Soil Salinity Saline Soil Solution Cation Exchange and Colloidal Phenomena Mineral Weathering Boron Chemistry Irrigation Water Quality

Saline Soil Solution Na +, K +, Ca 2+ and Mg 2+ CO 3 2-, SO 4 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 = log κ I for κ = 4 dS m -1 is 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. Φ T = Φ M + Φ O

So with Φ O very -, Φ M cannot be as negative and plant extract water from soil. Permanent wilt at -15 bar, 0 osmotic decrease in Φ T

Alkalinity = [HCO 3 - ] + 2[CO 3 2- ] + [H 2 PO 4 - ] + 2[HPO 4 2- ] + 3[PO 4 3- ] + [B(OH) 4 - ] + [L - ] + [OH - ] – [H + ] which is mostly from bicarbonate so can determine pH based on it, pH = log(HCO 3 - ) – log P CO2 pH = log γ - + log [HCO 3 - ] – log P CO2

Cation Exchange and Colloidal Phenomena Focusing on Na +, Ca 2+ and Mg 2+ consider this ternary system and the three binary exchange reactions involved, 2Na + ads + Ca 2+ = Ca 2+ ads + 2Na +V K Na Ca = N Ca (Na + ) 2 / (N Na ) 2 (Ca 2+ ) 2Na + ads + Mg 2+ = Mg 2+ ads + 2Na +V K Na Mg = Mg 2+ ads + Ca 2+ = Ca 2+ ads + Mg 2+V K Mg Ca = Most data on the Ca 2+ / Na + exchange, say that V K Na Ca ~ 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, V K Na Ca = Γ ([Na + ] 2 / [Ca 2+ ]) x (1 - E Na 2 ) / 4E Na 2 γ Na 2 / γ Ca

V K Na Ca = Γ ([Na + ] 2 / [Ca 2+ ]) x (1 - E Na 2 ) / 4E Na 2 Recall Sodium Adsorption Ratio defined = [Na + ] / ([Ca 2+ ] + [Mg 2+ ]) 1/2 with concentration units mM. Thus, if only Na + and Ca 2+ considered, SAR = 10 -3/2 ([Na + ] 2 / [Ca 2+ ]) 1/2 Also, Exchangeable Sodium Percentage defined = cmol(+) Na x 100 / CEC ESP = 100 E Na V K Na Ca = 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.

SAR = 13 ESP = 15 EC = 4 Sodic Saline and Sodic Not Affected Saline

V K Na Ca = 2.5 Γ (SAR / ESP) 2 [1 – (ESP / 100) 2 ] For V K Na Ca = 2.2 and Γ at I = 0.02 m ESP is approximately linear in SAR. Value for I is typical as is that for V K Na Ca. Also, since V K Mg Ca ~ 1, and using the strict definition of SAR as ([Na + ] 2 / [Ca 2+ + Mg 2+ ]) 1/2, ESP ≈ kSAR

If ignore complexes of Ca 2+ and Mg 2+, write the approximation SAR P = [Na + ] T / ([Ca 2+ ] T + [Mg 2+ ] T ) 1/2 Which is smaller than SAR by about 12 % due to complexes but given slope k slightly > 1 in ESP = kSAR, ESP ~ SAR P Recall effect of E Na on smectite coagulation (Ca 2+ and Na + system), Generally consistent with dispersion at ESP = 15 %. Generally true and though ESP ~ SAR P, onset of dispersion as set by SAR P is SAR P = 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 Concentration is positively related to EC. Ranges in EC-SAR P show that definition of sodic and saline-sodic soils do not exactly specify whether dispersion occurs. Not sodic but possibly dispersed at low solution concentration. Sodic but possibly not dispersed due to higher solution concentration but not saline concentration.

Mineral Weathering Source of Na +, K +, Ca 2+, Mg 2+, Cl -, SO 4 2- 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. A a B b = aA b+ + bB a- K so = (A b+ ) a (B a+ ) b = γ b+ a [A b+ ] a γ a- b [B a- ] b and since log γ i = (i) 2 {I 1/2 / (1 + I 1/2 ) - 0.3I), [A b+ ] and [B a- ] increase γ for monovalent at increasing I

Another important aspect of dissolution in arid region soils is behavior of calcite, CaCO 3. For small displacement from equilibrium, the kinetics of dissolution or precipitation can be described by, d[Ca 2+ ] / dt = k [K so – (Ca 2+ )(CO 3 2- )] where K so = (Ca 2+ )(CO 3 2- ) and k depends on pH and surface area. Recalling Ω = (Ca 2+ )(CO 3 2- ) / K so, d[Ca 2+ ] / dt = k K so [1 – Ω] Can introduce pH effect into this expression besides that implicit in k by CaCO 3 + H + = Ca 2+ + HCO 3 - K = d[Ca 2+ ] / dt = k [K so – (Ca 2+ )(HCO 3 - ) / K (H + )]

Define 10 -pHs = (Ca 2+ )(HCO 3 - ) / K K so to give d[Ca 2+ ] / dt = kK so [1 – 10 pH-pHs ]where pH – pHs is Langelier Index So, is d[Ca 2+ ] / dt = + / -, i.e., is dissolution / precipitation occurring based on Index value? If use data from Table 12.1 for (Ca 2+ ), (HCO 3 - ) and pH = 7.02, pH – pHs = 0.58 and [1 – 10 pH-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 HCO 3 - (CO 2 ) from respiration.

Boron Chemistry log (H 3 BO 3 ) = log K + 1/3 log (H + ) – 1/6 log (Ca 2+ pH = 7.6 and (Ca 2+ ) = = = -0.82; (H 3 BO 3 ) = 0.151

Value M >> M (Table 12.1). Therefore, B release by dissolution must be slow and solution concentration controlled by adsorption. While H 3 BO 3 does not dissociate, it hydrolyzes H 3 BO 3 + H 2 O = B(OH) H + K = 5.8 x But (B(OH) 4 - ) = (H 2 BO 3 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 + H 3 BO 3 = SOB(OH) 2 + H 2 O

Irrigation Water Quality Salinity, sodicity and toxicity are concerns. Sensitive Tolerant Correspond to I = 0.010m I = 0.044m What is relationship between EC water and EC soil ? Depends on leaching fraction, LF

Leaching fraction, LF LF = depth of water leached below root zone / depth of applied water What is EC drainage for set EC water and LF? EC drainage = EC water / LF From EC drainage x depth of water leached = EC water x depth of applied water. If EC water = 1 dS m -1 and LF = 0.15, EC drainage = 6.7 dS m -1 However, average salinity in soil not the same as EC drainage. Determine average EC soil 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 EC drainage at soil surface, …, and at bottom of root zone. Average these values to estimate EC soil. 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, EC drainage1 = EC water / LF = EC water. If EC water = 1.0 dS m -1, EC soil = 1.0 dS m -1. At bottom of first ¼ depth segment, LF = 1176 mm – 0.4 (1000 mm) / 1176 mm = 0.66 EC drainage2 = 1.0 dS m -1 / 0.66 = 1.5 dS m -1.

At bottom of second depth segment, LF = [1176 mm – ( ) 1000 mm] / 1176 mm = 0.40 EC drainage3 = 1.0 dS m -1 / 0.40 = 2.5 dS m -1. Proceed the same way to get, EC soil = (EC drainage1 + … EC drainage5 ) / 5 = 3.2 dS m -1. Thus, for a LF = 0.15, EC soil = 3.2 x EC water. Relationship, EC soil = X(LF) EC water depends on distribution of ET with depth.

Using the empirical relationship, EC soil = X(LF) EC water and Table 12.5, for LF = 0.3, EC soil = EC water for LF > 0.3, EC soil < EC water for LF EC water EC water < 4 dS m -1 goes to saline soil EC water > 4 dS m -1 does not go to saline soil

@ low SAR water soil may become sodic if EC water is low. Even at high SAR water soil may not become sodic if EC water is sufficiently high.

May also consider effect of [Ca 2+ ] and [HCO 3 - ] in irrigation water on tendency for CaCO 3 in soil to dissolve or precipitate. Can use an adjusted SAR, adjRNa = [Na + ] / [Ca 2+ eq + Mg 2+ ] 1/2 in V K Na Ca = 2.5 Γ (SAR / ESP) 2 [1 – (ESP / 100) 2 ] to calculate ESP, however [Ca 2+ eq ] must be calculated. Value in soil when (HCO 3 - ) / (Ca 2+ ) is the same as in the irrigation water and there is calcite equilibrium at the P CO2 of the soil.

γ 2+ [Ca 2+ eq ] = { K so / [(HCO 3 - ) water (Ca 2+ ) water ] 2 } 1/3 P CO2 1/3 Guidelines for toxicity with respect to Na, Cl and B.