Tabla de Contenidos
Normality , which is represented by the letter N , is a unit of chemical concentration that expresses the number of equivalents of a solute in each liter of solution. It is expressed in units of eq.L -1 or eq/L which is read “normal” (ie a concentration 0.1 eq/L is read 0.1 normal). It is a very useful unit of concentration, greatly facilitating stoichiometric calculations no matter what reagent is being used.
However, it is also a unit of concentration that can lead to a bit of confusion, especially since the same solution can have more than one normal concentration. This is because the concept of the number of equivalents depends on what the solute is used for or what types of chemical reactions it will participate in.
The following sections explain in detail how to calculate normality from different data, including other concentration units.
Formulas to calculate normality
The formulas for calculating normality are very similar to those for molarity. The mathematical form of the definition of normality is:
where n eq. solute represents the number of solute equivalents and V solution represents the volume of the solution expressed in liters. If the number of equivalents is not known beforehand but the mass of the solute (a very common situation) then we can take advantage of the fact that the number of equivalents is calculated as the mass divided by the equivalent weight. Substituting this into the formula above, you get:
Where PE solute (the equivalent weight of the solute) represents the weight in grams of 1 equivalent of solute.
The equivalent weight of a substance is given by its molar mass divided by an integer that represents the number of equivalents for each mole of the substance, and which we will call ω (the Greek letter omega). That is to say:
Combining this equation with the previous one, we get:
Which can be used to calculate normality from the mass of the solute, its molar mass (or molecular weight, although it is not strictly the same) and the volume of the solution. Furthermore, one needs to know ω for the solute, and this is where the main source of confusion regarding normality lies, since ω can have different values for the same solute.
The concept of the number of equivalents
The key to understanding the concept of the number of equivalents, and indeed the reason that “normal” concentration or normality is so called, lies in ω. This number depends on the use to which the solute is put or the chemical reaction in which it will participate.
For each type of major chemical reaction that involves at least two chemical substances, we can define what we will call the “Normal” reactant, which is nothing more than a generic term that we use to identify the reactant that participates in the simplest possible version of the type. particular reaction.
For example , if we are talking about an acid-base reaction , the simplest case would be one in which any monoprotic acid (HA) reacts with a monobasic base (B), to give the respective conjugate pairs according to the following reaction:
The monoprotic acid HA and the monobasic base B are what we would call a normal acid and base, respectively. This means that any acid such as HCl or HNO 3 is a normal acid, and any base such as NaOH or NH 3 would be an example of a normal base.
If we now consider an acid such as sulfuric acid (H 2 SO 4 ) that is diprotic, the reaction with a normal base would be:
As we can see, each mole of this acid is “equivalent” to 2 moles of a normal acid , since it consumes two moles of a normal base. Therefore, we say that the number of equivalents per mole of sulfuric acid is 2 (ω=2 eq/mol). For this reason, a 0.1 molar solution of H 2 SO 4 is equivalent to a 0.2 molar solution of a normal acid, so we say that the normality of said solution is 0.2.
In other words, we can redefine the concept of normality as the equivalent molar concentration that a normal reactant would have participating in the same type of chemical reaction as the solute .
Acid-base reactions are just one example of a typical chemical reaction. There are other reactions and for each of them there is a particular way of defining the normal reactant (that is, of defining ω). The following table shows how ω is determined for each type of solute, depending on the reaction in which it will be involved:
| type of chemical reaction | reagent type | Number of equivalents per mole (ω) |
| Salt metathesis reactions | ionic salts | ω is given by the total number of positive or negative charges in the neutral salt (both numbers are the same). It is calculated by multiplying the number of cations by their charge or the number of anions by theirs. |
| Acid Base Reactions | acids | ω is given by the number of hydrogens that give up in the reaction. |
| Bases | ω is given by the number of hydrogens that it can capture | |
| Redox reactions | oxidizing agents | ω is given by the number of electrons captured by each molecule of oxidizing agent in the balanced reduction half-reaction. |
| reducing agents | ω is given by the number of electrons that each molecule of reducing agent gives up in the balanced oxidation half-reaction. | |
| Solutes that do not participate in reactions | ——- | ω is worth 1eq/mol |
When is normality used?
Normality is mainly used in situations involving chemical reactions in solution, as they facilitate stoichiometric calculations without the need to write balanced or balanced chemical reactions.
Because of the way the number of equivalents per mole is defined, the number of equivalents of one reactant will always equal the number of equivalents of the other when they react in stoichiometric ratios.
Since the number of equivalents can be easily found from the normality and the volume of solution, we can carry out stoichiometric calculations very quickly without worrying about the details of the reaction.
This is particularly practical in volumetric titrations or titrations, since, at the equivalence point of the titration, it will always be true that:
And substituting the equivalents by the product of the normality by the volume, we obtain:
How to calculate normality from other units of concentration
Starting molarity (M)
Converting between molarity and normality is very easy, since the second is always an integer multiple of the first as shown below:
If we know the molarity of a solution, we can calculate its various normalities simply by multiplying the molarity by the respective number of equivalents per mole, ω.
From percentage m/V (%m/V)
The mass -volume percentage indicates the mass in grams of solute that is per 100 mL of solution. Taking this into account, the normality, in terms of the mass-volume percentage, is:
In this equation, the factor of 10 comes from the conversion factor from mL to L (1000) and 100% from the percent formula. To ensure unit consistency, percentage should be given units of g/mL and factor 10 should be given ml/L.
From percentage m/m (%m/m)
The only difference between converting the %m/V to normality and converting the %m/m is that you have to multiply by the density of the solution to be able to transform the 100 g of solution (of the %m/m) to volume. After rearranging the equation and making all the transformations, the formula remains:
where all factors have the same meaning as before and d solution is the density of the solution in g/mL.
Steps to Calculate Normality
Step 1: Obtain the necessary data
In this step, we analyze what data we have about the solution, the solute, or the solvent. This may include masses, number of equivalents, volumes, densities, or other units of concentration.
Step 2: Select the appropriate formula
Once we know what data we have, we can select which of the formulas we will use. For example, if we know the volume of the solution and the number of equivalents, we use the former formula, but if we know the percentage m/m and the density, we use the latter.
Step 3: Analyze the solute to determine ω
This involves first determining the type of reaction the solute will participate in to see if it will be assigned ω as a salt, an acid, a base, or an oxidizing or reducing agent. There are cases in which the same compound can react in different ways. For example, potassium dichromate (K 2 Cr 2 O 7 ) is both a basic salt and an oxidizing agent, so it could be assigned ω as if it were a base, a salt, or an oxidizing agent.
TIP: If you don’t have information about what it will be used for, the general rule is that salts are always treated as salts, even if they are acids, bases, oxidizing or reducing agents. The same with molecular (non-ionic) solutes, in which case ω=1 is taken.
Step 4: Apply the formula
Having ω and all the other information, all that remains is to apply the formula. The only detail to take into account is that we have to make sure that we have all the variables in the correct units so that our calculations are consistent.
Examples of normality calculation
Example 1
Determine the normality of a solution prepared by dissolving 350 mg of sodium sulfate (Na 2 SO 4 ) in 150 mL of solution.
SOLUTION:
Steps 1 and 2: In this case we have the mass of the solute (350mg) and the volume of the solution (150mL), so we will use equation 3:
Also, using the atomic masses of sodium, sulfur, and oxygen, the molar mass of the salt is determined to be 142 g/mol.
Step 3: Sodium sulfate is a salt made up of two Na + cations and one SO 4 2- anion . Therefore, ω in this case is worth 2x(1)=1x(2)=2 eq/mol.
Step 4: Finally, the data is substituted, the transformations to grams and liters are carried out and the normality is calculated:
Therefore, the solution has a normal 0.0329 concentration of sodium sulfate.
Example 2
Determine the normality of a solution prepared by diluting 10 mL of a 25% m/v concentrated phosphoric acid solution to a final volume of 250 mL.
SOLUTION:
Steps 1 and 2: In this case, you start with a concentrated solution that is diluted. We can calculate the normality of the first solution and then calculate the normality of the diluted solution, or carry out the dilution first and the conversion to normality later. In this example we will do it the latter way.
Since it is a dilution, the dilution formula is applied, which is:
From where the concentration of the diluted solution is cleared, which is the one that interests us:
We also need the molar mass of the solute (H 3 PO 4 ) which is 98.0 g/mol. With these, we can calculate the normality using the formula of equation 5:
Step 3: Phosphoric acid is an acid, so ω is given by the number of ionizable protons it contains. Since it is a triprotic acid, then ω=3 eq/mol.
Step 4: We apply the formula:
Therefore, the diluted solution has a normal 0.306 concentration of phosphoric acid.
Example 3
Determine the normality of a 0.05 molar solution of Ca 2+ ions .
SOLUTION:
This is a particular and considerably common case, since many times what matters is the concentration of a particular ion and not that of a complete salt. When this happens, everything is done in the same way, except that the number of equivalents per mole is simply taken to be the charge on the ion, in this case 2.
Since in this case the molarity is known, then we use equation 4:
Finally, the solution has a normal 0.1 concentration of calcium ions.
References
Chang, R., & Goldsby, K. (2013). Chemistry (11th ed.). McGraw-Hill Interamericana de España SL
Normality . (2020, June 12). Alicante server. https://glosarios.servidor-alicante.com/quimica/normalidad
quimicas.net. (n.d.). Examples of Normality . https://www.quimicas.net/2015/05/ejemplos-de-normalidad.html
UNAM CCH “East.” (2019, September 23). Normal concentration . Slideshare. https://es.slideshare.net/Amon_Ra_C/normal-concentration
