PO 37: Describe the long-term change in soil pH from applying N

This information can be used to calculate the amount of lime required to neutralize a given application of fertilizer. For example:

• Find the CaCO₃ equivalent (lime requirement) of 150 lb N/acre applied as urea: 150 lb N x 1.8 lb CaCO3/lb N = 270 lb lime requirement

• Find the CaCO₃ equivalent of 150 lb N/acre applied as ammonium sulfate: 150 lb N x 5.4 lb CaCO3/lb N = 810 lb lime requirement.

The following reactions explain why there are differences in the amount of acidity per pound of N for the different N fertilizer materials.

• Ammonia: NH₃ + H₂O → NH₄⁺ + OH^-
• The reaction produces 1 OH^- per N.  This OH^- neutralizes one of the two H⁺ that will be produced by nitrification – ½ of the acidity is neutralized.
• Urea: NH₂CONH₂ + H₂O → 2NH₃ + 2H₂O → 2NH₄⁺ + 2OH^-
• The reaction produces 1 OH^- per N.  This OH^- neutralizes one of the two H⁺ that will be produced by nitrification – ½ of the acidity is neutralized.
• Ammonium Nitrate: NH₄NO₃
• No OH^- is produced when this fertilizer reacts.  As only one N is NH₄⁺ (the other is already NO₃^- and does not need conversion), only one H⁺ is formed per N.
• Ammonium Sulfate: (NH₄)₂SO₃
• No OH^- is produced when this fertilizer reacts; therefore, all the acidity produced by nitrification remains and none is neutralized.
• Monoammonium Phosphate (MAP): NH₄H₂PO₄
• No OH^- is produced when this fertilizer reacts; therefore, all the acidity produced by nitrification remains and none is neutralized.
• Diammonium Phosphate (DAP): (NH₄)₂HPO₄ + H₂O → 2NH₄ + HPO₄^2- + H₂O → H₂PO₄^- + OH
• The reaction produces 1 OH^- per 2N.  This OH^- neutralizes one of the four H⁺ that will be produced by nitrification – ¼ of the acidity is neutralized.