How to make 10 from a 25 solution. Solutions of percentage concentration

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

SI units in clinical laboratory diagnostics.

In clinical laboratory diagnostics, the International System of Units is recommended to be used in accordance with the following rules.

1. Liters should be used as units of volume. It is not recommended to use fractional or multiples of a liter (1-100 ml) in the denominator.

2. The concentration of measured substances is indicated as molar (mol/l) or as mass (g/l).

3. Molar concentration is used for substances with a known relative molecular weight. The ionic concentration is indicated as a molar concentration.

4. Mass concentration is used for substances whose relative molecular weight is unknown.

5. Density is indicated in g/l; clearance - in ml / s.

6. The activity of enzymes on the amount of substances in time and volume is expressed as mol / (s * l); µmol/(s*l); nmol/(s*l).

When converting units of mass into units of quantity of a substance (molar), the conversion factor is K=1/Mr, where Mr is the relative molecular weight. In this case, the initial unit of mass (gram) corresponds to the molar unit of the amount of substance (mol).

General characteristics.

Solutions are homogeneous systems consisting of two or more components and products of their interaction. The role of a solvent can be played not only by water, but also by ethyl alcohol, ether, chloroform, benzene, etc.

The dissolution process is often accompanied by heat release (exothermic reaction - dissolution of caustic alkalis in water) or heat absorption (endothermic reaction - dissolution of ammonium salts).

Liquid solutions include solutions of solids in liquids (solution of salt in water), solutions of liquids in liquids (solution of ethyl alcohol in water), solutions of gases in liquids (CO 2 in water).

Solutions can be not only liquid, but also solid (glass, an alloy of silver and gold), as well as gaseous (air). The most important and common are aqueous solutions.

Solubility is the property of a substance to dissolve in a solvent. By solubility in water, all substances are divided into 3 groups - highly soluble, slightly soluble and practically insoluble. Solubility primarily depends on the nature of the substances. Solubility is expressed as the number of grams of a substance that can be maximally dissolved in 100 g of a solvent or solution at a given temperature. This amount is called the solubility coefficient or simply the solubility of the substance.

A solution in which no further dissolution of a substance occurs at a given temperature and volume is called saturated. Such a solution is in equilibrium with an excess of the solute, it contains the maximum possible amount of the substance under given conditions. If the concentration of the solution does not reach the saturation concentration under the given conditions, then the solution is called unsaturated. A supersaturated solution contains more than a saturated solution. Supersaturated solutions are very unstable. A simple shaking of the vessel or contact with the crystals of the solute results in instantaneous crystallization. In this case, the supersaturated solution becomes a saturated solution.

The concept of "saturated solutions" should be distinguished from the concept of "supersaturated solutions". A concentrated solution is a solution with a high solute content. Saturated solutions of different substances can vary greatly in concentration. In highly soluble substances (potassium nitrite), saturated solutions have a high concentration; in poorly soluble substances (barium sulfate), saturated solutions have a small concentration of the solute.

In most cases, the solubility of a substance increases with increasing temperature. But there are substances whose solubility increases slightly with increasing temperature (sodium chloride, aluminum chloride) or even decreases.

The dependence of the solubility of various substances on temperature is depicted graphically using solubility curves. Temperature is plotted on the abscissa axis, solubility is plotted on the ordinate axis. Thus, it is possible to calculate how much salt falls out of the solution when it is cooled. The release of substances from a solution with a decrease in temperature is called crystallization, while the substance is released in its pure form.

If the solution contains impurities, then the solution will be unsaturated with respect to them even with a decrease in temperature, and the impurities will not precipitate. This is the basis of the method of purification of substances - crystallization.

In aqueous solutions, more or less strong compounds of solute particles with water are formed - hydrates. Sometimes such water is so strongly associated with the solute that, when it is released, it enters the composition of the crystals.

Crystalline substances containing water in their composition are called crystalline hydrates, and the water itself is called crystallization. The composition of crystalline hydrates is expressed by a formula indicating the number of water molecules per molecule of the substance - CuSO 4 * 5H 2 O.

Concentration is the ratio of the amount of a solute to the amount of a solution or solvent. The concentration of the solution is expressed in weight and volume ratios. Weight percentages indicate the weight content of a substance in 100 g of a solution (but not in 100 ml of solution!).

Technique for preparing approximate solutions.

The necessary substances and the solvent are weighed in such ratios that the total amount is 100 g. If the solvent is water, the density of which is equal to one, it is not weighed, but a volume equal to the mass is measured. If the solvent is a liquid whose density is not equal to unity, it is either weighed or the amount of solvent expressed in grams is divided by the density index and the volume occupied by the liquid is calculated. Density P is the ratio of body mass to its volume.

The unit of density is the density of water at 4 0 C.

Relative density D is the ratio of the density of a given substance to the density of another substance. In practice, the ratio of the density of a given substance to the density of water, taken as a unit, is determined. For example, if the relative density of a solution is 2.05, then 1 ml of it weighs 2.05 g.

Example. How much 4 carbon chloride should be taken to prepare 100 g of a 10% fat solution? Weigh 10 g of fat and 90 g of CCl 4 solvent or, by measuring the volume occupied by the required amount of CCl 4 , divide the mass (90 g) by the relative density index D = (1.59 g/ml).

V = (90 g) / (1.59 g/ml) = 56.6 ml.

Example. How to prepare a 5% solution of copper sulfate from the crystalline hydrate of this substance (calculated as anhydrous salt)? The molecular weight of copper sulfate is 160 g, crystalline hydrate is 250 g.

250 - 160 X \u003d (5 * 250) / 160 \u003d 7.8 g

Therefore, you need to take 7.8 g of crystalline hydrate, 92.2 g of water. If the solution is prepared without conversion to anhydrous salt, the calculation is simplified. The given amount of salt is weighed and the solvent is added in such an amount that the total weight of the solution is 100 g.

Volume percentages show how much of a substance (in ml) is contained in 100 ml of a solution or mixture of gases. For example, a 96% ethanol solution contains 96 ml of absolute (anhydrous) alcohol and 4 ml of water. Volume percentages are used when mixing mutually soluble liquids, in the preparation of gas mixtures.

Weight-volume percentages (conditional way of expressing concentration). Indicate the weight amount of the substance contained in 100 ml of the solution. For example, a 10% NaCl solution contains 10 g of salt in 100 ml of solution.

Technique for preparing percentage solutions from concentrated acids.

Concentrated acids (sulfuric, hydrochloric, nitric) contain water. The ratio of acid and water in them is indicated in weight percentages.

The density of solutions in most cases is above unity. The percentage of acids is determined by their density. When preparing more dilute solutions from concentrated solutions, their water content is taken into account.

Example. It is necessary to prepare a 20% solution of sulfuric acid H 2 SO 4 from concentrated 98% sulfuric acid with a density D = 1.84 g / ml. Initially, we calculate how much concentrated solution contains 20 g of sulfuric acid.

100 - 98 X \u003d (20 * 100) / 98 \u003d 20.4 g

It is practically more convenient to work with volumetric rather than weight units of acids. Therefore, it is calculated what volume of concentrated acid occupies the desired weight amount of the substance. To do this, the number obtained in grams is divided by the density index.

V = M/P = 20.4/1.84 = 11 ml

You can also calculate in another way, when the concentration of the initial acid solution is immediately expressed in weight-volume percentages.

100 – 180 X = 11 ml

When special accuracy is not required, when diluting solutions or mixing them to obtain solutions of a different concentration, the following simple and quick method can be used. For example, you need to prepare a 5% solution of ammonium sulfate from a 20% solution.

Where 20 is the concentration of the solution taken, 0 is water, and 5 is the required concentration. Subtract 5 from 20, and write the resulting value in the lower right corner, subtracting 0 from 5, write the number in the upper right corner. Then the diagram will take the following form.

This means that you need to take 5 parts of a 20% solution and 15 parts of water. If you mix 2 solutions, then the scheme is preserved, only the initial solution with a lower concentration is written in the lower left corner. For example, by mixing 30% and 15% solutions, you need to get a 25% solution.

Thus, you need to take 10 parts of a 30% solution and 15 parts of a 15% solution. Such a scheme can be used when special accuracy is not required.

Accurate solutions include normal, molar, standard solutions.

A normal solution is a solution in which 1 g contains g - equivalent of a solute. The weight amount of a complex substance, expressed in grams and numerically equal to its equivalent, is called the gram equivalent. When calculating the equivalents of compounds such as bases, acids and salts, the following rules can be used.

1. The base equivalent (E o) is equal to the molecular weight of the base divided by the number of OH groups in its molecule (or by the valency of the metal).

E (NaOH) = 40/1=40

2. Acid equivalent (E to) is equal to the molecular weight of the acid divided by the number of hydrogen atoms in its molecule, which can be replaced by a metal.

E (H 2 SO 4) = 98/2 = 49

E (HCl) \u003d 36.5 / 1 \u003d 36.5

3. Salt equivalent (E s) is equal to the molecular weight of the salt divided by the product of the valency of the metal by the number of its atoms.

E (NaCl) \u003d 58.5 / (1 * 1) \u003d 58.5

In the interaction of acids and bases, depending on the properties of the reactants and the reaction conditions, not necessarily all hydrogen atoms present in the acid molecule are replaced by a metal atom, but acid salts are formed. In these cases, the gram equivalent is determined by the number of hydrogen atoms replaced by metal atoms in a given reaction.

H 3 PO 4 + NaOH = NaH 2 PO + H 2 O (gram equivalent equals gram molecular weight).

H 3 PO 4 + 2NaOH \u003d Na 2 HPO 4 + 2H 2 O (gram equivalent is equal to half a gram of molecular weight).

When determining the gram equivalent, knowledge of the chemical reaction and the conditions under which it occurs is required. If you need to prepare decinormal, centinormal or millinormal solutions, take, respectively, 0.1; 0.01; 0.001 grams is the equivalent of a substance. Knowing the normality of the solution N and the equivalent of the solute E, it is easy to calculate how many grams of the substance are contained in 1 ml of the solution. To do this, you need to divide the mass of the solute by 1000. The amount of solute in grams contained in 1 ml of the solution is called the titer of the solution (T).

T \u003d (N * E) / 1000

T (0.1 H 2 SO 4) \u003d (0.1 * 49) / 1000 \u003d 0.0049 g / ml.

A solution with a known titer (concentration) is called titrated. Using a titrated alkali solution, it is possible to determine the concentration (normality) of an acid solution (acidimetry). Using a titrated acid solution, it is possible to determine the concentration (normality) of an alkali solution (alkalimetry). Solutions of the same normality react in equal volumes. At different normalities, these solutions react with each other in volumes inversely proportional to their normalities.

N to / N u \u003d V u / V to

N to * V to \u003d N u * V u

Example. For titration of 10 ml of HCl solution, 15 ml of 0.5 N NaOH solution went. Calculate the normality of the HCl solution.

N to * 10 \u003d 0.5 * 15

N k \u003d (0.5 * 15) / 10 \u003d 0.75

N=30/58.5=0.5

Fixanals - pre-prepared and sealed in ampoules, accurately weighed amounts of reagent required to prepare 1 liter of 0.1 N or 0.01 N solution. Fixanals are liquid and dry. Dry ones have a longer shelf life. The technique for preparing solutions from fixanals is described in the appendix to the box with fixanals.

Preparation and testing of decinormal solutions.

Decinormal solutions, which are often used as initial solutions in the laboratory, are prepared from chemically frequent preparations. The required weight is weighed on technochemical scales or pharmaceutical scales. When weighing, an error of 0.01 - 0.03 g is allowed. In practice, an error can be made in the direction of some increase in the weight obtained by calculation. The sample is transferred to a volumetric flask, where a small amount of water is added. After complete dissolution of the substance and equalization of the temperature of the solution with the air temperature, the flask is topped up with water up to the mark.

The prepared solution requires verification. The check is carried out with the help of solutions prepared by their fixanals, in the presence of indicators, the correction factor (K) and titer are set. The correction factor (K) or correction factor (F) shows how much (in ml) of the exact normal solution corresponds to 1 ml of this (prepared) solution. To do this, 5 or 10 ml of the prepared solution is transferred into a conical flask, a few drops of the indicator are added and titrated with an exact solution. The titration is carried out twice and the arithmetic mean value is calculated. The results of the titration should be approximately the same (difference within 0.2 ml). The correction factor is calculated from the ratio of the volume of the exact solution V t to the volume of the test solution V n.

K \u003d V t / V n.

The correction factor can also be determined in the second way - by the ratio of the titer of the test solution to the theoretically calculated titer of the exact solution.

K = T practical / T theor.

If the left sides of an equation are equal, then their right sides are equal.

V t / V n. = T pract. / T theor.

If the practical titer of the test solution is found, then the weight content of the substance in 1 ml of the solution is determined. In the interaction of the exact and tested solution, 3 cases can occur.

1. The solutions interacted in equal volumes. For example, 10 ml of the test solution was used to titrate 10 ml of a 0.1 N solution. Therefore, the normality is the same and the correction factor is equal to one.

2. 9.5 ml of the test subject was used for interaction with 10 ml of the exact solution, the test solution turned out to be more concentrated than the exact solution.

3. 10.5 ml of the test subject went into interaction with 10 ml of the exact solution, the test solution is weaker in concentration than the exact solution.

The correction factor is calculated to the second decimal place, fluctuations from 0.95 to 1.05 are allowed.

Correction of solutions, the correction factor of which is greater than one.

The correction factor shows how many times a given solution is more concentrated than a solution of a certain normality. For example, K is 1.06. Therefore, 0.06 ml of water must be added to each ml of the prepared solution. If 200 ml of the solution remains, then (0.06 * 200) \u003d 12 ml - add to the remaining prepared solution and mix. This method of bringing solutions to a certain normality is simple and convenient. When preparing solutions, you should prepare them with more concentrated solutions, rather than dilute solutions.

Preparation of precise solutions, the correction factor of which is less than one.

In these solutions, some part of the gram equivalent is missing. This missing part can be identified. If you calculate the difference between the titer of a solution of a certain normality (theoretical titer) and the titer of this solution. The value obtained shows how much substance must be added to 1 ml of a solution to bring it to a solution concentration of a given normality.

Example. The correction factor for approximately 0.1 N sodium hydroxide solution is 0.9, the volume of the solution is 1000 ml. Bring the solution to exactly 0.1 N concentration. Gram - the equivalent of caustic soda - 40 g. Theoretical titer for a 0.1 N solution - 0.004. Practical caption - T theor. * K = 0.004 * 0.9 = 0.0036

T theor. - T pract. = 0.004 - 0.0036 = 0.0004

1000 ml of solution remained unused - 1000 * 0, 0004 \u003d 0.4 g.

The resulting amount of the substance is added to the solution, mixed well, and the titer of the solution is determined again. If the starting material for the preparation of solutions are concentrated acids, alkalis, and other substances, then it is necessary to make an additional calculation to determine how much of the concentrated solution contains the calculated value of this substance. Example. 4.3 ml of an exact 0.1 N NaOH solution was used to titrate 5 ml of approximately 0.1 N HCl solution.

K = 4.3/5 = 0.86

The solution is weak, it must be strengthened. We calculate T theor. , T practical and their difference.

T theor. = 3.65 / 1000 = 0.00365

T pract. = 0.00365 * 0.86 = 0.00314

T theor. - T pract. = 0.00364 - 0.00314 = 0.00051

200 ml of solution remained unused.

200*0.00051=0.102g

For a 38% HCl solution with a density of 1, 19, we make up a proportion.

100 - 38 X \u003d (0.102 * 100) / 38 \u003d 0.26 g

We convert weight units into volume units, taking into account the density of the acid.

V = 0.26 / 1.19 = 0.21 ml

Preparation of 0.01 N, 0.005 N from decinormal solutions, having a correction factor.

Initially, it is calculated what volume of a 0.1 N solution should be taken for preparation from a 0.01 N solution. The calculated volume is divided by the correction factor. Example. It is necessary to prepare 100 ml of a 0.01 N solution from 0.1 N with K = 1.05. Since the solution is 1.05 times more concentrated, you need to take 10 / 1.05 \u003d 9.52 ml. If K \u003d 0.9, then you need to take 10 / 0.9 \u003d 11.11 ml. In this case, take a slightly larger amount of the solution and bring the volume in the volumetric flask to 100 ml.

For the preparation and storage of titrated solutions, the following rules apply.

1. Each titrated solution has its own shelf life. During storage, they change their titer. When performing the analysis, it is necessary to check the titer of the solution.

2. It is necessary to know the properties of solutions. The titer of some solutions (sodium hyposulfite) change over time, so their titer is set no earlier than 5-7 days after preparation.

3. All bottles with titrated solutions must have a clear inscription indicating the substance, its concentration, correction factor, the time of preparation of the solution, the date of the titer check.

4. In analytical work, much attention should be paid to calculations.

T \u003d A / V (A - hitch)

N \u003d (1000 * A) / (V * g / eq)

T = (N * g/eq) / 1000

N = (T * 1000) / (g/eq)

A molar solution is one in which 1 liter contains 1 g * mol of a solute. A mole is a molecular weight expressed in grams. 1 molar solution of sulfuric acid - 1 liter of this solution contains 98 g of sulfuric acid. A centimole solution contains 0.01 mol in 1 liter, a millimolar solution contains 0.001 mol. A solution whose concentration is expressed as the number of moles per 1000 g of solvent is called molal.

For example, 1 liter of 1 M sodium hydroxide solution contains 40 g of the drug. 100 ml of solution will contain 4.0 g, i.e. solution 4/100 ml (4g%).

If the sodium hydroxide solution is 60/100 (60 mg%), its molarity must be determined. 100 ml of the solution contains 60 g of sodium hydroxide, and 1 liter - 600 g, i.e. 1 liter of 1 M solution should contain 40 g of sodium hydroxide. Molarity of sodium - X \u003d 600 / 40 \u003d 15 M.

Standard solutions are called solutions with precisely known concentrations used for the quantitative determination of substances by colorimetry, nephelometry. A sample for standard solutions is weighed on an analytical balance. The substance from which the standard solution is prepared must be chemically pure. standard solutions. Standard solutions are prepared in the volume required for consumption, but not more than 1 liter. The amount of substance (in grams) required to obtain standard solutions - A.

A \u003d (M I * T * V) / M 2

M I - Molecular weight of the solute.

T - Solution titer by analyte (g/ml).

V - Target volume (ml).

M 2 - Molecular or atomic mass of the analyte.

Example. It is necessary to prepare 100 ml of a standard solution of CuSO 4 * 5H 2 O for the colorimetric determination of copper, and 1 ml of the solution should contain 1 mg of copper. In this case, M I = 249.68; M 2 = 63, 54; T = 0.001 g/mL; V = 100 ml.

A \u003d (249.68 * 0.001 * 100) / 63.54 \u003d 0.3929 g.

A portion of the salt is transferred to a 100 ml volumetric flask and water is added up to the mark.

Control questions and tasks.

1. What is a solution?

2. What are the ways to express the concentration of solutions?

3. What is the titer of the solution?

4. What is a gram equivalent and how is it calculated for acids, salts, bases?

5. How to prepare a 0.1 N sodium hydroxide NaOH solution?

6. How to prepare a 0.1 N solution of sulfuric acid H 2 SO 4 from a concentrated one with a density of 1.84?

8. What is the way to strengthen and dilute solutions?

9. Calculate how many grams of NaOH are needed to prepare 500 ml of a 0.1 M solution? The answer is 2 years.

10. How many grams of CuSO 4 * 5H 2 O should be taken to prepare 2 liters of 0.1 N solution? The answer is 25 years.

11. 15 ml of 0.5 N NaOH solution was used for titration of 10 ml of HCl solution. Calculate - the normality of HCl, the concentration of the solution in g / l, the titer of the solution in g / ml. The answer is 0.75; 27.375 g/l; T = 0.0274 g/ml.

12. 18 g of a substance are dissolved in 200 g of water. Calculate the weight percent concentration of the solution. The answer is 8.25%.

13. How many ml of a 96% sulfuric acid solution (D = 1.84) should be taken to prepare 500 ml of a 0.05 N solution? The answer is 0.69 ml.

14. Titer of H 2 SO 4 solution = 0.0049 g/ml. Calculate the normality of this solution. The answer is 0.1 N.

15. How many grams of caustic soda should be taken to prepare 300 ml of a 0.2 N solution? The answer is 2.4 g.

16. How much do you need to take a 96% solution of H 2 SO 4 (D = 1.84) to prepare 2 liters of a 15% solution? The answer is 168 ml.

17. How many ml of a 96% sulfuric acid solution (D = 1.84) should be taken to prepare 500 ml of a 0.35 N solution? The answer is 9.3 ml.

18. How many ml of 96% sulfuric acid (D = 1.84) should be taken to prepare 1 liter of 0.5 N solution? The answer is 13.84 ml.

19. How much is the molarity of a 20% hydrochloric acid solution (D = 1.1). The answer is 6.03 M.

20 . Calculate the molar concentration of 10% nitric acid solution (D = 1.056). The answer is 1.68 M.

Quest Source: Decision 2446. USE 2017 Mathematics, I.V. Yashchenko. 36 options.

Task 11. By mixing 25% and 95% acid solutions and adding 20 kg of pure water, a 40% acid solution was obtained. If, instead of 20 kg of water, 20 kg of a 30% solution of the same acid were added, then a 50% acid solution would be obtained. How many kilograms of a 25% solution were used to make the mixture?

Decision.

Let us denote by x kg the mass of a 25% solution, and by y kg the mass of a 95% solution. It can be seen that the total mass of the acid in the solution after mixing them is equal to . The problem says that if you mix these two solutions and add 20 kg of pure water, you will get a 40% solution. In this case, the mass of the acid will be determined by the expression . Since the mass of acid remains the same after adding 20 kg of pure water, we have an equation of the form

By analogy, the second equation is obtained, when instead of 20 kg of water, 20 kg of a 30% solution of the same acid is added and a 50% acid solution is obtained:

We solve the system of equations, we get:

We multiply the first equation by -9, and the second by 11, we have.

In the process of preparing solutions by diluting concentrates, it is necessary to carry out quick and error-free calculations of the required amount of the initial concentrate and solvent combined into one solution.

When calculating the dilution of concentrates, in which the concentration is indicated as the ratio of the amount of solute to the amount of solution, the required amount of dry matter is multiplied by the dilution value, i.e. to the second digit of the concentration ratio.

For example, if the required amount of dry soluble substance is 5 g, and the concentrated solution has a concentration of 1: 10, then the required amount of concentrate solution will be: 5 x 10 = 50 (ml).

If the concentration of the blank solution is indicated as the ratio of the solute to the solvent reduced to unity (for example, 1 + 3), then, by analogy with the previous case of a concentrated solution, it is necessary to take:

5 x (1 + 3) = 20 (ml).

If the concentration of the semi-finished solution is expressed as a percentage and is equal, for example, to 10%, then under the same conditions it must be taken: 5 x 100 / 10 = 50 (ml).

In pharmacy practice, it is very often necessary to determine the required amount of a stock solution by its concentration (in percent), the amount of the prepared solution and its concentration (in percent), the amount of diluted solution prepared and its concentration (also in percent).

For example, there is an X% concentrated solution.

To determine the amount of this solution required to obtain A ml of a diluted solution with a concentration of Y% (let's denote it B), it is necessary to carry out the following calculations.

The amount of a solute in a concentrated solution is: X x B / 100, and in the resulting diluted solution - Y x A / 100. Since both values ​​\u200b\u200bare equal, then, respectively:

X x B / 100 = Y x A / 100.

From here we express the volume of X% concentrated solution required to obtain A ml of Y% diluted solution:

B \u003d Y x A / X (ml). And the amount of solvent necessary to dilute the workpiece, therefore, will be equal to A - B (ml).

Sometimes it is necessary to prepare solutions of a given concentration from two solutions (one with a higher and the other with a lower concentration). For example, there are two solutions with concentrations X and Y%. In order to determine in what ratio these solutions should be mixed in order to obtain C ml of a solution with a concentration of Z%, we carry out calculations. Let's denote the required amount of X-% solution as D, then Y-% solution will require (C - D) ml. Considering the previous calculations, we get:

X x D + Y x (C - D) = Z x C.

Hence: D \u003d C x (Z - Y) / (X - Y) (ml).

Very convenient for diluting concentrated solutions is the use of the so-called mixing rule. Let's assume that from two solutions with concentrations X and Y% it is necessary to prepare a Z% solution. Determine in what ratio you need to mix the initial solutions. Let the desired values ​​be equal: A (X% solution) and B (Y% solution) ml.

Therefore, the amount of the prepared Z% solution should be equal to: (A + B) ml.

Then: X x A + Y x B \u003d Z x (A + B), or A / B \u003d (Z - Y) / (X - Z).

Equating the corresponding members of the relations, we have:

A \u003d Z - Y, B \u003d X - Z.

Example 1

Let us calculate in what proportions it is necessary to mix 35% and 15% solutions in order to obtain a 20% solution.

After completing the necessary calculations, we get that you need to mix 5 parts of a 35% solution and 15 parts of a 15% solution. As a result of mixing, 20 parts of a 20% solution will be obtained.

Example 2

Let us calculate in what proportions it is necessary to mix water, i.e. 0% solution, and 25% solution to get 10% solution. After the calculations, we get that you need to mix 10 parts of a 25% solution and 15 parts of water. As a result, 25 parts of a 10% solution will be obtained.

Preparation of solutions. A solution is a homogeneous mixture of two or more substances. The concentration of a solution is expressed in different ways:

in weight percent, i.e. by the number of grams of the substance contained in 100 g of the solution;

in volume percent, i.e. by the number of volume units (ml) of the substance in 100 ml of solution;

molarity, i.e. the number of gram-moles of a substance in 1 liter of solution (molar solutions);

normality, i.e. the number of gram equivalents of a solute in 1 liter of solution.

Solutions of percentage concentration. Percentage solutions are prepared as approximate, while the sample of the substance is weighed on technochemical scales, and the volumes are measured with measuring cylinders.

Several methods are used to prepare percentage solutions.

Example. It is necessary to prepare 1 kg of a 15% sodium chloride solution. How much salt is needed for this? The calculation is carried out according to the proportion:

Therefore, water for this must be taken 1000-150 \u003d 850 g.

In those cases when it is necessary to prepare 1 liter of a 15% sodium chloride solution, the required amount of salt is calculated in a different way. According to the reference book, the density of this solution is found and, multiplying it by a given volume, the mass of the required amount of solution is obtained: 1000-1.184 \u003d 1184 g.

Then follows:

Therefore, the required amount of sodium chloride is different for the preparation of 1 kg and 1 liter of solution. In cases where solutions are prepared from reagents containing water of crystallization, it should be taken into account when calculating the required amount of the reagent.

Example. It is necessary to prepare 1000 ml of a 5% solution of Na2CO3 with a density of 1.050 from a salt containing water of crystallization (Na2CO3-10H2O)

The molecular weight (weight) of Na2CO3 is 106 g, the molecular weight (weight) of Na2CO3-10H2O is 286 g, from here the required amount of Na2CO3-10H2O is calculated to prepare a 5% solution:

Solutions are prepared by dilution method as follows.

Example. It is necessary to prepare 1 l of a 10% HCl solution from an acid solution with a relative density of 1.185 (37.3%). The relative density of a 10% solution is 1.047 (according to the reference table), therefore, the mass (weight) of 1 liter of such a solution is 1000X1.047 \u003d 1047 g. This amount of solution should contain pure hydrogen chloride

To determine how much 37.3% acid needs to be taken, we make up the proportion:

When preparing solutions by diluting or mixing two solutions, the diagonal scheme method or the "rule of the cross" is used to simplify calculations. At the intersection of two lines, the given concentration is written, and at both ends on the left is the concentration of the initial solutions, for the solvent it is equal to zero.