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The valence factor of a base is equal to its acidity. The acidity of a base is equal to furnishable hydroxyl ion (OH-) in its aqueous solution. With this background, we define equivalent weight of a base as :
$$\text{Equivalent weight},E=\frac{\text{Molecular weight of base}}{\text{Acidity}}$$
Acidity of KOH is 1, whereas acidity of $Ca{\left(OH\right)}_{2}$ is 2. Hence, equivalent weight of KOH is (39 + 16 + 1)/1 = 56/1 = 56. Similarly, equivalent weight of $Ca{\left(OH\right)}_{2}$ is {40 + 2X(16+1)}/2 = 74/2=37.
The valence factor of a compound depends on the manner a compound is involved in a reaction. The compounds of alkali metal salts and alkaline earth metal salts are, however, constant. These compounds are ionic and they dissociate in ionic components in aqueous solution. In this case, valence factor is equal to numbers of electronic charge on either cation or anion.
$$\text{Equivalent weight},E=\frac{\text{Molecular weight of compound}}{\text{Numbers of electronic charge on cation or anion}}$$
The numbers of electronic charge on cation of $NaHC{O}_{3}$ is 1. Hence, equivalent weight of $NaHC{O}_{3}$ is (23 + 1 + 12 + 3X16)/1 = 84.
If we look at the defining ratio of equivalent weight of a compound (AB) formed of two radicals (say A and B), then we can rearrange the ratio as :
$$\text{Equivalent weight, E}=\frac{\text{Molecular weight of Radical A}}{\text{Numbers of electronic charge}}+\frac{\text{Molecular weight of Radical B}}{\text{Numbers of electronic charge}}$$
Thus,
$$\Rightarrow \text{Equivalent weight of AB}=\text{Equivalent weight of A}+\text{Equivalent weight of B}$$
The valence factor of an ion is equal to numbers of electronic charge on the ion. Therefore, we define equivalent weight of an ion as :
$$\text{Equivalent weight},E=\frac{\text{Molecular weight of ion}}{\text{Numbers of electronic charge}}$$
The numbers of electronic charge on carbonate ion ( $C{O}_{3}^{2-}$ ) is 2. Hence, equivalent weight of carbonate ion is (12 + 3X16)/1 = 60/2 = 30. Similarly, equivalent weight of aluminum ion ( $A{l}^{3+}$ ) is 27/3 = 9.
In a redox reaction, one of the reacting entities is oxidizing agent (OA). The other entity is reducing agent (RA). The oxidizer is recipient of electrons, whereas reducer is releaser of electrons. The valence factor for either an oxidizing or reducing agent is equal to the numbers of electrons transferred from one entity to another.
$$\text{Equivalent weight},E=\frac{\text{Molecular weight of compound}}{\text{Numbers of electrons transferred in redox reaction}}$$
Alternatively,
$$\text{Equivalent weight},E=\frac{\text{Molecular weight of compound}}{\text{Change in oxidation number in redox reaction}}$$
Potassium dichromate in acidic medium is a strong oxidizer. It means it gains electrons during redox reaction. Potassium dichromate in acidic solution results in :
$${K}_{2}C{r}_{2}{O}_{7}+14{H}^{+}+6{e}^{-}\to 2{K}^{+}+2C{r}^{3+}+7{H}_{2}O$$
$$\text{Equivalent weight of}\phantom{\rule{1em}{0ex}}{K}_{2}C{r}_{2}{O}_{7}=\frac{294.2}{6}=49$$
Study of redox reaction is in itself an exclusive and extensive topic. We shall, therefore, discuss redox reaction separately.
It is equal to mass in grams numerically equal to equivalent weight. If the mass of a chemical entity is “g” grams, then the given mass contains gram equivalents given by :
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