Husein Ajwa, Emeritus Department of Plant Sciences UC Davis

Husein Ajwa, Emeritus Department of Plant Sciences UC Davis

HUSEIN AJWA Plant Sciences Department, UC Davis 1636 East Alisal St., Salinas, CA 93905 Office: (831) 755-2823, Fax: (831) 755-2898 email: haajwa@ucdavis.edu EDUCATION Ph.D., Soil Science/Soil Chemistry, Iowa State University (1993 ) M.S., Soil Science, University of Wisconsin, WI (1989) EMPLOYMENT Extension Soil Specialist, University of California, Davis (2001 2014, Emeritus) Soil Scientist/Chemist, USDA-ARS, Fresno (1995 to 2001)

Topics Chemical equilibrium Definitions and Basic Examples Fertilizers and Plant Nutrition Phosphorus Chemistry and Equilibrium Reactions Stability diagrams

Chemical Equilibrium

Chemical Equilibrium Equilibrium Constant The relative concentration of products and reactants at equilibrium is a constant. Equilibrium constant (K): For a general chemical reaction Equilibrium constant: Where: K [ C ] [ A] c a [ D] [ B] - small superscript letters are the stoichiometry coefficients - [A] concentration chemical species A relative to standard state d b

Interpretation of the Equilibrium Constant K [Products] [Reactants For K>1, we can state that the amount of product is greater than the amount of reactant. The equilibrium is favored to the right. when k>1, then Log K = positive value For K<1, the amount of reactants is greater than the products. The equilibrium is favored to the left. when K<1, then Log K = negative value ]

Because water is amphiprotic, one water molecule can react with another to form an OH ion and an H 3 O + ion in an autoionization process: 2H 2 O (l) H 3 O + (aq) + OH (aq) Equilibrium constant K for pure water can be written as: [H 3 O + ] [OH ] K c = [H 2 O] 2 [H 3 O + ][OH ] = 10 14 M = K w [H 3 O + ]=[OH ] = 10 7 M log [H 3 O + ] = Log K H = -7 ph = -log [H 3 O + ] = 7

Chemical Equilibrium Equilibrium Constant Example: Given the reactions and equilibrium constants: K w = 1.0 x 10-14 K NH3 = 1.8 x 10-5 Find the equilibrium constant for the reaction: Solution: K 1 = K w K 2 =1/K NH3 K 3 =K w *1/K NH3 =5.6x10-10

Solubility Equilibrium Precipitation NaCl(s) Na + (aq) + Cl - (aq); K sp = [Na + ][Cl - ] (K sp is called solubility product) PbCl 2 precipitate because the ion product is greater than K sp

Chemical Equilibrium Complex Formation 1.) High concentration of an ion may redissolve a solid Ion first causes precipitation Forms complex ions, consists of two or more simple ions bonded to each other ppt. formation Complex forms and redissolves solid

Crop Production Fertilizers

Phosphorus Chemistry Unlike N, P does not readily change oxidation states Unlike C, N, and S, no major gaseous forms Most common inorganic forms: HPO 2 4 and H 2 PO 1 4 Phosphate Pyrophosphate

Chemical Reactions

Phosphorus in Plant Nutrition Essential to plants and animals ATP DNA, RNA Phospholipid membranes Critical to energy flow in cells

Phosphorus and Soil Fertility Most of the P in soil is not available to plants Soluble P added to soils can be fixed by reaction with Fe and Al in acid soils and Ca and Mg in alkaline soils. Example: Fe 3+ (soln) + PO 4 3- (soln) FePO 4 iron phosphate precipitate This reaction is extremely favorable at low ph.

The Phosphorus Cycle P Component Input to soil Loss from soil Animal manures and biosolids Plant residues Crop harvest Fertilizers Nonlabile P Organic phosphorus Microbial Plant residue Humus Immobilization Mineralization Plant uptake Soil solution phosphorus HPO -2 4 Leaching H 2 PO -1 4 (usually minor) Weathering Dissolution Precipitation Nonlabile P Secondary minerals (CaP, FeP, MnP, AlP) Runoff and erosion Nonlabile P Primary minerals (apatite) Labile P (Active) Adsorbed on mineral surfaces (clays, Fe & Al oxides, carbonates) Adsorption Equilibrium Desorption

Chemical Equilibria in Soil Science and Crop Production Example: Phosphorus

Phosphoric acid, H 3 PO 4 Polyprotic acid: contains more than one ionizable proton H 3 PO 4 (aq) + H 2 O(l) H 2 PO 4 - (aq) + H3 O + (aq) K a1 = [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] = 7.2 x 10-3 Log K=-2.15 H 2 PO 4 - (aq) + H2 O(l) HPO 4 2- (aq) + H3 O + (aq) K a2 = [H 3 O+ ] [HPO 4 2- ] [H 2 PO 4 - ] = 6.3 x 10-8 Log K=-7.2 HPO 4 2- (aq) + H2 O(l) PO 4 3- (aq) + H3 O + (aq) K a3 = [H 3 O + ] [PO 4 3- ] [HPO 4 2- ] = 4.2 x 10-13 Log K=-12.35 K a1 > K a2 > K a3

Figure was adapted from Lindasy (Chemical Equilibria and Reaction Models, 1979) Effect of Ca, Mg, and Fe on P speciation in water at various ph values

Phosphorus in soil solution is governed by several processes of interactions with the soil solid phase (adsorption/desorption or precipitation/dissolution), all of which are controlled by major factors: Soil ph Concentrations of cations such as Ca, Fe, and Al Concentrations of competing inorganic anions (especially bicarbonate) and organic ligands (such as carboxylic anions).

Calcium Phosphates Equilibrium expression for dissolution reaction of Ca-P minerals Ca(H 2 PO 4 ) 2. H 2 O Ca 2+ + 2H 2 PO 4 - + H2 O Log K = -1.15 Mono calcium phosphate, K sp ~10-3 CaHPO 4. 2H 2 O + H + Ca 2+ + H 2 PO 4 - + 2H2 O Log K = 0.63 Brushite K sp ~10-7 + acid Ca 4 H(PO 4 ) 3. 2.5H 2 O + 5H + 4Ca 2+ + 3H 2 PO 4 - + 2.5H2 O Log K = 11.76 Octacalcium phosphate Ca 5 (PO 4 ) 3 OH + 7H + 5Ca 2+ + 3H 2 PO 4 - + H2 O Log K = 14.46 Hydroxyapatite ~10-30 Ca 5 (PO 4 ) 3 F + 6H + 5Ca 2+ + 3H 2 PO 4 - + F - Log K = -0.21 Fluoroapatite ~10-59

Calcium Phosphates Precipitation Neutral/alkaline solution

Effect of Al and Fe on P availability in Soil Acidic solution

Common P Minerals found in Soils (listed in order of decreasing solubility) Acidic Soils Variscite - AlPO 4.2H 2 O Strengite - FePO 4.2H 2 O Neutral and Calcareous Soils Dicalcium phosphate dihydrate (DCPD): CaHPO 4.2H 2 O Dicalcium phosphate (DCP): CaHPO 4 x Octacalcium phosphate (OCP): Ca 4 H(PO 4 ) 3.2.5H2O x Tricalcium phosphate (TCP): Ca 3 (PO 4 ) 2 x Hydroxyapatite (HA): Ca 5 (PO4) 3 OH x Fluorapatite (FA): Ca 5 (PO 4 ) 3 F (least soluble)

Calcium Phosphates Ca(H 2 PO 4 ) 2. H 2 O Ca 2+ + 2H 2 PO 4 - + H2 O Log K = -1.15 Mono calcium phosphate CaHPO 4. 2H 2 O + H + Ca 2+ + H 2 PO 4 - + 2H2 O Log K = 0.63 Brushite Ca 4 H(PO 4 ) 3. 2.5H 2 O + 5H + 4Ca 2+ + 3H 2 PO 4 - + 2.5H2 O Log K = 11.76 Octacalcium phosphate Ca 5 (PO 4 ) 3 OH + 7H + 5Ca 2+ + 3H 2 PO 4 - + H2 O Log K = 14.46 Hydroxyapatite Ca 5 (PO 4 ) 3 F + 6H + 5Ca 2+ + 3H 2 PO 4 - + F - Log K = -0.21 Fluoroapatite

Figure was adapted from Lindasy (Chemical Equilibria and Reaction Models, 1979)

Calcium Phosphates Ca(H 2 PO 4 ) 2. H 2 O Ca 2+ + 2H 2 PO 4 - + H2 O Log K = -1.15 Mono calcium phosphate CaHPO 4. 2H 2 O + H + Ca 2+ + H 2 PO 4 - + 2H2 O Log K = 0.63 Brushite Ca 4 H(PO 4 ) 3. 2.5H 2 O + 5H + 4Ca 2+ + 3H 2 PO 4 - + 2.5H2 O Log K = 11.76 Octacalcium phosphate Ca 5 (PO 4 ) 3 OH + 7H + 5Ca 2+ + 3H 2 PO 4 - + H2 O Log K = 14.46 Hydroxyapatite Ca 5 (PO 4 ) 3 F + 6H + 5Ca 2+ + 3H 2 PO 4 - + F - Log K = -0.21 Fluoroapatite

A B ph 4.5 Event Precipitate as: A. add fertilizer soluble P added B. From A to B soluble P decreases DCP C. From B to C DCP dissolves FA D. From C to D FA dissolves Variscite C D Figure was adapted from Lindasy (Chemical Equilibria and Reaction Models, 1979)

Example

Example Figure was adapted from Lindasy (Chemical Equilibria and Reaction Models, 1979) Solubility of Ca, Fe, and Al in soil

Example 50 lbs P fertilizer/ac 200 lbs P fertilizer/ac ph + ph 2 PO 4 - DCPD 1 hr 1 hr 5 d 20 d 5 d 20 d 40 d 40 d ph ½pCa

Olsen P test: closed vs open containers

Thank you very much!