STOICHIOMETRY n FACTOR

                                       N- FACTOR (advanced batch)

Dear students ,
I think most of you must be going through revision .I thought of giving a thrust to your revision .
I would like to brief you about stoichiometry which is the stepping stone of physical chemistry .As you know important term  in stoichiometry is n-factor which is the back bone of equivalent concept. To begin one must be very clear with the definition of  n-factor w.r.t Acid / Base , it is not  the no. of  Proton or hydroxide present but it is actually  the no. ofreplaceable or replaced  ions.
For salt,it depends whether the reaction is undergoing  red-ox reaction or displacement reaction. In displacement reaction one should always remember no. of  moles of cation /anion actually taking part in the reaction .

On converting sodium carbonate to sodium bicarbonate or to carbonic acid  n-factor is different even though no. of moles of cation and its oxidation state remains same.
Moving on to red-ox reaction, n-factor for disproportionate reaction is not defined. 

We do have few tricks to solve disproportionate reaction n-factor for e.g decomposition  of hydrogen peroxide  where if 8 gm of  oxygen is  produced then weight of hydrogen peroxide taken must be equal to its equivalent weight.
In few compounds like potassium permanganate there will be fixed  oxidation state change,in acidic medium 5,in neutral medium 3and in basic medium 1. At times just by mugging  up the changes gives wrong answer as in case of permanganate (in presence of sulfite) n-factor is 2.
One should always remember to understand the reaction as it is core of stoichiometry, one can develop by following minute details of inorganic chemistry 

for e.g in Iodimetry  titration  we need to follow neutral medium i.e pH around 6 to 8.
If we go for high pH say around 10 we may end up converting iodine to iodate ion.
Similarly  if we try low pH then starch gets hydrolyzed.
For better practice i advice you to go through old IITJEE question specially like 92 ,97 ,98 ,95 etc.
Before finishing off with my first  interaction ,i would like to ask you a simple question and lets see who can answer?     
What will be the n-factor for Permanganate ion in presence of HF ?

• Equal equivalents or Meq of reactants react to give same number of Eq.or Meq of products.
•  Moles and milli-moles react according to stoichiometery of equation.
• The sum of mole fraction of components in a two components system is equal to unity.

Normality (N) = Equivalents of solute / vol in lit. 
Equivalent = N X V (in lit) = Weight / Equivalent wt
Milli -Equivalent = N X  V in ml = wt./Eq .wt  X 1000 
                                                  Equivalent weight
The no of parts by mass of the substance which combine or displace directly or in directly 1.008 parts by mass  of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine or 108 parts by mass of silver.
Eq wt wt of an element may vary with change of valency

•  Eq .wt of an element = Mass of element  X 11200 / vol of hydrogen displaced(NTP)
•   Equivalent weight of element = atomic weight /valence.
•   Eq wt of compound = mol . wt / total charge on cation or anion
•   Eq  wt of acid = mol .wt / Basicity in a particular reaction.
•   Eq wt of base = mol .wt / Acidity  in a particular reaction.
•   Eq wt of acid salt = formula wt / Replaceable H atom in acid salt.
•   Eq wt of an ion = formula wt. / charge on ion

Moles = wt / mol. wt
Molarity (M) = Moles of solute / volume in litre
Milli- moles =M X V in ml. = wt / mol. wt. X 1000.
Molarity =  Normality X  1 / x-factor.      : N = M X n-factor