The normal it is a measure of concentration used, less and less frequently, in the chemistry of solutions. It indicates how reactive the solution of the dissolved species is, rather than how high or dilute its concentration is. It is expressed as gram-equivalents per liter of solution (Eq / L).
Much confusion and debate have arisen in the literature regarding the term 'equivalent', as it varies and has its own value for all substances. Likewise, the equivalents depend on the chemical reaction under consideration; therefore, normality cannot be used arbitrarily or globally.
For this reason, the IUPAC has advised to stop using it to express concentrations of solutions..
However, it is still used in acid-base reactions, widely used in volumetry. This is partly because, considering the equivalents of an acid or a base, it makes calculations much easier; Furthermore, acids and bases always behave in the same way in all scenarios: they release or accept hydrogen ions, H+.
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Although normality by its mere definition can generate confusion, in short, it is nothing more than molarity multiplied by an equivalence factor:
N = nM
Where n is the equivalence factor and depends on the reactive species, as well as the reaction in which it participates. Then, knowing its molarity, M, its normality can be calculated by simple multiplication.
If, on the other hand, only the mass of the reagent is available, its equivalent weight will be used:
PE = PM / n
Where MW is the molecular weight. Once you have PE, and the mass of the reactant, just apply a division to obtain the equivalents available in the reaction medium:
Eq = g / PE
And finally, the definition of normality says that it expresses the gram-equivalents (or equivalents) per one liter of solution:
N = g / (PE ∙ V)
What is equal to
N = Eq / V
After these calculations, it is obtained how many equivalents the reactive species has per 1L of solution; or, how many mEq are there per 1mL of solution.
But what are the equivalents? They are the parts that have in common a set of reactive species. For example, to acids and bases, what happens to them when they react? Free or accept H+, regardless of whether it is a hydracid (HCl, HF, etc.), or an oxacid (HtwoSW4, HNO3, H3PO4, etc.).
Molarity does not discriminate the number of H that the acid has in its structure, or the amount of H that a base can accept; just consider the whole in molecular weight. However, normality takes into account how the species behave and, therefore, the degree of reactivity.
If an acid releases an H+, molecularly only a base can accept it; in other words, an equivalent always reacts with another equivalent (OH, in the case of bases). Likewise, if one species donates electrons, another species must accept the same number of electrons..
From here comes the simplification of the calculations: knowing the number of equivalents of one species, it is known exactly how many are the equivalents that react from the other species. Whereas with the use of moles, one must adhere to the stoichiometric coefficients of the chemical equation.
Starting with the HF and H pairtwoSW4, For example, to explain the equivalents in its neutralization reaction with NaOH:
HF + NaOH => NaF + HtwoOR
HtwoSW4 + 2NaOH => NatwoSW4 + 2HtwoOR
To neutralize HF, one mole of NaOH is needed, while HtwoSW4 requires two moles of base. This means that HF is more reactive since it needs a smaller amount of base for its neutralization. Why? Because HF has 1H (an equivalent), and HtwoSW4 2H (two equivalents).
It is important to emphasize that, although HF, HCl, HI and HNO3 they are "just as reactive" according to normality, the nature of their bonds and, therefore, their strength of acidity, are totally different.
So, knowing this, the normality for any acid can be calculated by multiplying the number of H by its molarity:
1 ∙ M = N (HF, HCl, CH3COOH)
2 ∙ M = N (HtwoSW4, HtwoSeO4, HtwoS)
With the H3PO4 you have 3H, and therefore have three equivalents. However, it is a much weaker acid, so it does not always release all of its H+.
Furthermore, in the presence of a strong base, not all its H's necessarily react+; This means that attention should be paid to the reaction where you are participating:
H3PO4 + 2KOH => KtwoHPO4 + 2HtwoOR
In this case, the number of equivalents is equal to 2 and not 3, since only 2H react+. While in this other reaction:
H3PO4 + 3KOH => K3PO4 + 3HtwoOR
It is considered that the normality of H3PO4 is three times its molarity (N = 3 ∙ M), since this time all its hydrogen ions react.
For this reason, it is not enough to assume a general rule for all acids, but also, it must be known exactly how many H+ participate in the reaction.
A very similar case occurs with the bases. For the following three bases neutralized with HCl we have:
NaOH + HCl => NaCl + HtwoOR
Ba (OH)two + 2HCl => BaCltwo + 2HtwoOR
Al (OH)3 + 3HCl => AlCl3 + 3HtwoOR
El Al (OH)3 it needs three times more acid than NaOH; that is, the NaOH needs only a third of the amount of added base to neutralize the Al (OH)3.
Therefore, NaOH is more reactive, since it has 1OH (one equivalent); the Ba (OH)two has 2OH (two equivalents), and the Al (OH)3 three equivalents.
Although it lacks OH groups, NatwoCO3 is capable of accepting up to 2H+, and therefore, it has two equivalents; but if you accept only 1H+, then participate with an equivalent.
When a cation and anion join together to precipitate into a salt, the number of equivalents for each is equal to its charge:
Mgtwo+ + 2Cl- => MgCltwo
Thus, the Mgtwo+ has two equivalents, while Cl- has only one. But what is the normality of MgCltwo? Its value is relative, it can be 1M or 2 ∙ M, depending on whether the Mg is considered.two+ or Cl-.
The number of equivalents for the species involved in redox reactions is equal to the number of electrons gained or lost during the same..
3CtwoOR4two- + CrtwoOR7two- + 14H+ => 2Cr3+ + 6COtwo + 7HtwoOR
What will be normal for the CtwoOR4two- and CrtwoOR7two-? For this, the partial reactions where electrons participate as reactants or products must be taken into account:
CtwoOR4two- => 2COtwo + 2e-
CrtwoOR7two- + 14H+ + 6e- => 2Cr3+ + 7HtwoOR
Each CtwoOR4two- releases 2 electrons, and each CrtwoOR7two- accepts 6 electrons; and after balancing, the resulting chemical equation is the first of the three.
So, the normality for CtwoOR4two- is 2 ∙ M, and 6 ∙ M for CrtwoOR7two- (remember, N = nM).
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