Notes on the Gypsum Requirement (GR) of Soil

This article provides a notes on the Gypsum Requirement (GR) of soil.

The main principle for the reclamation of sodic or alkali soils is to replace exchangeable Na by another cation calcium (Ca²⁺). Of all calcium compounds, gypsum (CaSO₄.2H₂O) is considered the best and cheapest for the reclamation purpose. Calcium (Ca²⁺) solubilized from gypsum replaces sodium (Na⁺) leaving soluble sodium sulphate (Na₂SO₄) in the water, which is then leached out. Gypsum requirement (GR) is expressed as me of Ca²⁺ per 100 gm. soil.

The gypsum requirement (GR) is the calculated amount of gypsum necessary to add to reclaim the soil. For calculating the dosage of gypsum needed to reclaim a particular alkali soil, advantage is taken of the assumption that through the use of a calcareous amendment, the desired level of quantitative replacement of exchangeable sodium from the soil by calcium is possible.

Since an exchangeable sodium percentage (ESP) of 10 and below is considered safe for tolerable physical condition of the soil, replacement by calcium to this level (ESP, 10) is all that is attempted in practice, though for certain tolerant crops like paddy, wheat and barley the replacement may be attained to higher ESP level than 10. Gypsum requirement is determined from the formula

Gypsum requirement (GR) i.e. me of Ca²⁺/100 g soil = ESP (initial)-ESP (final) × CEC/100

ESP (initial) is obtained from the analysis of soil before reclamation or application of gypsum; ESP (final) is usually kept at 10 and CEC is the cation exchange capacity in me/ 100 g or Cmol (pp⁺) kg^-1 of the soil.

For example, the gypsum requirement of a soil having initial ESP 60, final ESP 10 and CEC 30 Cmole (p⁺) kg^-1 will be

me of Ca²⁺/100 g soil = [60 – 10]/100 × 30

= 50 × 30/100 = 15

Since one hectare of soil to a depth of 15 cm (6 inches) weighs approximately 2 × 10⁶ kg and 1 me of replaceable gypsum as CaSO₄.2H₂O equals 860 ppm of amendment, the theoretical amount of gypsum required per hectare will be:

Theoretical amount of gypsum (kg) ha^-1

= 860/106 × 2 × 10⁶× 15= 1720 × 15

= 25,800 kg = 25.80 tons

This has to be multiplied by the purity factor of gypsum to arrive at the field requirement of the amendment. Once, the gypsum requirement in tons per hectare is known the equivalent quantities of other chemical ameliorants can be calculated from the conversion Table 15.3.

It is evident that quantitative replacement of sodium by calcium from the soil exchange complex is never possible.

Two factors which govern the level of equilibrium in soils and soil solutions between sodium and calcium ions are:

(i) Initial exchangeable sodium percentage of soil and

(ii) The total cation concentration of the soil solution.

Usually to compensate for these factors, the U.S. Salinity Laboratory staff recommends that the gypsum requirement obtained by calculation may be multiplied by 1.25 to obtain the desired level of sodium replacement.

The efficiency of gypsum and other amendments will depend on the fineness as well as purity of the material. Usually, fineness with particle size of 2 mm or 0.59 mm (10 or 30 mesh sieve) has been found optimum for the amelioration purpose. Thorough incorporation and mixing of the amendment in the soil is an another important factor in determining its beneficial effect.

However, the calculation for the actual amount of gypsum for field application is not possible because various other side factors interferes with the determination of exchangeable sodium percentage (ESP) viz. texture, surface area and type of clay minerals, organic matter and exchangeable magnesium.

In India, it is evident that even 50 per cent of the theoretical gypsum requirement for replacement of exchangeable sodium in sodic soils has improved their physical conditions and assisted responses to management practices. The amount of gypsum (CaSO₄.2H₂O) and sulphur (S) required to ameliorate the sodic soil on the basis of exchangeable sodium is given in Table 15.4.

Problem:

There are various mathematical formulae which can be used for the determination of gypsum requirement to ameliorate the sodic soils taking into consideration different interference of soil properties. Using a soil bulk density of 1.34 g/c.c. for an average soil, the gypsum requirement is described by the following equation:

GR = metric tons of gypsum required/weight of soil per hectare to some fixed depth

= (Na_x) 4.50 metric tons of gypsum per hectare (30 cm).

= (Na_x) 2.25 metric tons of gypsum per hectare furrow slice (15 cm)

where Na_x is the number of milliequivalents of exchangeable sodium to be replaced by calcium from the added gypsum.

A sodic soil has an average exchangeable sodium percentage (ESP) of 24 per cent in the top 45 cm (18 inches) and a cation exchange capacity (CEC) of 18 C mol (p⁺) kg^-1of soil. The average exchangeable sodium to be left in the top 30 cm is 6 per cent. Calculate (i) the amount of gypsum, and (ii) the amount of sulphur required to reclaim the soil for the top 30 cm.

Solution:

The gypsum requirement factor is calculated as follows, using a soil bulk density of 1.34 g/c.c.

A hectare area 30 cm deep has a volume of about

100 m × 100 m × 0.30 m or 10,000 cm × 10,000 cm × 30 cm = 3 × 10⁹ c.c.

Therefore, the weight of soil per hectare.

30 cm depth = (3 × 10⁹c.c.) × (1.34 g/c.c. bulk density)

= 4.02 × 10⁹g of soil per hectare 30 cm

= 4.02 × 10⁹/10³kg of soil per hectare 30 cm

= 4.02 × 10⁶ kg of soil per hectare 30 cm

= 40,20,000 kg of soil per hectare 30 cm.

Again,

1 me of Na⁺ weighs 23 mg of Na⁺,

Kilograms of Na + per hectare, 15 cm is calculated using 2 × 10⁶ kg per hectare 15 cm of soil weight

Na⁺ /100g soil = 23 mg Na⁺/100 g of soil

= 23 mg Na⁺/100,000 mg of soil

= 460 mg Na⁺ /2000,000 mg of soil

= 460 kg Na⁺ /2000,000 kg of soil

Thus 1 me of Na⁺/100 g soil is equivalent to 460 kg of Na⁺ per hectare taken to a depth of 15 cm. So 1 me of Na⁺/100 g soil is equivalent to 920 kg (460 × 2) of Na⁺ per hectare 30 cm depth soil.

= (Na_x) (86/23)(920 kg Na/me/hectare30 cm)(1metric ton/1000 kg)

= (Na_x) (3.44) metric tons per hectare 30 cm depth per milliequivalent replaced

If the gypsum is not pure, or for other reasons is not 100 per cent efficient, more than the amount calculated must be added. Experience suggests that the gypsum is only 75-80 per cent efficient, so the amount added must be increased accordingly. Adding 30 per cent more gives these equations.

GR = (Na_x) 4.50 metric tons of gypsum per hectare foot soil.

= (Na_x) 2.25 metric tons of gypsum per hectare furrow slice (15 cm)

The milliequivalents of Na⁺ required to be replaced are calculated as the total exchangeable Na minus the exchangeable Na to be left.

CEC × ESP/100 = meq of exchangeable Na⁺ per 100 g soil

18 × 24/100

= 4.32 meq of exchangeable Na⁺ per 100 g soil

Again,

18 × 6/100 = 1.08 meq of exchangeable Na⁺ per 100 g soil to be left

Therefore, milliequivalents of exchangeable Na⁺ needs to be replaced per 100 g soil (Na_x)

= (4.32 – 1.08) = 3.24

So the amount of gypsum (CaSO₄.2H₂O) required to reclaim the soil metric tons/ha 30 cm

= (Na_x) (4.50)

= (3.24 me Na⁺/100 g) 4.50 = 14.58

Again amount of gypsum (CaSO₄.2H₂O) required to reclaim in metric tons per hectare furrow slice

= (Na_x) (2.25)

= 3.24 × 2.25 = 7.29

(ii) It has been found that only 0.18 times as much sulphur by weight is needed compared to gypsum, so the needed sulphur is:

14.58 tons gypsum × 0.18 = 2.62 metric tons of sulphur per hectare 30 cm soil

Again, 7.29 tons gypsum × 0.18 = 1.31 metric tons per hectare furrow slice (15 cm).