Soils 205-90

Lecture 12- Salt-affected soils

Videos                                                            Pages in Text.

18,19

412-446

A. Excess salt content = saline soils

Saline soils have accumulations of Ca, Mg, K, Na, Cl, SO4 (Na is relatively low)

defined as having an electrical conductivity of > 4 dS/m

Problems:  osmotic potential

 

Excess Na⁺ content = sodic soils

salt-affected agricultural soil in showing decreased productivity

B. Soil Salinity

1. High soluble salt content = low osmotic potential (y_o)

2. EC of saturation extract > 4 ds/m

(a) add water to soil to saturation

(b) extract solution with vacuum

= "saturation extract"

(c) measure electrical conductivity (EC) in decisiemen/meter (ds/m)

- EC proportional to salt content

- ds/m same as mmho/cm (millimhos/centimeter)

3. Salt at EC > 4 ds/m = problem with many plants

(a) salt sensitive plants (EC = 2 ds/m)

ü bean, onion, potato, raspberry, carrot, dogwood,
larch, linden, peach, rose, tomato

to

(b) salt tolerant plants (EC = 10 ds/m)

ü sugarbeets, barley, cotton, elm, locust, oak,
willow,rosemary, wheat grass, wild rye

(see table 10.2)

C. Soil Sodicity

1. High Na⁺ content

2. Related to flocculation-dispersion tendancy

(a) flocculation = good physical condition

(b) dispersion = poor physical condition

(c)

flocculation

dispersion

è attraction ç	ç repulsion è
Ca++ & Mg++	Na+

 

(d) both Na⁺ and salt content

most soils of medium salt content:
Na⁺ = dispersion
Ca⁺² = flocculation

3. ESP = exchangeable sodium percentage

ESP = (exchangeable Na^+/exchangeable Na⁺⁾ x 100

example: Na⁺ = 5 cmol_c/kg

CEC = 15 cmol_c/kg

ESP = (5/15) x 100 = 33 %

4. When ESP > 15 % = potential dispersion

(a) high sand = more difficult to disperse

- tolerate higher ESP

(b) high clay + low salt = disperse easily

- keep ESP lower

5. Predicted from saturation extract or irrigation water

(a) sodium adsorption ratio (SAR)

SAR = (Na⁺) /(Ca⁺² + Mg⁺²)^1/2

where (Na⁺), (Ca⁺²), (Mg⁺²)
= concentration of ions in the solution (mmol/L)

(b) SAR approximately equals ESP

SAR is measured Þ Þ ESP is estimated
in water or extract for soil solids

(c) good quality irrigation water:

- for salt hazard = EC < 2 ds/m

- for Na⁺ hazard = SAR < 15

D. Soil Properties

1. Saline (nonsodic) soils

(a) EC > 4 ds/m

Þ high Ca⁺² & Mg⁺² salts

(b) ESP (SAR) < 15

(c) pH < 8.5

· "neutral" salts

· often calcareous (lime-containing)

- then pH = 8.2 - 8.5

(d) "white alkali" soils

= white surface crust

2. Saline-Sodic soils

(a) EC > 4 ds/m

Þ high Na⁺ salts

(b) ESP (SAR) > 15

(c) pH < 8.5

· controlled by salts and lime

· pH> 8.5, if Na₂CO₃ salts

(d) also called "white alkali"

· but subject to dispersion

· if leached with good water, then ê

3. Sodic (nonsaline) soils

(a) EC < 4 ds/m

(b) ESP (SAR) > 15

(c) highly dispersed

· low salt + high Na⁺

· poor physical condition

(d) pH > 8.5

· hydrolysis of Na⁺

Na-x + H₂O ¬¾® H-x + Na⁺ + OH^- Þ high pH

(e) high pH disperses organic matter Þ dark color

= "black alkali"

(f) worst soil of the salt-affected

· high dispersion = water-logging

· can result from leaching saline-sodic soil

 

Comparisons of Salt-affected Soils

E. Management of salt-affected soils

1. Drainage - remove cause of condition

2. Leaching

(a) remove soluble salts

(b) leach soil only if it is nonsodic

- exchangeable Na⁺ + leaching = dispersion

(c) requires

x good drainage and drainage area

- results in degraded water

x source of "good" water

- low EC and low SAR

3. Na⁺ removal (for sodic soils)

(a) replace Na⁺ with Ca⁺² and Mg⁺²

(b) commonly use gypsum = CaSO₄^·2H₂O

2 Na-X + CaSO₄ ¬¾¾® Ca-X + Na₂SO₄

(c) S^o and H₂SO₄ in calcareous soils

S^o ¾® H₂SO₄ by microorganisms

CaCO₃ + H₂SO₄ ¾® CaSO₄ + CO2 + H₂O

lime

gypsum

(d) after Ca⁺² (Mg⁺²) replace Na⁺

- leach out the Na⁺ salts

(e) must know if Na⁺ is present before reclamation

4. Control

- maintain a salt balance by irrigation management
and/or crops grown

F. Gypsum requirement

?? gypsum if CEC is 12 cmol_c/kg and ESP is 20 %?

(20/100)(12 cmol_c/100 kg soil) = 2.4 cmol_c/kg soil

(mol_c/100 cmol_c)(mol CaSO₄/2 mol_c)( 136 g CaSO₄/mol CaSO₄)
= 1.6 g CaSO₄/kg soil

(kg soil/1000 g soil)( 454 g soil/lb soil)( lb CaSO₄/454 g CaSO₄)
(4 x 10⁶ lb soil/A ft)(T CaSO₄/2000 lb CaSO₄)
= 3.2 T CaSO₄/A ft

back to contents