Dorogov's drug ASD-2 is widely used for therapy various diseases humans and animals. It is intended for both indoor and outdoor use. But most often not pure form, but in solutions. Today we’ll talk about how to prepare a 1% solution.

How to make a 1% solution of ASD-2 for douching, skin treatment and compresses?

The schemes and methods of using the composition are simple. Scientist A.V. Dorogov has developed several protocols for taking the drug for therapy various pathologies. It is according to these schemes that patients are treated. The product is also recommended for external use: lotions, microenemas and vaginal rinsing.

For douching, use a 1% solution. It is very easy to prepare. Need to mix required quantity drops or milliliters of medicine with boiled, slightly cooled water. The ratio of components is 1:100.

If we take 1 ml of medicine, then it must be mixed with 99 ml of water. How to do it easier and more correctly:

1. take 100 ml of boiled water into a measuring cup;
2. use a syringe to take 1 ml (cube) of water from the glass, 99 ml remains;
3. With another syringe, through a puncture of the rubber stopper, according to the instructions for the drug kit, we collect 1 cube of ASD-2;
4. dip the needle of the syringe with the medicine into water;
5. carefully squeeze out the drug;
6. no additional mixing is required, the medicine itself quickly mixes with water;
7. We use the prepared solution immediately, do not store it, otherwise its healing qualities will be lost.

Attention! Do not open the bottle while taking the medicine. When the adaptogen interacts with air, they are lost medicinal properties composition, and it simply becomes inactive.

Since the stimulant has a specific, rather unpleasant aroma, it is preferable to mix it with water near an open window and try not to inhale the vapors of the drug.

In what cases should it be used?

External use of an antiseptic stimulant helps cure a wide variety of ailments, including gynecological and skin ones. The medicine has powerful anti-inflammatory, wound-healing, antibacterial and antiseptic effects. Using the solution will help in:

• cure skin diseases: psoriasis, neurodermatitis, trophic ulcers, eczema;
• therapy of pathologies skin fungal origin;
• accelerating the wound healing process;
• treatment of gynecological ailments: thrush, endometriosis, cervical erosion, uterine fibroids.

Douching with diluted liquid should be done two to three times a day. The duration of the therapeutic course is until complete recovery.

Dorogov’s drug is very effective and unique. Use it as directed and in the correct dosage, and it will help in curing many ailments.

(get a less concentrated solution from a more concentrated solution)

1 action:

Number of ml of a more concentrated solution (which must be diluted)

Required volume in ml (to be prepared)

Concentration of the less concentrated solution (the one you want to obtain)

Concentration of a more concentrated solution (the one we are diluting)

Action 2:

Number of ml of water (or diluent) = or water up to (ad) required volume ()

Task No. 6. A bottle of ampicillin contains 0.5 dry medicine. How much solvent do you need to take so that 0.5 ml of solution contains 0.1 g of dry matter?

Solution: when diluting the antibiotic per 0.1 g of dry powder, take 0.5 ml of solvent, therefore, if,

0.1 g dry matter – 0.5 ml solvent

0.5 g dry matter - x ml solvent

we get:

Answer: In order for 0.5 ml of solution to contain 0.1 g of dry matter, it is necessary to take 2.5 ml of solvent.

Task No. 7. A bottle of penicillin contains 1 million units of dry medicine. How much solvent do you need to take so that 0.5 ml of solution contains 100,000 units of dry matter?

Solution: 100,000 units of dry matter – 0.5 ml of dry matter, then 100,000 units of dry matter – 0.5 ml of dry matter.

1000000 units – x

Answer: In order for 0.5 ml of solution to contain 100,000 units of dry matter, it is necessary to take 5 ml of solvent.

Task No. 8. A bottle of oxacillin contains 0.25 dry medicine. How much solvent do you need to take so that 1 ml of solution contains 0.1 g of dry matter?

Solution:

1 ml of solution – 0.1 g

x ml - 0.25 g

Answer: In order for 1 ml of solution to contain 0.1 g of dry substance, you need to take 2.5 ml of solvent.

Problem No. 9. The price of dividing an insulin syringe is 4 units. How many divisions of the syringe correspond to 28 units? insulin? 36 units? 52 units?

Solution: In order to find out how many divisions of the syringe correspond to 28 units. insulin required: 28:4 = 7 (divisions).

Similar: 36:4=9 (divisions)

52:4=13(divisions)

Answer: 7, 9, 13 divisions.

Problem No. 10. How much do you need to take a 10% solution of clarified bleach and water (in liters) to prepare 10 liters of a 5% solution.

Solution:

1) 100 g – 5 g

(G) active substance

2) 100% – 10g

(ml) 10% solution

3) 10000-5000=5000 (ml) water

Answer: you need to take 5000 ml of clarified bleach and 5000 ml of water.

Problem No. 11. How much do you need to take a 10% solution of bleach and water to prepare 5 liters of a 1% solution.

Solution:

Since 100 ml contains 10 g of active substance,

1) 100g – 1ml

5000 ml – x

(ml) active substance

2) 100% – 10ml

00 (ml) 10% solution

3) 5000-500=4500 (ml) water.

Answer: you need to take 500 ml of a 10% solution and 4500 ml of water.

Problem No. 12. How much do you need to take a 10% solution of bleach and water to prepare 2 liters of a 0.5% solution.

Solution:

Since 100 ml contains 10 ml of active substance,

1) 100% – 0.5ml

0 (ml) active substance

2) 100% – 10 ml

(ml) 10% solution

3) 2000-100=1900 (ml) water.

Answer: you need to take 10 ml of a 10% solution and 1900 ml of water.

Problem No. 13. How much chloramine (dry substance) per g and water is needed to prepare 1 liter of a 3% solution.

Solution:

1) 3g – 100 ml

G

2) 10000 – 300=9700ml.

Answer: To prepare 10 liters of a 3% solution, you need to take 300 g of chloramine and 9700 ml of water.

Problem No. 14. How much chloramine (dry) should be taken in g and water to prepare 3 liters of a 0.5% solution.

Solution:

Percentage is the amount of substance in 100 ml.

1) 0.5 g – 100 ml

G

2) 3000 – 15 = 2985 ml.

Answer: to prepare 10 liters of a 3% solution you need to take 15g of chloramine and 2985ml of water

Problem No. 15 . How much chloramine (dry) should be taken in g and water to prepare 5 liters of a 3% solution.

Solution:

Percentage is the amount of substance in 100 ml.

1) 3 g – 100 ml

G

2) 5000 – 150= 4850 ml.

Answer: To prepare 5 liters of a 3% solution, you need to take 150 g of chloramine and 4850 ml of water.

Problem No. 16. To apply a warm compress from a 40% solution ethyl alcohol you need to take 50ml. How much 96% alcohol do you need to use to apply a warm compress?

Solution:

According to formula (1)

ml

Answer: To prepare a warming compress from a 96% ethyl alcohol solution, you need to take 21 ml.

Problem No. 17. Prepare 1 liter of 1% bleach solution for treating equipment from 1 liter of 10% stock solution.

Solution: Calculate how many ml of 10% solution you need to take to prepare a 1% solution:

10g – 1000 ml

Answer: To prepare 1 liter of a 1% bleach solution, you need to take 100 ml of a 10% solution and add 900 ml of water.

Problem No. 18. The patient should take the medicine 1 mg in powders 4 times a day for 7 days, then how much of this medicine needs to be prescribed (calculation is in grams).

Solution: 1g = 1000mg, therefore 1mg = 0.001g.

Calculate how much medication the patient needs per day:

4* 0.001 g = 0.004 g, therefore, for 7 days he needs:

7* 0.004 g = 0.028 g.

Answer: This medicine must be prescribed 0.028 g.

Problem No. 19. The patient needs to be administered 400 thousand units of penicillin. Bottle of 1 million units. Dilute 1:1. How many ml of solution should be taken?

Solution: When diluted 1:1, 1 ml of solution contains 100 thousand action units. 1 bottle of penicillin, 1 million units each, diluted in 10 ml of solution. If the patient needs to administer 400 thousand units, then it is necessary to take 4 ml of the resulting solution.

Answer: you need to take 4 ml of the resulting solution.

Problem No. 20. Inject the patient with 24 units of insulin. The syringe division price is 0.1 ml.

Solution: 1 ml of insulin contains 40 units of insulin. 0.1 ml of insulin contains 4 units of insulin. To administer 24 units of insulin to a patient, you need to take 0.6 ml of insulin.

To prepare solutions of molar and normal concentrations, a sample of the substance is weighed on an analytical balance, and the solutions are prepared in a volumetric flask. When preparing acid solutions, the required volume of concentrated acid solution is measured with a burette with a glass stopcock.

The weight of the solute is calculated to the fourth decimal place, and the molecular weights are taken with the accuracy with which they are given in the reference tables. The volume of concentrated acid is calculated accurate to the second decimal place.

Example 1. How many grams of barium chloride are needed to prepare 2 liters of 0.2 M solution?

Solution. The molecular weight of barium chloride is 208.27. Therefore. 1 liter of 0.2 M solution should contain 208.27-0.2= = 41.654 g BaCl 2 . To prepare 2 liters you will need 41.654-2 = 83.308 g of BaCl 2.

Example 2. How many grams of anhydrous soda Na 2 C0 3 are required to prepare 500 ml of 0.1 N. solution?

Solution. The molecular weight of soda is 106.004; equivalent unit mass 5 N a 2 C0 3 =M: 2 = 53.002; 0.1 eq. = 5.3002 g

1000 ml 0.1 n. solution contain 5.3002 g Na 2 C0 3
500 »» » » » X " Na 2 C0 3

5,3002-500
x=—— Gooo—- = 2-6501 g Na 2 C0 3.

Example 3. How much concentrated sulfuric acid (96%: d=l.84) is required to prepare 2 liters of 0.05 N. sulfuric acid solution?

Solution. The molecular weight of sulfuric acid is 98.08. Equivalent mass of sulfuric acid 3h 2 so 4 = M: 2 = 98.08: 2 = 49.04 g. Mass 0.05 eq. = 49.04-0.05 = 2.452 g.

Let's find how much H 2 S0 4 should be contained in 2 liters of 0.05 n. solution:

1 l-2.452 g H 2 S0 4

2"- X » H 2 S0 4

X = 2.452-2 = 4.904 g H 2 S0 4.

To determine how much 96.% H 2 S0 4 solution needs to be taken for this, let’s make a proportion:

\ in 100 g of conc. H 2 S0 4 -96 g H 2 S0 4

U» » H 2 S0 4 -4.904 g H 2 S0 4

4,904-100
U=——— §6—— = 5.11 g H 2 S0 4 .

We recalculate this amount to volume: ,. R 5,11

K = 7 = TJ = 2 ‘ 77 ml -

Thus, to prepare 2 liters of 0.05 N. solution you need to take 2.77 ml of concentrated sulfuric acid.

Example 4. Calculate the titer of a NaOH solution if it is known that its exact concentration is 0.0520 N.

Solution. Let us recall that the titer is the content in 1 ml of a solution of a substance in grams. Equivalent mass of NaOH = 40 01 g Let's find how many grams of NaOH are contained in 1 liter of this solution:

40.01-0.0520 = 2.0805 g.

1 liter of solution: -n=- =0.00208 g/ml. You can also use the formula:

9N

Where T- titer, g/ml; E- equivalent mass; N- normality of the solution.

Then the titer of this solution:

f 40,01 0,0520

“NaOH =——— jooo—— 0.00208 g/ml.

„ “Rie P 5 - Calculate the normal concentration of the HN0 3 solution if it is known that the titer of this solution is 0.0065. To calculate, we use the formula:

T ■ 1000 63,05

5hno 3 = j- = 63.05.

The normal concentration of nitric acid solution is:

- V = 63.05 = 0.1030 n.

Example 6. What is normal concentration solution, if it is known that 200 ml of this solution contains 2.6501 g of Na 2 C0 3

Solution. As was calculated in example 2, Zma 2 co(=53.002.
Let's find how many equivalents are 2.6501 g of Na 2 C0 3: G
2.6501: 53.002 = 0.05 eq. /

In order to calculate the normal concentration of a solution, we create a proportion:

1000 "" X "

1000-0,05
x = —————— =0.25 eq.

1 liter of this solution will contain 0.25 equivalents, i.e. the solution will be 0.25 N.

For this calculation you can use the formula:

R- 1000

Where R - amount of substance in grams; E - equivalent mass of the substance; V - volume of solution in milliliters.

Zia 2 with 3 = 53.002, then the normal concentration of this solution

2.6501-10С0 N = 53.002-200

Determine what you know and what you don't. In chemistry, dilution usually means obtaining small quantity a solution of known concentration, followed by diluting it with a neutral liquid (for example, water) and thus obtaining a less concentrated solution of a larger volume. This operation is very often used in chemical laboratories, so reagents are stored in concentrated form for convenience and diluted if necessary. In practice, as a rule, you know the initial concentration, as well as the concentration and volume of the solution you want to obtain; wherein The volume of the concentrated solution that needs to be diluted is unknown.

• In another situation, for example, when solving a school problem in chemistry, another quantity may act as an unknown quantity: for example, you are given the initial volume and concentration, and you need to find the final concentration of the final solution with its known volume. In any case, it is useful to write down the known and unknown quantities before starting a problem.
• Let's look at an example. Let us say we need to dilute a solution with a concentration of 5 M to obtain a solution with a concentration of 1 mmM. In this case, we know the concentration of the initial solution, as well as the volume and concentration of the solution that needs to be obtained; Not the volume of the original solution that must be diluted with water is known.
  • Remember: in chemistry, M serves as a measure of concentration, also called molarity, which corresponds to the number of moles of a substance per 1 liter of solution.

Substitute the known values ​​into the formula C 1 V 1 = C 2 V 2. In this formula, C 1 is the concentration of the initial solution, V 1 is its volume, C 2 is the concentration final solution, and V 2 is its volume. From the resulting equation you can easily determine the desired value.

• Sometimes it is useful to put a question mark in front of the quantity you are trying to find.
• Let's return to our example. Let’s substitute the values ​​we know into the equation:
  • C 1 V 1 = C 2 V 2
  • (5 M)V 1 = (1 mM)(1 l). Concentrations have different units of measurement. Let's look at this in a little more detail.

Please account for any differences in units of measurement. Because dilution leads to a decrease in concentration, often significantly, concentrations are sometimes measured in different units. If you miss this, you could be off by several orders of magnitude. Before solving the equation, convert all concentration and volume values ​​to the same units.

• In our case, two concentration units are used, M and mM. Let's convert everything to M:
  • 1 mM × 1 M/1.000 mM
  • = 0.001 M.

Let's solve the equation. When you have reduced all quantities to the same units, you can solve the equation. To solve it, knowledge of simple algebraic operations is almost always sufficient.

• For our example: (5 M)V 1 = (1 mM)(1 l). Reducing everything to the same units, we solve the equation for V 1.
  • (5 M)V 1 = (0.001 M)(1 L)
  • V 1 = (0.001 M)(1 l)/(5 M).
  • V 1 = 0.0002 l, or 0.2 ml.

Think about applying your results in practice. Let's say you have calculated the desired value, but you are still finding it difficult to prepare a real solution. This situation is quite understandable - the language of mathematics and pure science is sometimes far from real world. If you already know all four quantities included in the equation C 1 V 1 = C 2 V 2, proceed as follows:

• Measure the volume V 1 of a solution with concentration C 1 . Then add dilution liquid (water, etc.) so that the volume of the solution becomes equal to V 2. This new solution will have the required concentration (C 2).
• In our example, we first measure out 0.2 ml of the original solution with a concentration of 5 M. Then we dilute it with water to a volume of 1 l: 1 l - 0.0002 l = 0.9998 l, i.e. add 999.8 ml of water to it. The resulting solution will have the concentration we need 1 mM.

Approximate solutions. When preparing approximate solutions, the amounts of substances that must be taken for this purpose are calculated with little accuracy. To simplify calculations, the atomic weights of elements can sometimes be taken rounded to whole units. So, for a rough calculation, the atomic weight of iron can be taken equal to 56 instead of the exact -55.847; for sulfur - 32 instead of the exact 32.064, etc.

Substances for preparing approximate solutions are weighed on technochemical or technical balances.

In principle, the calculations when preparing solutions are exactly the same for all substances.

The amount of the prepared solution is expressed either in units of mass (g, kg) or in units of volume (ml, l), and for each of these cases the amount of solute is calculated differently.

Example. Let it be required to prepare 1.5 kg of 15% sodium chloride solution; We first calculate the required amount of salt. The calculation is carried out according to the proportion:

i.e. if 100 g of solution contains 15 g of salt (15%), then how much of it will be required to prepare 1500 g of solution?

The calculation shows that you need to weigh out 225 g of salt, then take 1500 - 225 = 1275 g of iuzhio water.

If you are asked to obtain 1.5 liters of the same solution, then in this case you will find out its density from the reference book, multiply the latter by the given volume and thus find the mass of the required amount of solution. Thus, the density of a 15% noro sodium chloride solution at 15 0C is 1.184 g/cm3. Therefore, 1500 ml is

Therefore, the amount of substance for preparing 1.5 kg and 1.5 liters of solution is different.

The calculation given above is applicable only for the preparation of solutions of anhydrous substances. If an aqueous salt is taken, for example Na2SO4-IOH2O1, then the calculation is slightly modified, since the water of crystallization must also be taken into account.

Example. Let you need to prepare 2 kg of 10% Na2SO4 solution based on Na2SO4 * 10H2O.

The molecular weight of Na2SO4 is 142.041, and Na2SO4*10H2O is 322.195, or rounded to 322.20.

The calculation is carried out first using anhydrous salt:

Therefore, you need to take 200 g of anhydrous salt. The amount of salt decahydrate is calculated from the calculation:

In this case, you need to take water: 2000 - 453.7 = 1546.3 g.

Since the solution is not always prepared in terms of anhydrous salt, the label, which must be stuck on the container with the solution, must indicate what salt the solution is prepared from, for example, a 10% solution of Na2SO4 or 25% Na2SO4 * 10H2O.

It often happens that a previously prepared solution needs to be diluted, i.e., its concentration must be reduced; solutions are diluted either by volume or by weight.

Example. It is necessary to dilute a 20% solution of ammonium sulfate so as to obtain 2 liters of a 5% solution. We carry out the calculation in the following way. From the reference book we find out that the density of a 5% solution of (NH4)2SO4 is 1.0287 g/cm3. Therefore, 2 liters of it should weigh 1.0287 * 2000 = 2057.4 g. This amount should contain ammonium sulfate:

Considering that losses may occur during measuring, you need to take 462 ml and bring them to 2 liters, i.e. add 2000-462 = 1538 ml of water to them.

If the dilution is carried out by mass, the calculation is simplified. But in general, dilution is carried out based on volume, since liquids, especially in large quantities, it is easier to measure by volume than to weigh.

It must be remembered that in any work with both dissolution and dilution, you should never pour all the water into the vessel at once. The container in which the required substance was weighed or measured is rinsed several times with water, and each time this water is added to the solution vessel.

When special precision is not required, when diluting solutions or mixing them to obtain solutions of a different concentration, you can use the following simple and quick method.

Let's take the already discussed case of diluting a 20% solution of ammonium sulfate to 5%. First we write like this:

where 20 is the concentration of the solution taken, 0 is water and 5" is the required concentration. Now subtract 5 from 20 and write the resulting value in the lower right corner, subtracting zero from 5, write the number in the upper right corner. Then the diagram will look like this :

This means that you need to take 5 volumes of a 20% solution and 15 volumes of water. Of course, such a calculation is not very accurate.

If you mix two solutions of the same substance, the scheme remains the same, only the numerical values ​​change. Suppose that by mixing a 35% solution and a 15% solution, you need to prepare a 25% solution. Then the diagram will look like this:

i.e. you need to take 10 volumes of both solutions. This scheme gives approximate results and can be used only when special accuracy is not required. It is very important for every chemist to cultivate the habit of accuracy in calculations when necessary, and to use approximate figures in cases where this will not affect the results work. When greater accuracy is needed when diluting solutions, the calculation is carried out using formulas.

Let's look at a few of the most important cases.

Preparation of a diluted solution. Let c be the amount of solution, m% the concentration of the solution that needs to be diluted to a concentration of p%. The resulting amount of diluted solution x is calculated using the formula:

and the volume of water v for diluting the solution is calculated by the formula:

Mixing two solutions of the same substance different concentrations to obtain a solution of a given concentration. Let by mixing a parts of an m% solution with x parts of a p% solution we need to obtain a /% solution, then:

Precise solutions. When preparing accurate solutions, the calculation of the quantities of the required substances will be checked with a sufficient degree of accuracy. Atomic weights of elements are taken from the table, which shows their exact values. When adding (or subtracting) use exact value the term with the fewest decimal places. The remaining terms are rounded, leaving one decimal place after the decimal place than in the term with the smallest number of decimal places. As a result, as many digits after the decimal point are left as there are in the term with the smallest number of decimal places; in this case, the necessary rounding is performed. All calculations are made using logarithms, five-digit or four-digit. The calculated quantities of the substance are weighed only on an analytical balance.

Weighing is carried out either on a watch glass or in a weighing bottle. The weighed substance is poured into a clean, washed volumetric flask through a clean, dry funnel. in small portions. Then, from the washing machine, the glass or watch glass in which the weighing was carried out is washed several times with small portions of water over the funnel. The funnel is also washed several times from the washing machine with distilled water.

To pour solid crystals or powders into a volumetric flask, it is very convenient to use the funnel shown in Fig. 349. Such funnels are made with a capacity of 3, 6, and 10 cm3. You can weigh the sample directly in these funnels (non-hygroscopic materials), having previously determined their mass. The sample from the funnel is very easily transferred to a volumetric flask. When the sample is poured, the funnel, without removing it from the neck of the flask, is washed well with distilled water from the rinse.

As a rule, when preparing accurate solutions and transferring the solute into a volumetric flask, the solvent (for example, water) should occupy no more than half the capacity of the flask. Stopper the volumetric flask and shake it until the solid is completely dissolved. After this, the resulting solution is added to the mark with water and mixed thoroughly.

Molar solutions. To prepare 1 liter of a 1 M solution of a substance, 1 mole of it is weighed out on an analytical balance and dissolved as indicated above.

Example. To prepare 1 liter of 1 M solution of silver nitrate, find the molecular weight of AgNO3 in the table or calculate it, it is equal to 169.875. The salt is weighed out and dissolved in water.

If you need to prepare a more dilute solution (0.1 or 0.01 M), weigh out 0.1 or 0.01 mol of salt, respectively.

If you need to prepare less than 1 liter of solution, then dissolve a correspondingly smaller amount of salt in the corresponding volume of water.

Normal solutions are prepared in the same way, only by weighing out not 1 mole, but 1 gram equivalent of the solid.

If you need to prepare a half-normal or decinormal solution, take 0.5 or 0.1 gram equivalent, respectively. When preparing not 1 liter of solution, but less, for example 100 or 250 ml, then take 1/10 or 1/4 of the amount of substance required to prepare 1 liter and dissolve it in the appropriate volume of water.

Fig. 349. Funnels for pouring the sample into the flask.

After preparing a solution, it must be checked by titration with a corresponding solution of another substance of known normality. The prepared solution may not correspond exactly to the normality that is specified. In such cases, an amendment is sometimes introduced.

In production laboratories, exact solutions are sometimes prepared “according to the substance being determined.” The use of such solutions facilitates calculations during analysis, since it is enough to multiply the volume of the solution used for titration by the titer of the solution in order to obtain the content of the desired substance (in g) in the amount of any solution taken for analysis.

When preparing a titrated solution for the analyte, calculations are also carried out using the gram equivalent of the soluble substance, using the formula:

Example. Suppose you need to prepare 3 liters of potassium permanganate solution with an iron titer of 0.0050 g/ml. The gram equivalent of KMnO4 is 31.61, and the gram equivalent of Fe is 55.847.

We calculate using the above formula:

Standard solutions. Standard solutions are solutions with different, precisely defined concentrations used in colorimetry, for example, solutions containing 0.1, 0.01, 0.001 mg, etc. of dissolved substance in 1 ml.

In addition to colorimetric analysis, such solutions are needed when determining pH, for nephelometric determinations, etc. Sometimes standard solutions are stored in sealed ampoules, but more often they have to be prepared immediately before use. Standard solutions are prepared in a volume of no more than 1 liter, and more often - less. Only with a large consumption of the standard solution can you prepare several liters of it, and then only on the condition that the standard solution will not be stored for a long time.

The amount of substance (in g) required to obtain such solutions is calculated using the formula:

Example. It is necessary to prepare standard solutions of CuSO4 5H2O for the colorimetric determination of copper, and 1 ml of the first solution should contain 1 mg of copper, the second - 0.1 mg, the third - 0.01 mg, the fourth - 0.001 mg. First, prepare a sufficient amount of the first solution, for example 100 ml.