146. Concentration units – Normality (4).



Salts are ionic compounds, consisting of cations (positively charged ions) and anions (negatively charged ions).

e.g.- CuSO4 ⇒ Copper sulfate is a salt that contains Cu2+ ions and SO42- ions.
NaCl ⇒  Sodium Chloride contains Na+ and Cl ions.


The net charge is written with the magnitude before the sign. e.g- a doubly charged cation will be denoted as 2+ and NOT +2. For a single charge, the number 1 is not written.

As mentioned in post 143, for salts we have two variations-

i)where the oxidation number(ON) does not change after the reaction-  For such a reaction, the n-factor is the modulus of total charge on the cation/anion.

Modulus means the absolute numeric value or co-efficient. For an ion, modulus of total charge means the numeric value of charge(not the + or – sign). Modulus is indicated by writing the quantity between two straight lines.
e.g.- |Mg2+| = 2 , |SO42-| = 2.

Consider the following reaction –

NaOH +HCl → NaCl + H2O

In the above reaction, the oxidation number of sodium is +1 before and after the reaction. The oxidation number of chlorine remains the same too. For such reactions, the n-factor is the absolute numeric value of the charge. In this example, it will be 1.

ii)where oxidation number(ON) changes after the reaction (REDOX REACTION) – In redox reactions, the oxidation number changes as one element gets oxidized (increase in ON) and the other gets reduced(decrease in ON).In these type of reactions, the n-factor is the change in oxidation number per mole of the substance.

CuSO4 +Zn→ ZnSO4 +Cu.

Before the reaction, the oxidation number of copper (in copper sulfate) is +2 i.e it exists as Cu2+  . The oxidation number of elemental zinc is zero. After the reaction,  elemental copper is formed with ON =0, and zinc exists as Zn2+ ion in zinc sulfate. So, in this reaction-

Change in Oxidation numberModulus of change
Cu2+ → Cu2-0=2|2| =2
Zn →Zn 2+0-2=-2|-2| = 2

As the change in ON for both copper and zinc is 2, the nf=2.

The molecular weight of CuSO4 is 160 g/mol. What will be its equivalent weight?
Equivalent weight = Molecular weight /nf = 160/2 = 80g/eq.
Thus, if we dissolve 80g of CuSO4 in 1 litre of solution, we will get a 1N CuSO4 solution.

Similarly, molecular weight of ZnSO4 is 162g/mol.
∴ Equivalent weight of ZnSO4 = 162/2= 81g/eq.
Thus, if we dissolve 81g of ZnSO4 in 1 litre of solution, we will get a 1N ZnSO4 solution.

In the next post we will learn more about normality. Till then,

Be a perpetual student of life and keep learning….

Good Day !

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