × 10 = 8%

Result: The amount of free carbonate in soil sample 8%.

29 Exp. No: 13 Date: 24.07.06 Name of the Experiment: Determination of Manganese from supplied sample. Principle: The usual and most6 satisfactory method of determining manganese in solution or extraction is the oxidation of Mn2+ to MnO4- by K or Na- meta-period ate in a strongly acid solution and to measure calorimetrically the permanganate form. 2 Mn 2 + + 5 IO4− + 3H 2 O = 2 MnO4− + 5 IO3− + 6 H + This reaction will proceed smoothly over a wide range of Mn concentration. The little interference encountered can usually be readily eliminated. Permanganate is stable in the presence of excess period ate. The intensity of the permanganate color developed will remain unchanged in the absence of reducing agents and is readily reproduced. Organic matter and chloride is the principal interfering agent that must be eliminated in analysis of soil. So, reducing substances must be absent during the period ate oxidation of manganese and these are destroyed with nitric or sulfuric acid and H2O2, before development of permanganate. Chloride can be tolerated to some extent particularly if phosphoric acid is used; if perchloric acid is used to evaporate the solution there will be little interference from chloride. Apparatus: 1. 2. 3. 4. 5. 6. 7.

Conical flask Filter paper (whatman no.40) Beaker pipette Hot plate Spectrophotometer Asbestos

Reagent: 1. 2. 3. 4. 5. 6.

H2SO4 H2O2, 30% H3PO4, 85% KIO4 NH4OAc Standard Mn-solution

Procedure: 1. 2. 3. 4. 5.

Take 5 ml total extract in beaker. Add 2ml H2SO4, 5ml of 30% H2O2, 5ml H3PO4 and 0.2gm KIO4. Heat the solution in digestion chamber. Then cool the beaker to room temperature. Then transfer it to a 50ml volumetric flask and bring the volume to a 100 ml H2O.

30 6. Read the absorbance at 540nm. Data: ppm 0 0.5 1.0 1.5 2.0 3.0

Reading for standard solution: Absorbance 0 0.018 0.034 0.052 0.068 0.108

Reading for sample Absorbance (concentrated) 0.006

Calculation: ppm from the Curve × 50 ml volume × 100 Amount of soil × amount of extract 0.1625 × 50 × 100 = 0.5 × 5 = 325 ppm = 0.0325 %

ppm S =

Result: The total soil Mn of the soil sample was 0.0325 %.

Absorbance (concentrated) 0 0.018 0.034 0.052 0.068 0.108 ppm from graph 0.1625

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