Tiêu đề 4.SOME BASIC CONCEPTS IN CHEMISTRY_06.03.14 - FINAL ( 123 - 153 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 123 NAME OF THE TOPIC Jr Chemistry E/M JR/SR SUBJECT- JEE - NIT - MAINS - VOL - No. GENERAL CHEMISTRY SYNOPSIS CALCULATION OFn-FACTOR For calculating the n-factor of any reactant in any reac- tion. One must know the kind of reaction it is, the reac- tions may be classified into the following three types. 1. Acid-Base Reactions/Neutralization reactions 2. Redox Reactions 3. Precipitation Reactions/Double decomposition re- actions Acid-Base Reaction: According, to the Arrhenius, an acid is a substance that furnishes H+ ion(s) in solution, a base is a substance that furnishes OH- ion(s) in solution and neutralization is a reaction in which H+ ion furnished by acid combines with OH- ions furnished by base. The number of H+ ion(s) furnished per molecule of the acid is its n-factor also called basicity. Similarly the number of OH- ion(s) furnished by the base per molecule is its n-factor, also called acidity. Some Examples (n 1) HCl H Cl      2 4 4 (n 1) H SO H HSO      2 2 4 4 (n 2) H SO 2H SO      3 4 2 4 (n 1) H PO H H PO      3 3 4 4 (n 3) H PO 3H PO      3 3 2 3 (n 1) H PO H H PO      3 3 3 (n 2) H PO 2H HPO      The n-factor of H3 PO3 cannot be 3 as it has only two dissociable H+ ions. So, its n-factor or dissociable pro- tons is 1 or 2 as one of the H-atoms is linked with P atom directly. P O HO OH H Similarly,    CH COOH CH3COO  H (n 1) 3 n-factor of CH3 COOH is 1, because it contains only one dissociable H+ ion. Now, we will consider the n-factor of some bases. (n 1) NaOH Na OH      2 (n 1) Ba(OH) [Ba(OH)] OH      2 2 (n 2) Ba(OH) Ba 2OH      3 3 (n 3) Al(OH) Al 3OH      Similarly, n-factor of Al(OH)3 can also be 1 or 2 or 3, depending upon the number of OH– released. REDOX REACTIONS Those reactions which involve the exchange of electrons are called redox reactions. For the calculation of n- factor of oxidising agent or reducing agent, the method depends upon the change in oxidation state of the species considered. We will discuss them one by one. i) When only one atom undergoing either reduction or oxidation e.g. 7 2 H 2 4 n 5 MnO Mn        In such a case, we consider the change in oxidation state of atom undergoing oxidation or reduction change per molecule as the n-factor of the species.
Jr Chemistry E/M 124 NISHITH Multimedia India (Pvt.) Ltd., JR/SR SUBJECT- JEE - NIT - MAINS - VOL - No. STIOCHIOMETRY n-factor = |(+2) ×1 – (+7)  1| = 5 2 3 2 3 n 1 Fe Fe       n-factor = |(+3) × 1 – (+2) × 1| = 1 3 2 4 2 2 2 4 2 n 2 C O 2CO        n-factor = |(+4) × 2 – (+3) × 2| = 2 6 2 3 2 2 3 2 7 n 6 Cr O 2Cr         n-factor = |(+3) × 2 – (+6)× 2| = 6 ii) Salts which reacts in such a way that only one atom undergoes change in oxidation state but appears in two products with the same oxidation state: In such a case the method of calculation of n-factor remains the same i.e., we will calculate the change in oxidation state of the atom per mole of that substance (reactant). 6 2 3 3 2 3 3 Cr O Cr Cr 2 7          In this example, oxidation state of Cr changes from +6 to +3 in both the products. So n-factor = |(+6) × 2 – (+3) × 2| = 6 iii) Salts which react in such a way that only one atom undergoes change in oxidation state but goes in two products with different oxidation state as a result of either only oxidation or only reduction. 6 6 2 2 7 3MnO4 2Mn Mn         In such a case, it is impossible to calculate the n-factor until and unless one knows that how much of  MnO4 is changing to Mn2+ and how much to Mn6+ and if one knows the balanced equation then there is no need of calculation of n-factor. Nevertheless in such case the n-factor can be calculated by deducing the total change in oxidation state divided by total number of atom undergoing reduction/oxidation change. So, for the calculation of n-factor in the above example, out of three moles of  MnO4 , two moles are being converted to Mn2+ and one mole changes to Mn6+. So total decrease in oxidation state of Mn. = | [2 × (+2) – 2 × (+7)] | + | [1 × (+6) – 1 × (+7)]| = | 4 – 14 | + | 6 – 7 | = 11 So, n-factor = 3 11 iv) Salts which react in such a way that only one atom undergoes change in oxidation state in two product, in one product with changed oxidation state and in other product with same oxidation state as that of reactant. In such case also one cannot calculate the n-factor without knowing the balanced chemical equation because one must know how much of atom has changed its oxidation state. For example. K2 Cr2O7 + 14HCl  2KCl + 2CrCl3 + 3Cl2 + 7H2O Let us calculate the n-factor of HCl. Out of 14 moles of Cl– (in HCl) only 6 moles of Cl– are changing its oxidation state from –1 to 0 in the product Cl2 and the oxidation state of remaining 8 Cl– ions remains same in KCl and CrCl3 . So, total no. of moles of electrons lost by 14 moles of HCl is 6. So each mole of HCl takes up 6/14 i.e., 3/7 moles of electrons and hence n-factor of HCl is 3/7. v) Salts which react in such a way that two or more atoms in the salt undergoes change in oxidation states as a result of either oxidation or reduction. Let us consider the following example, FeC2O4  Fe3+ + 2CO2 In this case, the oxidation of both Fe2+ and C3+ are changing from + 2 and +3 to +3 and +4 respectively. In such a case we will calculate the n-factor of the salt as the total increase or decrease in oxidation state per mole of the salt. As one can see that one mole of FeC2O4 contains one mole of Fe2+ and one mole of C2O4 2– (i.e. 2 carbon atoms per mole of C­2O4 2– . Total change in oxidation state = | 1 × (+2) – 1 × (+3) | + | 2 × (+3) – 2 × (+4) | = 1 + 2 = 3 So, n-factor of FeC2O4 is 3 vi) Salts which react in such a way that two atoms undergoing change in oxidation state but one undergoing oxidation and other reduction reaction. In such a case one has to calculate the change in oxidation state of either the atom being oxidized or the atom being reduced. For example. 3 2 6 2 0 2 3 2 2 4 2 7 2 2 3 2 (N H ) Cr O N Cr O 4H O          
NISHITH Multimedia India (Pvt.) Ltd., 125 NAME OF THE TOPIC Jr Chemistry E/M JR/SR SUBJECT- JEE - NIT - MAINS - VOL - No. In this reaction, the oxidation state of N is increasing by 6 units and that of Cr is decreasing by 6 unit. So, we can consider either oxidation or reduction product for the calculation of n-factor and it will be the same. n-factor of (NH4 ) 2 Cr2O7 considering oxidation = |(-3) ×2 – (0) × 2| = 6 n-factor of (NH4 ) 2 Cr2O7 considering reduction = |(+6) × 2 – (+3) × 2| = 6 vii) Species which undergoes disproportionation reaction: Those reaction in which oxidant and reductant are the same species or the same element from the species is getting oxidized as well as reduced. When the number of moles of atoms being oxidized is equal to the number of moles of atoms being reduced. The n-factor can be calculated by knowing the balanced chemical equation and considering any of the change taking place. Say for example, 2H2O2  2H2O + O2 Out of 2 moles of H2O2 consumed in the reaction, one mole of H2O2 is being oxidized (H2O2  O2 ) and one mole of H2O2 is being reduced (H2O2  2H2O). First consider the oxidation reaction 2 0 2 ( 1) 2 H2O2 O     n-factor = |2 × 0 – (–1) × 2| = 2 Again, considering reduction reaction ( 2) 2 2 ( 1) 2 H2O2 2H O      n-factor = |(-2) × 2 – (-1) × 2| = 2 So, n-factor of H2O2 either considering oxidation or reduction reaction is same i.e. 2. P R E C I P I T A T I O N / D O U B L E DECOMPOSITION REACTIONS: In such reaction, there is no change in oxidation state of any atom. The n-factor of the salt can be calculated by multiplying the oxidation state of the cation/anion by total no. of atoms per molecule of the salt. For example BaCl Na SO BaSO 2NaCl 4 n 2 2 4 n 2 2      for BaCl2 n-factor = Oxidation state of Ba atom in BaCl2 × number of Ba atoms in 1 molecule of BaCl2 = (+2) × 1 = 2 for Na2 SO4 n-factor = Oxidation state of Na × number of Na-atoms in 1 molecule of Na2 SO4 = (+1) × 2 = 2 TITRATIONS The process of determination of concentration of a solution with the help of a solution of known concentration (standard solution) is called titration. Titration is divided into following four categories. i) Simple Titration ii) Double Titration iii) Back Titration iv) Iodimetric and Iodometric Titration Simple Titration: A known volume of the solution of unknown concentration is taken in a flask and required reagents are added to it. The solution of known concentration is added from the burette in the solution of unknown concentration till the latter reacts completely. This process is called titration. At the end point (equivalence point) the equivalents or milliequivalents of the two reacting substances are equal. Volume of solution (A) = VA litres Normality of solution (A) = NA Equivalents of substance (A) = NAVA Similarly, equivalents of substance (B) = NBVB At the equivalence point (end point) the equivalents (not the moles) of the two substance are equal. NAVA (litre) = NB  VB (litre) Illustration 19: 40 ml N/10 HCl and 60 ml N/20 KOH are mixed together. Calculate the normality of the acid or base left. The normality of the salt formed in the solution will be (A) 0.06 (B) 0.09 (C) 1 (D) 0.03
Jr Chemistry E/M 126 NISHITH Multimedia India (Pvt.) Ltd., JR/SR SUBJECT- JEE - NIT - MAINS - VOL - No. STIOCHIOMETRY Solution: (D) Milli equivalents of HCl = N × V (ml) = 1 40 4 10   Milli equivalents of KOH = N × V (ml) = 1 60 3 20   One milli equivalent of an acid neutralizes one milli equivalent of a base Milli equivalent of HCl left = 4 – 3 = 1 Total volume of the solution = 40 + 60 = 100 ml Milli equivalents of HCl = N × V (ml) 1 = N × 100 Normality (N) of HCl left in solution = 0.01 Salt formed = Milli equivalent of acid or base neutralized Milli equivalents of the salt formed = N × V (ml) 3 = N × 100 Normality (N) of salt formed = 0.03 Double Titration: If an aqueous solution contains a mixture of any two of the three NaOH, NaHCO3 and Na2 CO3 and it has to be titrated against an acid HCl or H2 SO4 , it will require two indicators to determine the strength of the bases present. The two indicators used are phenophthalein and methyl orange. Method: A given volume of the aqueous solution of the bases is taken and phenophthalein indicator is added to it. This solution is titrated with an acid of known normality to the end point the volume of the acid used is noted. This end point is called first end point. Now in the same solution methyl orange is added and gain titrated with an acid of known strength to the end point. It is called second end point. The volume of acid, used in the second end point is also noted. The normality of the bases present is then calculated. Principle: In the presence of phenolphthalein indicator i) NaOH is completely neutralized by the acid. ii) Only half of the milliequivalents of Na2 CO3 present are titrated as the phenolphthalein will show the colour change when only NaHCO3 (weak base) and neutral species are left in the reaction mixture. The following reactions take place, NaOH + HCl  NaCl + H2O Na2 CO3 + HCl  weak base NaHCO3 + NaCl Since phenolphthalein is a weak organic acid, and it changes its colour in weakly basic medium (NaHCO3 ), so as soon as the Na2 CO3 is converted to NaHCO3 phenophthalein shows the colour change indicating the completion of the reaction. In the presence of methyl orange, all the basic substances left in the mixture will be neutralized by acid and methyl orange will show the colour change when the medium is weakly acidic (H2O + CO2 i.e. H2 CO3 ). Titration of the solution containing both NaOH and Na2 CO3 : A given volume of the aqueous solution is titrated with an acid of known normality using phenophthalein indicator. Suppose ‘a’ milli equivalents of acid are used in the first end point then, milli equivalent of NaOH + 1⁄2 milli equivalent of Na2 CO3 = milli equivalent of acid = a ...(1) Now in the same already titrated solution methyl orange indicator is added and again titrated to the end point. Suppose ‘b’ milli equivalents of the acid are used at the second end point. 1⁄2 milli equivalents of Na2 CO3 = milli equivalents of acid = b ...(2) From equation (1) and (2) Milli equivalents of acid used by Na2 CO3 = 2b milli equivalents of Na2 CO3 Milli equivalents of acid used by NaOH = a – b = milli equivalent of NaOH Knowing the milli equivalents of Na2 CO3 or NaOH and the volume of the solution titrated, their normality can be calculated.