Preparation of titrated solutions. Calculations in the preparation of solutions of molar and normal concentrations

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 (such as 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, you know the initial concentration, as well as the concentration and volume of the solution that you want to receive; wherein the volume of the concentrated solution to be diluted is unknown.

• In another situation, for example, when solving a school problem in chemistry, another quantity may act as an unknown: for example, you are given an initial volume and concentration, and you need to find the final concentration of the final solution with a known volume. In any case, it is useful to write down the known and unknown quantities before starting the problem.
• Consider an example. Suppose we need to dilute a solution with a concentration of 5 M in order to obtain a solution with a concentration of 1 mm. In this case, we know the concentration of the initial solution, as well as the volume and concentration of the solution to be obtained; not the volume of the initial solution to be diluted with water is known.
  • Remember: in chemistry, M is a measure of concentration, also called molarity, which corresponds to the number of moles of a substance per 1 liter of solution.

To prepare solutions of molar and normal concentrations, a sample of the substance is weighed on an analytical balance, and the solutions are prepared in a volumetric flask. When preparing acid solutions, the required volume of a concentrated acid solution is measured with a burette with a glass tap.

The weight of the solute is counted to the fourth decimal place, and the molecular weights are taken with the accuracy with which they are given in the reference tables. The volume of concentrated acid is calculated to the second decimal place.

Example 1. How many grams of barium chloride is needed to prepare 2 liters of a 0.2 M solution?

Decision. The molecular weight of barium chloride is 208.27. Hence. 1 liter of 0.2 M solution should contain 208.27-0.2 = = 41.654 g BaCl 2 . To prepare 2 liters, 41.654-2 \u003d 83.308 g of BaCl 2 will be required.

Example 2. How many grams of anhydrous soda Na 2 C0 3 will be required to prepare 500 ml of 0.1 n. solution?

Decision. The molecular weight of soda is 106.004; equivalent share weight 5 N a 2 C0 3 \u003d M: 2 \u003d 53.002; 0.1 eq. = 5.3002 g.

1000 ml 0.1 n. solution contain 5.3002 g of Na 2 C0 3
500 »» » » » X » Na 2 C0 3

5,3002-500
x=—— Gooo-- = 2-6501 g Na 2 C0 3.

Example 3 How much concentrated sulfuric acid (96%: d=1.84) is required to prepare 2 liters of 0.05N. sulfuric acid solution?

Decision. The molecular weight of sulfuric acid is 98.08. Equivalent mass of sulfuric acid 3h 2 so 4 \u003d M: 2 \u003d 98.08: 2 \u003d 49.04 g. Weight 0.05 equiv. \u003d 49.04-0.05 \u003d 2.452 g.

Let's find how much H 2 S0 4 should be contained in 2 l 0.05 n. solution:

1 l-2.452 g H 2 S0 4

2"- X » H 2 S0 4

X \u003d 2.452-2 \u003d 4.904 g H 2 S0 4.

In order to determine how much a 96% solution of H 2 S0 4 should be taken for this, we compose the proportion:

\ in 100 g conc. H 2 S0 4 -96 g H 2 S0 4

At» » H 2 S0 4 -4.904 g H 2 S0 4

4,904-100
At=——— §6—— = 5.11 g H 2 S0 4 .

Convert this amount to volume: ,. R 5,11

K \u003d 7 \u003d TJ \u003d 2 "77 ml -

Thus, to prepare 2 liters of 0.05 N. solution should take 2.77 ml of concentrated sulfuric acid.

Example 4. Calculate the titer of a NaOH solution if its exact concentration is known to be 0.0520 N.

Decision. Recall that the titer is the content in 1 ml of a solution of a substance in grams. Equivalent mass of NaOH \u003d 40 01 g Find how many grams of NaOH are contained in 1 liter of this solution:

40.01-0.0520 = 2.0805 g

1 liter of solution: -u = - = 0.00208 g / ml. You can also use the formula:

9 N

where T- titer, g/ml; E- equivalent weight; N- the normality of the solution.

Then the titer of this solution is:

f 40,01 0,0520

“NaOH =——— jooo—— 0.00208 g/ml.

„ “Rie P 5 - Calculate the normal concentration of a solution of HN0 3, if it is known that the titer of this solution is 0.0065. To calculate, we use the formula:

T ■ 1000 63,05

5hno 3 = j- = 63.05.

The normal concentration of a nitric acid solution is:

- V \u003d 63.05 \u003d 0.1030 n.

Example 6. What is the normal concentration of a solution if it is known that 200 ml of this solution contains 2.6501 g of Na 2 C0 3

Decision. As was calculated in example 2, Zma 2 co(=53.002.
Let's find how many equivalents are 2.6501 g of Na 2 C0 3: G
2.6501: 53.002 = 0.05 equiv. /

In order to calculate the normal concentration of the solution, we compose the proportion:

1000 » » X "

1000-0,05
x = —————— =0.25 equiv.

1 liter of this solution will contain 0.25 equivalents, i.e. the solution will be 0.25 n.

For this calculation, you can use the formula:

R- 1000

where R - amount of substance in grams; E - equivalent mass of the substance; V is the volume of the solution in milliliters.

Zia 2 co 3 \u003d 53.002, then the normal concentration of this solution

2.6501-10С0 N = 53.002-200

Salt solution may be needed for a variety of purposes, for example, it is part of some traditional medicine. So how to prepare a 1% solution if there are no special beakers at home to measure the amount of product? In general, even without them, you can make a 1% salt solution. How to cook it is detailed below. Before proceeding with the preparation of such a solution, you should carefully study the recipe and accurately determine the necessary ingredients. The thing is that the definition of "salt" can refer to different substances. Sometimes it turns out to be ordinary edible salt, sometimes rock or even sodium chloride. As a rule, in a detailed recipe it is always possible to find an explanation of which substance is recommended to be used. Folk recipes often also indicate magnesium sulfate, which has the second name "epsom salt".

If a substance is required, for example, for gargling or relieving pain from a tooth, then most often in this case it is recommended to use a saline solution of sodium chloride. In order for the resulting product to have healing properties and not harm the human body, only high-quality ingredients should be selected for it. For example, rock salt contains a lot of extra impurities, so instead of it it is better to use ordinary fine salt (you can also use iodized salt for rinsing). As for water, at home you should use filtered or at least boiled water. Some recipes recommend using rainwater or snow. But, given the current ecological state, this is not worth doing. Especially for residents of large cities. It is better to just thoroughly clean the tap water.

If there was no special filter at home, then the well-known "old-fashioned" method can be used to purify water. It involves freezing tap water in the freezer. As you know, in the process, it is the purest liquid that first turns into ice, and all harmful impurities and dirt sink to the bottom of the container. Without waiting for the freezing of the entire glass, you should remove the upper ice part and then melt it. Such water will be as pure and safe as possible for health. It can be used to prepare saline solution.

Now it is worth deciding on the units of measurement for liquid and solid matter. For salt, it is most convenient to use a teaspoon. As you know, it contains 7 grams of the product, if the spoon is with a slide, then 10. The latter option is more convenient to use to calculate the percentage. It is easy to measure water with an ordinary faceted glass if there are no special beakers in the house. It contains 250 milliliters of water. The mass of 250 milliliters of pure fresh water is 250 grams. It is most convenient to use half a glass of liquid or 100 grams. Next is the most difficult stage of preparing the saline solution. It is worth once again carefully studying the recipe and deciding on the proportions. If it is recommended to take a 1% salt solution in it, then in every 100 grams of liquid, 1 gram of solid will need to be dissolved. The most accurate calculations will suggest that it will be necessary to take 99 grams of water and 1 gram of salt, but such accuracy is unlikely to be required.

It is quite possible to allow some error and, for example, add one heaping teaspoon of salt to one liter of water to get a 1% saline solution. Currently, it is often used, for example, in the treatment of colds and especially sore throats. You can also add soda or a few drops of iodine to the finished solution. The resulting gargle mixture will be an excellent effective and effective remedy for sore throat. Unpleasant sensations will go away after just a few procedures. By the way, such a solution is not prohibited for use by the smallest family members. The main thing is not to overdo it with additional ingredients (especially with iodine), otherwise you can damage the oral mucosa and only aggravate the condition of a sore throat.

Also, a saline solution can be used to relieve a pulling aching toothache. True, it is more efficient to use a more saturated one, for example, 10 percent. Such a mixture is really able to relieve painful discomfort in the oral cavity for a short time. But it is not a drug, so you should never postpone a visit to the dentist after relief.

approximate solutions. When preparing approximate solutions, the amounts of substances that must be taken for this are calculated with little accuracy. Atomic weights of elements to simplify calculations can be taken rounded sometimes to whole units. So, for a rough calculation, the atomic weight of iron can be taken equal to 56 instead of the exact -55.847; for sulfur - 32 instead of the exact 32.064, etc.

Substances for the preparation of approximate solutions are weighed on technochemical or technical scales.

Fundamentally, the calculations in the preparation of solutions are exactly the same for all substances.

The amount of the prepared solution is expressed either in units of mass (g, kg) or in units of volume (ml, l), and for each of these cases, the calculation of the amount of the dissolved substance is carried out differently.

Example. Let it be required to prepare 1.5 kg of a 15% sodium chloride solution; pre-calculate the required amount of salt. The calculation is carried out according to the proportion:

i.e. if 100 g of the solution contains 15 g of salt (15%), then how much will it take to prepare 1500 g of the solution?

The calculation shows that you need to weigh 225 g of salt, then take 1500 - 225 = 1275 g. ¦

If it is given to obtain 1.5 liters of the same solution, then in this case, according to the reference book, its density is found out, the latter is multiplied by the given volume and thus the mass of the required amount of solution is found. Thus, the density of a 15%-horo solution of sodium chloride at 15 0C is 1.184 g/cm3. Therefore, 1500 ml is

Therefore, the amount of substance for preparing 1.5 kg and 1.5 l of solution is different.

The calculation given above is applicable only for the preparation of solutions of anhydrous substances. If an aqueous salt is taken, for example Na2SO4-IOH2O1, then the calculation is somewhat modified, since crystallization water must also be taken into account.

Example. Let it be necessary to prepare 2 kg of 10% Na2SO4 solution starting from Na2SO4 *10H2O.

The molecular weight of Na2SO4 is 142.041 and Na2SO4*10H2O is 322.195, or rounded 322.20.

The calculation is carried out first on anhydrous salt:

Therefore, you need to take 200 g of anhydrous salt. The amount of decahydrate salt is found from the calculation:

Water in this case must be taken: 2000 - 453.7 \u003d 1546.3 g.

Since the solution is not always prepared in terms of anhydrous salt, then on the label, which must be stuck on the vessel with the solution, it is necessary to indicate from which salt the solution is prepared, for example, 10% Na2SO4 solution or 25% Na2SO4 * 10H2O.

It often happens that the previously prepared solution needs to be diluted, i.e., its concentration should be reduced; solutions are diluted either by volume or by weight.

Example. It is necessary to dilute a 20% solution of ammonium sulfate so as to obtain 2 liters of a 5% solution. We carry out the calculation in the following way. We learn from the reference book that the density of a 5% solution of (NH4) 2SO4 is 1.0287 g/cm3. Therefore, 2 liters of it should weigh 1.0287 * 2000 = 2057.4 g. This amount should contain ammonium sulfate:

Considering that losses may occur during measuring, you need to take 462 ml and bring them to 2 liters, i.e. add 2000-462 = 1538 ml of water to them.

If the dilution is carried out by weight, the calculation is simplified. But in general, dilution is carried out on a volume basis, since liquids, especially in large quantities, are easier to measure by volume than to weigh.

It must be remembered that in all work, both with dissolution and dilution, one should never pour all the water into the vessel at once. Rinse with water several times the dishes in which the weighing or measuring of the desired substance was carried out, and each time this water is added to the vessel for the solution.

When special accuracy is not required, when diluting solutions or mixing them to obtain solutions of a different concentration, you can use the following simple and quick method.

Let us take the already analyzed case of diluting a 20% solution of ammonium sulphate to 5%. First we write like this:

where 20 is the concentration of the solution taken, 0 is water and 5 "is the required concentration. Now we subtract 5 from 20 and write the resulting value in the lower right corner, subtracting zero from 5, we write the number in the upper right corner. Then the circuit will look like this :

This means that you need to take 5 volumes of a 20% solution and 15 volumes of water. Of course, such a calculation is not accurate.

If you mix two solutions of the same substance, then the scheme remains the same, only the numerical values ​​\u200b\u200bare changed. Let a 25% solution be prepared by mixing a 35% solution and a 15% solution. Then the diagram will look like this:

i.e. you need to take 10 volumes of both solutions. This scheme gives approximate results and can be used only when special accuracy is not required. It is very important for any chemist to cultivate the habit of accuracy in calculations when necessary, and to use approximate figures in cases where this will not affect the results. work. When greater accuracy is needed when diluting solutions, the calculation is carried out using formulas.

Let's look at some of the most important cases.

Preparing a diluted solution. Let c be the amount of solution, m% is the concentration of the solution to be diluted to a concentration of n%. The resulting amount of dilute solution x is calculated by the formula:

and the volume of water v for diluting the solution is calculated by the formula:

Mixing two solutions of the same substance of different concentration to obtain a solution of a given concentration. Let by mixing a parts of an m% solution with x parts of a n% solution, you need to obtain a /% solution, then:

precise solutions. When preparing exact solutions, the calculation of the quantities of the required substances will be checked already with a sufficient degree of accuracy. The atomic weights of the elements are taken from the table, which shows their exact values. When adding (or subtracting), the exact value of the term with the fewest decimal places is used. The remaining terms are rounded off, leaving one more decimal place after the decimal point than in the term with the least number of digits. As a result, as many digits after the decimal point are left as there are in the term with the least number of decimal places; while doing the necessary rounding. All calculations are made using logarithms, five-digit or four-digit. The calculated amounts of the substance are weighed only on an analytical balance.

Weighing is carried out either on a watch glass or in a bottle. The weighed substance is poured into a cleanly washed volumetric flask through a clean, dry funnel in small portions. Then, from the washer, several times with small portions of water, the bnzhe or the watch glass in which the weighing was carried out is washed over the funnel. The funnel is also washed several times with distilled water.

For pouring solid crystals or powders into a volumetric flask, it is very convenient to use the funnel shown in Fig. 349. Such funnels are made with a capacity of 3, 6, and 10 cm3. You can weigh the sample directly in these funnels (non-hygroscopic materials), having previously determined their mass. The sample from the funnel is very easily transferred to the volumetric flask. When the sample is poured, the funnel, without removing the flask from the throat, is well washed with distilled water from the wash bottle.

As a rule, when preparing accurate solutions and transferring the solute to a volumetric flask, the solvent (for example, water) should occupy no more than half the capacity of the flask. Stopper the volumetric flask and shake it until the solid dissolves completely. The resulting solution is then filled up to the mark with water and mixed thoroughly.

molar solutions. To prepare 1 liter of a 1 M solution of a substance, 1 mol of it is weighed on an analytical balance and dissolved as described above.

Example. To prepare 1 liter of 1 M solution of silver nitrate, find in the table or calculate the molecular weight of AgNO3, it is equal to 169.875. Salt is weighed and dissolved in water.

If you need to prepare a more dilute solution (0.1 or 0.01 M), weigh out respectively 0.1 or 0.01 mol of salt.

If you need to prepare less than 1 liter of solution, then dissolve a correspondingly smaller amount of salt in the corresponding volume of water.

Normal solutions are prepared in a similar way, only weighing not 1 mole, but 1 gram equivalent of a solid.

If you need to prepare a semi-normal or decinormal solution, take 0.5 or 0.1 gram equivalent, respectively. When preparing not 1 liter of solution, but less, for example 100 or 250 ml, then take 1/10 or 1/4 of the amount of the substance required to prepare 1 liter and dissolve in the appropriate volume of water.

Fig 349. Funnels for pouring a sample into a flask.

After preparing the solution, it must be checked by titration with an appropriate solution of another substance with a known normality. The prepared solution may not correspond exactly to the normality that is given. In such cases, an amendment is sometimes introduced.

In production laboratories, accurate solutions are sometimes prepared “by the substance to be determined”. The use of such solutions facilitates calculations in analyzes, since it is enough to multiply the volume of the solution used for titration by the titer of the solution in order to obtain the content of the desired substance (in g) in the amount of any solution taken for analysis.

When preparing a titrated solution for the analyte, the calculation is also carried out according to the gram equivalent of the dissolved substance, using the formula:

Example. Let it be necessary to prepare 3 liters of potassium permanganate solution with an iron titer of 0.0050 g / ml. The gram equivalent of KMnO4 is 31.61 and the gram equivalent of Fe is 55.847.

We calculate according to the above formula:

standard solutions. Standard solutions are called solutions with different, precisely defined concentrations used in colorimetry, for example, solutions containing 0.1, 0.01, 0.001 mg, etc. of a solute in 1 ml.

In addition to colorimetric analysis, such solutions are needed when determining pH, for nephelometric determinations, etc. Sometimes standard solutions are stored in sealed ampoules, but more often they have to be prepared immediately before use. Standard solutions are prepared in a volume of no more than 1 liter, and more often - less.Only with a large consumption of the standard solution, it is possible to prepare several liters of it, and then on condition that the standard solution will not be stored for a long time.

The amount of substance (in g) required to obtain such solutions is calculated by the formula:

Example. It is necessary to prepare standard solutions of CuSO4 5H2O for the colorimetric determination of copper, and 1 ml of the first solution should contain 1 mg of copper, the second - 0.1 mg, the third - 0.01 mg, the fourth - 0.001 mg. First prepare a sufficient amount of the first solution, for example 100 ml.

Alkali solutions. Caustic alkalis and their solutions actively absorb moisture and carbon dioxide from the air, so it is difficult to prepare accurate titer solutions from them. It is best to make such solutions from fixanals. To do this, take a test tube with a fixanal of the required normality and a 1-liter volumetric flask. A glass funnel is inserted into the flask with a glass striker inserted into it, the sharp end of which is turned upwards.

When the striker is correctly placed in the funnel, the fixanal ampoule is allowed to fall freely so that the thin bottom of the ampoule breaks when it hits the sharp end of the striker. After that, the lateral recess of the ampoule is pierced 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 an amount that after cooling the solution to 20 ° C, only a few drops should be added to the mark. The titrated alkali solution should be stored under conditions that exclude the possibility of its contact with air.

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

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

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

Table 31

To prepare 1 liter of 0.1 N. sodium hydroxide solution, weigh 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 well mixed and placed in a bottle protected from carbon dioxide. After that, the titer is established, that is, the exact concentration of the solution.

The titer can be set according to 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 one. If the molecular weight of oxalic acid is 126.05 g, then its gram equivalent will be 126.05: 2 = 63.025 g.

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

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

Depending on the nature of the crystallization of the substance, the solution saturated in the 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 allowed to stand at ambient temperature.

If very small crystals have fallen 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 still for 12-15 hours.

Then the crystals are separated from the mother liquor by filtering under vacuum (Buchner funnel), carefully 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 it is necessary to take 63.025: 10 \u003d 6.3025 g per 1 liter of solution. But to set the titer of such an amount of oxalic acid solution, there is a lot; enough to prepare 100 ml. To do this, about 0.63 g of recrystallized oxalic acid is weighed on an analytical balance to the fourth decimal place, for example, 0.6223 g. A 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 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?

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

To calculate the normality of an alkali, we use the relation VN=ViNt, i.e., the product of volume and normality of a known solution is equal to the product of volume and normality for an unknown solution. We get: 20-0.09873 \u003d 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 alkali and the resulting product divided by 1000. Then the alkali titer will be

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

When using succinic acid to set the titer, 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, then its gram equivalent 59.02 g.

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

Titer setting 0.1 N NaOH solution by weight 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). We dissolve the sample in distilled water (about 100 ml), then add 3-5 drops of phenolphthalein and titrate with alkali (NaOH). Assume that 28 ml of NaOH is used for titration. We calculate the NaOH titer and correct it 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, making up the proportion, we find out how much NaOH is 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 practical titer to theoretical

Checking the normality of an alkali solution by a titrated acid solution. 20-25 ml of a titrated acid solution (HC1 or H 2 S0 4) are measured with a burette into three conical flasks and titrated with a NaOH solution until the color of methyl orange changes.

Assume that for the titration of three samples of 20 ml of 0.1015 N. HC1 solution consumed an average of 19.50 ml of NaOH solution. The alkali normality 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 recognizable by density.

In analytical work we use chemically pure acids. To prepare a solution of one or another acid, the amount of concentrated acids is usually taken by volume, calculated from the density.

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

How much by volume do you need to take H 2 S0 4 with a density of 1.84, so that, diluting it to 1 liter, you 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 volume units, we get

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

For example, during titration, it was found that 1 ml of 0.1 N. a solution of H 2 S0 4 does not contain 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 was taken from the initial solution for titration, which is 0.02, then less water should be taken, i.e. 41-(41-0.02) \u003d 41-0.8 \u003d 40.2 ml . This is the amount of water and add from the burette to the flask with the solution.

The above work is quite painstaking when performed, so it is possible to prepare approximately accurate solutions by introducing a correction factor that is applied in the work for each titration. In this case, the spent number of milliliters of the solution is multiplied by the correction factor.

The correction factor is calculated by the formula

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

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

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 amount of starting substances for preparing 1 liter of solutions of different normality (Table 32).

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