Equivalent weight uses and how to calculate it (with examples)

3894
Alexander Pearson

The equivalent weight (PE) of a substance is one that participates in a chemical reaction, and is used as the basis for a titration. Depending on the type of reaction, it can be defined in one way or another.

For acid-base reactions, PE is the weight in grams of the substance required to supply or react with one mole of H+ (1.008 g); for redox reactions, the weight in grams of the substance that is required to supply or react with one mole of electrons.

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For precipitation or complex formation reactions, the weight of substance required to supply or react with one mole of a monovalent cation, 1/2 mole of a divalent cation, 1/3 mole of a trivalent cation . And so on.

Although it may seem a bit complicated at first, some substances always behave chemically in the same way; therefore, it is not difficult to learn the PE values ​​given the cases.

Article index

  • 1 Origin of equivalent weight
  • 2 Uses
    • 2.1 Use in general chemistry 
    • 2.2 Use in volumetric analysis
    • 2.3 Use in gravimetric analysis
    • 2.4 Uses in polymer chemistry
  • 3 How to calculate it? Examples
    • 3.1 -Equivalent weight of a chemical element
    • 3.2 -Equivalent weight of an oxide
    • 3.3 -Equivalent weight of a base
    • 3.4 -Equivalent weight of an acid
    • 3.5 -Equivalent weight of a salt
  • 4 References

Origin of equivalent weight

John Dalton (1808) proposed the equivalent weight of hydrogen as a unit of mass. However, a number of objections to this approach arose. For example, it was noted that most elements did not react directly with hydrogen to form simple compounds (XH)..

Furthermore, elements with various oxidation states, for example permanganate, have more than one equivalent weight. This made it difficult to accept the equivalent weight as a unit of mass..

The presentation by Dimitri Mendeleev (1869) of his periodic table, in which the chemical properties of the elements were related to the ordered order of their atomic weights, constituted a strong argument from those who objected to the use of the equivalent weight as a unit of mass.

Actually, there is no need to use the term "equivalent", since any stoichiometric calculation can be done in terms of moles. However, this term is often used and should not be ignored..

For convenience, the term "equivalent" was introduced: an equivalent of any acid reacts with an equivalent of any base; one equivalent of any oxidizing agent reacts with one equivalent of any reducing agent, etc..

Applications

Use in general chemistry

Metals

The use of PE in elements and chemical compounds has been replaced by the use of its molar mass. The main reason is the existence of elements and compounds with more than an equivalent weight.

For example, iron (Fe), an element with an atomic weight of 55.85 g / mol, has two valences: +2 and +3. Therefore, it has two equivalent weights: when it works with valence +2 its equivalent weight is 27.93 g / eq; while, when using the valence +3, its equivalent weight is 18.67 g / eq.

Of course, it is not possible to speak of the existence of an equivalent weight of Fe, but the existence of an atomic weight of Fe can be pointed out..

Acids

Phosphoric acid has a molecular weight of 98 g / mol. This acid when dissociated into H+ + HtwoPO4-,  has an equivalent weight of 98 g / eq, since it releases 1 mole of H+. If phosphoric acid dissociates into H+ +  HPO4two-, its equivalent weight is (98 g.mol-1) / (2eq / mol-1) = 49 g / eq. In this dissociation the H3PO4 releases 2 moles of H+.

Although it is not titratable in an aqueous medium, the H3PO4 can dissociate in 3 H+  +   PO43-. In this case, the equivalent weight is (98 g.mol-1) / (3 mol eq.-1) = 32.7 g / eq. The H3PO4 delivers in this case 3 moles of H+.

So, phosphoric acid has up to 3 equivalent weights. But this is not an isolated case, so for example, sulfuric acid has two equivalent weights and carbonic acid also has.

Use in volumetric analysis

-To reduce the errors that can be made during the action of weighing substances, in analytical chemistry the use of a substance of greater equivalent weight is preferred. For example, in the titration of a sodium hydroxide solution with acids of different equivalent weights. The use of the acid with the highest equivalent weight is recommended.

-When using the mass of a solid acid that can react with sodium hydroxide, you have the option of choosing between three solid acids: oxalic acid dihydrate, potassium acid phthalate and potassium hydrogeniodate, with equivalent weights respectively. of 63.04 g / eq, 204.22 g / eq and 389 g / eq.

In this case, it is preferred to use potassium hydrogeniodate acid in the titration of sodium hydroxide, since as it has a greater equivalent weight, the relative error made when weighing it is less..

Use in gravimetric analysis

Equivalent weight is defined in its own way in this substance analysis technique. Here, it is the mass of precipitate that corresponds to one gram of the analyte. This is the element or compound of interest in the study or analysis that is being carried out.

In gravimetry it is common to quote the results of the analyzes as a fraction of the mass of the analyte, often expressed as a percentage..

The equivalence factor is explained as a numerical factor by which the mass of the precipitate must be multiplied to obtain the mass of the analyte, usually expressed in grams..

Gravimetric determination of nickel

For example, in the gravimetric determination of nickel, the precipitate that contains it is bis (nickel dimethylglyoximate) with a molar mass of 288.915 g / mol. The molar mass of nickel is 58.6934 g / mol.

The molar mass of the precipitate divided by the molar mass of nickel produces the following result:

288.915 g.mol-1/ 58.6934 g.mol-1 = 4.9224. This means that 4.9224 g of the compound is equivalent to 1 g of nickel; or in other words, 4.9224 g of the precipitate contains 1 g of nickel.

The equivalence factor is calculated by dividing the molar mass of nickel by the molar mass of the precipitate that contains it: 58.693 g.mol-1/ 288.915 g.mol-1 = 0.203151. This tells us that per gram of the nickel-containing precipitate there are 0.203151 g of nickel.

Uses in polymer chemistry

In polymer chemistry, the equivalent weight of a polymerization reagent is the mass of the polymer that has one equivalent of reactivity.

It is particularly important in the case of ion exchange polymers: one equivalent of an ion exchange polymer can exchange one mole of monocharged ions; but only half a mole of doubly charged ions.

It is common to express the reactivity of a polymer as the inverse of the equivalent weight, which is expressed in units of mmol / g or meq / g.

How to calculate it? Examples

-Equivalent weight of a chemical element

It is obtained by dividing its atomic weight by its valence:

Peq = Pa / v

There are items that have only one equivalent weight and items that can have 2 or more.

Equivalent weight of calcium

Atomic weight = 40 g / mol

Valencia = +2

Peq = 40 g.mol-1/2eq.mol-1

20 g / eq

Equivalent weight of aluminum

Atomic weight = 27 g / mol

Valencia = +3

Peq = 27 g.mol-1/ 3 eq.mol-1

9 g / eq

Nickel equivalent weight

Atomic weight = 58.71 g / mol

Valencia = +2 and +3

Nickel has two equivalent weights corresponding to when it reacts with valence +2 and when it reacts with valence +3.

Peq = 58.71 g.mol-1/ 2 eq. Mol-1

29.35 g / eq

Peq = 58.71 g.mol-1/ 3 eq.mol-1

19.57 g / eq

-Equivalent weight of an oxide

One way to calculate the equivalent weight of an oxide is by dividing its molecular weight by the valence product of the metal and the subscript of the metal..

Peq = Pm / V S

Pm = molecular weight of oxide.

V = valence of metal

S = subscript of the metal

The product V S is called the total or net charge of the cation.

Equivalent weight of aluminum oxide (AltwoOR3)

Molecular weight = Al (2 x 27 g / mol) + O (3 x 16 g / mol)

102 g / mol

Valencia = +3

Subscript = 2

Peq AltwoOR3 = Pm / V · S

Peq AltwoOR3 = 102 g.mol-1/ 3 eqmol-1. two

17 g / eq

There is another way to solve this problem based on stoichiometry. In 102 g of aluminum oxide there are 54 grams of aluminum and 48 grams of oxygen.

Peq del Al = Atomic weight / Valencia

27 g mol-1/ 3 eq.mol-1

9 g / eq

Based on the equivalent weight of aluminum (9 g / eq), it is calculated that in 54 g of aluminum there are 6 equivalents of aluminum.

Then from the properties of the equivalents: 6 equivalents of aluminum will react with 6 equivalents of oxygen to give 6 equivalents of aluminum oxide.

In 102 g. of aluminum oxide there are 6 equivalents.

Therefore:

Al PeqtwoOR3 = 102 g / 6 eq

17 g / eq

-Equivalent weight of a base

The equivalent weight is obtained by dividing its molecular weight by the number of oxyhydryl groups (OH).

Equivalent Weight of Ferrous Hydroxide, Fe (OH)two

Molecular weight = 90 g / mol

OH number = 2

Small Fe (OH)two = 90 g.mol-1/ 2 eq. Mol-1

45 g / eq

-Equivalent weight of an acid

In general, it is obtained by dividing its molecular weight by the number of hydrogen it gives up or releases. However, polyrotonic acids can dissociate or release their H in various ways, so they can have more than one equivalent weight..

Equivalent weight of hydrochloric acid, HCl

Equivalent weight HCl = molecular weight / number of hydrogen

Small HCl = g.mol-1/ 1 eq.mol-1

36.5 g / eq

Equivalent weight of sulfuric acid

Sulfuric acid (HtwoSW4) can be dissociated in two ways:

HtwoSW4 => H+   +    HSO4-

HtwoSW4 => 2 H+   +    SW4two-

When he releases an H+ your PE is:

Molecular weight = 98 g / mol

Peq = 98 g.mol-1/ 1 eq.mol-1

98 g / small

And when it releases 2H+:

Molecular weight = 98 g / mol

Peq = 98 g.mol-1/ 2 eq. Mol-1

49 g / eq

For the same reason, phosphoric acid (H3PO4) of molecular weight 98 g / mol, it can have up to three equivalent weights: 98 g / eq, 49 g / eq and 32.67 g / eq.

-Equivalent weight of a salt

And finally, the equivalent weight of a salt can be calculated by dividing its molecular weight by the product of the valence of the metal by the subscript of the metal..

PE = PM / V S

Ferric Sulfate Fetwo(SW4)3

Molecular weight = 400 g / mol

Iron valence = +3 eq / mol

Iron subscript = 2

Peq = 400 g.mol-1/ 3 eq.mol-1 x 2

66.67 g / eq

References

  1. Day, R. A. JR. And Underwood, A. L. Quantitative Analytical Chemistry. Translation of the 5to English Edition. Editorial Prentice Hall Interamericana
  2. Inorganic chemistry. (s.f.). Determination of equivalent weights of Oxides. Recovered from: fullquimica.com
  3. Wikipedia. (2018). Equivalent weight. Recovered from: en.wikipedia.org
  4. The Editors of Encyclopaedia Britannica. (2016, September 26). Equivalent weight. Encyclopædia Britannica. Recovered from: britannica.com
  5. Ori, Jack. (April 30, 2018). How to Calculate Equivalent Weight. Sciencing. Recovered from: sciencing.com
  6. The Equivalent Weight of an Acid Part 2: Titration of the Unknown Acid Sample. (s.f.). Recovered from: faculty.uml.edu
  7. Bergstresser M. (2018). Equivalent Weight: Definition & Formula. Study. Recovered from: study.com

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