1. Soil Reaction Chapter 9

3. Here are some relations and terms you need: H 2 O = H + + OH - Water dissociates as above and the Equilibrium constant for it is K w = [H + ][OH - ] = 10 -14 So, log[H + ] + log[OH - ] = -14 -log[H + ] – log[OH - ] =14 pH + pOH = 14 Keep in mind, low pH means high [H + ] and visa versa.

5. Remember why Al 3+, Fe 3+, etc. are acidic, they hydrolyze, Al 3+ + H 2 O = AlOH 2+ + H +, AlOH 2+ + H 2 O = Al(OH) 2 + + H +, Al(OH) 2 + + H 2 O = Al(OH) 3 + H +

6. Sources of H + and OH - Moderately acid soil 5 < pH < 6.5 Exchangeable H + and Al(OH) x (3-x)+ Higher %BS so fewer acid cations

8. Sources of H + and OH - Neutral to alkaline soils pH > 6.5 High %BS so few acidic cations

9. Negligible exchangeable H + and Al(OH) x (3-x)+ but some acids remain that are strongly adsorbed.

10. Sources of H + and OH - Carbonate and bicarbonate salts are present at alkaline pH CaCO 3 → Ca 2+ + CO 3 2- CO 3 2- + H 2 O → H 2 CO 3 + 2OH -

12. Pools of Acidity

13. Residual is adsorbed H + and Al(OH) x (3-x)+ that is unextractable Can be determined by extraction at alkaline pH █2H + + Ba 2+ → █Ba 2+ + 2H + 2H + + 2OH - → 2H 2 O

15. Buffering Well, H + is a very reactive chemical species and it reacts, consuming it. Thus, the concentration of H + is less than would otherwise be the case. Sure, adding acid at a concentration greater than exists in the soil solution increases the concentration of H + in the soil solution but not nearly a much as would be the case without these various reactions. See next slide.

16. Buffering Al(OH) 3 + 3H +  Al 3+ + 3H 2 O

17. Buffering

18. CO 3 2- + 2H +  H 2 CO 3 H 2 CO 3  H 2 O + CO 2 ↑ So what happens? It depends on how much acid you add and how much base is present. If you add more acid than base, then the base is consumed and you make the soil acidic. Usually, you are interested in adding base to an acid soil but in some cases you may wish to decrease the basicity.

19. Buffering

21. Consider the matter of raising the pH by adding a base, like CaCO 3, calcite. Two things are going on: 1) reaction of the base, CO 3 2-, with H +, consuming the latter and producing H 2 O and CO 2 ; and 2) Ca 2+ replaces acidic cations that are adsorbed on soil colloids, forcing them into solution where they react with CO 3 2- or HCO 3 - or OH - (the latter two produced from CO 3 2- reaction with water). The soil solution pH increases ([H + ] decreases) and the fraction of the CEC made up of acidic cations decreases, i.e., the %BS increases.

23. Two soils have the same shape buffer curve, they have the same pH (= 5.5) but the CEC of soil A is 10 cmole(+) / kg and the CEC of soil B is 20 cmole(+) / kg. Which soil would require more CaCO 3 (lime) to raise its pH to 6.5? How much more?

24. ∆ % BS = (90 % – 50 %) = 40 % 40 % of 10 cmol (+) / kg = 4 cmol (+) Ca / kg 40 % of 20 cmol (+) / kg = 8 cmol (+) Ca / kg

26. Why pH Changes Respiration in the soil produces, CO 2, which reacts with water to form H 2 CO 3, which dissociates giving H +. Carbonic acid is weak but it’s more or less constantly produced. Further, when microbes decompose organic matter, the N and S that it contains is effectively released as nitric and sulfuric acids, HNO 3 and H 2 SO 4, which are strong acids. Various organic acids (-COOH groups) are also released. Of course, these acids yield H + to the soil solution.

27. Why pH Changes And this is what happens due to this natural production of H + in the soil --

28. Why pH Changes Acidifying fertilizers Microbial oxidation of NH 4 + NH 4 + + 2O 2 → NO 3 - + 2H + + H 2 O Fertilizers that contain ammonium or from which ammonium is derived (like, NH 3 + H + = NH 4 + or urea, (H 2 N) 2 CO + H 2 O = 2NH 3 + CO 2 ), are oxidized by certain soil microorganisms, generating H+. You can count on the above reaction to occur, and can figure how much acid is generated based on how much ammonium-containing fertilizer is added.

29. Removal of a lot of biomass Let’s say 10 Mg (= 10000 kg) of biomass is removed from 1 ha and it contains 1 % Ca. Therefore, 100 kg Ca removed. 100 kg = 100000 gAW Ca = 40 g / mole So, 100000 g / (40 g / mole) = 2500 mole Ca removed Or 2500 x (100 cmole / mole) x 2 = 500,000 cmol (+) removed 1 HFS = 2000 Mg or 2,000,000 kg 500,000 cmole (+) / 2,000,000 kg = 0.25 cmol (+) / kg

30. The previous calculation may not look like much, 0.25 cmol(+) kg -1, however, figure that other base cations are removed in harvested biomass and sum that over several harvests –the effect can be substantial, particularly if the CEC is low, i.e., the effect on decreasing %BS is more pronounced.

31. Why pH Changes The phenomenon is like natural acidification but by adding organic matter respiration and decomposition of organic matter are substantially increased.

32. Why pH Changes

34. CaNa ↓ █ ↓ Ca Na HCa Na H Ca-saturatedNa-saturated

35. CaNa ↓ █ ↓ Ca Na HCa Na H Ca-saturatedNa-saturated % BS and pH increase in both cases but If dominated by Ca, there will also exist a CaCO 3 phase which controls pH at ≈ 8.4 Na 2 CO 3 is soluble so there is no pH control Unlike all the foregoing, this effect raises the pH.

36. The below data are for an acid sandy loam: Base Cations Acid Cations ----------mmol (+) / 100 g or cmol (+) / kg ---------- Ca Mg K Na Al H 2.0 0.6 0.3 0.1 6.4 0.6 CEC = ? and % BS = ? CEC = sum of all cationic charge, 10.0 cmol (+) kg -1 %BS = (sum of base cationic charge x 100%) / CEC, or 70%

37. You can do it with pH sensitive dyes (like a colorimetric titration or like with Duplex indicator in lab) or you can do it with an ion (H + ) selective electrode, right?

38. Sulfur lowers pH 2S + 3O 2 + 2H 2 O → 2H 2 SO 4 Lime raises pH To a minor extent, the above reaction occurs, as they say, chemically. However, sulfur oxidizing microbes are principally responsible.

39. CaCO 3 Calcite CaMg(CO 3 ) 2 Dolomite CaOBurned lime (quicklime) Ca(OH) 2 Hydrated lime ∆ CaCO 3 → CaO + CO 2 ↑ These are the types of lime materials. The advantage of the non-limestones, the oxides and hydroxides, is that they are more soluble and so react faster to consume H +. To get good results using limestone, it should be finely ground to increase surface area and rate of dissolution.

41. It’s worth pointing out that fixing topsoil acidity is an easy matter, just till in some lime. However, fixing subsoil acidity is more difficult because unless the lime is placed down there in the first place, its reactivity is largely confined to where it was applied --it is mostly immobile. Subsoil acidity can be a real problem in some cases if it curtails deep root development, thereby restricting the root system to a smaller portion of the profile (access to fewer potential nutrients, and increased drought risk, especially. A way around the expense of deep tillage and liming is to use another source of Ca 2+ (replaces acidic cations on soil colloids, putting them into solution where they may leach deeper and away) that is much more soluble and mobile is gypsum, CaSO 4. Besides the effect of Ca 2+ from it, the sulfate tends to bond to soil colloids, and in doing so releases a tad of OH -, a pseudo-liming effect, however, gypsum is not considered a lime material.

43. Which will neutralize more acidity, 1 kg of 90 % purity CaCO 3 or 0.74 kg of Ca(OH) 2 ? You have to have a common basis of comparison, like moles –one mole of Either will neutralize two moles of H+, Ca(OH) 2 + 2H + → 2H 2 O + Ca 2+ or CaCO 3 + 2H + → H 2 O + CO 2 + Ca 2+ MW Ca(OH)2 = 74 g mole -1 and MW CaCO3 = 100 g mole -1 Therefore, 1000 g x 0.90 / 100 g mole -1 = 9 moles CaCO 3 740 g / 74 g mole -1 = 10 moles Ca(OH) 2

44. Based on a buffer curve for a certain soil, 1.0 millimoles of OH - are required to raise the pH of 10 g (oven-dry mass) of this soil from an initial value of 5.0 to 6.5. How many Mg per hectare furrow slice (HFS) of CaCO 3 are required to raise the pH of this soil from 5.0 to 6.5? One Mg = 1000 kg. Assume 2000 Mg / HFS. (1.0 mmole OH / 10 g soil ) x (50 mg CaCO3 / 1 mmole OH ) x (0.001 g / 1 mg) = 0.005 g CaCO3 / 1 g soil just a ratio of mass to mass so multiply it by mass of soil, 2000 Mg / HFS to give,10 Mg CaCO3 / HFS The key, besides seeing the sense in the multiplied ratios, is knowing what mass of lime material is chemically equivalent to 1 mmole of OH -1. This means figuring molecular weight, MW (and milli-MW, mMW), and recognizing that a mmole of any lime, CaO, etc., will neutralize 2 mmoles H +.