Calculations for preparing aqueous solutions. Preparation of titrated solutions

Alkali solutions. Caustic alkalis and their solutions actively absorb moisture and carbon dioxide from the air, so preparing solutions of precise titre from them is difficult. It is best to make such solutions from fixanals. To do this, take a test tube with a fixation of the required normality and a 1 liter volumetric flask. A glass funnel with a glass striker inserted into it, the sharp end of which faces upward, is inserted into the flask.

When the striker is properly positioned in the funnel, the ampoule containing the fixanal is allowed to fall freely so that the thin bottom of the ampoule breaks when it hits the sharp end of the striker. After this, the side recess of the ampoule is punched and the contents are allowed to flow out. Then, without changing the position of the ampoule, it is thoroughly washed with well-boiled distilled water, cooled to a temperature of 35-40°C and taken in such quantity that after cooling the solution to 20°C, only a few drops need to be added to the mark. The titrated alkali solution should be stored in conditions that exclude the possibility of its contact with air.

If there is no fixanal, titrated solutions are prepared from sodium hydroxide (or potassium hydroxide) preparations. The molecular weight of NaOH is 40.01. This number is also its gram equivalent.

To prepare 1 liter 1 i. NaOH solution, you need to take 40 g of chemically pure caustic soda, and to prepare 1 liter 0.1 N. solution - ten times less, i.e. 4 g.

For the convenience of calculating the required amount of starting substances for the preparation of 1 liter of titrated solutions of alkalis of different normalities, we recommend using the data given in Table 31.

Table 31

To prepare 1 liter 0.1 n. sodium hydroxide solution, weigh out a little more than 4 g (4.3-4.5 g) of the drug and dissolve in a small volume of distilled water (about 7 ml).

After settling, the solution is carefully poured (without sediment) into a liter volumetric flask and brought to the mark with distilled freshly boiled water.

The prepared solution is mixed well and placed in a bottle protected from carbon dioxide. After this, the titer is established, i.e. the exact concentration of the solution.

The titer can be set using oxalic or succinic acid. Oxalic acid (C g H 2 0 4 -2H 2 0) is dibasic, and, therefore, its gram equivalent will be equal to half the molecular weight. If the molecular weight of oxalic acid is 126.05 g, then its gram equivalent will be 126.05: 2 = 63.025 g.

The existing oxalic acid should be recrystallized once or twice and only then used to set the titer.

Recrystallization is carried out as follows: take an arbitrary amount of the substance intended for recrystallization, dissolve it by heating, trying to obtain the highest possible concentration of the solution or saturated solution. If necessary, filter this solution through a hot filter funnel. The filtrate is collected in an Erlenmeyer flask, porcelain cup or glass.

Depending on the nature of crystallization of the substance, the solution, saturated in a hot state, is cooled. To quickly cool the solution during recrystallization, the crystallizer is placed in cold water, snow or ice. With slow cooling, the solution is left to stand at ambient temperature.

If very small crystals fall out, they are dissolved again by heating; the vessel in which the dissolution was carried out is immediately wrapped in several layers with a towel, covered with a watch glass and left to stand completely undisturbed for 12-15 hours.

Then the crystals are separated from the mother liquor, filtered under vacuum (Buchner funnel), thoroughly squeezed, washed and dried.

Preparing 0.1 n. NaOH solution, it is necessary to have a solution of oxalic acid of the same normality; for this, for 1 liter of solution you need to take 63.025: 10 = 6.3025 g. But to set the titer, this amount of oxalic acid solution is too much; It is enough to prepare 100 ml. To do this, about 0.63 g of recrystallized oxalic acid is weighed on an analytical balance, accurate to the fourth decimal place, for example, 0.6223 g. The taken sample of oxalic acid is dissolved in a volumetric flask (per 100 ml). Knowing the mass of the substance taken and the volume of the solution, it is easy to calculate its exact concentration, which in this case is not equal to 0.1 n., but somewhat less.

From the prepared solution, take 20 ml with a pipette, add a few drops of phenolphthalein and titrate with the prepared alkali solution until a faint pink color appears.

Let 22.05 ml of alkali be used for titration. How to determine its titer and normality?

0.6223 g of oxalic acid was taken instead of the theoretically calculated amount of 0.6303 g. Consequently, its normality will not be exactly 0.1

To calculate the normality of alkali, we use the relation VN=ViNt, i.e. the product of volume and normality known solution is equal to the product of volume and normality for an unknown solution. We get: 20-0.09873 =22.05-a:, from where

To calculate the titer or content of NaOH in 1 ml of solution, the normality should be multiplied by the gram equivalent of the alkali and the resulting product divided by 1000. Then the titer of the alkali will be

But this titer does not correspond to 0.1 n. NaOH solution. To do this, they resort to the coefficient To, i.e., the ratio of the practical titer to the theoretical one. In this case it will be equal

When used to set the title succinic acid its solution is prepared in the same order as oxalic acid, based on the following calculation: the molecular weight of succinic acid (C 4 H 6 0 4) is 118.05 g, but since it is dibasic, its gram equivalent is 59.02 g .

To prepare 1 liter of decinormal solution of succinic acid, it must be taken in an amount of 59.02: 10 = 5.902, and for 100 ml of solution -0.59 g.

Setting the titer to 0.1 n. NaOH solution by gravimetric method. To set the titer to 0.1 N. NaOH solution, we take a sample of succinic acid with an accuracy of 0.0001 g (for example, 0.1827 g). Dissolve the sample in distilled water (about 100 ml), then add 3-5 drops of phenolphthalein and titrate with alkali (NaOH). Let's assume that 28 ml of NaOH was used for titration. The calculation of the NaOH titer and the correction to it are carried out as follows: since the gram equivalent of NaOH, equal to 40.01 g, corresponds to the gram equivalent of succinic acid, equal to 59.02 g, then, by making up the proportion, we find out what amount of NaOH contained in weighed amount of succinic acid: 40.01-59.02

We calculate the titer of NaOH, i.e. the content of NaOH in 1 ml of solution. It is equal to: 0.1238: 28 = 0.00442. The correction to the NaOH titer is equal to the ratio of the practical to theoretical titer

Checking the normality of an alkali solution against a titrated acid solution. 20-25 ml of titrated acid solution (HC1 or H2S04) are measured into three conical flasks with a burette and titrated with NaOH solution until the color changes to methyl orange.

Let us assume that for the titration of three samples of 20 ml each, 0.1015 N. HC1 solution consumed an average of 19.50 ml of NaOH solution. The normality of alkali will be

Acid solutions. In most cases, the laboratory has to deal with sulfuric, hydrochloric and nitric acids. They are in the form of concentrated solutions, the percentage of which is determined by density.

For analytical work we use chemically pure acids. To prepare a solution of a particular acid, we usually take the amount of concentrated acids by volume, calculated by density.

For example, you need to prepare 0.1 N. solution H 2 S0 4 . This means that 1 liter of solution should contain

How much volume do you need to take H 2 S0 4 with a density of 1.84 in order to dilute it to 1 liter to get 0.1 N? solution?

An acid with a density of 1.84 contains 95.6% H 2 S0 4. Therefore, for 1 liter of solution it must be taken in grams:

Expressing the mass in volumetric units, we get

Having measured exactly 2.8 ml of acid from the burette, dilute it to 1 liter in a volumetric flask, then titrate with alkali to check the normality.

For example, during titration it was found that 1 ml of 0.1 N. H 2 S0 4 solution contains not 0.0049 g of H 2 S0 4, but 0.0051 g. To calculate the amount of water that needs to be added to 1 liter of acid, we make up the proportion:

Therefore, 41 ml of water must be added to this solution. But considering that 20 ml of the original solution was taken for titration, which is 0.02, then less water needs to be taken, i.e. 41-(41-0.02) = 41-0.8 = 40.2 ml . This amount of water is added from the burette to the flask with the solution.

The above work is quite painstaking when performed, so you can prepare approximately accurate solutions by introducing a correction factor, which is used in the work for each titration. In this case, we multiply the consumed number of milliliters of solution by a correction factor.

The correction factor is calculated using the formula

Where V - volume of the test solution taken for titration;

k t- correction factor of an alkali solution of known normality, by which the titer of the newly prepared acid solution is established;

Y x is the volume of an alkali solution of known normality used for titration of the test acid.

Table 32

To facilitate the process of preparing titrated solutions of acids, we offer a table of the amounts of starting substances for preparing 1 liter of solutions of different normalities (Table 32).

It must be kept in mind that when dissolving acids, the acid should be added to the water, and not vice versa.

Simple chemical solutions can be easily prepared in a variety of ways at home or at work. Whether you are making a solution from a powder material or diluting a liquid, you can easily determine the correct amount of each component. When preparing chemical solutions, do not forget to use personal protective equipment to avoid damage.

Steps

Calculation of percentages using the formula for weight/volume

Determine the percentage by weight/volume of the solution. Percentages show how many parts of a substance are present in one hundred parts of a solution. In application to chemical solutions this means that if the concentration is 1 percent, then 100 milliliters of solution contains 1 gram of substance, that is, 1 ml/100 ml.

• For example, by weight: a 10 percent solution by weight contains 10 grams of the substance dissolved in 100 milliliters of solution.
• For example, by volume: a 23 percent solution by volume contains 23 milliliters of liquid compound in every 100 milliliters of solution.

1. Determine the volume of solution you want to prepare. To find out the required mass of a substance, you must first determine the final volume of the solution you need. This volume depends on how much solution you will need, how often you will use it, and the stability of the finished solution.

   • If you need to use fresh solution each time, prepare only the amount needed for one use.
   • If the solution retains its properties for a long time, you can prepare large quantity to use it in the future.

2. Calculate the number of grams of substance required to prepare the solution. To calculate the number of grams required, use the following formula: number of grams = (percentage required)(volume required/100 ml). In this case, the required percentages are expressed in grams, and the required volume - in milliliters.

   • Example: you need to prepare a 5% NaCl solution with a volume of 500 milliliters.
   • number of grams = (5g)(500ml/100ml) = 25 grams.
   • If NaCl is given as a solution, simply take 25 milliliters of NaCl instead of the number of grams of powder and subtract that volume from the final volume: 25 milliliters of NaCl per 475 milliliters of water.

3. Weigh the substance. After you calculate the required mass of the substance, you should measure this amount. Take a calibrated scale, place the pan on it and set it to zero. Weigh required amount substance in grams and pour it out.

   • Before continuing to prepare the solution, be sure to clean the scale from any remaining powder.
   • In the example above, you need to weigh out 25 grams of NaCl.

4. Dissolve the substance in the required amount of liquid. Unless otherwise specified, water is used as a solvent. Take a measuring beaker and measure out the required amount of liquid. After this, dissolve the powder material in the liquid.

   • Label the container in which you will store the solution. Clearly indicate the substance and its concentration on it.
   • Example: Dissolve 25 grams of NaCl in 500 milliliters of water to obtain a 5 percent solution.
   • Remember that if you are diluting a liquid substance, to obtain the required amount of water, you must subtract the volume of the added substance from the final volume of the solution: 500 ml - 25 ml = 475 ml water.

   Preparation of molecular solution

   1. Determine the molecular weight of the substance used using the formula. The formula molecular weight (or simply molecular weight) of a compound is written in grams per mole (g/mol) on the side of the bottle. If you can't find the molecular weight on the bottle, look it up online.

      • The molecular weight of a substance is the mass (in grams) of one mole of that substance.
      • Example: The molecular weight of sodium chloride (NaCl) is 58.44 g/mol.

   2. Determine the volume of the required solution in liters. It is very easy to prepare one liter of solution, since its molarity is expressed in moles/liter, but you may need to make more or less than a liter, depending on the purpose of the solution. Use the final volume to calculate the number of grams needed.

      • Example: it is necessary to prepare 50 milliliters of a solution with a mole fraction of NaCl of 0.75.
      • To convert milliliters to liters, divide them by 1000 and get 0.05 liters.

   3. Calculate the number of grams required to prepare the required molecular solution. To do this, use the following formula: number of grams = (required volume)(required molarity)(molecular weight according to the formula). Remember that the required volume is expressed in liters, molarity in moles per liter, and molecular weight according to the formula in grams per mole.

      • Example: If you want to prepare 50 milliliters of a solution with a mole fraction of NaCl of 0.75 (molecular weight according to the formula: 58.44 g/mol), you should calculate the number of grams of NaCl.
      • number of grams = 0.05 L * 0.75 mol/L * 58.44 g/mol = 2.19 grams NaCl.
      • By reducing the units of measurement, you get grams of the substance.

   4. Weigh the substance. Using a properly calibrated scale, weigh out the required amount of substance. Place the pan on the scale and set it to zero before weighing. Add the substance to the bowl until you get the required mass.

      • Clean the scale pan after use.
      • Example: Weigh out 2.19 grams of NaCl.

   5. Dissolve the powder in the required amount of liquid. Unless otherwise noted, most solutions are made using water. In this case, the same volume of liquid is taken that was used to calculate the mass of the substance. Add the substance to the water and stir until completely dissolved.

      • Label the container with the solution. Clearly label the solute and molarity so that the solution can be used later.
      • Example: Using a beaker (an instrument for measuring volume), measure 50 milliliters of water and dissolve 2.19 grams of NaCl in it.
      • Stir the solution until the powder is completely dissolved.

   Diluting solutions of known concentration

   1. Determine the concentration of each solution. When diluting solutions, you need to know the concentration of the original solution and the solution you want to obtain. This method is suitable for diluting concentrated solutions.

      • Example: You need to prepare 75 milliliters of a 1.5 M NaCl solution from a 5 M solution. The original solution has a concentration of 5 M, and you need to dilute it to a concentration of 1.5 M.

   2. Determine the volume of the final solution. You need to find the volume of the solution you want to obtain. You will have to calculate the amount of solution that will be needed to dilute this solution to the required concentration and volume.

      • Example: You need to prepare 75 milliliters of a 1.5 M NaCl solution from a starting solution of 5 M. In this example, the final volume of the solution is 75 milliliters.

   3. Calculate the volume of solution that will be needed to dilute the initial solution. To do this, you will need the following formula: V 1 C 1 = V 2 C 2, where V 1 is the volume of the required solution, C 1 is its concentration, V 2 is the volume of the final solution, C 2 is its concentration.

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 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.

For cooking percent solutions use several techniques.

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 using the dilution method 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.

Usually, when the name “solution” is used, true solutions are meant. In true solutions, the solute in the form of individual molecules is distributed among the solvent molecules. Not all substances dissolve equally well in any liquid, i.e. The solubility of various substances in certain solvents is different. Typically, the solubility of solids increases with increasing temperature, so when preparing such solutions, in many cases it is necessary to heat them.

No more than a certain amount of a given substance can be dissolved in a certain amount of each solvent. If you prepare a solution containing per unit volume greatest number a substance that can dissolve at a given temperature, and add at least a small amount of soluble substance to it, then it will remain undissolved. Such a solution is called saturated.

If you prepare a concentrated solution that is close to saturated by heating, and then quickly but carefully cool the resulting solution, a precipitate may not form. If you throw a salt crystal into such a solution and stir it or rub it with a glass rod against the walls of the vessel, then salt crystals will fall out of the solution. Consequently, the cooled solution contained more salt than corresponded to its solubility at a given temperature. Such solutions are called supersaturated.

The properties of solutions always differ from the properties of the solvent. The solution boils at more than high temperature than pure solvent. The solidification temperature, on the contrary, is lower for solutions than for solvents.

Based on the nature of the solvent taken, solutions are divided into aqueous and non-aqueous. The latter include solutions of substances in organic solvents (alcohol, acetone, benzene, chloroform, etc.). The solvent for most salts, acids and alkalis is water. Biochemists rarely use such solutions; they often work with aqueous solutions of substances.

In each solution, the substance content is different, so it is important to know the quantitative composition of the solution. Exist various ways expressions for solution concentrations: in mass fractions of the dissolved substance, moles per 1 liter of solution, equivalents per 1 liter of solution, grams or milligrams per 1 ml of solution, etc.

The mass fraction of the dissolved substance is determined as a percentage. Therefore these solutions are called percent solutions.

The solute mass fraction (ω) expresses the ratio of the mass of the solute (m 1) to the total mass of the solution (m).

ω = (m 1 /m) x 100%

The mass fraction of the dissolved substance is usually expressed per 100 g of solution. Therefore, a 10% solution contains 10 g of substance in 100 g of solution or 10 g of substance and 100-10 = 90 g of solvent.

Molar concentration is determined by the number of moles of a substance in 1 liter of solution. The molar concentration of a solution (M) is the ratio of the amount of dissolved substance in moles (ν) to a certain volume of this solution (V).

The volume of solution is usually expressed in liters. In laboratories, the value of molar concentration is usually denoted by the letter M. Thus, a monomolar solution is denoted 1 M (1 mol/l), a decimolar solution - 0.1 M (0.1 mol/l), etc. In order to determine how many grams of a given substance are in 1 liter of a solution of a given concentration, it is necessary to know its molar mass (see periodic table). It is known that the mass of 1 mole of a substance is numerically equal to its molar mass, for example, the molar mass of sodium chloride is 58.45 g/mol, therefore, the mass of 1 mole of NaCl is equal to 58.45 g. Thus, 1 M NaCl solution contains 58.45 g sodium chloride in 1 liter of solution.

Molar concentration equivalent(normal concentration) is determined by the number of equivalents of the dissolved substance in 1 liter of solution.

Let's look at the concept of “equivalent”. For example, HCl contains 1 mole of atomic hydrogen and 1 mole of atomic chlorine. We can say that 1 mole of atomic chlorine is equivalent (or equivalent) to 1 mole of atomic hydrogen, or the equivalent of chlorine in the HCl compound is 1 mole.

Zinc does not combine with hydrogen, but displaces it from a number of acids:

Zn + 2HC1 = Zn C1 2 + H 2

From the reaction equation it is clear that 1 mole of zinc replaces 2 moles of atomic hydrogen in hydrochloric acid. Therefore, 0.5 mol of zinc is equivalent to 1 mol of atomic hydrogen, or the zinc equivalent for this reaction will be 0.5 mol.

Complex compounds can also be equivalents, for example in the reaction:

2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O

1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. It follows that 1 mol of sodium hydroxide is equivalent in this reaction to 0.5 mol of sulfuric acid.

It must be remembered that in any reaction the substances react in equivalent quantities. To prepare solutions containing a certain number of equivalents of a given substance, it is necessary to be able to calculate the molar mass of the equivalent (equivalent mass), i.e., the mass of one equivalent. The equivalent (and therefore the equivalent mass) is not a constant value for a given compound, but depends on the type of reaction in which the compound enters.

Equivalent mass of acid equal to its molar mass divided by the basicity of the acid. Thus, for nitric acid HNO 3 the equivalent mass is equal to its molar mass. For sulfuric acid, the equivalent mass is 98:2 = 49. For tribasic phosphoric acid, the equivalent mass is 98:3 = 32.6.

In this way, the equivalent mass for reactions is calculated complete exchange or complete neutralization. For reactions incomplete neutralization and incomplete exchange the equivalent mass of a substance depends on the course of the reaction.

For example, in react:

NaOH + H 2 SO 4 = NaHSO 4 + H 2 O

1 mole of sodium hydroxide is equivalent to 1 mole of sulfuric acid, so in this reaction the equivalent mass of sulfuric acid is equal to its molar mass, i.e. 98 g.

Equivalent base mass equal to its molar mass divided by the oxidation state of the metal. For example, the equivalent mass of sodium hydroxide NaOH is equal to its molar mass, and the equivalent mass of magnesium hydroxide Mg(OH) 2 is equal to 58.32:2 == 29.16 g. This is how the equivalent mass is calculated only for the reaction complete neutralization. For reaction incomplete neutralization this value will also depend on the course of the reaction.

Equivalent mass of salt equal to the molar mass of the salt divided by the product of the oxidation state of the metal and the number of its atoms in the salt molecule. So the equivalent mass of sodium sulfate is equal to 142: (1x2) = 71 g, and the equivalent mass of aluminum sulfate Al 2 (SO 4) 3 is equal to 342: (3x2) = 57 g. However, if salt is involved in an incomplete exchange reaction, then only the number of metal atoms participating in the reaction is taken into account.

Equivalent mass of a substance participating in a redox reaction, is equal to the molar mass of a substance divided by the number of electrons accepted or given up by a given substance. Therefore, before making calculations, it is necessary to write the reaction equation:

2CuSO 4 + 4KI = 2CuI + I 2 + 2K 2 SO 4

Cu 2+ + e - à Cu +

I - - e - à I o

The equivalent mass of CuSO 4 is equal to the molar mass (160 g). In laboratory practice, the name “normal concentration” is used, which is denoted in various formulas by the letter N, and when the concentration of a given solution is designated by the letter “n”. A solution containing 1 equivalent in 1 liter of solution is called one-normal and is designated 1 N, containing 0.1 equivalent - decinormal (0.1 N), 0.01 equivalent - centinormal (0.01 N).

The titer of a solution is the number of grams of a substance dissolved in 1 ml of solution. In the analytical laboratory, the concentration of working solutions is recalculated directly to the substance being determined. Then the titer of the solution shows how many grams of the substance being determined corresponds to 1 ml of the working solution.

The concentration of solutions used in photometry of the so-called standard solutions, are usually expressed by the number of milligrams in 1 ml of solution.

When preparing acid solutions a concentration of 1:x is often used, indicating how many parts by volume of water (X) are per part of concentrated acid.

To approximate solutions These include solutions whose concentration is expressed as a percentage, as well as solutions of acids, the concentration of which is indicated by the expression 1:x. Before preparing solutions, prepare dishes for preparing and storing them. If you are preparing a small amount of solution that will be used during the day, then it does not have to be poured into a bottle, but can be left in the flask.

On the flask it is necessary to write special wax pencil(or marker) the formula of the solute and the concentration of the solution, for example HC1 (5%). At long-term storage A label must be placed on the bottle in which the solution will be stored indicating what solution is in it and when it was prepared.

Dishes for preparing and storing solutions must be washed and rinsed with distilled water.

To prepare solutions, only pure substances and distilled water should be used. Before preparing the solution, it is necessary to calculate the amount of solute and the amount of solvent. When preparing approximate solutions, the amount of solute is calculated to the nearest tenth, molecular weight values ​​are taken rounded to whole numbers, and when calculating the amount of liquid, fractions of a milliliter are not taken into account.

The technique for preparing solutions of various substances is different. However, when preparing any approximate solution, a sample is taken on a technochemical balance, and the liquids are measured with a measuring cylinder.

Preparation of salt solutions. You need to prepare 200 g of a 10% solution of potassium nitrate KNO3.

The required amount of salt is calculated according to the proportion:

100 g - 10 g KNO 3

200 g - X g KNO 3 X = (200 x 10) / 100 = 20 g KNO 3

Amount of water: 200-20=180 g or 180 ml.

If the salt from which the solution is prepared is contains water of crystallization, then the calculation will be slightly different. For example, you need to prepare 200 g of a 5% CaCl 2 solution based on CaCl 2 x 6H 2 O.

First, the calculation is made for anhydrous salt:

100 g - 5 g CaCl 2

200 g - X g CaCl 2 X = 10 g CaCl 2

The molecular weight of CaCl 2 is 111, the molecular weight of CaCl 2 x 6H 2 O is 219, therefore, 219 g of CaCl 2 x 6H 2 O contains 111 g of CaCl 2.

Those. 219 - 111

X - 10 X = 19.7 g CaCl 2 x 6H 2 O

To obtain the required solution, it is necessary to weigh out 19.7 g of CaCl 2 x 6H 2 O salt. The amount of water is 200-19.7 = 180.3 g, or 180.3 ml. Water is measured using a measuring cylinder, so tenths of a millimeter are not taken into account. Therefore, you need to take 180 ml of water.

The salt solution is prepared as follows. The required amount of salt is weighed out on a technical chemical scale. Carefully transfer the sample into a flask or glass where the solution will be prepared. Measure out required quantity water with a graduated cylinder and pour approximately half of the measured amount into a flask with a sample of water. By vigorous stirring, complete dissolution of the sample is achieved, and sometimes this requires heating. After dissolving the sample, add the remaining amount of water. If the solution is cloudy, it is filtered through a folded filter.

Preparation of alkali solutions. Calculation of the amount of alkali required to prepare a solution of a given concentration is carried out in the same way as for salt solutions. However, solid alkali, especially not very well purified, contains many impurities, so it is recommended to weigh out alkali in an amount greater than the calculated amount by 2-3%. The technique for preparing alkali solutions has its own characteristics.

When preparing alkali solutions, the following rules must be observed:

1. Pieces of alkali should be taken with tongs, tweezers, and if you need to take them with your hands, then be sure to wear rubber gloves. Granulated alkali in the form of small cakes is poured with a porcelain spoon.

2. You cannot weigh lye on paper; For this you should use only glass or porcelain dishes.

3. Alkali cannot be dissolved in thick-walled bottles, since during dissolution the solution becomes very hot; the bottle may burst.

The amount of alkali weighed on a technochemical scale is placed in a large porcelain cup or glass. Such an amount of water is poured into this container so that the solution has a concentration of 35-40%. Stir the solution with a glass rod until all the alkali has dissolved. Then the solution is left to stand until it cools and a precipitate forms. The precipitate consists of impurities (mainly carbonates) that do not dissolve in concentrated alkali solutions. The remaining alkali is carefully poured into another vessel (preferably using a siphon), where the required amount of water is added.

Preparation of acid solutions. Calculations for preparing acid solutions are different than for preparing solutions of salts and alkalis, since the concentration of acid solutions is not 100% due to the water content; The required amount of acid is not weighed out, but measured using a measuring cylinder. When calculating acid solutions, standard tables are used, which indicate the percentage of the acid solution, the density of this solution at a certain temperature and the amount of this acid contained in 1 liter of a solution of a given concentration.

For example, you need to prepare 1 liter of 10% HCl solution based on the available 38.0% acid with a density of 1.19. From the table we find that a 10% acid solution at room temperature has a density of 1.05, therefore, its mass of 1 liter is 1.05 x 1000 == 1050 g.

For this amount, the content of pure HCl is calculated:

100 g - 10 g HCl

1050 g - X g HCl X = 105 g HCl

An acid having a density of 1.19 contains 38 g of HCl, therefore:

X = 276 g or 276: 1.19 = 232 ml.

Amount of water: 1000 ml - 232 ml = 768 ml.

Acid solutions are often used the concentration of which is expressed 1:x, where x is an integer indicating how many volumes of water should be taken per volume of concentrated acid. For example, a 1:5 acid solution means that when preparing the solution, 5 volumes of water were mixed with 1 volume of concentrated acid.

For example, prepare 1 liter of sulfuric acid solution 1:7. There will be 8 parts in total. Each part is equal to 1000:8 = 125 ml. Therefore, you need to take 125 ml of concentrated acid and 875 ml of water.

When preparing acid solutions, the following rules must be observed:

1. The solution cannot be prepared in a thick-walled bottle, since when diluting acids, especially sulfuric acid, strong heating occurs. Acid solutions are prepared in flasks.

2. When diluting, do not pour water into the acid. The calculated amount of water is poured into the flask, and then the required amount of acid is added in a thin stream, gradually, with stirring. Acid and water are measured using graduated cylinders.

3. After the solution has cooled, pour it into a bottle and stick a label; the paper label is waxed; You can make a label with special paint directly on the bottles.

4. If the concentrated acid from which the dilute solution will be prepared is stored for a long time, then it is necessary to clarify its concentration. To do this, measure its density and use the table to find the exact acid content in the solution.

Concentration of precise solutions expressed as molar or normal concentration or titer. These solutions are usually used in analytical work; They are rarely used in physicochemical and biochemical studies.

Weights for the preparation of precise solutions are calculated with an accuracy of up to the fourth decimal place, and the accuracy of molecular masses corresponds to the accuracy with which they are given in the reference tables. The sample is taken on an analytical balance; the solution is prepared in a volumetric flask, i.e. the amount of solvent is not calculated. The prepared solutions should not be stored in volumetric flasks; they are poured into a bottle with a well-selected stopper.

If the exact solution needs to be poured into a bottle or another flask, proceed as follows. The bottle or flask into which the solution will be poured is thoroughly washed, rinsed several times with distilled water and allowed to stand upside down to drain the water, or dried. Rinse the bottle 2-3 times with small portions of the solution that you are going to pour, and then pour the solution itself. Each precise solution has its own shelf life.

Cooking calculations molar and normal solutions carried out as follows.

Example 1.

It is required to prepare 2 liters of 0.5 M Na 2 CO 3 solution. The molar mass of Na 2 CO 3 is 106. Therefore, 1 liter of 0.5 M solution contains 53 g of Na 2 CO 3. To prepare 2 liters you need to take 53 x 2 = 106 g Na 2 CO 3. This amount of salt will be contained in 2 liters of solution.

Another way to visualize the calculation:

1 liter of 1M Na 2 CO 3 solution contains 106 g of Na 2 CO 3

(1L - 1M - 106 g)

2 l of 1M Na 2 CO 3 solution contains x g Na 2 CO 3

(2L - 1M - x g);

when counting, “the hand closes” the central part of the expression (1M)

We find that 2 liters of 1M Na 2 CO 3 solution contains 212 g of Na 2 CO 3

(2L - 1M - 212 g)

And 2 liters of 0.5M Na 2 CO 3 solution (“the left side closes”) contains x g Na 2 CO 3 (2 l - 0.5 M - x g)

Those. 2 l of 0.5M Na 2 CO 3 solution contains 106 g of Na 2 CO 3

(2 L - 0.5 M - 106 g).

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

1 action:

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

Required volume in ml (to be prepared)

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

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

Action 2:

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

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

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

0.1 g dry matter – 0.5 ml solvent

0.5 g dry matter - x ml solvent

we get:

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

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

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

1000000 units – x

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

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

Solution:

1 ml of solution – 0.1 g

x ml - 0.25 g

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

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

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

Similar: 36:4=9 (divisions)

52:4=13(divisions)

Answer: 7, 9, 13 divisions.

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

Solution:

1) 100 g – 5 g

(d) active substance

2) 100% – 10g

(ml) 10% solution

3) 10000-5000=5000 (ml) water

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

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

Solution:

Since 100 ml contains 10 g of active substance,

1) 100g – 1ml

5000 ml – x

(ml) active substance

2) 100% – 10ml

00 (ml) 10% solution

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

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

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

Solution:

Since 100 ml contains 10 ml of active substance,

1) 100% – 0.5ml

0 (ml) active substance

2) 100% – 10 ml

(ml) 10% solution

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

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

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

Solution:

1) 3g – 100 ml

G

2) 10000 – 300=9700ml.

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

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

Solution:

Percentage is the amount of substance in 100 ml.

1) 0.5 g – 100 ml

G

2) 3000 – 15 = 2985 ml.

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

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

Solution:

Percentage is the amount of substance in 100 ml.

1) 3 g – 100 ml

G

2) 5000 – 150= 4850 ml.

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

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

Solution:

According to formula (1)

ml

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

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

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

10g – 1000 ml

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

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

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

Calculate how much medication the patient needs per day:

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

7* 0.004 g = 0.028 g.

Answer: This medicine must be prescribed 0.028 g.

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

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

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

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

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