" APPLICATION OF POAC ( PRINCIPAL OF ATOMIC CONSERVATION) AND LAW OF CHEMICAL EQUIVALENCE IN NUMERICALS OF PHYSICAL CHEMISTRY OF JEE / ISEET.

1.    Mole concept

2.    POAC

3.    Equivalent method ( law of chemical equivalence)

POAC & equivalent method does not depend on stoichiometric coefficient, but one necessary condition is that we must have given reactant from which products are to be produced. i.e. balancing of reaction is not necessary for applying POAC and eq. method. POAC is used for simpler reaction.

For saving time Eq. Method should apply for solving complex/ redox reaction rather than POAC method.

POAC method:- principle of atomic conservation

Through the chemical reaction, total no. of atoms of reactant must be equal to products. If atoms are conserved, moles of those atoms shall also be conserved. Examples:-

NaNO₃     →   NaNO₂ + O₂

Reactant & product contains only 1 Na atom. So we can write as

1 * mole of NaNO₃ = 1 * mole of NaNO₂

1*[ wt. Of NaNO₃/ m.wt of NaNO₃] = 1*[ wt. Of NaNO₂/m.wt. of NaNO₂]

Similarly we can apply POAC on N & O atoms

1* mole of NaNO₃    = 1* mole of NaNO₂   (for N atoms)

3* mole of NaNO₃    = 2* mole of NaNO₂ + 2 * mole of O₂   (for O atoms)

Since one mole of NaNO₃, NaNO₂ & O₂ contain 3, 2, and 2 moles of O atoms respectively.

KClO₃ → KCl + O₂

For   K atom

1* mole of KClO₃ = 1* mole of KCl

3 * mole of KClO₃ = 2 * mole of O₂

Q.1 calculates the no. of O & P atoms in 1420 gm of P₄O₁₀. M.wt of P₄O₁₀  = 568

Mole of P₄O₁₀ = 1420/ 568 = 5 moles

1* mole of P₄O₁₀ = 4* mole of P

5 moles of   P₄O₁₀ = 20 moles of P

20 moles of P   =   20 * 6.022* 10 ²³ atoms of P = 12.044 * 10 ²⁴ atoms of P

Similarly O atom

1* mole of P₄O₁₀ = 10* mole of O

5 mole of P₄O₁₀ = 50 mole of O

50 mole of O = 50* 6.022 * 10²³ atoms of O

= 30.11 * 10²⁴ atoms of O

Equivalent method:-

Sum of equivalents( Eqs.) or milliequivalents ( meqs.) of all oxidising agents is equal to sum of Eqs. Or  meqs. Of all reducing agents.

i.e. sum of equivalents of oxidising agents = sum of equivalents of reducing agents.

Example{ 1} Suppose we have a mixture of  FeO+ Fe₃O₄ ,in this mixture we add acidic  KMnO₄.

Here Fe is in oxidation state of +2 in FeO and +8/3 in Fe₃O₄ i.e. less than the maximum Oxidation state of Fe (less than +3). Hence in the presence of oxidising agent, it has a tendency to oxidise itself. In KMnO_4,   Mn has +7 oxidation state {max. Oxd. State} so it can not be further oxidise. It can be only reduce to itself in the presence of any reducing agent so 

Meq. of FeO + meq. of Fe₃O₄ = meq. Of KMnO₄

Or n₁ * mole of Fe₃O₄ + n₂ * mole of FeO = n₃ * mole of KMnO₄

Or           N₁V₁       +     N₂V₁      =      N₃V₂

Example 2. If a mixture contains FeO+ Fe₃O₄ + Fe₂O₃ (which is not reacted to each other) and we add KMnO₄  in it initially, which oxidise Fe of mixture to completely to its higher oxidation state ( +3) then we write

Meq. Of FeO+ meq. of Fe₃O₄ = meq. Of KMnO₄

Because Fe is in max. Oxidation state(+3)  in  Fe₂O₃. So it can’t be further oxidise. so Fe₂O₃ does not react with  KMnO₄ .

If after this oxidation process we add excess of KI (reducing agent), it will reduce Fe₂O₃ present in the initial solution. In this case Fe₂O₃  is react as oxidising agent because of max. Oxd. State (+3) and  KI as reducing agent

How to prepare for ISEET 2014:- “Learn the fundamental concepts of Physics, Chemistry and Mathematics” what is the meaning of learning the concepts.

How to prepare for ISEET 2014

“Learn the fundamental concepts of Physics, Chemistry and Mathematics”: - everybody says that “learn fundamental concepts to crack any competitive exams, learn basic concepts. Don’t just remember, or mugup the things.” but do u know what is exactly meaning of “learning the fundamentals and basics” Let’s see what is the meaning of learning the concepts.

For example if you are learning about equivalent no., equivalent wt. Normally most students think that always equivalent wt. Of H₂SO₄ is 49. If you are thinking the same, means your concepts also not clear. It is not always 49.

Let’s clear the concept

 Equivalent wt = molecular wt. / n factor

Now n factor for acids = basicity of acids (no. Of total replaceable H⁺ ion)

Because there are two H in H₂SO₄ are replaceable. Its n factor is 2.

So eq. Wt = 98/2 =49

But it is always not true. n factor also depend on chemical reaction. If in any reaction H₂SO₄ only donate 1 H⁺ ion and gets converted into HSO₄^- for example :-

NaOH + H₂SO₄ →NaHSO₄ + H₂O

In the above reaction H₂SO₄  only donating 1 H⁺. So here n factor for H₂SO₄ is one 1. So eq.wt = 98/1= 98 in this reaction

But in the following reaction

2 NaOH + H₂SO₄ →Na₂SO₄ + 2H₂O

H₂SO₄ donate its all 2 H⁺ ion and gets converted into SO₄^-2 ion. So in this reaction n factor for H₂SO₄ is 2. So eq. Wt of H₂SO₄ in this reaction = 98/2 = 49.

Because H₂SO₄ have total 2 H⁺ to donate. So generally we assume that it is donating its all two H⁺. Hence generally we consider n factor for H₂SO₄ is two. But if in the question there is any reaction of H₂SO₄ is given then carefully see the reaction whether it is donating 1 or two H⁺. According to that your ans. of eq.wt. of H₂SO₄ would be 98 or 49. But if there is no such reaction of H₂SO₄ is given then we assume that H₂SO₄  is donating its all two H⁺ ion so n factor is assumed two and eq.wt is 49.

VOLUMETRIC CALCULATIONS:- NUMERICALS FOR IIT JEE / ISEET

VVVOLUMETRIC CALCULATION :- TRY TO SOLVE FOLLOWING NUMERICALS FOR IIT JEE/ ISEET

1.  A solution contains 1.2046* 10²⁴  HCl molecule in one dm³ of the solution. What is the normality of the solution?
2.   A solution containing 0.275 gm of NaOH required 35.4 ml of HCl for neutralization. What was the normality of the HCl?
3.   The formula mass of an acid is 82.0. In a titration, 100 cm³ of a solution of this acid containing 39.0 gm of this acid per litre were completely neutralized by 95.0 c.c. of NaOH per litre. What is the basicity of the acid?
4.  A solution contains Na₂CO₃ and NaHCO₃. 10 ml. of the solution required 2.5 ml of 0.1 M H₂SO₄ for neutralization using phenolphthalein as indicator. Methyl orange is than added when a further 2.5 ml of 0.2 M H₂SO₄ was required. Calculate the amount of   Na₂CO₃ in one litre of the solution.
5.    500 ml of a solution contains 2.65 gm of Na₂CO₃ and 4 gm of caustic soda; 20 ml. of this solution have been titrated each time against N/ 10 H₂SO_4.  Find the titer values if (i) methyl orange is taken as indicator, and also if (ii) phenolphthalein is taken as an indicator.
6.  25 ml. of a solution containing NaOH and Na₂CO₃ when titrated against N/2 HCl using phenolphthalein as indicator required 40 ml. of HCl. The same vol. of mix. When titrated against N/2 HCl using methyl orange as indicator required 45 cc of HCl. Calculate the amounts of NaOH and Na₂CO₃ in 1 litre of the mixture.
7.     Acidified KMnO₄   oxidizes oxalic acid to CO_2. what vol. in the litres of 10^-4 M KMnO₄ required to completely oxidize 0.5 litre of 10^-2 M oxalic acid in acid medium?                                                   

MOLECONCEPT AND LAW OF CHEMICAL EQUIVALENCE

If you are preparing for ISEET 2014 or  JEE OR NEET 2014 or any other medical or engineering competitive exam u have to start your preparation from physical chemistry "MOLE CONCEPT and LAW OF CHEMICAL EQUIVALENCE "  because its basic concept and also used in a no. of other topic also like in Faraday's law (electro chemistry), chemical and ionic equilibrium etc.