## A New Look at Calculating Calcium Requirements, in metric units

##### 01 April 2012

We have worked with PACE Turf to develop the Minimum Levels for Sustainable Nutrition (MLSN). These new guidelines ensure ample amounts of mineral elements are present in the soil to meet the requirements of turfgrass. The target for calcium is 360 ppm using the Mehlich 3 extractant, meaning that if calcium is present in the soil at 360 ppm or more, none is needed as fertilizer, because the soil can supply all the calcium that the plant requires.

Calcium is also used to replace sodium in the soil. For that use of calcium, it is not a fertilizer, but rather a soil amendment. And the MLSN guideline for sodium is to have no more than 110 ppm in the soil, measured in a Mehlich 3 extract.

What if you have sodium at more than 110 ppm in the soil, perhaps from salt spray, or more likely, from sodium in the irrigation water? A new document from PACE Turf explains how to calculate the calcium requirement. That document, however, uses pounds and acres and thousands of square feet. I make the calculations here in metric units.

- First we look at how much sodium is in the soil. Our goal in this calculation is to find out how much calcium we need to apply to match the amount of excess sodium. If we can apply calcium to match the sodium, then it will be easier to leach the sodium from the soil.
- Let's pretend we have just received a soil test report and the sodium is at 150 ppm. This means that there are 150 mg of sodium for each kg of soil, and our excess sodium is 40 ppm (or 40 mg) more than the 110 ppm maximum on the MLSN guidelines. 150-110 = 40 mg/kg
- We can take that 40 mg of sodium and find out how many ions it is. One millimole of sodium has a mass of 23 mg, so we have 40/23 = 1.7 millimoles of sodium per kg. Let's write millimole as mmol for simplicity. Now we are ready to calculate how much calcium we require.
- Sodium is a monavalent cation, meaning it has one positive charge for each ion. We have 1.7 mmol of sodium, and also 1.7 mmol of positive charge. We need 1.7 mmol of positive charge from calcium. But calcium is a divalent cation, with two positive charges for each ion. So we only need 0.85 mmol of calcium. One mmol of calcium has a mass of 40.1 mg, so we need 40.1*0.85 = 34 mg of calcium. We've almost got it: 34 mg of calcium will provide the same amount of charge as will 40 mg of sodium.
- We now need to consider how much area and volume of soil we will apply that calcium to, so that we can get an application rate. Let's assume we have a rootzone 15 cm deep and with a bulk density of 1.33 g/cm
^{3}. In that case, 1 m^{2}of the soil has a mass of 200 kg. Remember, we've previously been making our calculations on a per kg basis. Now we want to change to square meters. To replace the excess sodium, we have already calculated that we need 34 mg of calcium per kg of soil. And we have 200 kg soil in 1 m^{2}, so we need 34*200 = 6,800 mg calcium per square meter (6.8 g/m^{2}). - That's how much calcium we require in this example. If you want to simplify this and not work through the calculation every time, we can simply apply 0.17 mg of Ca for every 1 ppm sodium in excess of 110 ppm. If sodium were at 120 ppm on the soil test, the calcium requirement would be (120-110)*0.17 = 1.7 g/m
^{2}. If sodium were at 204 on the soil test, the calcium requirement would be (204-110)*0.17 = 16 g/m^{2}.

Of course, if you have more than 360 ppm Ca and less than 110 ppm Na on the soil test, then there is no calcium requirement. The soil already has enough, and the sodium does not require treatment.