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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-, SO42- 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 that
they could not thrive, even survive at this concentration. The main problem
is extraction of water from the soil.
ΦT = ΦM + ΦO
Relationship between matric potential and
water content --soil moisture characteristic
ΦT = Φ M + ΦO
So with ΦO very -, ΦM cannot
be as negative and plant extract
water from soil.
Permanent wilt at
-15 bar, 0 osmotic
decrease in ΦT
Matric Potential
Tolerance to varies with species
Alkalinity = [HCO3-] + 2[CO32-] + [H2PO4-] + 2[HPO42-] + 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 the
three binary exchange reactions involved,
2Na+ads + Ca2+ = Ca2+ads + 2Na+
VK Ca
Na
= NCa (Na+)2 / (NNa)2(Ca2+)
2Na+ads + Mg2+ = Mg2+ads + 2Na+
VK Mg
Na
=
Mg2+ads + Ca2+ = Ca2+ads + Mg2+
VK Ca
Mg
=
Most data on the Ca2+ / Na+ exchange, say that VKNaCa ~ 2.2 ≠ 1
Rewriting the selectivity coefficient in terms of solution phase concentrations
and expressing mole fractions on the exchanger in terms of equivalent
charge fractions gives,
VK Ca
Na
= Γ ([Na+]2 / [Ca2+]) x (1 - ENa2) / 4ENa2
γNa2 / γCa
VK Ca
Na
= Γ ([Na+]2 / [Ca2+]) x (1 - ENa2) / 4ENa2
Recall Sodium Adsorption Ratio defined = [Na+] / ([Ca2+] + [Mg2+])1/2 with
concentration 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
VK Ca
Na
= 2.5 Γ (SAR / ESP)2 [1 – (ESP / 100)2]
Also, recall significance of ESP > 15 % as defining a sodic soil if
the EC < 4 dS m-1.
Otherwise, if EC > 4 dS m-1, it is a saline-sodic soil.
Sodic
Not Affected
Saline and Sodic
Saline
SAR = 13
ESP = 15
EC = 4
VK Ca
Na
= 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 typical
as is that for VKNaCa.
Also, since VKMgCa ~ 1,
and using the strict
definition 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 given
slope 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 substantially
reduced hydraulic conductivity at ESP > 15 % and SAR > 13 is the basis
for defining sodic soils.
However, the dispersive effects occurs only when total salinity is sufficiently
low, i.e., below the ccc of the colloid.
Generally consistent.
Increasing salinity offsets effect of Na+
on disintegration of soil structure.
Effect of solution
concentration on
structural stability
shown in Fig. 12.3.
Concentration is
positively related to
EC.
Not sodic but possibly
dispersed at low
solution concentration.
Ranges in EC-SARP
show that definition of
sodic and saline-sodic
soils do not exactly
specify whether dispersion
occurs.
Sodic but possibly
not dispersed due to
higher solution concentration
but 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 infiltration
and percolation of rainwater (low EC) would have on tendency of
colloids 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
0.0001
0.001
0.01
0.1
-0.00161
-0.00505
-0.01554
-0.04501
-0.10765
1.00
0.99
0.96
0.90
0.78
γ for monovalent at increasing I
Another important aspect of dissolution in arid region soils is
behavior of calcite, CaCO3.
For small displacement from equilibrium, the kinetics of dissolution
or 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, instead
the 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 thought
due to ligand exchange
effectively involving loss of
H+,
SOH + H3BO3 =
SOB(OH)2 + H2O
Irrigation Water Quality
Salinity, sodicity and toxicity are concerns.
Sensitive Tolerant
Correspond to
I = 0.010m
I = 0.044m
What is relationship
between ECwater and
ECsoil?
Depends on
leaching 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 series
of soil depths down to the bottom of the root zone. This approach requires
an 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 the
bottom 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 with
depth.
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 soil
ECwater > 4 dS m-1
does not go to saline soil
@ low SARwater
soil may become
sodic if ECwater is
low.
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
VK
Na
Ca
= 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 calcite
equilibrium 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.