Soil Salinity

Saline Soil Solution

Cation Exchange and Colloidal Phenomena

Mineral Weathering

Boron Chemistry

Irrigation Water Quality

Saline Soil Solution

Na+, K+, Ca2+ and Mg2+

CO32-, SO4

2- and Cl- dominant species

Soils of arid regions most likely to be considered saline

Defined as having an EC > 4 dS m-1

Thus, from Marion-Babcock, log I = 1.159 + 1.009 log κ

I for κ = 4 dS m-1 is 0.059 molal

However, many plants (crops) are sufficiently sensitive to salinity thatthey could not thrive, even survive at this concentration. The main problemis extraction of water from the soil.

ΦT = ΦM + ΦO

ΦT = ΦM + ΦO

So with ΦO very -, ΦM cannotbe as negative and plant extractwater from soil.

Permanent wilt at -15 bar, 0 osmoticdecrease in ΦT

Tolerance to varies with species

Alkalinity = [HCO3-] + 2[CO3

2-] + [H2PO4-] + 2[HPO4

2-] + 3[PO43-] +

[B(OH)4-] + [L-] + [OH-] – [H+]

which is mostly from bicarbonate so can determine pH based on it,

pH = 7.8 + log(HCO3-) – log PCO2

pH = 7.8 + log γ- + log [HCO3-] – log PCO2

Cation Exchange and Colloidal Phenomena

Focusing on Na+, Ca2+ and Mg2+ consider this ternary system and thethree binary exchange reactions involved,

2Na+ads + Ca2+ = Ca2+

ads + 2Na+ VKNaCa = NCa (Na+)2 / (NNa)2(Ca2+)

2Na+ads + Mg2+ = Mg2+

ads + 2Na+ VKNaMg =

Mg2+ads + Ca2+ = Ca2+

ads + Mg2+ VKMgCa =

Most data on the Ca2+ / Na+ exchange, say that VKNaCa ~ 2.2 ≠ 1

Rewriting the selectivity coefficient in terms of solution phase concentrationsand expressing mole fractions on the exchanger in terms of equivalentcharge fractions gives,

VKNaCa = Γ ([Na+]2 / [Ca2+]) x (1 - ENa

2) / 4ENa2

γNa2 / γCa

VKNaCa = Γ ([Na+]2 / [Ca2+]) x (1 - ENa

2) / 4ENa2

Recall Sodium Adsorption Ratio defined = [Na+] / ([Ca2+] + [Mg2+])1/2 withconcentration units mM.

Thus, if only Na+ and Ca2+ considered, SAR = 10-3/2 ([Na+]2 / [Ca2+])1/2

Also, Exchangeable Sodium Percentage defined = cmol(+)Na x 100 / CEC

ESP = 100 ENa

VKNaCa = 2.5 Γ (SAR / ESP)2 [1 – (ESP / 100)2]

Also, recall significance of ESP > 15 % as defining a sodic soil ifthe EC < 4 dS m-1.

Otherwise, if EC > 4 dS m-1, it is a saline-sodic soil.

SAR = 13ESP = 15

EC = 4

Sodic Saline and Sodic

Not Affected Saline

VKNaCa = 2.5 Γ (SAR / ESP)2 [1 – (ESP / 100)2]

For VKNaCa = 2.2 and Γ at I = 0.02 m ESP is approximately linear in SAR.

Value for I is typicalas is that for VKNa

Ca.

Also, since VKMgCa ~ 1,

and using the strictdefinition of SAR as([Na+]2 / [Ca2+ + Mg2+])1/2,

ESP ≈ kSAR

If ignore complexes of Ca2+ and Mg2+, write the approximation

SARP = [Na+]T / ([Ca2+]T + [Mg2+]T)1/2

Which is smaller than SAR by about 12 % due to complexes but givenslope k slightly > 1 in ESP = kSAR,

ESP ~ SARP

Recall effect of ENa on smectite coagulation (Ca2+ and Na+ system),

Generally consistent with dispersion at ESP = 15 %.

Generally true and though ESP ~ SARP, onset of dispersion as set by SARP is SARP = 13.

Effect of Na+ on dispersion, clogging of pores and swelling to substantiallyreduced hydraulic conductivity at ESP > 15 % and SAR > 13 is the basisfor defining sodic soils.

However, the dispersive effects occurs only when total salinity is sufficientlylow, i.e., below the ccc of the colloid.

Generally consistent.

Increasing salinity offsets effect of Na+

on disintegration of soil structure.

Effect of solutionconcentration onstructural stabilityshown in Fig. 12.3.

Concentration ispositively related to EC.

Ranges in EC-SARP

show that definition ofsodic and saline-sodicsoils do not exactlyspecify whether dispersionoccurs.

Not sodic but possibly dispersed at lowsolution concentration.

Sodic but possiblynot dispersed due tohigher solution concentrationbut not saline concentration.

Mineral Weathering

Source of Na+, K+, Ca2+, Mg2+, Cl-, SO42- in arid region soils is

mineral dissolution. This tends to counteract the effect that infiltrationand percolation of rainwater (low EC) would have on tendency ofcolloids to disperse.

Interestingly, high salinity increases the solubility of minerals.

AaBb = aAb+ + bBa-

Kso = (Ab+)a(Ba+)b = γb+a[Ab+]a γa-

b[Ba-]b

and since log γi = -0.512 (i)2 {I1/2 / (1 + I1/2) - 0.3I), [Ab+] and [Ba-] increase

0.00001 -0.00161 1.000.0001 -0.00505 0.99

0.001 -0.01554 0.960.01 -0.04501 0.90

0.1 -0.10765 0.78

γ for monovalent at increasing I

Another important aspect of dissolution in arid region soils isbehavior of calcite, CaCO3.

For small displacement from equilibrium, the kinetics of dissolutionor precipitation can be described by,

d[Ca2+] / dt = k [Kso – (Ca2+)(CO32-)]

where Kso = (Ca2+)(CO32-) and k depends on pH and surface area.

Recalling Ω = (Ca2+)(CO32-) / Kso,

d[Ca2+] / dt = k Kso [1 – Ω]

Can introduce pH effect into this expression besides that implicit in k by

CaCO3 + H+ = Ca2+ + HCO3- K = 1010.3

d[Ca2+] / dt = k [Kso – (Ca2+)(HCO3-) / K (H+)]

d[Ca2+] / dt = k [Kso – (Ca2+)(HCO3-) / K (H+)]

Define 10-pHs = (Ca2+)(HCO3-) / K Kso

to give

d[Ca2+] / dt = kKso [1 – 10pH-pHs] where pH – pHs is Langelier Index

So, is d[Ca2+] / dt = + / -, i.e., is dissolution / precipitation occurring based on Index value?

If use data from Table 12.1 for (Ca2+), (HCO3-) and pH = 7.02,

pH – pHs = 0.58 and [1 – 10pH-pHs] < 0 so calcite is precipitating.

Alternatively, Ω > 1.

Interestingly, this is typically the case. Equilibrium does not exist, insteadthe soil solution is supersaturated with respect to calcite.

Why?

Rate of precipitation is slowed by surface adsorption of organics.

Another possibility is constant input of HCO3- (CO2) from respiration.

Boron Chemistry

log (H3BO3) = log K + 1/3 log (H+) – 1/6 log (Ca2+); @ pH = 7.6 and (Ca2+) = 0.0027

= 1.28 - 2.53 + 0.43 = -0.82; (H3BO3) = 0.151

Value 0.151 M >> 0.000038 M (Table 12.1).

Therefore, B release by dissolution must be slow and solution concentration controlled by adsorption.

While H3BO3 does not dissociate, it hydrolyzes

H3BO3 + H2O = B(OH)4- + H+ K = 5.8 x 10-10

But (B(OH)4-) = (H2BO3) @ pH = 9.23 and does not contribute much to (B)T

Nevertheless, adsorption envelope behavior thoughtdue to ligand exchangeeffectively involving loss ofH+,

SOH + H3BO3 =

SOB(OH)2 + H2O

Irrigation Water Quality

Salinity, sodicity and toxicity are concerns.

Sensitive Tolerant

Correspond to

I = 0.010mI = 0.044m

What is relationshipbetween ECwater andECsoil?

Depends onleaching fraction, LF

Leaching fraction, LF

LF = depth of water leached below root zone / depth of applied water

What is ECdrainage for set ECwater and LF?

ECdrainage = ECwater / LF

From ECdrainage x depth of water leached = ECwater x depth of applied water.

If ECwater = 1 dS m-1 and LF = 0.15, ECdrainage = 6.7 dS m-1

However, average salinity in soil not the same as ECdrainage.

Determine average ECsoil by applying this procedure stepwise to a seriesof soil depths down to the bottom of the root zone. This approach requiresan assumption on how much soil water is used in ET in each depth segment.

Assume 40% of the water in the upper ¼ depth segment meets the crop ET,30% comes from the second ¼, 20% from the third ¼ and 10% from thebottom ¼.

Determine ECdrainage at soil surface, …, and at bottom of root zone.Average these values to estimate ECsoil.

For LF = 0.15 and annual ET = 1000 mm,

Applied water = 1000 mm / (1 – LF) = 1176 mm

Assume that at soil surface, all applied water leaches below the surface.This gives LF = 1.0. Therefore,

ECdrainage1 = ECwater / LF = ECwater. If ECwater = 1.0 dS m-1, ECsoil = 1.0 dS m-1.

At bottom of first ¼ depth segment,

LF = 1176 mm – 0.4 (1000 mm) / 1176 mm = 0.66

ECdrainage2 = 1.0 dS m-1 / 0.66 = 1.5 dS m-1.

At bottom of second depth segment,

LF = [1176 mm – (0.4 + 0.3) 1000 mm] / 1176 mm = 0.40

ECdrainage3 = 1.0 dS m-1 / 0.40 = 2.5 dS m-1.

Proceed the same way to get,

ECsoil = (ECdrainage1 + … ECdrainage5) / 5 = 3.2 dS m-1.

Thus, for a LF = 0.15, ECsoil = 3.2 x ECwater.

Relationship, ECsoil = X(LF) ECwater depends on distribution of ET withdepth.

Using the empirical relationship,

ECsoil = X(LF) ECwater

and Table 12.5,

for LF = 0.3, ECsoil = ECwater

for LF > 0.3, ECsoil < ECwater

for LF < 0.3, ECsoil > ECwater

ECwater < 4 dS m-1

goes to saline soilECwater > 4 dS m-1

does not go to saline soil

@ low SARwater

soil may becomesodic if ECwater islow.

Even at high SARwater

soil may not become sodic if ECwater is sufficiently high.

May also consider effect of [Ca2+] and [HCO3-] in irrigation water on

tendency for CaCO3 in soil to dissolve or precipitate.

Can use an adjusted SAR,

adjRNa = [Na+] / [Ca2+eq + Mg2+]1/2

in VKNaCa = 2.5 Γ (SAR / ESP)2 [1 – (ESP / 100)2]

to calculate ESP, however [Ca2+eq] must be calculated.

Value in soil when (HCO3-) / (Ca2+) is the same

as in the irrigation water and there is calciteequilibrium at the PCO2 of the soil.

γ2+[Ca2+eq] = {102.5Kso / [(HCO3

-)water(Ca2+)water]2}1/3 PCO21/3

Guidelines for toxicity with respect to Na, Cl and B.