Determination of equivalent masses of elements
Equivalent masses can be determined by the following methods:
1. Hydrogen displacement method
2. Oxide method
3. Chloride method
4. Metal displacement method
1.Hydrogen displacement method
This method is used to determine the equivalent mass of those metals such as magnesium, zinc and aluminium which react with dilute acids and readily displace hydrogen.
Mg + 2HCl -> MgCl2 + H2
Zn + H2SO4 - > ZnSO4 + H2
2Al + 6HCl - > 2AlCl3 + 3H2
From the mass of the metal and the volume of hydrogen displaced, the equivalent mass of the metal can be calculated.
This method is employed to determine the equivalent mass of those elements which readily form their oxides eg. magnesium, copper etc. Oxide of an element can be formed by direct method or by indirect method.
Magnesium forms its oxide directly on heating
2Mg + O2 ® 2 MgO
In the case of copper, its oxide is obtained in an indirect manner i.e. copper is dissolved in concentrated nitric acid and the copper(II) nitrate formed after evaporation is strongly heated to give copper (II) oxide.
Cu + 4HNO3 ® Cu(NO3)2 + 2H2O + 2 NO2
2Cu(NO3)2 ® 2CuO + 4 NO2 + O2
Mass of the element taken = w1 g
Mass of the oxide of the element = w2 g
Mass of oxygen = (w2-w1) g
(w2 - w1) g of oxygen has combined with w1 g of the metal.
8 g of oxygen will combine with w1 / (w2 - w1) x 8
This value represents the equivalent mass of the metal.
The equivalent mass of those elements which readily form their chlorides can be determined by chloride method. For example, a known mass of pure silver is dissolved completely in dilute nitric acid. The resulting silver nitrate solution is treated with pure hydrochloric acid when silver chloride is precipitated. It is then filtered, dried and weighed. Thus from the masses of the metal and its chloride, the equivalent mass of the metal can be determined as follows :
Mass of the metal = w1 g
Mass of the metal chloride = w2 g
Mass of chlorine = (w2 - w2) g
(w2 - w1) g of chlorine has combined with w1 of the metal
35.46 g of chlorine will combine with 35.46 x w1 / (w2 - w1) g of the metal
This value gives the equivalent mass of the metal.
Uses of volumetric laws
If the volume of the acid is represented as V1, the normality of the acid as N1, volume of base as V2 the normality of the base as N2, then from the law of volumetric analysis it follows that
V1 x N1 = V2 x N2
All volumetric estimations are based on the above relationship. If any three quantities are known, the fourth one can readily be calculated using the above expression.
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