Ability to leave and mix nutrients is critical to mastering hydroponics. But it should be understood that this is a very laborious and time-consuming process. Even seasoned hydroponicists prefer ready-made feeding systems like the Hydroponics Kit. This will be a more suitable option for a beginner and save you from storing an abundance of raw materials and laboratory equipment.
Next, we will consider an example of composing and preparing a solution of a given concentration of nutrients.
Before starting to formulate the nutrient mixture, it is necessary to select the solution that is most suitable for the culture, climate and growing method. Recommendations for feeding nutrient mixtures are described in the article “Selection of nutrient solutions”, and the base of solutions for selection is described in the section of the site “Nutrient mix recipes for hydroponics”.
For an example of calculation, let’s take the very popular “Nutrient solution of Chesnokov and Bazyrina”. The concentration of nutrients is expressed in milligrams of an element in 1 liter of solution [mg / l]. On the website, it looks like this:
For convenience, I recommend compiling a table. It is convenient to use Microsoft Excel, or specialized programs for calculating nutritional mixtures. The table will look like this:
Blend component
We make, g / l
N
P
K
Ca
Mg
…
In the finished solution, mg / l
Specified solution, mg / l
Let’s transfer the concentration of elements from the base to the table (we do not take into account the content of ammonium and nitrate nitrogen in this example).
Blend component
We make, g / l
N
P
K
Ca
Mg
…
In the finished solution, mg / l
Specified solution, mg / l
140
38.5
190
165
30
Next, you need to select the substances from which the future mixture will be prepared. As a rule, these are mineral fertilizers.
It is worth starting from a component, which we want to add with only one substance. I prefer to start with calcium or magnesium. Let’s start with magnesium.
Magnesium
It is known that for 1 mass unit of MgO, there are 0,603 mass units of Mg. This figure is calculated on the basis of the molar masses of compounds and simple substances, and can be found under the term “Oxide factor”. These figures for different substances are given in the article “Oxide factor”, or can be calculated by the online calculator of molar masses on the page of the article “Molar mass of chemical compounds”.
Let’s make a simple proportion:
- Let’s take 16,2 g of MgO as 1 mass unit,
- then x g Mg is 0,603 mass units.
x = 16.2 · 0.603 = 9.77 g
We get the following result: 100 grams of fertilizer contains 9,77 grams of pure magnesium. Let’s enter the data into the table:
Blend component
We make, g / l
N
P
K
Ca
Mg
MgO
Magnesium sulfate, wt%
9.77
16.2
…
In the finished solution, mg / l
Specified solution, mg / l
140
38.5
190
165
30
Now you need to calculate the amount of fertilizer that needs to be applied to obtain a solution with a magnesium concentration of 30 mg / l. A concentration of 30 mg / L tells us that 1 liter of solution contains 30 milligrams of magnesium, or, translated into grams, 0,03 grams of magnesium. For the calculation, we again compose a simple proportion:
- 100 grams of fertilizer apply 9,77 grams of Mg,
- then x gram of fertilizer will add 0,03 gram of Mg.
x = 100 0.03 / 9.77 = 0.31
Conclusion: it is necessary to apply 0,31 grams of magnesium sulfate fertilizer per 1 liter of solution. Let’s fill in the table:
Blend component
We make, g / l
N
P
K
Ca
Mg
MgO
Magnesium sulfate, wt%
0,31
9.77
16.2
…
In the finished solution, mg / l
30
Specified solution, mg / l
140
38.5
190
165
30
Calcium
The oxide factor for CaO is 0,715. Fertilizer pure calcium content:
Ca = 27 · 0.715 = 19.3%
It is necessary to add 165 mg of calcium per 1 liter to the solution.
- 100 grams of fertilizer introduces 19,3 grams of Ca,
- then x gram of fertilizer will add 0,165 gram of Ca.
x = 100 0.165 / 19.3 = 0.855
Conclusion: it is necessary to apply 0,855 grams of calcium nitrate fertilizer per 1 liter of solution.
Important! It is worth considering that calcium nitrate, in addition to calcium, also introduces nitrogen into the solution. The nitrogen content in the fertilizer is 14,9%. It turns out that 0,855 grams of fertilizer additionally with calcium will add 0,127 grams of nitrogen (0,855 14,9 / 100), which is equal to 127 mg. Let’s fill in the table:
Blend component
We make, g / l
N
P
K
Ca
Dog
Mg
MgO
Magnesium sulfate, wt%
0,31
9.77
16.2
Calcium nitrate, wt%
0,855
14,9
19,3
27
…
In the finished solution, mg / l
127
165
30
Specified solution, mg / l
140
38.5
190
165
30
Phosphorus
Let’s start the calculation for potassium. Similar to previous calculations:
Oxide factor for K2O is 0.83. Fertilizer pure potassium content:
It is necessary to add 190 mg of potassium per 1 liter to the solution.
- 100 grams of fertilizer apply 27,39 grams of K,
- then x gram of fertilizer will add 0,190 gram K.
x = 100 0.190 / 27.39 = 0.69
Conclusion: it is necessary to make 0,69 grams of “potassium monophosphate” fertilizer per 1 liter of solution.
Important! Together with potassium, phosphorus is also introduced.
Oxide factor for P2O5 is 0.436. Fertilizer pure phosphorus content:
P = 50 · 0.436 = 21.8 %
We add 0,69 grams of “potassium monophosphate” fertilizer to the solution, and therefore 0,15 grams of phosphorus (0,69 · 21,8 / 100). 0,15 grams = 150 mg, which is significantly more than we need. Conclusion: we are calculating starting with phosphorus.
Calculation for phosphorus. Similar to previous calculations:
Oxide factor for P2O5 is 0.436. Fertilizer pure phosphorus content:
P = 50 · 0.436 = 21.8 %
It is necessary to add 38,5 mg of phosphorus per 1 liter to the solution.
- 100 grams of fertilizer applies 21,8 grams of P,
- then x gram of fertilizer will add 0,0385 gram P.
x = 100 0.0385 / 21.8 = 0.177
Conclusion: it is necessary to make 0.177 grams of “potassium monophosphate” fertilizer per 1 liter of solution.
Important! Together with phosphorus, potassium is also added.
Oxide factor for K2O is 0.83. Fertilizer pure potassium content:
K = 33 · 0.83 = 27.39%
We add 0.177 grams of “potassium monophosphate” fertilizer to the solution, and therefore 0,048 grams of potassium (0.177 · 27.39 / 100). Let’s fill in the table:
Blend component
We make, g / l
N
P
P2O5
K
K2O
Ca
Dog
Mg
MgO
Magnesium sulfate, wt%
0,31
9.77
16.2
Calcium nitrate, wt%
0,855
14,9
19,3
27
Potassium monophosphate, wt%
0.177
21.8
50
27.39
33
…
In the finished solution, mg / l
127
38.5
48
165
30
Specified solution, mg / l
140
38.5
190
165
30
Nitrogen
The nitrogen content in the fertilizer is 13.6%. It is necessary to add 13 mg of nitrogen (140-127 mg). Let’s make the proportion:
- 100 grams of fertilizer apply 13,6 grams of N,
- leaves x gram of fertilizer 0,013 gram of N is applied.
x = 100 0.013 / 13.6 = 0.096
Conclusion: it is necessary to apply 0.096 grams of potassium nitrate fertilizer per 1 liter of solution.
Important! Potassium is added together with nitrogen.
Oxide factor for K2O is 0.83. Fertilizer pure potassium content:
K = 46 · 0.83 = 38,18%
We add 0.096 grams of potassium nitrate fertilizer to the solution, and therefore 0,037 grams of potassium (0.096 · 38,18 / 100). In total, 85 mg of potassium (37 + 48 g) in solution. Let’s fill in the table:
Blend component
We make, g / l
N
P
P2O5
K
K2O
Ca
Dog
Mg
MgO
Magnesium sulfate, wt%
0,31
9.77
16.2
Calcium nitrate, wt%
0,855
14,9
19,3
27
Potassium monophosphate, wt%
0.177
21.8
50
27.39
33
Potassium nitrate, wt%
0,096
13,6
38,18
46
…
In the finished solution, mg / l
140
38.5
85
165
30
Specified solution, mg / l
140
38.5
190
165
30
potassium
Oxide factor for K2O is 0.83. Fertilizer pure potassium content:
K = 50 · 0.83 = 41.5%
It is necessary to add 105 mg of potassium per 1 liter (190-85 g) to the solution.
- 100 grams of fertilizer apply 41,5 grams of K,
- then x gram of fertilizer will add 0,105 gram K.
x = 100 0.105 / 41,5 = 0.253
Conclusion: it is necessary to apply 0,253 grams of potassium sulfate fertilizer per 1 liter of solution. Let’s fill in the table:
Blend component
We make, g / l
N
P
P2O5
K
K2O
Ca
Dog
Mg
MgO
Magnesium sulfate, wt%
0,31
9.77
16.2
Calcium nitrate, wt%
0,855
14,9
19,3
27
Potassium monophosphate, wt%
0.177
21.8
50
27.39
33
Potassium nitrate, wt%
0,096
13,6
38,18
46
Potassium sulfate, wt%
0,253
41,5
50
In the finished solution, mg / l
140
38.5
190
165
30
Specified solution, mg / l
140
38.5
190
165
30
The ready-made solution corresponds to the finished one – the solution is composed correctly. To prepare a larger amount of solution, we make a simple recalculation by multiplying the applied quantities by the required volume in liters. Calculation example for 5 liters:
Blend component
We make, g / l
We make, g / 5l
Magnesium sulfate
0,31
1,55
Calcium nitrate
0,855
4,275
Potassium monophosphate
0.177
0,885
Potassium nitrate
0,096
0,48
Potassium sulphate
0,253
1,265
As you know, water for the preparation of nutrient solutions may contain a certain amount of dissolved salts, which should be taken into account when compiling nutrient solutions. Let’s say water has the following composition:
Name Ca Mg K Content, mg / l 50 25 30
All that needs to be done is to correct the composition of the solution before starting the calculation. It looks like this:
Blend component
We make, g / l
N
P
K
Ca
Mg
…
In the finished solution, mg / l
Specified solution taking into account the composition of water, mg / l
140
38,5
160
115
5
Specified solution, mg / l
140
38.5
190
165
30
Water, mg / l
30
50
25
Next, we carry out the calculation, similar to the instructions posted above.
Weighing small and minimal amounts of substances can be difficult if an analytical balance is not available. Using household scales for this purpose, one can never be sure of the accuracy of weighing at least up to 0,5 g. There is a simple way of preparing solutions without having accurate scales. Let’s consider the example of a solution of trace elements according to Hoagland.
Let’s prepare in distilled water a 0,5% solution of all compounds of trace elements that we need only in small quantities (for example, tin chloride, potassium iodide, cobalt nitrate, etc.). So, we will dissolve, for example, 5 g of potassium iodide in 1 liter of distilled water. If we need only 0,5 g, then we simply take 100 cubic meters from this solution. cm, which contain exactly 0,5 g. The required quantity of cubic centimeters is measured with an accurate, albeit cheap, pipette, syringe or beaker. Using this method, one should not forget that, according to the recipe for the preparation of Hoagland’s solution, all quantities are indicated per 18 liters of water. Therefore, having dissolved all concentrates prepared by us separately in about 10 liters of water, we only then bring the total amount of liquid to 18 liters with water.
Acidification of the nutrient solution
The nutrient solution usually needs to be acidified. The absorption of ions by plants causes a gradual alkalinization of the solution. Any solution having a pH of 7 or higher will most often need to be adjusted to the optimum pH. Various acids can be used to acidify the nutrient solution, but sulfuric acid is usually used because it is always available and cheap.
When adjusting the pH with both acids and alkalis, rubber gloves should be worn to avoid skin burns. An experienced chemist is adept at handling concentrated sulfuric acid, adding the acid drop by drop to the water. But for novice hydroponists, perhaps it is better to turn to an experienced chemist and ask him to prepare a 25% sulfuric acid solution. While the acid is being added, the solution is stirred and its pH is determined. Having learned the approximate amount of sulfuric acid, in the future it can be added from a graduated cylinder.
Sulfuric acid must be added in small portions so as not to acidify the solution too much, which will then have to be alkalinized again. In an inexperienced worker, acidification and alkalinization can go on indefinitely. In addition to wasting time and reagents, such regulation unbalances the nutrient solution due to the accumulation of ions unnecessary to plants.
Alkalinization of the nutrient solution
Too acidic solutions are made alkaline with caustic sodium (sodium hydroxide). As its name suggests, it is corrosive, so rubber gloves should be worn. It is recommended to purchase sodium hydroxide in pill form. In household chemical stores, sodium hydroxide can be purchased as a pipe cleaner, such as Mole. Dissolve one pellet in 0,5 L of water and gradually add the alkaline solution to the nutrient solution with constant stirring, frequently checking its pH. No mathematical calculations can calculate how much acid or alkali needs to be added in a given case.
If you want to grow several crops in the same pallet, you need to select them so that not only their optimal pH coincides, but also the needs for other growth factors. For example, yellow daffodils and chrysanthemums need a pH of 6,8, but different moisture conditions, so they cannot be grown on the same pallet. If you give daffodils as much moisture as chrysanthemums, the daffodil bulbs will rot. In experiments, rhubarb reached its maximum development at pH 6,5, but it could grow even at pH 3,5. Oats, which prefer a pH of about 6, give good yields at pH 4, if the nitrogen dose in the nutrient solution is greatly increased. Potatoes grow in a fairly wide pH range, but they thrive best at a pH of 5,5. Below this pH, high yields of tubers are also obtained, but they acquire a sour taste. To obtain maximum high quality yields, the pH of the nutrient solutions must be precisely adjusted.
