N factor here we mean a conversion factor by which we divide molar mass of substance to get equivalent mass and it depends on nature of substance which varies from one condition to another condition. We can divide n-factor calculation is two category.
In case of non-redox reaction.
(a) n factor of acid =Basicity of the acid
Basicity: Number of replaceable H+ ion.
Ex.1
n factor of HCl = 1
n factor of CH3COOH=1
n factor of H2SO4=2
(b)n factor of base= acidity of the base
Acidity: Number of replaceable OH– ion.
Ex.2
n factor of NaOH=1
n factor of Ca(OH)2=2
n factor of Al(OH)3=3
n factor of B(OH)3=1 (because it is a mono basic acid)
(c) n factor Salt: Total number of positive or negative charge.
Ex.3
n factor of NaCl=1
n factor of Na2SO4=2
n factor of K2SO4. Al2 (SO4)3. 24H2O = 8
Ex.4
Find the n factor of H3PO4 in the following reaction.
H3PO4 + Ca (OH)3 → CaHPO4 + 2H2O
Sol.
Basicity of H3PO4 in the above reaction is 2
∴ the n factor of H3PO4 is 2
In case of redox reaction
(a) From oxidation number
n factor of oxidizing or reducing agent=change in oxidation number per molecule.
⇒ consider a salt AxBy in which the O.S. of A is +c. It changes to a compound AdE in which the O.S. of A is +f. Here we are assuming that B does not undergo any change in O.S. A+cxBy → A+fdE (Obviously AxBy must have reacted with some other substance to produce the product AdE. That means other substance has the atom E in it.) The ‘n’ factor is = |xc-xf|.
Ex.5
Find the n factor of KMnO4 in different medium.
Sol.
(i) In acidic medium
KMnO4→Mn+2
Change in oxidation number of Mn=+7-2=5
∴ the n factor of KMnO4=5
(ii) Basic medium
KMnO4→K2MnO4
N factor of KMnO4 = +7 – 6 = 1
(iii) Neutral medium
KMnO4 → MnO2
N factor of MnO2 = +7 – 4 = 3
(b)From ion electron method:
N factor = total number of electrons transferred per mole of the reactant
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