Sodium bicarbonate is most often used in the form of a soda solution. O useful properties soda became known to mankind not so long ago, but soda solution is already actively used in many areas of human life, once again showing its positive effect.
Recipes for soda solutions that help overcome illnesses are very simple and accessible to everyone.
How to prepare and where to use the remedy
Soda solution has been widely used in various fields of human activity. Dry sodium bicarbonate itself is used both in industry and in cooking, but soda solution is most in demand in horticulture, medicine and cosmetology.
It is very easy to prepare the remedy - required amount white powder is added to the liquid and mixed until the particles are completely dissolved. Most often, soda is stirred in water, but for the treatment of certain ailments, sodium bicarbonate is more useful with milk than with water. But in cosmetology, a soda solution is made on the basis of a shampoo that is used to wash hair.
Despite the fact that it is easy to make a soda solution, when preparing it, it is important to correctly observe the proportions of the recommended substances.
Otherwise, the tool can become not only useless, but also harmful.
Application in horticulture
For a garden and a kitchen garden solution baking soda irreplaceable. With it, you can:
- get rid of powdery mildew - a disease of many cultivated plants that destroys young shoot leaves;
- rejuvenate rose bushes by adding a little ammonia to the bicarbonate solution;
- eliminate small grass that makes its way in the crevices of garden paths;
- defeat caterpillars that eat young cabbage leaves;
- feed the tomatoes, after which their fruits will become even meatier and sweeter;
- acidify the soil for growing certain types of cultivated plants;
- protect grapes from gray rot and make berries more sugary;
- clean your hands after working in the garden from the dirt firmly ingrained in them.
Application in medicine
Soda solution is a real savior from many diseases. Some doctors even insist that sodium bicarbonate can cure cancer.
Gargling with a solution of soda very quickly and noticeably relieves discomfort in the throat. However, while research on the effect of soda on oncopathology is still ongoing, it is safe to say that this substance is able to cope with many ailments:
- heartburn - the remedy neutralizes hyperacidity;
- colds - sodium bicarbonate is able to relieve the first symptoms of a cold and prevent the development of the disease;
- diaper rash in infants - the solution promotes the rapid healing of wounds on the surface of the skin;
- cystitis - soda in a dissolved state helps the body fight pathogenic microorganisms;
- burns - a solution of sodium bicarbonate applied to the affected area relieves pain, and wounds heal faster;
- hypertension - sodium bicarbonate helps to remove excess water and significantly reduce arterial pressure;
- runny nose - a simple solution of soda and salt perfectly replaces expensive pharmaceutical preparations, and you can rinse your nose with it as much as you like without harm to health;
- thrush - with the help of sodium bicarbonate, you can get rid of the thrush hated by women, since the Candida fungus is afraid of an alkaline environment;
- prolonged cough - with the help of soda and honey dissolved in milk, you can get rid of a long dry cough, promote sputum discharge;
- sore throat - gargling with soda solution softens the manifestations of the disease, promotes the discharge of purulent plugs and cleansing of the throat, also when pus is discharged, the body temperature drops significantly and the person becomes much better;
- dermatitis and psoriasis - soda lotions have an extremely beneficial effect on the affected skin surface;
- calluses, boils and corns, on the surface of which a cotton swab dipped in the prepared product is applied;
- smoking - with the help of sodium bicarbonate solution, smokers rinse their mouths;
- slags and toxins that soda dissolved in water successfully removes from the body;
- manifestations of motion sickness in transport.
Application in cosmetology
To give split ends beauty and strength, you can use a product with bicarbonate, made right before the shampooing procedure. For this, 2 tsp. soda should be added to 3 tbsp. l. shampoo to get a fairly strong concentrated product. They wash their hair with this shampoo once a week, and at other times they use the usual detergents. After a month, you can notice that the hair is gaining strength, splitting less, becoming thick and shiny.
Also, sodium bicarbonate can be mixed with body gel to make a kind of scrub and peel off dead particles of the epidermis from the body. This will help make your skin look healthier.
With the help of sodium bicarbonate, you can restore the acid balance of the skin, and thus eliminate acne on its surface. To do this, a cotton swab is dipped in soda and wiped on the face twice a week. Wipe the skin of the face very gently, massaging movements, avoiding the area under the eyes. If you do the procedure regularly, then acne on the surface of the skin will not appear for a long time.
To successfully apply soda, it is very important to pay attention to how to make the solution correctly. You should not mix proportions by eye and assume that such a remedy will become an assistant - in many cases, it was the excess of the dose of the main substance that caused allergic reactions or worsened the patient's condition. And in gardening, the solution is too strong soda ash and completely destroy the crop.
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), the exact value of the term with the fewest decimal places is used. 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 during analysis, since it is enough to multiply the volume of the solution used for titration by the titer of the solution 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.
To the question, tell me how to make a 1% solution with copper sulfate and lime (or soda ash) given by the author Larysa lymar the best answer is Bordeaux liquid is not prepared with soda ash. Soda ash is sodium carbonate, and you need calcium hydroxide (slaked lime). Otherwise, you will get malachite. How to cook - read:
Answer from Mujtahid.[guru]
Answer from spacer[guru]
Answer from I-beam[guru]
Answer from Special[guru]
to prepare a 1% copper solution, dilute 100 g of vitriol in 10 liters of water and neutralize 100 g of slaked lime ==>>
Answer from Natali Natali[guru]
Answer from Zhanna S[guru]
1-% BORDEAUX mixture.
and also diluted with 5 liters of water.
(on green leaves) .
1-% BURGUNDE liquid
Add 50 g. soap.
COPPER_SOAP SOLUTION.
+ Soap in the same amount.
Success in the fight against diseases)
Answer from Mujtahid.[guru]
Buy better ready-made Bordeaux liquid.
Answer from Galina Russova (Churkina) GALJ[guru]
PROCESS PLANTS FROM PHYTOFTORA OF OTHER DISEASES
Answer from Elena Akentieva[guru]
Do not suffer with the Bordeaux mixture, a very inconvenient preparation in terms of preparation (does not mix well) and processing (clogs the sprayer). Buy Ordan or Abiga Peak, wonderful fungicides, no hassle.
Answer from Kostenko Sergey[guru]
to prepare a 1% copper solution, dilute 100 g of vitriol in 10 liters of water and neutralize 100 g of slaked lime ==>> 10 liters of Bordeaux with a copper concentration of 1%
Answer from Natali Natali[guru]
in 5 liters warm water dilute 100 g of copper sulfate and separately dilute 100 g of lime in 5 liters of water. Then pour the solution of vitriol into the solution of lime - NOT vice versa and get a 1% solution of Bordeaux liquid. In other words: for 10 liters of water - 100 g of vitriol and lime
Answer from Zhanna S[guru]
Bordeaux mixture is prepared on the basis of copper sulphate and lime.
BURGUNDY - from copper sulphate and soda (food can also be used) + soap.
1-% BORDEAUX mixture.
100 g of quicklime and slaked in a small amount water, diluted with water to 5 liters, milk of lime is obtained.
In another container (non-metal)
dissolve 100 g of copper sulfate in hot water
and also diluted with 5 liters of water.
A solution of copper sulphate is poured into milk of lime and mixed well.
Copper to lime, not the other way around!
You can dissolve 100 g of copper sulfate in 1 liter hot water
and pour, gradually stirring, into 9 liters of milk of lime.
But mixing both concentrated solutions,
and then diluted with water up to 10 liters is unacceptable.
It turns out a mixture of poor quality.
Properly prepared liquid has a turquoise, sky blue color and a neutral or slightly alkaline reaction.
Acidity is tested with litmus paper.
which is dyed blue.
You can lower any clean (not rusty) piece of iron into the solution.
In an acidic environment, copper actively settles on iron.
The acidic mixture will burn the leaves.
It is neutralized by adding milk of lime.
1% mixture is applied to vegetative plants
(on green leaves).
1-% BURGUNDE liquid
100 g of copper sulfate and 100 g of soda ash per 10 liters of water. Dissolve separately.
merge together, check the acidity,
i.e., they are prepared in the same way as the Bordeaux mixture, only the lime is replaced with soda.
Add 50 g. soap.
COPPER_SOAP SOLUTION.
10 g of copper sulphate are dissolved in 0.5 l of hot water.
Separately, 100 g of soap is diluted in 10 liters of water (preferably warm).
A solution of copper sulphate is poured into a soap solution in a thin stream with constant stirring.
The drug is prepared before spraying.
Copper-soap preparation (emulsion) can be prepared in higher concentrations
(20 g of copper sulfate and 200 g of soap
or 30 g of vitriol and 300 g of soap per 10 liters of water).
A properly prepared emulsion should have a greenish color and not form flakes.
To avoid coagulation of the drug in cases of its preparation in hard water,
reduce the amount of copper sulfate
or add 0.5% (50 g per 10 l of water) soda ash (linen) to the water.
Can be used in conjunction with karbofos (20 g per 10 l of emulsion)
for simultaneous control of aphids and spider mites.
Soda ash, or laundry, soda (sodium carbonate) is a white crystalline powder, soluble in water.
Used to control powdery mildew
at a concentration of 0.5% (50 g per 10 liters of water).
+ Soap in the same amount.
To prepare a working solution, dilute soap in soft water and add soda, previously dissolved in a small amount of water.
Success in the fight against diseases)
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?
Solution. 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?
Solution. 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?
Solution. 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 eq. \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.
Solution. 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 the normal concentration of a 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 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
Not everyone remembers what “concentration” means and how to properly prepare a solution. If you want to get 1- percentage 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.
