Advice 1: How to calculate percentage concentration of the solution

Tasks for the calculation of percentage concentration of the solutions must not only meet the study section chemistry. The ability to carry out the relevant calculations can be helpful in everyday life, for example, when recalculating the concentration of a solution of acetic acid in the period of preserving vegetables. You will need
• calculator.
Instruction
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Any solution consists of a solute and solvent. In most cases the solvent is water. To calculate the percentage concentration (or mass fraction of the solute), you must use the formula:W = m (solute) / m (solution) x 100% W – mass fraction of solute (or percent concentration), %From the same formula, we can deduce the mass of solute if you know the mass of the solution and percent concentration of a solution.
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Example No. 1. Calculate the mass fraction (percentage) of salt (NaCl) if the mass of (NaCl) 5 g, and the mass of the solution (NaCl) 100 g. In this problem, it remains only to substitute in the formula proposed in the condition parameters:W = m (b.-VA) / m (R-RA) x 100 % W (NaCl) = m (NaCl) / m (solution NaCl) x 100 % W (NaCl) = 5 g / 100 g x 100 % = 5 %Answer: W (NaCl) = 5 %
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Example No. 2. Calculate the mass percent ( % ) potassium bromide (KBr), if the mass of salt (KBr) 10 g, and the weight of water, 190 g Before using formula to calculate percentage concentration, calculate the mass of the solution, which consists of water and dissolved substances:m (solution) = m (solute) + m (water) Consequently:m (R-RA KBr) = 10 grams + 190 grams = 200 godstate in the basic formula are found and specified in the condition parameters:W = m (b.-VA) / m (R-RA) x 100 % W (KBr) = m (KBr) / m (solution KBr) x 100 % W (KBr) = 10 g / 200 g x 100 % = 5% Answer: W (KBr) = 5 %
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Example No. 3. Calculate the percentage concentration of acetic acid (CH3COOH), if the mass of the acid (CH3COOH) and 30 g, and the mass of water 170 g. Calculate the mass of the solution, which consists of water and acetic acid:m (solution) = m (solute) + m (water) Consequently:m (R-RA CH3COOH) = 30 g + 170 g = 200 godstate in the formula of required parameters:W = m (b.-VA) / m (R-RA) x 100 %W (CH3COOH) = m (CH3COOH) / m (solution CH3COOH) x 100 % W (CH3COOH) = 30 g / 200 g x 100 % = 15% Answer: W (CH3COOH) = 15 %

Advice 2 : How to calculate the concentration of a substance

The concentration of the solution is a measure of how much mass or other substances contained in a certain volume or mass of solution. Even the most remote from chemistry, one is faced with this concept at every step: for example, buying in the store 9% vinegar for home canning, or 20% cream to add to coffee. How is calculated the concentration of the solution? Instruction
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Suppose 200 or 300 milliliters of water dissolve 58.5 grams of sodium chloride, that is familiar to all of salt. Then, Prilepa water brought the total weight of the solution up to one kilogram. It is easy to guess that the solution in this case will contain 58.5 grams of salt and 941,5 grams of water. What will be the mass fraction of salt?
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To calculate it is easy, for this amount of salt, divide by the total mass of solution and multiply by 100%, it will look like the following:(58,5/1000) * 100% = 5,85%.
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Formulate the task a little differently. The same amount of salt dissolved in water, then brought the solution volume up to one liter. What will be the molar concentration of the solution?
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Remember the definition of molar concentration. This is the number of moles of dissolved substancecontained in one liter of solution. What is the mol of salt? Its formula is NaCl, the molar mass is approximately 58.5 per. In other words, in one liter of solution you have contains exactly one mole of salt. You will receive a 1.0 molar solution.
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Now go back to the original conditions of the problem where the total weight of the solution amounted to exactly one kilogram. How do you find the molality of this solution?
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And here, too, there is nothing difficult. Above you have already figured out that 58.5 grams of common salt have 941, 5 grams of water. Substituting the known values into the formula m = v/M, where M is the value of the molality and v is the number of moles of the substance in the solution, and M is the mass of solvent in kilograms, we get:1,0/0,9415 = 1,062 Moralny solution.
The concentration can be expressed in many different ways. For example, using the concepts of "mass fraction", "molarity" (i.e., how many moles of a substance is in 1 liter solution), molality (number of moles of a substance is in 1 kilogram of solvent), etc.

Advice 3 : How to determine the concentration of the acid

Concentration - dimensional quantity, by expressing the solution composition (in particular, the content of dissolved substances). Sometimes it happens that the value is unknown. For example, in lab top bottles may be one, just signed - HCl (hydrochloric acid). To conduct many experiments information requires much more than just a name. Therefore it is necessary to use experimental techniques such as titration or density determination. You will need
• -the alkali solution, the exact concentration
• -burette
• -volumetric pipettes
• indicator
• -a set of hydrometers
Instruction
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One of the most simple ways to determine the concentration of acid involves the direct titration (the process of gradual addition of solution of known concentration(titrant) to a solution of the analyte with the aim to fix the point of equivalence (end of reaction)). In this case it is convenient to use the neutralization with an alkali. Conclusion it can be easily identified by adding the indicator (for example, in acid phenolphthalein clear, and upon addition of alkali becomes crimson; methyl orange in acidic medium pink, and in alkaline - orange).
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Take the burette (volume 15-20 ml), install it in the tripod using the legs. It should be clearly stated, otherwise, swinging of the tip can drop a few extra drops, which will spoil the whole process. Sometimes one drop changes the color of the indicator. This point need to be detected.
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Please be ware and reagents conical flasks for titration (4-5 pieces of small volume), several pipettes (as Mora - without division, and dimensional), volumetric flask of 1 l, fiksanala alkali, indicator, distilled water.
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Prepare a solution of precise concentration of alkali (e.g., NaOH). To do this, use fixanal (vial sealed it with a substance, which when diluted in 1 liter of water is obtained with a 0.1 normal solution). Of course, you can use the exact linkage. But the first option is more precise and more reliable.
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Next, fill the burette with alkali. In a conical flask place 15 ml of unknown concentration of acid (possibly HCl) add 2-3 drops of the indicator. And proceed directly to the titration. As soon as the indicator changes color and approximately 30 will to remain so, stop the process. Write down how much was spent alkali (for example, 2.5 ml).
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Then follow the course of the work another 2-3 times. This is done to obtain whiter the exact result. After calculate the average volume of alkali. Vav = (V1+V2+V3)/3, V1 is the result of the first titration, in ml V2 - is the result of the second, in ml, V3 - the third volume, ml 3 - number of performed reactions. For example, Vav = (2,5+2,7+2,4)/3 = 2,53 ml.
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After the experiment, you can proceed to the main estimates. In this situation the relation is valid: C1*V1 = C2*V2 where C1 is the concentration of the alkali solution, normal (n), V1 is the average volume used in the reaction of alkali ml, C2 is the concentration of the acid solution, n, V2 - volume of acid involved in the reaction, ml. of the C2 - value is unknown. So, it must be expressed through the known data. C2 = (C1*V1)/V2, i.e., S2 = (0,1 * 2,53)/ 15 = 0,02 B. Conclusion: during the titration of HCl solution 0.1 n NaOH, was clarified by the concentration of the acid - 0.02 N.
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Another common way to find out the concentration of the acid is, for starters, to know its density. To do this, purchase a set of hydrometers (in chemical or specialty store, you can also order online or visit terms of trade of facilities for motorists).
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Pour acid into beaker and place in it the hydrometers as long as they will continue to sink or to be ejected to the surface. When the device will be like a float, select a numeric value on it. This figure is the density of the acid. Next, using the relevant literature (reference Lurie), is not difficult to determine in the table the desired concentration.
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Regardless of which method you select, do not forget about safety.

Advice 4 : How to calculate the initial concentration

In the course of a reaction some substances are transformed into others while changing its composition. Thus, the "source concentration"is the concentration of substances prior to the occurrence of a chemical reaction, that is, converting them into other substances. Of course, this transformation is accompanied by a decrease in their number. Accordingly, it reduced the concentration of the starting materials, until a zero value if the reaction has proceeded to the end, irreversibly, and the components were taken in equal amounts. Instruction
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Say you have the following task. Was leaking some kind of chemical reaction in which a starting material taken as A and B, turned into products, for example, relatively To and G. That is, the reaction took place according to the following scheme: A + B = B + G. If the concentration of substance B is equal to 0.05 mol/l, and the substance G to 0.02 mol/l, was established a kind of chemical equilibrium. You need to determine what the initial concentrations A0 and B0, if the equilibrium constant Kr is equal to the amount of 0.04?
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To solve this problem take the equilibrium concentration of the substance And the value of "x" and the concentration of a substance In for "y". And also remember that the equilibrium constant Kr is calculated by the following formula: [V][G]/[A][B].
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In the course of solving we get the following calculation: 0,04 = 0,02 y /x of 0.05. That is, by simple calculation you will get that y = 0.1 x.
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Now once again look closely at the above chemical equation. From this it follows that one mole of the substances A and B were formed on one pray compounds In G. Based on this, the initial molar concentration of a can be represented as follows:A0 = x + 0,02 A0 = x + y
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Remember that the value of "y" as you just defined, is equal to one level of 0.1 x. Transforming these equations in the future, you will get: x + 0,02 = 1.1 x. From this it follows that x = 0.2 mol/l, and then the initial concentration [A0] is equal to 0,2 + 0,02 = 0.22 mol/l.
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What substance B? The initial concentration B0 is much easier. To determine the equilibrium concentration of the substance necessary to allow the equilibrium concentration of the product-substance of G. That is, [B0] = 0,05 + 0,02 = 0,07 mol/l. the Answer will be: [A0] = 0.22 mol/l, [B0] = 0,07 mol/l. the task is solved.

Advice 5 : How to calculate molar concentration

Molar concentration is the value which indicates how many moles of the substance is 1-m liter of solution. For example, it is known that in a liter of solution is exactly of 58.5 grams of common salt – chloride of sodium. Molar because this substance is 58,5 g/mol, we can say that in this case you have odnokolernyh salt solution. (Or, in record, 1M solution). You will need
• - table of solubility of substances.
Instruction
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The solution of this problem depends on the specific conditions. If you know the exact material mass and the exact volume of the solution, the solution is very simple. For example, 15 grams of barium chloride contains 400 milliliters of solution. What is its molar concentration?
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Start with what you remember the exact formula of the salt: BaCl2. On the periodic table, determine the atomic masses of the elements included in its composition. And, given the index 2 from chlorine, get molecular weight: 137 + 71 = 208. Therefore, the molar mass of barium chloride 208 g/mol.
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And in terms of the tasks, in the solution contains 15 grams of this substance. How much is it in moles? Dividing 15 by 208, will receive: approximately 0,072 mole.
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Now you need to take into account that the solution volume to 1 liter, and 0.4. 0,072 dividing by 0.4, we get the answer: 0,18. That is, you have approximately 0.18-molar solution of barium chloride.
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Let's build the solution. Suppose you began to dissolve in 100 milliliters of water at room temperature already mentioned, familiar to you, table salt is sodium chloride. Added you in small portions, stirring constantly and waiting until complete dissolution. And then came the moment when a small portion is not dissolved until the end, despite the intense mixing. You need to determine what is the molar concentration of the resulting solution.
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First of all, you need to find the table of solubility of substances. They are in the majority of chemical directories, you can also find this information on the Internet. You can easily determine that at room temperature the saturation limit (i.e., solubility limit) of sodium chloride is 31.6 grams/100 grams of water.
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For this task, you dissolved salt in 100 milliliters of water, but its density is almost equal 1. So we conclude: in the resulting solution contains approximately 31.6 grams of sodium chloride. Small undissolved excess, as well as some volume change upon dissolution of the salt can be neglected, the error will be small.
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Respectively, in 1 liter of solution would contain 10 times more salt – 316 grams. Given that the molar mass of sodium chloride, as mentioned in the beginning, is 58.5 g/mol, you will easily find the answer: 316/58,5 = a 5.4-molar solution.

Advice 6 : How to calculate the concentration

With the concept of the concentration of people is found not only in science but also in everyday life. For example, the mass fraction of fat listed on food products (milk, butter etc.) are not that other, as percentage concentration. Also it has another molar, normal and molalla concentration. And any of them easily calculated. You will need
• - handle;
• paper;
• - the periodic table;
• calculator.
Instruction
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To find the mass fraction (percentage concentration) of a substance, divide its mass by the total mass of the solution (mixture). The result will be in fractions of a unit, which can then convert into a percentage that will also be true. For example, given the task: to prepare the solution took 150 g of water and 50 g sugar. You must calculate the percentage concentration of the solute. For the solution first, write the formula and then find the optimum value:ω (sugar)= m(sugar)/m(solution) = 50/ (150+50) =0,25 * 100% = 25%the solution contains 25% sugar.
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When calculating the molar concentration, you need the amount of substance divided by the total volume of the solution. The unit of measure in this case would be mol/L. the Formula for calculation is as follows: C = n(solute)/V, where C is the molar concentration (mol/l);n – amount of substance (mol);V is the total volume of the mixture (liter).
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Normal concentration is expressed in gram equivalent/liter and represents the number of equivalents of a certain substance in 1 l of solution, which is, in chemical reactions, 1G of hydrogen or 8 g of oxygen. Suppose you want to calculate the normality of 70% sulfuric acid, the density of which is equal to 1,615 g/l. the conditions of the problem it is clear that 100 g of solution contains 70 grams of acid. So first find the volume of a given solution: V = 100/1,615 = accounting period 61.92 (ml). Then calculate the mass of H2SO4 in 1 liter of solution: m(H2SO4) = 1000*70/61,92 = 1130,49 (g). And then calculate the normality, considering that the acid is dibasic:SN = m*z/ M =1130,49*2/98 = N of 23.06.
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If you want to calculate molalou the concentration of the solution (molality), use the following formula: Cm = n/m, gdes – molalla concentration is measured in mol/kg;n – number of a certain substance in moles;m is the total mass of the solution in kilograms.Unlike molarity molalla concentration depends on the temperature conditions of the reaction.
In analytical chemistry are often used instead of concentration titer. It shows the amount of a certain substance in each milliliter of solution.

Advice 7 : How to determine the equilibrium constant

Constant equilibrium characterizes the displacement of the reversible chemical reaction toward the formation of the reaction products or starting materials. To calculate the constant of equilibrium in a variety of ways, depending on the conditions of the problem. You will need
• - handle;
• paper for records;
• calculator.
Instruction
1
A constant equilibrium can be expressed using equilibrium concentrations of the reaction participants, i.e. the concentration of substances at a time when the speed of the forward reaction equals the rate of reverse. Given the reversible reaction of substances A and b In certain conditions with the formation of substances: na+mB ↔ zС, where n, m, z – the coefficients in the equation reactions. A constant equilibrium can be expressed by: Kc = [C]^z/ ([A]^n*[B]^m), where [C], [A], [B] is the equilibrium concentration of substances..
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In the first type of task you want to define the constant of equilibrium of the equilibrium concentrations. Equilibrium concentrations may not be asked directly. When solution first write down the reaction equation, place coefficients.
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Example: nitrogen monoxide under certain conditions, reacts with oxygen c NO2 formation. The initial concentrations of NO and O2 – 18 mol/l and 10 mol/L. it is Known that methane 60% O2. You want to find the constant of equilibrium reaction. 4
Write down the reaction equation, place coefficients. Please note, the proportions in which substances react. Calculate the concentration of O2, which entered into reaction: 10моль*0,6 = 6 mol/L. From the reaction equations, find the concentration of unreacted NO – 12 mol/L. the NO2 concentration is 12 mol/L.
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Determine the amount of unreacted NO: 18-12 = 6 mol. And unreacted oxygen: 10-6 = 4 mol. Calculate the constant of equilibrium: KS = 12^2/(6^2*4) = 1.
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If the clause specified the rate constants of direct and reverse reactions, find the constant of equilibrium from the relation: K = k1/k2 where k1, k2 – the rate constants of the forward and reverse chemical reactions.
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In an isothermal process and Isobaric process a constant equilibrium can be found from the equation of the standard changes of Gibbs energy: ΔGр- = - RT*lnKc = -8,31 T*2,3 lgKc, where R is the universal gas constant, equal to 8,31; T – reaction temperature, K; lnKc – the natural logarithm of the constants of equilibrium. For convenience, convert it into decimal lgKc by multiplying by a factor of 2.3.
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To determine the change in standard Gibbs energy of reaction, you can from the equation for isothermal Isobaric process: ΔG = ΔH – T ΔS, where T is the reaction temperature, K; ∆ H - enthalpy, kJ/mol; ΔS is entropy, j/(mol-deg). The value of enthalpy and entropy for 1 mole of basic chemical compounds at a temperature of 25 ° C are given in reference literature. If the reaction temperature differs from 25 ° C, the values of enthalpy and entropy should be given in the problem statement.
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Magnitude of the ΔG of the reaction at 25 ° C you can find the folding potentials of education ΔGобр each of the reaction products and subtracting from the sum ΔGобр starting materials. The potential values of education 25oC for 1 mole of various substances are given in reference tables. Note
In the case that the participants in the reaction are in different States of aggregation, in the formula for determination of the equilibrium constant includes the concentrations of substances in more a mobile (gas or liquid) state.
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