(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
(G) 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 substances then,
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 should be prescribed of this medicine(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.
Determine what you know and what you don't. In chemistry, dilution usually means taking a small amount of a solution of known concentration and then diluting it with a neutral liquid (such as water) to create a larger, less concentrated solution. This operation is very often used in chemical laboratories, so reagents are stored in concentrated form for convenience and diluted if necessary. In practice, as a rule, the initial concentration is known, as well as the concentration and volume of the solution that needs to be obtained; at the same time the volume of concentrated solution that needs to be diluted is unknown.
Substitute known values into the formula C 1 V 1 = C 2 V 2. In this formula, C 1 is the concentration of the initial solution, V 1 is its volume, C 2 is the concentration of the final solution, and V 2 is its volume. From the resulting equation you can easily determine the desired value.
Please account for any differences in units of measurement. Because dilution leads to a decrease in concentration, often significantly, concentrations are sometimes measured in different units. If you miss this, you could be off by several orders of magnitude. Before solving the equation, convert all concentration and volume values to the same units.
Let's solve the equation. When you have reduced all quantities to the same units, you can solve the equation. To solve it, knowledge of simple algebraic operations is almost always sufficient.
Think about applying your results in practice. Let's say you have calculated the desired value, but you are still finding it difficult to prepare a real solution. This situation is quite understandable - the language of mathematics and pure science is sometimes far from the real world. If you already know all four quantities included in the equation C 1 V 1 = C 2 V 2, proceed as follows:
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.
For cooking percent solutions use several techniques.
Example. It is necessary to prepare 1 kg of 15% solution sodium chloride. 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 a 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 a 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.
SI units in clinical laboratory diagnostics.
In clinical laboratory diagnostics International system units are recommended to be used in accordance with the following rules.
1. The unit of volume should be liter. It is not recommended to use submultiples or multiples of a liter (1-100 ml) in the denominator.
2. The concentration of the measured substances is indicated as molar (mol/l) or mass (g/l).
3. Molar concentration is used for substances with known relative molecular weight. Ionic concentration is reported as molar concentration.
4. Mass concentration is used for substances whose relative molecular weight is unknown.
5. Density is indicated in g/l; clearance – in ml/s.
6. Enzyme activity on the amount of substances in time and volume is expressed as mol/(s*l); µmol/(s*l); nmol/(s*l).
When converting units of mass into units of quantity of a substance (molar), the conversion factor is K=1/Mr, where Mr is the relative molecular mass. In this case, the initial unit of mass (gram) corresponds to the molar unit of the amount of substance (mol).
General characteristics.
Solutions are homogeneous systems consisting of two or more components and products of their interaction. Not only water, but also ethyl alcohol, ether, chloroform, benzene, etc. can act as a solvent.
The dissolution process is often accompanied by the release of heat (exothermic reaction - dissolution of caustic alkalis in water) or absorption of heat (endothermic reaction - dissolution of ammonium salts).
Liquid solutions include solutions of solids in liquids (a solution of salt in water), solutions of liquids in liquids (a solution of ethyl alcohol in water), solutions of gases in liquids (CO 2 in water).
Solutions can be not only liquid, but also solid (glass, an alloy of silver and gold), as well as gaseous (air). The most important and common are aqueous solutions.
Solubility is the property of a substance to dissolve in a solvent. Based on their solubility in water, all substances are divided into 3 groups - highly soluble, slightly soluble and practically insoluble. Solubility primarily depends on the nature of the substances. Solubility is expressed by the number of grams of a substance that can be maximally dissolved in 100 g of solvent or solution at a given temperature. This amount is called the solubility coefficient or simply the solubility of the substance.
A solution in which, at a given temperature and volume, no further dissolution of the substance occurs is called saturated. Such a solution is in equilibrium with an excess of the solute; it contains the maximum amount of substance possible under the given conditions. If the concentration of a solution does not reach the saturation concentration under given conditions, then the solution is called unsaturated. A supersaturated solution contains more substance than a saturated solution. Supersaturated solutions are very unstable. Simple shaking of the vessel or contact with crystals of the dissolved substance leads to instantaneous crystallization. In this case, the supersaturated solution turns into a saturated solution.
The concept of “saturated solutions” should be distinguished from the concept of “supersaturated solutions”. A solution with a high solute content is called concentrated. Saturated solutions different substances can vary greatly in concentration. For highly soluble substances (potassium nitrite), saturated solutions have a high concentration; For poorly soluble substances (barium sulfate), saturated solutions have a low concentration of solute.
In the vast majority of cases, the solubility of a substance increases with increasing temperature. But there are substances whose solubility increases slightly with increasing temperature (sodium chloride, aluminum chloride) or even decreases.
Solubility Dependence various substances on temperature is depicted graphically using solubility curves. Temperature is plotted on the abscissa axis, and solubility is plotted on the ordinate axis. Thus, it is possible to calculate how much salt falls out of the solution as it cools. The release of substances from solution as the temperature decreases is called crystallization, and the substance is released in its pure form.
If the solution contains impurities, then the solution will be unsaturated in relation to them even when the temperature decreases, and the impurities will not precipitate. This is the basis for the method of purifying substances – crystallization.
IN aqueous solutions more or less strong compounds of dissolved substance particles with water are formed - hydrates. Sometimes such water is so tightly bound to the dissolved substance that when it is released, it becomes part of the crystals.
Crystalline substances containing water are called crystalline hydrates, and the water itself is called crystallization water. The composition of crystal hydrates is expressed by a formula indicating the number of water molecules per molecule of substance - CuSO 4 * 5H 2 O.
Concentration is the ratio of the amount of solute to the amount of solution or solvent. The concentration of the solution is expressed in weight and volume ratios. Weight percentages indicate the weight content of the substance in 100 g of solution (but not in 100 ml of solution!).
Technique for preparing approximate solutions.
Weigh out necessary substances and the solvent in such ratios that the total amount is 100 g. If the solvent is water, the density of which is equal to unity, it is not weighed, but the volume is measured, equal to mass. If the solvent is a liquid whose density is not equal to unity, it is either weighed, or the amount of solvent expressed in grams is divided by the density indicator and the volume occupied by the liquid is calculated. Density P is the ratio of body mass to its volume.
The density of water at 4 0 C is taken as the unit of density.
Relative density D is the ratio of the density of a given substance to the density of another substance. In practice, they determine the ratio of the density of a given substance to the density of water, taken as a unit. For example, if the relative density of a solution is 2.05, then 1 ml of it weighs 2.05 g.
Example. How much carbon 4 chloride should be taken to prepare 100 g of 10% fat solution? Weigh out 10 g of fat and 90 g of CCl 4 solvent or, measuring the volume occupied required quantity CCl 4, divide the mass (90 g) by the relative density D = (1.59 g/ml).
V = (90 g) / (1.59 g/ml) = 56.6 ml.
Example. How to prepare a 5% solution of copper sulfate from crystalline hydrate of this substance (calculated as anhydrous salt)? The molecular weight of copper sulfate is 160 g, crystal hydrate is 250 g.
250 – 160 X = (5*250) / 160 = 7.8 g
Therefore, you need to take 7.8 g of crystalline hydrate, 92.2 g of water. If the solution is prepared without converting to anhydrous salt, the calculation is simplified. Weigh out the specified amount of salt and add the solvent in such an amount that the total weight of the solution is 100 g.
Volume percentages show how much of a substance (in ml) is contained in 100 ml of a solution or mixture of gases. For example, a 96% ethyl alcohol solution contains 96 ml of absolute (anhydrous) alcohol and 4 ml of water. Volume percentages are used when mixing mutually soluble liquids and when preparing gas mixtures.
Weight-volume percentage ratios (a conventional way of expressing concentration). Indicate the weight amount of the substance contained in 100 ml of solution. For example, a 10% NaCl solution contains 10 g of salt in 100 ml of solution.
Technique for preparing percentage solutions from concentrated acids.
Concentrated acids (sulfuric, hydrochloric, nitric) contain water. The ratio of acid and water in them is indicated in weight percentages.
The density of solutions is in most cases above unity. The percentage of acids is determined by their density. When preparing more dilute solutions from concentrated solutions, the water content in them is taken into account.
Example. It is necessary to prepare a 20% solution of sulfuric acid H 2 SO 4 from concentrated 98% sulfuric acid with a density D = 1.84 g/ml. Initially, we calculate how much of the concentrated solution contains 20 g of sulfuric acid.
100 – 98 X = (20*100) / 98 = 20.4 g
In practice, it is more convenient to work with volumetric rather than weight units of acids. Therefore, they calculate what volume of concentrated acid occupies the required weight amount of the substance. To do this, the number obtained in grams is divided by the density indicator.
V = M/P = 20.4 / 1.84 = 11 ml
It can be calculated in another way, when the concentration of the initial acid solution is immediately expressed in weight-volume percentages.
100 – 180 X = 11 ml
When special precision is not required, when diluting solutions or mixing them to obtain solutions of different concentrations, you can use the following simple and in a fast way. For example, you need to prepare a 5% solution of ammonium sulfate from a 20% solution.
Where 20 is the concentration of the solution taken, 0 is water, and 5 is the required concentration. We subtract 5 from 20, and write the resulting value in the lower right corner, subtracting 0 from 5, write the number in the upper right corner. Then the diagram will take the following form.
This means that you need to take 5 parts of a 20% solution and 15 parts of water. If you mix 2 solutions, the diagram remains the same, only the original solution with a lower concentration is written in the lower left corner. For example, by mixing 30% and 15% solutions you need to get a 25% solution.
Thus, you need to take 10 parts of a 30% solution and 15 parts of a 15% solution. This scheme can be used when special accuracy is not required.
Precise solutions include normal, molar, and standard solutions.
A solution is called normal if 1 g contains g – equivalent of a dissolved substance. The weight amount of a complex substance, expressed in grams and numerically equal to its equivalent, is called a gram equivalent. When calculating the equivalents of compounds such as bases, acids and salts, you can use the following rules.
1. The base equivalent (E o) is equal to the molecular weight of the base divided by the number of OH groups in its molecule (or by the valence of the metal).
E (NaOH) = 40/1 = 40
2. The acid equivalent (Ek) is equal to the molecular weight of the acid divided by the number of hydrogen atoms in its molecule that can be replaced by the metal.
E(H 2 SO 4) = 98/2 = 49
E(HCl) = 36.5/1=36.5
3. The salt equivalent (E s) is equal to the molecular weight of the salt divided by the product of the valence of the metal and the number of its atoms.
E(NaCl) = 58.5/(1*1) = 58.5
When acids and bases interact, depending on the properties of the reacting substances and the reaction conditions, not all hydrogen atoms present in the acid molecule are necessarily replaced by a metal atom, and acid salts are formed. In these cases, the gram equivalent is determined by the number of hydrogen atoms replaced by metal atoms in a given reaction.
H 3 PO 4 + NaOH = NaH 2 PO + H 2 O (gram equivalent is equal to gram molecular weight).
H 3 PO 4 + 2NaOH = Na 2 HPO 4 + 2H 2 O (gram equivalent is equal to half a gram molecular weight).
When determining the gram equivalent, knowledge is required chemical reaction and the conditions in which it occurs. If you need to prepare decinormal, centinormal or millinormal solutions, take 0.1, respectively; 0.01; 0.001 grams is the equivalent of the substance. Knowing the normality of the solution N and the equivalent solute E, it is easy to calculate how many grams of the substance are contained in 1 ml of solution. To do this, you need to divide the mass of the dissolved substance by 1000. The amount of dissolved substance in grams contained in 1 ml of solution is called the titer of the solution (T).
T = (N*E) / 1000
T (0.1 H 2 SO 4) = (0.1 * 49) / 1000 = 0.0049 g/ml.
A solution with a known titer (concentration) is called titrated. Using a titrated alkali solution, you can determine the concentration (normality) of an acid solution (acidimetry). Using a titrated acid solution, you can determine the concentration (normality) of an alkali solution (alkalimetry). Solutions of the same normality react in equal volumes. At different normalities, these solutions react with each other in volumes inversely proportional to their normalities.
N k / N sh = V sh / V k
Nk * Vk = N sch * V sch
Example. To titrate 10 ml of HCl solution, 15 ml of 0.5 N NaOH solution was used. Calculate the normality of the HCl solution.
Nk * 10 = 0.5 * 15
Nk = (0.5 * 15) / 10 = 0.75
N=30/58.5=0.5
Fixanals are pre-prepared and sealed in ampoules, precisely weighed amounts of reagent required to prepare 1 liter of 0.1 N or 0.01 N solution. Fixanales come in liquid and dry forms. Dry ones have a longer shelf life. The technique for preparing solutions from fixanals is described in the appendix to the box with fixanals.
Preparation and testing of decinormal solutions.
Decinormal solutions, which are often starting materials in the laboratory, are prepared from chemically frequent medications. The required sample is weighed on a technical chemical scale or pharmaceutical scale. When weighing, an error of 0.01 - 0.03 g is allowed. In practice, you can make an error in the direction of slightly increasing the calculated weight. The sample is transferred to a volumetric flask, where a small amount of water is added. After the substance is completely dissolved and the temperature of the solution is equalized with the air temperature, the flask is topped up with water to the mark.
The prepared solution requires checking. The test is carried out using solutions prepared from their fixatives, in the presence of indicators, and the correction factor (K) and titer are established. The correction factor (K) or correction factor (F) shows what amount (in ml) of an exact normal solution corresponds to 1 ml of a given (prepared) solution. To do this, transfer 5 or 10 ml of the prepared solution into a conical flask, add a few drops of indicator and titrate with the exact solution. Titration is carried out twice and the arithmetic mean is calculated. The titration results should be approximately the same (difference within 0.2 ml). The correction factor is calculated based on the ratio of the volume of the exact solution Vt to the volume of the test solution Vn.
K = V t / V n.
The correction factor can also be determined in a second way - by the ratio of the titer of the test solution to the theoretically calculated titer of the exact solution.
K = T practical / T theor.
If the left sides of an equation are equal, then their right sides are equal.
V t / V n. = T practical / T theor.
If the practical titer of the test solution is found, then the weight content of the substance in 1 ml of solution has been determined. When the exact solution and the solution being tested interact, 3 cases can occur.
1. The solutions interacted in equal volumes. For example, the titration of 10 ml of a 0.1 N solution required 10 ml of the test solution. Therefore, normality is the same and the correction factor is equal to one.
2. 9.5 ml of the test solution was used to interact with 10 ml of the exact solution; the test solution turned out to be more concentrated than the exact solution.
3. 10.5 ml of the test solution was used to interact with 10 ml of the exact solution; the test solution is weaker in concentration than the exact solution.
The correction factor is calculated accurate to the second decimal place; fluctuations from 0.95 to 1.05 are allowed.
Correction of solutions whose correction factor is greater than one.
The correction factor shows how many times a given solution is more concentrated than a solution of a certain normality. For example, K is 1.06. Therefore, 0.06 ml of water must be added to each ml of the prepared solution. If 200 ml of solution remains, then (0.06*200) = 12 ml - add to the remaining prepared solution and mix. This method of bringing solutions to a certain normality is simple and convenient. When preparing solutions, you should prepare them with more concentrated solutions, rather than dilute solutions.
Preparation of accurate solutions, the correction factor of which is less than one.
In these solutions, some part of the gram equivalent is missing. This missing part can be identified. If you calculate the difference between the titer of a solution of a certain normality (theoretical titer) and the titer of a given solution. The resulting value shows how much substance must be added to 1 ml of solution to bring it to the solution concentration of a given normality.
Example. The correction factor for approximately 0.1 N sodium hydroxide solution is 0.9, the volume of the solution is 1000 ml. Bring the solution to exactly 0.1 N concentration. Gram equivalent of sodium hydroxide – 40 g. Theoretical titer for a 0.1 N solution – 0.004. Practical titer - T theory. * K = 0.004 * 0.9 = 0.0036 g.
T theor. - T pract. = 0.004 – 0.0036 = 0.0004 g.
1000 ml of solution remained unused - 1000 * 0.0004 = 0.4 g.
The resulting amount of the substance is added to the solution, mixed well, and the titer of the solution is determined again. If the starting material for preparing solutions are concentrated acids, alkalis, and other substances, then it is necessary to make an additional calculation to determine how much of the concentrated solution contains the calculated amount of this substance. Example. The titration of 5 ml of approximately 0.1 N HCl solution required 4.3 ml of an exact 0.1 N NaOH solution.
K = 4.3/5 = 0.86
The solution is weak, it needs to be strengthened. We calculate T theory. , T practice and their difference.
T theor. = 3.65 / 1000 = 0.00365
T pract. = 0.00365 * 0.86 = 0.00314
T theor. - T pract. = 0.00364 – 0.00314 = 0.00051
200 ml of solution remained unused.
200 * 0.00051 = 0.102 g
For a 38% HCl solution with a density of 1.19, we make up a proportion.
100 – 38 X = (0.102 * 100) / 38 = 0.26 g
We convert weight units into volume units, taking into account the density of the acid.
V = 0.26 / 1.19 = 0.21 ml
Preparation of 0.01 N, 0.005 N from decinormal solutions, having a correction factor.
Initially, calculate what volume of 0.1 N solution should be taken to prepare from a 0.01 N solution. The calculated volume is divided by the correction factor. Example. It is necessary to prepare 100 ml of 0.01 N solution from 0.1 N with K = 1.05. Since the solution is 1.05 times more concentrated, we need to take 10/1.05 = 9.52 ml. If K = 0.9, then you need to take 10/0.9 = 11.11 ml. IN in this case take a slightly larger amount of solution and adjust the volume in the volumetric flask to 100 ml.
The following rules apply for the preparation and storage of titrated solutions.
1. Each titrated solution has its own shelf life. During storage they change their titer. When performing an analysis, it is necessary to check the titer of the solution.
2. It is necessary to know the properties of solutions. The titer of some solutions (sodium hyposulfite) changes over time, so their titer is established no earlier than 5-7 days after preparation.
3. All bottles with titrated solutions must have a clear label indicating the substance, its concentration, correction factor, time of preparation of the solution, and date of titration check.
4. During analytical work, great attention must be paid to calculations.
T = A / V (A – sample)
N = (1000 * A) / (V * g /eq)
T = (N * g/eq) / 1000
N = (T * 1000) / (g/eq)
A solution is called molar if 1 liter contains 1 g*mol of solute. Mole is molecular weight expressed in grams. 1-molar solution of sulfuric acid - 1 liter of such a solution contains 98 g of sulfuric acid. A centimolar solution contains 0.01 mol in 1 liter, a millimolar solution contains 0.001 mol. A solution whose concentration is expressed by the number of moles per 1000 g of solvent is called molal.
For example, 1 liter of 1 M sodium hydroxide solution contains 40 g of the drug. 100 ml of solution will contain 4.0 g, i.e. solution 4/100 ml (4g%).
If the sodium hydroxide solution is 60/100 (60 mg%), you need to determine its molarity. 100 ml of solution contains 60 g of sodium hydroxide, and 1 liter - 600 g, i.e. 1 liter of 1 M solution should contain 40 g of sodium hydroxide. The molarity of sodium is X = 600 / 40 = 15 M.
Standard solutions are solutions with precisely known concentrations used for quantification substances using colorimetry and nephelometry. Samples for standard solutions are weighed on an analytical balance. The substance from which the standard solution is prepared must be chemically pure. Standard solutions. Standard solutions are prepared in the volume required for consumption, but not more than 1 liter. The amount of substance (in grams) required to obtain standard solutions – A.
A = (M I * T * V) / M 2
M I – Molecular mass of the solute.
T – Titer of the solution for the substance being determined (g/ml).
V – Set volume (ml).
M 2 – Molecular or atomic mass analyte.
Example. Need to prepare 100 ml standard solution CuSO 4 * 5H 2 O for the colorimetric determination of copper, and 1 ml of solution should contain 1 mg of copper. In this case, M I = 249.68; M2 = 63.54; T = 0.001 g/ml; V = 100 ml.
A = (249.68*0.001*100) / 63.54 = 0.3929 g.
Transfer a sample of salt to a 100 ml volumetric flask and add water to the mark.
Security questions and tasks.
1. What is a solution?
2. What ways are there to express the concentration of solutions?
3. What is the titer of the solution?
4. What is a gram equivalent and how is it calculated for acids, salts, bases?
5. How to prepare a 0.1 N solution of sodium hydroxide NaOH?
6. How to prepare a 0.1 N solution of sulfuric acid H 2 SO 4 from concentrated acid with a density of 1.84?
8. What is the method for strengthening and diluting solutions?
9. Calculate how many grams of NaOH are needed to prepare 500 ml of 0.1 M solution? The answer is 2 years.
10. How many grams of CuSO 4 * 5H 2 O do you need to take to prepare 2 liters of 0.1 N solution? The answer is 25 g.
11. To titrate 10 ml of HCl solution, 15 ml of 0.5 N NaOH solution was used. Calculate the normality of HCl, the concentration of the solution in g/l, the titer of the solution in g/ml. The answer is 0.75; 27.375 g/l; T = 0.0274 g/ml.
12. 18 g of a substance are dissolved in 200 g of water. Calculate the weight percent concentration of the solution. The answer is 8.25%.
13. How many ml of 96% sulfuric acid solution (D = 1.84) do you need to take to prepare 500 ml of 0.05 N solution? The answer is 0.69 ml.
14. Titer of H 2 SO 4 solution = 0.0049 g/ml. Calculate the normality of this solution. The answer is 0.1 N.
15. How many grams of sodium hydroxide do you need to take to prepare 300 ml of 0.2 N solution? The answer is 2.4 g.
16. How much do you need to take a 96% solution of H 2 SO 4 (D = 1.84) to prepare 2 liters of a 15% solution? The answer is 168 ml.
17. How many ml of 96% sulfuric acid solution (D = 1.84) do you need to take to prepare 500 ml of 0.35 N solution? The answer is 9.3 ml.
18. How many ml of 96% sulfuric acid (D = 1.84) do you need to take to prepare 1 liter of 0.5 N solution? The answer is 13.84 ml.
19. What is the molarity of a 20% solution hydrochloric acid(D = 1.1). The answer is 6.03 M.
20. Calculate the molar concentration of a 10% nitric acid solution (D = 1.056). The answer is 1.68 M.
Task 11. Mixing 25% and 95% acid solutions and adding 20 kg clean water, received a 40% acid solution. 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.
