Formula for solving problems for diluting solutions. Preparation and use of soda solution
(obtain a less concentrated solution from a more concentrated solution)
1 action:
Number of ml of a more concentrated solution (to be diluted)
Required volume in ml (to be prepared)
The concentration of a less concentrated solution (the one that needs to be obtained)
The concentration of a more concentrated solution (the one that we dilute)
2 action:
Number of ml of water (or diluent) = or water up to (ad) the required volume ()
Task number 6. In a vial of ampicillin is 0.5 dry drug. How much solvent should be taken in order to have 0.1 g of dry matter in 0.5 ml of solution.
Decision: when diluting the antibiotic to 0.1 g of dry powder, 0.5 ml of the solvent is taken, therefore, if,
0.1 g dry matter - 0.5 ml solvent
0.5 g of dry matter - x ml of solvent
we get:
Answer: in order to have 0.1 g of dry matter in 0.5 ml of the solution, 2.5 ml of the solvent must be taken.
Task number 7. In a vial of penicillin is 1 million units of a dry drug. How much solvent should be taken in order to have 100,000 units of dry matter in 0.5 ml of solution.
Decision: 100,000 units of dry matter - 0.5 ml of dry matter, then in 100,000 units of dry matter - 0.5 ml of dry matter.
1000000 U - x
Answer: in order to have 100,000 units of dry matter in 0.5 ml of the solution, it is necessary to take 5 ml of the solvent.
Task number 8. In a vial of oxacillin is 0.25 dry drug. How much solvent do you need to take in order to have 0.1 g of dry matter in 1 ml of solution
Decision:
1 ml of solution - 0.1 g
x ml - 0.25 g
Answer: in order to have 0.1 g of dry matter in 1 ml of the solution, 2.5 ml of the solvent must be taken.
Task #9. The price of division of an insulin syringe is 4 units. How many divisions of the syringe corresponds to 28 units. insulin? 36 units? 52 units?
Decision: In order to find out how many divisions of the syringe corresponds to 28 units. insulin needed: 28:4 = 7 (divisions).
Similarly: 36:4=9(divisions)
52:4=13(divisions)
Answer: 7, 9, 13 divisions.
Task number 10. How much you need to take a 10% solution of clarified bleach and water (in liters) to prepare 10 liters of a 5% solution.
Decision:
1) 100 g - 5g
(d) active substance
2) 100% - 10g
(ml) 10% solution
3) 10000-5000=5000 (ml) water
Answer: it is necessary to take 5000 ml of clarified bleach and 5000 ml of water.
Task number 11. How much you need to take a 10% solution of bleach and water to prepare 5 liters of a 1% solution.
Decision:
Since 100 ml contains 10 g of the active substance,
1) 100g - 1ml
5000 ml - x
(ml) active substance
2) 100% - 10ml
00 (ml) 10% solution
3) 5000-500=4500 (ml) of water.
Answer: it is necessary to take 500 ml of a 10% solution and 4500 ml of water.
Task number 12. How much you need to take a 10% solution of bleach and water to prepare 2 liters of a 0.5% solution.
Decision:
Since 100 ml contains 10 ml of the active substance,
1) 100% - 0.5 ml
0 (ml) active ingredient
2) 100% - 10 ml
(ml) 10% solution
3) 2000-100=1900 (ml) of water.
Answer: it is necessary to take 10 ml of a 10% solution and 1900 ml of water.
Task number 13. How much chloramine (dry matter) should be taken in g and water to prepare 1 liter of a 3% solution.
Decision:
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.
Task number 14. How much chloramine (dry) should be taken in g and water to prepare 3 liters of a 0.5% solution.
Decision:
Percentage - the amount of a 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 15 g of chloramine and 2985 ml of water
Task number 15 . How much chloramine (dry) should be taken in g and water to prepare 5 liters of a 3% solution.
Decision:
Percentage - the amount of a substance in 100 ml.
1) 3 g - 100 ml
G
2) 5000 - 150= 4850ml.
Answer: to prepare 5 liters of a 3% solution, you need to take 150 g of chloramine and 4850 ml of water.
Task number 16. For setting a warm compress from a 40% solution ethyl alcohol you need to take 50 ml. How much 96% alcohol should I take to apply a warm compress?
Decision:
According to formula (1)
ml
Answer: To prepare a warming compress from a 96% solution of ethyl alcohol, you need to take 21 ml.
Task number 17. Prepare 1 liter of 1% bleach solution for inventory processing from 1 liter of stock 10% solution.
Decision: 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, take 100 ml of a 10% solution and add 900 ml of water.
Task number 18. The patient should take the medicine 1 mg in powders 4 times a day for 7 days, then how much it is necessary to prescribe this medicine (calculation is carried out in grams).
Decision: 1g = 1000mg, therefore 1mg = 0.001g.
Calculate how much the patient needs medication per day:
4 * 0.001 g \u003d 0.004 g, therefore, for 7 days he needs:
7* 0.004 g = 0.028 g.
Answer: of this medicine, it is necessary to write out 0.028 g.
Task number 19. The patient needs to enter 400 thousand units of penicillin. Bottle of 1 million units. Dilute 1:1. How many ml of solution should be taken.
Decision: When diluted 1:1, 1 ml of the solution contains 100 thousand units of action. 1 bottle of penicillin 1 million units diluted with 10 ml of solution. If the patient needs to enter 400 thousand units, then you need to take 4 ml of the resulting solution.
Answer: you need to take 4 ml of the resulting solution.
Task number 20. Give the patient 24 units of insulin. The division price of the syringe is 0.1 ml.
Decision: 1 ml of insulin contains 40 units of insulin. 0.1 ml of insulin contains 4 units of insulin. To enter the patient 24 units of insulin, you need to take 0.6 ml of insulin.
Salt solution may be needed for a variety of purposes, for example, it is part of some products. traditional medicine. So how to cook 1- percentage solution if there are no special beakers at home to measure the amount of the product? In general, even without them, you can make a 1% salt solution. How to cook it is detailed below. Before proceeding with the preparation of such a solution, you should carefully study the recipe and accurately determine the necessary ingredients. The thing is that the definition of "salt" can refer to different substances. Sometimes it turns out to be normal edible salt, sometimes stone or even sodium chloride. As a rule, in detailed recipe it is always possible to find an explanation of which substance is recommended to be used. AT folk recipes often magnesium sulfate is also indicated, which has the second name "epsom salt".
If a substance is required, for example, for gargling or relieving pain from a tooth, then most often in this case it is recommended to use it saline solution sodium chloride. In order for the resulting product to have healing properties and did not harm the human body, it should be selected for it exclusively quality ingredients. For example, rock salt contains a lot of extra impurities, so instead of it it is better to use ordinary fine salt (you can also use iodized salt for rinsing). As for water, at home you should use filtered or at least boiled water. Some recipes recommend using rain water or snow. But, given the current ecological state, this is not worth doing. Especially for residents of large cities. It is better to just thoroughly clean the tap water.
If there was no special filter at home, then the well-known "old-fashioned" method can be used to purify water. It involves freezing tap water in the freezer. As you know, in the process, it is the purest liquid that first turns into ice, and all harmful impurities and dirt sink to the bottom of the container. Without waiting for the freezing of the entire glass, you should remove the upper ice part and then melt it. Such water will be as pure and safe as possible for health. It can be used to prepare saline solution.
Now it is worth deciding on the units of measurement for liquid and solid matter. For salt, it is most convenient to use a teaspoon. As you know, it contains 7 grams of the product, if the spoon is with a slide, then 10. The latter option is more convenient to use to calculate the percentage. It is easy to measure water with an ordinary faceted glass if there are no special beakers in the house. It contains 250 milliliters of water. Weight 250 milliliters net fresh water is equal to 250 grams. It is most convenient to use half a glass of liquid or 100 grams. Next is the most difficult stage of preparing the saline solution. It is worth once again carefully studying the recipe and deciding on the proportions. If it is recommended to take a 1% salt solution in it, then in every 100 grams of liquid, 1 gram of solid will need to be dissolved. The most accurate calculations will suggest that it will be necessary to take 99 grams of water and 1 gram of salt, but such accuracy is unlikely to be required.
It is quite possible to allow some error and, for example, add one heaping teaspoon of salt to one liter of water to get a 1% saline solution. Currently, it is often used, for example, in the treatment colds and especially sore throats. You can also add soda or a few drops of iodine to the finished solution. The resulting rinse mixture will be an excellent effective and effective tool against sore throat. Unpleasant sensations will go away after just a few procedures. By the way, such a solution is not prohibited for use by the smallest family members. The main thing is not to overdo it with additional ingredients(especially with iodine), otherwise you can damage the mucous membrane of the oral cavity and only aggravate the condition of the sore throat.
Also, a saline solution can be used to relieve a pulling aching toothache. True, it is more efficient to use a more saturated one, for example, 10 percent. Such a mixture is really able to relieve painful discomfort in the oral cavity for a short time. But she is not medicine Therefore, it is in no case possible to postpone a visit to the dentist after relief.
approximate solutions. When preparing approximate solutions, the amounts of substances that must be taken for this are calculated with little accuracy. Atomic weights of elements to simplify calculations can be taken rounded sometimes 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 the preparation of approximate solutions are weighed on technochemical or technical scales.
Fundamentally, the calculations in the preparation of 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 calculation of the amount of the dissolved substance is carried out differently.
Example. Let it be required to prepare 1.5 kg of a 15% sodium chloride solution; pre-calculate the required amount of salt. The calculation is carried out according to the proportion:
i.e. if 100 g of the solution contains 15 g of salt (15%), then how much will it take to prepare 1500 g of the solution?
The calculation shows that you need to weigh 225 g of salt, then take 1500 - 225 = 1275 g. ¦
If it is given to obtain 1.5 liters of the same solution, then in this case, according to the reference book, its density is found out, the latter is multiplied by the given volume and thus the mass of the required amount of solution is found. Thus, the density of a 15%-horo solution of sodium chloride 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 l 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 somewhat modified, since crystallization water must also be taken into account.
Example. Let it be necessary to prepare 2 kg of 10% Na2SO4 solution starting from Na2SO4 *10H2O.
The molecular weight of Na2SO4 is 142.041 and Na2SO4*10H2O is 322.195, or rounded 322.20.
The calculation is carried out first on anhydrous salt:
Therefore, you need to take 200 g of anhydrous salt. The amount of decahydrate salt is found from the calculation:
Water in this case must be taken: 2000 - 453.7 \u003d 1546.3 g.
Since the solution is not always prepared in terms of anhydrous salt, then on the label, which must be stuck on the vessel with the solution, it is necessary to indicate from which salt the solution is prepared, for example, 10% Na2SO4 solution or 25% Na2SO4 * 10H2O.
It often happens that the previously prepared solution needs to be diluted, i.e., its concentration should 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. We learn from the reference book 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 weight, the calculation is simplified. But in general, dilution is carried out on a volume basis, since liquids, especially in large quantities It is easier to measure by volume than to weigh.
It must be remembered that in all work, both with dissolution and dilution, one should never pour all the water into the vessel at once. Rinse with water several times the dishes in which the weighing or measuring of the desired substance was carried out, and each time this water is added to the vessel for the solution.
When special accuracy 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 us take the already analyzed case of diluting a 20% solution of ammonium sulphate 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 we subtract 5 from 20 and write the resulting value in the lower right corner, subtracting zero from 5, we write the number in the upper right corner. Then the circuit 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 accurate.
If you mix two solutions of the same substance, then the scheme remains the same, only the numerical values \u200b\u200bare changed. Let a 25% solution be prepared by mixing a 35% solution and a 15% 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 any 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 some of the most important cases.
Preparing a diluted solution. Let c be the amount of solution, m% is the concentration of the solution to be diluted to a concentration of n%. The resulting amount of dilute solution x is calculated by the formula:
and the volume of water v for diluting the solution is calculated by the formula:
Mixing two solutions of the same substance of different concentration to obtain a solution of a given concentration. Let by mixing a parts of an m% solution with x parts of a n% solution, you need to obtain a /% solution, then:
precise solutions. When preparing exact solutions, the calculation of the quantities of the required substances will be checked already with a sufficient degree of accuracy. The atomic weights of the elements are taken from the table, which shows their exact values. When adding (or subtracting) use exact value term with the fewest decimal places. The remaining terms are rounded off, leaving one more decimal place after the decimal point than in the term with the least number of digits. As a result, as many digits after the decimal point are left as there are in the term with the least number of decimal places; while doing the necessary rounding. All calculations are made using logarithms, five-digit or four-digit. The calculated amounts of the substance are weighed only on an analytical balance.
Weighing is carried out either on a watch glass or in a bottle. The weighed substance is poured into a cleanly washed volumetric flask through a clean, dry funnel in small portions. Then, from the washer, several times with small portions of water, the bnzhe or the watch glass in which the weighing was carried out is washed over the funnel. The funnel is also washed several times with distilled water.
For pouring 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 the volumetric flask. When the sample is poured, the funnel, without removing the flask from the throat, is well washed with distilled water from the wash bottle.
As a rule, when preparing accurate solutions and transferring the solute to 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 dissolves completely. The resulting solution is then filled up to the mark with water and mixed thoroughly.
molar solutions. To prepare 1 liter of a 1 M solution of a substance, 1 mol of it is weighed on an analytical balance and dissolved as described above.
Example. To prepare 1 liter of 1 M solution of silver nitrate, find in the table or calculate the molecular weight of AgNO3, it is equal to 169.875. Salt is weighed and dissolved in water.
If you need to prepare a more dilute solution (0.1 or 0.01 M), weigh out respectively 0.1 or 0.01 mol of salt.
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 a similar way, only weighing not 1 mole, but 1 gram equivalent of a solid.
If you need to prepare a semi-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 the substance required to prepare 1 liter and dissolve in the appropriate volume of water.
Fig 349. Funnels for pouring a sample into a flask.
After preparing the solution, it must be checked by titration with an appropriate solution of another substance with a known normality. The prepared solution may not correspond exactly to the normality that is given. In such cases, an amendment is sometimes introduced.
In production laboratories, accurate solutions are sometimes prepared “by the substance to be determined”. The use of such solutions facilitates calculations in analyzes, 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, the calculation is also carried out according to the gram equivalent of the dissolved substance, using the formula:
Example. Let it be necessary 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 according to the above formula:
standard solutions. Standard solutions are called solutions with different, precisely defined concentrations used in colorimetry, for example, solutions containing 0.1, 0.01, 0.001 mg, etc. of a solute 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, it is possible to prepare several liters of it, and then on 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 by 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.
Not everyone remembers what “concentration” means and how to properly prepare a solution. If you want to get a 1% solution of any substance, then dissolve 10 g of the substance in a liter of water (or 100 g in 10 liters). Accordingly, a 2% solution contains 20 g of the substance in a liter of water (200 g in 10 liters), and so on.
If it is difficult to measure a small amount, take a larger one, prepare the so-called stock solution and then dilute it. We take 10 grams, prepare a liter of a 1% solution, pour 100 ml, bring them to a liter with water (we dilute 10 times), and a 0.1% solution is ready.
How to make a solution of copper sulfate
To prepare 10 liters of copper-soap emulsion, you need to prepare 150-200 g of soap and 9 liters of water (rain is better). Separately, 5-10 g of copper sulfate are dissolved in 1 liter of water. After that, a solution of copper sulphate is added in a thin stream to soap solution while continuing to stir well. The result is a greenish liquid. If you mix poorly or rush, then flakes form. In this case, it is better to start the process from the very beginning.
How to prepare a 5% solution of potassium permanganate
To prepare a 5% solution, you need 5 g of potassium permanganate and 100 ml of water. First of all, pour water into the prepared container, then add the crystals. Then mix all this until a uniform and saturated purple color of the liquid. Before use, it is recommended to strain the solution through cheesecloth to remove undissolved crystals.
How to prepare a 5% urea solution
Urea is a highly concentrated nitrogen fertilizer. In this case, the granules of the substance are easily dissolved in water. To make a 5% solution, you need to take 50 g of urea and 1 liter of water or 500 g of fertilizer granules per 10 liters of water. Add granules to a container with water and mix well.
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 a concentrated acid solution is measured with a burette with a glass tap.
The weight of the solute is counted 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 to the second decimal place.
Example 1. How many grams of barium chloride is needed to prepare 2 liters of a 0.2 M solution?
Decision. The molecular weight of barium chloride is 208.27. Hence. 1 liter of 0.2 M solution should contain 208.27-0.2 = = 41.654 g BaCl 2 . To prepare 2 liters, 41.654-2 \u003d 83.308 g of BaCl 2 will be required.
Example 2. How many grams of anhydrous soda Na 2 C0 3 will be required to prepare 500 ml of 0.1 n. solution?
Decision. The molecular weight of soda is 106.004; equivalent share weight 5 N a 2 C0 3 \u003d M: 2 \u003d 53.002; 0.1 eq. = 5.3002 g.
1000 ml 0.1 n. solution contain 5.3002 g of 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=1.84) is required to prepare 2 liters of 0.05N. sulfuric acid solution?
Decision. The molecular weight of sulfuric acid is 98.08. Equivalent mass of sulfuric acid 3h 2 so 4 \u003d M: 2 \u003d 98.08: 2 \u003d 49.04 g. Weight 0.05 equiv. \u003d 49.04-0.05 \u003d 2.452 g.
Let's find how much H 2 S0 4 should be contained in 2 l 0.05 n. solution:
1 l-2.452 g H 2 S0 4
2"- X » H 2 S0 4
X \u003d 2.452-2 \u003d 4.904 g H 2 S0 4.
In order to determine how much a 96% solution of H 2 S0 4 should be taken for this, we compose the proportion:
\ in 100 g conc. H 2 S0 4 -96 g H 2 S0 4
At» » H 2 S0 4 -4.904 g H 2 S0 4
4,904-100
At=——— §6—— = 5.11 g H 2 S0 4 .
Convert this amount to volume: ,. R 5,11
K \u003d 7 \u003d TJ \u003d 2 "77 ml -
Thus, to prepare 2 liters of 0.05 N. solution should take 2.77 ml of concentrated sulfuric acid.
Example 4. Calculate the titer of a NaOH solution if its exact concentration is known to be 0.0520 N.
Decision. Recall that the titer is the content in 1 ml of a solution of a substance in grams. Equivalent mass of NaOH \u003d 40 01 g 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: -u = - = 0.00208 g / ml. You can also use the formula:
9 N
where T- titer, g/ml; E- equivalent weight; N- the normality of the solution.
Then the titer of this solution is:
f 40,01 0,0520
“NaOH =——— jooo—— 0.00208 g/ml.
„ “Rie P 5 - Calculate the normal concentration of a solution of HN0 3, 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 a nitric acid solution is:
- V \u003d 63.05 \u003d 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
Decision.
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 equiv. /
In order to calculate the normal concentration of the solution, we compose the proportion:
1000 » » X "
1000-0,05
x =
—————— =0.25 equiv.
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 is the volume of the solution in milliliters.
Zia 2 co 3 \u003d 53.002, then the normal concentration of this solution
2.6501-10С0 N = 53.002-200
