# How much Sulfur is needed to lower the soil pH? It depends…



## wors (Feb 2, 2019)

There are many posts about lowering the soil pH. This doesn't need to be a guess and can be calculated. Send a soil sample to a lab and test the CCE% (Calcium Carbonate Equivalency). Once the CCE% is known some simple math will tell the amount of Sulfur required to lower the pH. Then calculate the years based on the recommended 10lbs sulfur / year.


Acre furrow slice at 6" is 2,000,000 lbs.

43,560 ft2 in 1 acre

32lbs of elemental S will neutralize 100 lbs of CaCO3[/sub]

Example:

Lab test CCE% = 3.0%

Step 1: Calculate free CaCO3 per acre furrow slice

(2,000,000 * .03) = 60,000 lbs of free CaCO3 per acre furrow slice​
Step 2: Calculate lbs of free CaCO[sub]3[/sub] / M

(60,000 / 43.5) = 1,380 lbs of free CaCO[sub]3 [/sub]/ M​
Step 3: Calculate lbs of elemental S needed to neutralize lbs of free CaCO[sub]3 / M

(1,380 * (32/100) ) = 442 lbs of Sulfur​
Step 4: Calculate years required (assuming 2 x 5 lb apps of Sulfur)

(442 / 10) = 44 years​
CCE% -> Years rounded
0.5% = 7 years
1.0% = 15 years
2.0% = 30 years
3.0% = 45 years
4.0% = 60 years
5.0% = 75 years


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## gasdoc (Jul 24, 2019)

Thank you wors.

Is the 5lb/app only applicable to turf areas where the is concern for burn?

I'm establishing an orchard and ph is elevated. I calculated I need ~8lbs S/1000sqft. If I'm not concerned with turf can I drop the 8lbs, cover in mulch and be done with it?


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## wors (Feb 2, 2019)

How did you calculate needing ~8lbs S/1000sqft?


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## gasdoc (Jul 24, 2019)

I already did my soil test so can't get the CCE% unfortunately

I have sandy loam texture, cec 11, Ph 7.3.

I looked at a few different university websites that give charts such as this.



6-6.5 would be a good target

Using the "silt" column gets me ~8-14lbs/1M


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## wors (Feb 2, 2019)

I'd recommend sending in a sample to PSU for $25 to get your CCE% instead of guessing. Your screen capture cut off a very important point "respectively; soils are not calcareous". Sands can also be calcareous.


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## Ridgerunner (May 16, 2017)

The Covid lockdown has provided you with way too much free time. Good information though. :thumbup:

I think you have a typo:


> Step 2: Calculate lbs of free CaCO3 / ft2
> 
> (60,000 / 43.5) = 1,380 lbs of free CaCO3 / ft2
> 
> ...


I think hat should be per thousand (M) square feet rather than each square foot.


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## wors (Feb 2, 2019)

Good catch, thanks Ridgerunner.


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