I sampled the soil in the orchard from 0-12" depth and came up with the following results:

%CaCO3 = 2.8%

pH=7.4

Micros. in PPM
Mn= 9.5
Cu= 3.0
Zn= 10.7
Fe= 12.9
B = 0.09

Macros. in PPM

NO3= 4.2
P= 13.4
K= 720
% O.M. = 1.4

This is a sandy clay loam

It appears to me that the micro. levels are fine and that I'll be fertilizing with P and N, no problem there. However, that lime, looking at the leaves looks to be binding up the Fe and possibly Zn. If the trees were older and larger I'd have already done a leaf tissue sample to confirm which is the problem or both.

for now it makes sense to get the pH down to about 6.5 in the top foot of soil (or more) where most of the roots are so the micros. can be used by the trees. Increasing the OM is a good idea too, started doing that 3 years ago and will continue.

Now for the question, knowing the actual % lime level, the desired soil depth and the new target pH of 6.5, how does one calculate how much material to use?

BTW, I'll be adding about 1 lb P2O5/1000 sq. ft. + 0.3 lb N/100 sq. ft. to get those up where they need to be using one or more of the following - 10-34-0 liquid, 28-0-0 UAN, Urea. The purpose of those particular fertilizers is 3-fold, I have access to them all, they can be fertigated in and the urea in particular can be particularly useful to lower the soil pH. In addition, by fertigating in the pH dropping ferts. they can be carried into the root zone and begin acting much quicker than using S. The other drawback of S is that it has to be incorporated, to do so would chew up the root systems.

Methinks I'll be best using the 10-34-0 + urea as it will give me the N and P I need but I have no idea how to calculate a theoretical pH drop in all that darned lime which must be neutralized first before the pH is going anywhere. Perhaps some surface applied Ag. S would help over time with overhead water from the sky or hose, beats me.

What sayith thou?

I agree about the manure and probably be heading south a mile to a friend's winter feed lot to get some manure one of these months. the wood chips are used both as a mulch (can be very dry out here) and a source of O.M.. In the last 3 years I've added about 1 1/2 ft. of wood chips, the soils critter have reduced that to about 1" deep now. if I can swing it I'll be getting my pH lowering amendment(s) applied first this spring followed by manure and another 6" of wood chips.

About half of the trees started producing a few fruit for the first time last year, Wagener, Golden Russet and Roxbury Russet, I plan to use most fruit eventually for sweet cider. I must be doing the pruning right because there are a lot more fruit buds now than there were this time last year and the years before and the tree forms are getting very good.

Kimm: son of a gun, I just realized that Mg wasn't analyzed too, guess I'll have to call the lab and see what the story is! I did do asampling about 8 years ago about 400 ft. away in very similar soils and the Mg there was 205 ppm via amm. acetate extraction, don't imagine in this bottom land the orchard is a whole lot different.

One thing for certain, I'll have to calculate how much Ca CO3 is in the top foot on whatever measurement I choose (per 100 or 1000 sq. ft.) and don't see any problem doing that once I just get started. then will come the chemistry problem and I stink there.

I calculated the amount of lime in an a.f.s as follows based on 2.8% from the lab.

Weight of an a.f.s of soil = 2M lb. so,0.028 x 2M = 56,000 lb. lime in that a.f.s.. 56,000/43.56 = 1286 lb. lime/1000 ft.sq.. Truly, a load of lime assuming my math is correct!!!

Knowing how much lime is in my 1000 sq. ft. I don't know how to calculate how much S would be required to neutralize it, my target is pH 6.5 and that probably shouldn't be reached for with straight S this year alone. This Spring I'll also be fertigating in some Amm. sulfate for the N, just realizing that it's got the most punch of all N sources for neutralizing the lime. How do I calculate it's lime neutralizing effect? I other words, how do I calculate the amount of lime neutralized by given amounts of amm. sulfate and S?

CaCO3:
Molecular weight Ca+C+O(*3)=100
100 / valence (2) = 50
1/1000*50 = .05g CaCO3 = 1 meq CaCO3

Sulfur
Molecular Weight S = 16
16 / 2 = 8
1/1000*8 = .008g S

So, for every .05g of lime you have, you need .008g of sulfur to neutralize it. Or, to put it another way, for every 1g of CaCO3, you need .16g S.

From there, we can use the 1 to .16 ratio on any unit of weight that we want:
1286 (CaCO3)*.16 = 205.76 lbs. of sulfur to neutralize the lime in 1000 sq. ft. to a depth of 12 inches.

Of course, applying that much sulfur all at once would fry your rees.

I think the name of the game here is that it's possible to neutralize that much lime, just not feasible in your situation. Remember also that lime has a very low solubility and goes into solution slowly.

If it were me, I would just stick with the urea or ammonium sulfate and in that way, treat the soil solution to help free up the micros. If you can get some sulfur in there, that would be great because as the lime goes into solution, there will be sulfur present to neutralize it to some extent. That, along with acidifying fertilizers and judicious applications of chelates when necessary should do the trick. That's how I'd do it anyway.

Even without the sulfur, I would think that you could get a decent to good response with acidifying fertilizers and chelates. Either way, I'm afraid you may just have to make your peace with the fact that you will always be fighting lime.

(I swear, one of these days I'll proof read a post before submitting.)

Is all this CaCO3 natural or was some introduced?

All that free Lime is native, just less than a mile from here are some formations called chalk bluffs.

I saw in an OSU publication where part of Ohio has some calcareous soils, guess that technically is the midwest as opposed to the east. Must have something to do with all that under ground rock.

Horticultural types have extreme difficulty with this. I have been accused of 'sleep liming' since how else could I possibly end up with a soil pH over 7 without knowingly doing something to create it. I have pretty much given up on local recommendations for shade shrubs since the first half of the discussion is always taken up by explaining that I *CANNOT* grow the usual ericaceous suspects. The oddest part of that is that the local high end nursery sits on similar soil. One of these years they will figure out why the extra rhodos they've thrown in the ground are yellow. I really thought I was losing it until I started talking to the Soil & Water people.

Oh yeah, seeing which thread this is, one of the big local crops is apples. It's been that way for about 200 years now. The soils aren't an issue at all, and never have been.

Your apple crops probably never need irrigation, do they? I have to irrigate my trees and the water is high in bi-carbonates making the calcareous soil problem even worse. The good news, I figured that out a year ago and started acidifying the irrigation water to deal with the bi-carbonates.

Curious to know the concentration of lime in those apple growing regions. It could well be that my rootstocks don't do well at getting micros. in this soil whereas a different one would, don't think I'll live long enough to test that hypothesis. The growers in your area likely figured out long ago what rootstocks worked best in their calcareous soils, I've only been trying apples for about 8 years now.

I've been driving myself nuts trying to figure out where you derived the valence numbers for S and calcium carbonate and have been unsuccessful so far. On the other hand, I did find the valence for CaCO3 in a table, it agrees with yours (2) but am still in the dark as to where one derives valence numbers for elements and compounds. My sneaking suspicion is that with CaCO3, it is the Ca that is relevant for what I am trying to do to the soil, therefore; the valence of Ca is what matters, not the vaalence of CaCO3. am I correct?

Anyway, you're right. Actually, I didn't get the 16 from oxygen though. I was thinking of the atomic number for sulfur rather than the atomic weight. So, with a quick re-calculation, it would be 411.52 lbs. of sulfur per 1000 sq.ft.

First, the easiest way is to look at the groups at the top of columns in the periodic table and the Roman numeral will give you the valence. For example, everything in group IIA has 2 valence electrons, so a charge of +2. Everything in Group VB will have 5 valence electrons, which gives you a valence of -2.
Understanding that gets a little sticky. The important things are the number of electrons there are, how many protons there are and how many electrons are in the outer most shell (valence) shell.

Calcium:

You'll notice that there are 2 electrons in the valence shell. With the exception of the first valence shell (which holds only 2 electrons), atoms want 8 electrons in each shell "shell". (To make it easier to visulaize, they represent those shells as circular orbits here but the pattern actually changes depending on the number of electrons and any bonds attached to the atom.) Having 2 electrons in the valence shell means that in an ionic bond, calcium will "give up" those 2 electrons so that the next shell down will be the valence shell, complete with 8 electrons. If it "gives up" 2 electrons it would have 2 more protons than electrons (protons being positively charged and electrons being negatively charged. Atoms start with the same number of each). If calcium were to give up 2 electrons then it would be left with 18 electrons and 20 protons and so we think of it as having a charge of +2.

Sulfur:

You'll notice that sulfur has 6 valence electrons so it would accept 2 to fill its valence shell up to 8, giving it 18 electrons and 16 protons. We think of it as a -2.

For polyatomic ions like carbonate, you typically go to your references acids. For example, the reference acid for carbonate (CO3) is carbonic acid, which is H2CO3. If we know that there are 2 hydrogens in that acid, each with a charge of +1, then carbonate must have a charge of -2. It's also a way to "back door" some of the cations. After figuring out that carbonate is a -2 and there is 1 calcium in calcium carbonate, then calcium must be a +2.

Also, keep in mind that these are general guides. Many elements have multiple valences. (I'm trying to cram a lot of chemistry into a small space without getting too wordy)

Calcium can hook up with a sulfur to make the mineral calcium sulfide. "strongly attracted" is a relative term though. When talking about ionic bonds, the strength of the bond has largely to do with the electronegativity of the two "halves" of the salt. When looking at the periodic table, the farther to the right that you move, the greater the electronegativity.

Another thing that affects the strength of the attraction is the distance from the electrons in the outer shells of both "halves" and their respective nuclei. This is dictated by the number of electrons in the atoms involved. Remember, once you have 8 electrons in a shell it fills up and the next shell farther out becomes populated. The farther out you have to go, the farther the valence shell will be from the nucleus (where the protons are) and, consequently, the weaker the "pull".

Curious how S actually neutralizes carbonate when added to soil. i.e. what acid form is it in? If H2S then S is +2. If sulfuric acid (H2SO4) then S is in the +6 form. What I am getting at is that the acid neutralization potential (parts S to parts CO3) may not be the same ratio as the ratio of the valences. Just trying to understand that part.

Then the chemistry comes in and those 4H+ can neutral 2 CaCO3. Since it took 2S to result in 4H+ then for the purposes of calculating, it takes 1S to make 2H+ to neutralize 1 CaCO3. In that way, it just happens to work out that 1 meq S will neutralize 1 meq CaCO3.

Sorry, slow day.

Crazy little critters, that eat sulfur and pee out sulfuric acid. What a world.

My chemistry knowledge is nil (missed it along with anything else that bored me in school). I guess I need to look up what these various endings mean: "ate", "ite", "ide" etc. Not to mention how they vary in terms of their interaction with plants and microflora.

There's a misconception that sulfate lowers pH because some of the common sulfate salts are used to lower pH, most notably iron sulfate and aluminum sulfate. However, the pH change in that case is chemical rather than microbial and has to do with more complex reactions that rely more on the iron and aluminum involved.

As far as the types of Ca and sulfur that plants need, they require elemental calcium and they use sulfur in the sulfate form so between gypsum and lime, you've got them both covered.

1) from Colo. State Univ. - "If the free lime in Western Colorado soils could be neutralized, any additional applications of sulfur or sulfuric acid would bring about a pH change. Six tons of sulfur (or 20 tons of sulfuric acid) is required per acre to neutralize each percentage of calcium carbonate." I calculated out my 2.8% lime soil and determined that on a sq. ft. basis for my orchard. Not planning, as you stated earlier, to try and do it all at once but, maybe a tenth of a percent this Spring. In addition -

2) Apple trees uptake NH4+ from the soil (I didn't know if apple trees would before). why is that significant? -

3) when apple roots uptake NH4+, they exude H+ and acidify the rhizosphere. My hope is that in using amm sulfate for my N this Spring, the NH4+ will do it's thing and the micros. will become available enough to at least reduce the micro deficiency problem. The soil test indicated that there is more than adequate Fe and Zn there already. Based on the soil test, my amm sulfate rate will be 15.24 lb/1000 sq. ft. and deliver 3.2 lb N/1000 sq. ft..

4) between the small amount of S and amm sulfate for the N I think I may be on the right track, finally. Whaddya think? I can't calculate how much lime would be neutralized by the amm. sulfate, unfortunately, that would be good to know assuming it is a significant amount.

5)for sure, I'll be sampling for pH and %lime in the future to keep track of what happens.

Yes, apples do take up NH4 and you'll get some H+'s back that way but they also take up NO3 so as NH4 is oxidized to NO3 before being taken up (which happens relatively quickly, especially once the soil warms up), you'll get your H+'s that way too. I know you already know this but just remember to get your ammonium sulfate down into the soil as quickly as possible to avoid volatilization.

Theoretically, it would take about 733 lbs. per 1000 sq. ft. of ammonium sulfate to neutralize the lime in the top 6 inches. But soils and soil reactions are so dynamic and variable that we're just talking about good guesses using very precise chemistry to calculate ideals and applying them to an imprecise world. Either way, it would take a whole lotta time.

I think in a real world sense you'll be addressing the soil solution more than the lime. That's not a problem though. Like I said in an earlier post, I think you'll always be fighting the lime so addressing the soil solution is the way to go. You'll get some pH drop, you'll free up some micros and then the lime will buffer the pH back up. Lather, rinse, repeat.

If I'm following you correctly, are you looking at about 77 lbs. of sulfur per 1000 sq.ft. applied to the top? I would think that's an awful lot. My own personal rule is not to exceed 30 lbs. per 1000 sq. ft. tilled to 6 inches. Somewhere in that neighborhood is where I start to see plants burning. Add to that the fact that sulfur doesn't move down very well and would be even more concentrated in the top. I would be concerned about getting a real hot zone and frying some surface roots. That wouldn't be the end of the world but I think you could get some burning and put a pretty good chunk of the root system out of commission. I'm also afraid it wouldn't have much effect on lime or pH down beyond an inch or two in the profile. Over all, it would seem a little counterproductive to me.

Personally, if I were to go the sulfur route in this situation, I would probably stick with about 5 to 10 lbs. per 1000 sq. ft., scratch it into the surface and water it in well. That should give you a little extra punch in the top couple of inches anyway.