How to dilute the solution. Preparation of solutions from standard and concentrated liquids Compress on the throat

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 per 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 mother liquor and then dilute it. We take 10 grams, prepare a liter of a 1 percent solution, pour 100 ml, bring it to a liter with water (dilute 10 times), and the 0.1 percent 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 (preferably rainwater). Separately, dissolve 5-10 g of copper sulfate in 1 liter of water. After this, the solution of copper sulfate is added in a thin stream to the soap solution, while continuously stirring well. The result will be a greenish liquid. If you mix poorly or rush, flakes will form. In this case, it is better to start the process from the very beginning.

How to prepare a 5 percent 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 it all until the liquid has a uniform and rich purple color. Before use, it is recommended to strain the solution through cheesecloth to remove undissolved crystals.

How to prepare a 5 percent 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.

In the process of preparing solutions by diluting concentrates, you should carry out quick and error-free calculations of the required amount of the original 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 soluble substance 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 the concentrated solution will be equal to: 5 x 10 = 50 (ml).

If the concentration of the stock solution is indicated in the form of the ratio of solute to solvent reduced to one (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 to, for example, 10%, then under the same conditions it should be taken: 5 x 100 / 10 = 50 (ml).

In pharmacy practice, it is very often necessary to determine the required amount of a reserve solution by its concentration (in percent), the amount of the prepared solution and its concentration (in percent), the amount of the 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 dissolved substance in a concentrated solution is equal to: X x B / 100, and in the resulting dilute solution - Y x A / 100. Since both quantities are equal, then accordingly:

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

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

B = Y x A / X (ml). And the amount of solvent required to dilute the workpiece will therefore 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 to obtain C ml of a solution with a concentration of Z%, we carry out calculations. Let us denote the required amount of X-percent solution by D, then Y-percent solution will require (C – D) ml. Taking into account the previous calculations, we get:

X x D + Y x (C – D) = Z x C.

Hence: D = C x (Z – Y) / (X – Y) (ml).

Very convenient for diluting concentrated solutions is the use of the so-called mixing rule. Suppose that from two solutions with concentrations X and Y% you need to prepare a Z% solution. Let us determine in what ratio the initial solutions should be mixed. Let the required 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 = Z x (A + B), or A / B = (Z – Y) / (X – Z).

Equating the corresponding terms of the relations, we have:

A = Z – Y, B = X – Z.

Example 1

Let's calculate the ratios in which 35% and 15% solutions need to be mixed to obtain a 20% solution.

Having performed the necessary calculations, we find that you need to mix 5 parts of a 35% solution and 15 parts of a 15% solution. The result of mixing is 20 parts of a 20% solution.

Example 2

Let's calculate in what proportions the water needs to be mixed, i.e. 0% solution, and 25% solution to get a 10% solution. After carrying out the calculations, we find 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.

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, are 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 with water several times, 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 n%. 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 of 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 the exact value of the term with the least number of 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 in small portions through a clean, dry funnel. 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 into 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.

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 substance contained in 100 g of solution;

in volume percentage, 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 contained in 1 liter of solution (molar solutions);

normality, i.e. the number of gram equivalents of the dissolved substance in 1 liter of solution.

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

To prepare percentage solutions, several methods are used.

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

Therefore, for this you need to take 1000-150 = 850 g of water.

In cases where it is necessary to prepare 1 liter of 15% sodium chloride solution, the required amount of salt is calculated in a different way. Using the reference book, find the density of this solution and, multiplying it by the given volume, obtain the mass of the required amount of solution: 1000-1.184 = 1184 g.

Then it follows:

Therefore, the required amount of sodium chloride is different for preparing 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 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 as follows.

Example. It is necessary to prepare 1 liter of 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 = 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 - the concentration of the initial solutions; for the solvent it is equal to zero.

Job source: Solution 2446. Unified State Exam 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 we added 20 kg of a 30% solution of the same acid, we would get a 50% acid solution. How many kilograms of the 25% solution were used to prepare the mixture?

Solution.

Let us denote by x kg the mass of the 25% solution, and by y kg the mass of the 95% solution. It can be noted that the total mass of 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 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% solution of acid is obtained:

We solve the system of equations and get:

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