
Determine the equivalent weight of each given below, if formula weight of these compounds are X, Y and Z respectively.
(a) $ N{a_2}S{O_4} $
(b) $ N{a_3}P{O_4}.12{H_2}O $
(c) $ C{a_3}{\left( {P{O_4}} \right)_2} $
Answer
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Hint: Equivalent weight is also known as gram equivalent. It is the mass of one element. It can also be defined as the mass of a given substance which will combine with or displace a fixed quantity of other substances. It is used in volumetric analysis and gravimetric analysis also. The complete solution of this question is given in the below section in brief.
Complete answer:
(a)The equivalent weight of $ N{a_2}S{O_4} $ is $ = \dfrac{{Molecular{\text{ weight}}}}{{Total{\text{ charge on cation or anion}}}} $
$ {E_{N{a_2}S{O_4}}} = \dfrac{X}{2} $ . Because there are two sodium atoms and each of them has a charge of $ + 1 $ . So there is a total of $ + 2 $ charge and the charge on $ S{O_4} $ is $ - 2 $ .
(b)The equivalent weight of $ N{a_3}P{O_4}.12{H_2}O $ is $ = \dfrac{{Molecular{\text{ weight}}}}{{Total{\text{ charge on cation or anion}}}} $
$ {E_{N{a_3}P{O_4}.12{H_2}O}} = \dfrac{Y}{3} $ . Because the charge on $ P{O_4} $ is $ - 3 $ .
(c)The equivalent weight of $ C{a_3}{\left( {P{O_4}} \right)_2} $ is $ = \dfrac{{{\text{Molecular weight}}}}{{{\text{Total charge on cation or anion}}}} $
$ {E_{C{a_3}{{\left( {P{O_4}} \right)}_2}}} = \dfrac{Z}{6} $ . Because there are three calcium atoms and each of them has a charge of $ + 2 $ . So there is a total of $ + 6 $ charge and there are two atoms of compound $ P{O_4} $ and each of them having charge $ - 3 $ . So there is a total of $ - 6 $ charge.
Note:
Sodium sulfate is an inorganic compound. It’s all forms are white solids and are highly soluble in water. It is mainly used in the Kraft process of paper pulping and also used for the manufacture of detergents. It is ionic in nature. It does not react with the oxidizing as well as reducing agents. It also can be converted to sodium sulfide by carbo-thermal reduction at high temperature.
Complete answer:
(a)The equivalent weight of $ N{a_2}S{O_4} $ is $ = \dfrac{{Molecular{\text{ weight}}}}{{Total{\text{ charge on cation or anion}}}} $
$ {E_{N{a_2}S{O_4}}} = \dfrac{X}{2} $ . Because there are two sodium atoms and each of them has a charge of $ + 1 $ . So there is a total of $ + 2 $ charge and the charge on $ S{O_4} $ is $ - 2 $ .
(b)The equivalent weight of $ N{a_3}P{O_4}.12{H_2}O $ is $ = \dfrac{{Molecular{\text{ weight}}}}{{Total{\text{ charge on cation or anion}}}} $
$ {E_{N{a_3}P{O_4}.12{H_2}O}} = \dfrac{Y}{3} $ . Because the charge on $ P{O_4} $ is $ - 3 $ .
(c)The equivalent weight of $ C{a_3}{\left( {P{O_4}} \right)_2} $ is $ = \dfrac{{{\text{Molecular weight}}}}{{{\text{Total charge on cation or anion}}}} $
$ {E_{C{a_3}{{\left( {P{O_4}} \right)}_2}}} = \dfrac{Z}{6} $ . Because there are three calcium atoms and each of them has a charge of $ + 2 $ . So there is a total of $ + 6 $ charge and there are two atoms of compound $ P{O_4} $ and each of them having charge $ - 3 $ . So there is a total of $ - 6 $ charge.
Note:
Sodium sulfate is an inorganic compound. It’s all forms are white solids and are highly soluble in water. It is mainly used in the Kraft process of paper pulping and also used for the manufacture of detergents. It is ionic in nature. It does not react with the oxidizing as well as reducing agents. It also can be converted to sodium sulfide by carbo-thermal reduction at high temperature.
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