This shortened overview serves as an aid in estimating the magnitude of the analytical technology required when analyzing and controlling the nutrients with which the plants are fertilized.

The quality of analysis in chemistry has already reached a level of precision that is unnecessary for our purposes of controlled fertilization. In order not to shoot at sparrows when selecting the various analysis methods and analysis devices, we have listed here a very shortened overview of the necessary accuracies that are sufficient for checking the individual additives. The technology used in the chosen analysis method has a major influence on the overall operating costs.

In addition to checking the necessary substances, monitoring is also necessary to prevent over-fertilization. The nutrients produced by fish farming must not exceed a certain concentration, otherwise this will impair the optimal growth of the plants.

There are now a very large number of analysis methods on the market, which differ greatly in both the technology used and the on-site application. This overview will help you, even without our advice , to obtain offers from different manufacturers that exactly meet your needs. Here is a random selection of manufacturers.

Here you will find the essential compounds required for plant growth. Depending on the plant and/or growth phase, the form of administration, the chemical compound in which the desired “substance” is bound, can or must vary. In the previous cultivation method (in the soil), the microorganisms and fungi caused the necessary compounds to be broken down. Since no microorganisms take on this task in hydroponics, this is still a current area of ​​basic research.

Compounds and trace elements / orders of magnitude in nutrient solutions

K

potassium

0.5 - 10 mmol/L

Approx

calcium

0.2 - 5 mmol/L

S

sulfur

0.2 - 5 mmol/L

P

phosphorus

0.1 - 2 mmol/L

Mg

magnesium

0.1 - 2 mmol/L

Fe

iron

2 - 50 µmol/L

Cu

copper

0.5 - 10 µmol/L

Zn

zinc

0.1 - 10 µmol/L

Mn

manganese

0 - 10 µmol/L

b

boron

0 - 0.01 ppm

Mo

molybdenum

0 - 100 ppm

NO2

nitrite

0 – 100 mg/L

NO3

nitrate

0 – 100 mg/L

NH4

ammonia

0.1 - 8 mg/L

KNO3

Potassium nitrate

0 - 10 mmol/L

Ca(NO3)2

Calcium nitrate

0 - 10 mmol/L

NH4H2PO4

Ammonium dihydrogen phosphate

0 - 10 mmol/L

(NH4)2HPO4

Diammonium hydrogen phosphate

0 - 10 mmol/L

MgSO4

Magnesium sulfate

0 - 10 mmol/L

Fe-EDTA

Ethylenediaminetetraacetic acid

0 – 0.1 mmol/L

H3BO3

Boric acid

0 – 0.01 mmol/L

KCl

Potassium chloride

0 – 0.01 mmol/L

MnSO4

Manganese (II) sulfate

0 – 0.001 mmol/L

ZnSO4

Zinc sulfate

0 – 0.001 mmol/L

FeSO4

Iron(II) sulfate

0 – 0.0001 mmol/L

CuSO4

Copper sulfate

0 - 0.0002 mmol/L

MoO3

Molybdenum oxide

0 – 0.0002 mmol/L

Conversion: Mol and PPM

A technical definition of ppm

What is ppm? And how can something called "parts per million" be represented by mg/L? Parts per million indicates the number of "parts" of something in a million "parts" of something else. The "part" can be any unit, but when mixing solutions, ppm usually represents units of weight. In this context, ppm indicates how many grams of a solute there are per million grams of solvent (e.g. water).

1 g dissolved / 1,000,000 g solvent

When dealing with water at room temperature, it is common to assume that the density of the water is equal to 1 g/ml. Therefore we can describe the relationship as follows:

1 g dissolved in 1,000,000 ml of water

Then we divide ml by 1000 ml:

1 g dissolved in 1,000 L water

By dividing both units by 1000, the ratio becomes:

1 mg dissolved in 1 L water

Therefore, one can say 1 mg in 1 L of water is the same as 1 mg in 1,000,000 mg of water, or 1 part per million (assuming both room temperature and an atmospheric pressure of 1 atmosphere).

How do you convert ppm to moles?

To convert ppm to molarity or molarity to ppm, you only need to know the molar mass of the dissolved element or molecule. Here is a periodic table for the molar masses (top left: the atomic weight).

Take the molarity mol/L and multiply by its molar mass
g/mol to get g/L. Multiply by 1000 again to convert grams to milligrams and you have mg/L for aqueous solutions.

Example: Prepare a NaOH solution

You have a stock solution of 1 molar NaOH. How do you go about creating a 1L solution of 200 ppm NaOH? NaOH has a molar mass of 39,997 g/mol.

1. Convert 200 ppm to molarity.

First let's assume 200 ppm = 200 mg/L. Then divide the result by 1000 and you get g/L:  200 mg/L divided by 1000 mg/g equals 0.2 g/L.

Next, divide 0.2 g/L by the molar mass of NaOH (Na=22.9 O=16 H=1) to get the molarity: 0.2 g/L divided by 39,997 g/mol which is 0.005 mole /L.

2. Calculate the dilution recipe.

From step 1 we know the target molarity of 0.005 mol/L. To calculate the dilution we use the dilution equation:  m1⋅v1=m2⋅v2

where:
• m1— the concentration of the stock solution;
• m2— the concentration of the diluted solution;
• v1—the volume of the stock solution; and
• v2 - The volume of the diluted solution

We can enter the numbers for all variables except the volume of the stock solution:

1 M ⋅ v1 = 0.005 M ⋅ 1 L

By rearranging the equation, we find the required volume of the stock solution:
v1 = 0.005 M / 1 M  ⋅ 1 L = 0.005 L

Therefore we need to dilute 0.005 L (or 5 ml) stock solution to a final volume of
1 L and so we get 200 ppm NaOH solution.

How do I calculate ppm from volume concentration?

How to get volume ppm:

Take the molar concentration of the solutions in mol/L.
Multiply it by the molar mass in g/mol.
Divide it by the density of the solute in g/cm³.
Multiply everything by 1000 mg/g.
The resulting ppm volume unit is typically μL/L.

You can find a slightly more detailed example here for both conversion directions:

Convert moles to grams

Convert grams to moles

HowTos and measuring devices