Equivalent Concept and Volumentric Analysis

EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS 1. INTRODUCTION (a) All chemical combinations are followed by the law of equivalence. (b) As per the law of equivalence, for the two reacting substances A and B, the number of equivalents of A reacted = the number of equivalents of B reacted. (c) Let us consider x factor for A in the reaction is xA and x factor for B = xB. Let the amounts A and B reacted are wA and wB respectively. Let the formula wt. of A and B are MA and MB respectively. wA MA xA  the no. equivalents of A reacted =  the no. of equivalents of B reacted =  w A xA MA = wB MB xB w B xB MB (d) for a couple of reaction aA + bB  dD + eE x factor = x1 dD + eE  gG + hH x factor = x2 If x1  x2 , the law of equivalence is followed, but the entire calculation by equivalent concept becomes complicated as equivalent wt. of D in first reaction and second reaction are not same. (e) To, solve the problems of stoichiometry based on coupling reactions, by equivalent concept, individual step of the reactions are to be studied carefully. (f) If balanced chemical equations for the reactions given or if balancing of the chemical equations can be done easily, it is advisable to follow the mole concept rather than equivalent concept to eliminate mistakes. (g) For acid-base reactions or for single step reacti on, one can solve the probl em by calculating the equivalent wts of the reactants and products involved, with the aid of x factors. 2. DEFINITION OF EQUIVALENT WEIGHT (a) The equivalent weight of an element is that weight of the element that will combine with or replace directly or indirectly 1.0 gm (1 gm- 1 atom) of H, 35.5 gm (1 gm atom) of Cl or 8.0 gm. (½ gm-atom) of O or 108 gm (1 gm-atom) of Ag. (b) In the reaction Mg + Cl2  MgCl 2 1 atom of Mg loses 2 electrons to become Mg2+ ion. If we start with 1 gm-atom or 1 mole or 24 gm of Mg, we have NA (6.023 × 10 23) number of Mg atoms which would lose 2NA number of electrons and form NA number of Mg2+ ions. Therefore, we get 2NA number of electrons from 24 gm of Mg. So, NA number of 24 electrons can be obtained from = 12 gm of 2 Mg. Thus the equivalent weight of Mg = 12. (c) So we can also define equivalent weight of an element in otherway. Thus equivalent weight of an element is that weight of the element which loses or gained Avogadro number (NA) of electrons. (d) From the concept of the determination of oxidation state of the elements, you can easily calculate the number of electron transferred in a particular process. So, you can also calculate the amount of reactant which transferred NA number of electrons. Thereby, you can find the equivalent weight of the element. (e) For example, in the reaction 0 1 4 Na + O2 = 2 N a2 O O.S. of Na in pure state = 0 O.S. of Na in Na2O = +1  change in O.S. per gm-atom of Na = 1  no.of elect. released per g-atom of Na = NA Hence, equi. wt. of Na = atomic wt of Na = 23 (f) For example, in the reaction 0 4 S + O2 = S O 2 O.S. of S in pure state = 0 O.S. of S in SO2 = + 4 change in O.S. per gm-atom of S = 4 no.of elect. released per gm-atom of S =   4NA Hence equivalent weight of S in the given reaction atomic wt of S 32 = = 8 4 4 EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS (g) Let us find the equivalent wt.of Na and S as per definition. As per the equation given in (e),32 gm Oxygen combines with 4 × 23 gm of Na    4  23 8 gm Oxygen combines with ×8 gm of Na 32 = 23 gm of Na the equivalent weight of Na = 23 As per the equation given in (f) 32 gm sulphur combines with 2 ×16 gm of oxygen 2 × 16 gm of oxygen combines with 32 gm of S. 8 gm of oxygen combines with 32 × 8 gm of S 2  16 = 8 gm of S  the equivalent weight of S = 8 So, you must realise the importance of change in O.S. while calculating the equivalent weight. (h) Earlier we have discussed about "x" factor, whi ch can be ap plied confi dentl y wh il e calculating the equival ent weight of the substances in the redox reactions. Equivalent weight of element = Atomic wt of the element x factor Equivalent weight of compound = Formula wt of the compound x factor (i) Let us apply the formula mentioned above. For example, in the reaction 7 2 K MnO 4  M n the value of x in the given equation = 7 – 2 = 5  equiv. wt. of KMnO4 = 158 mol. wt. of KMnO 4 = = 31.6 5 5 For example, in the reaction 7 4 K M nO 4  Mn O2  mol. wt. of KMnO 4 3 = 2.1 The gram-equivalent weight or gram equivalent or simply equivalent and Determination of equivalent weights for various substances (a) Equivalent weight when expressed in grams the amount we get is known as gram-equivalent weight. (b) For the sake of simplicity, we introduce a term gram-equivalent weight, gram-equivalent or simply 1 equivalent which means equivalent of given substance weight in grams. Gramequivalent is also abbreviated as gm-equiv. (c) For example, 1 equivalent of chlorine = 35.5 gm 1 equivalent of oxygen = 8 gm  71 gm chlorine = 2 equivalent of chlorine  32 gm oxygen = 4 equivalent of oxygen. (d) The no. of equiv. of any substance = wt. of the subs tan ce in grams. Equiv. wt. of the subs tan ce in grams w [where w and E represent the weight E and equiv. wt. of the substance] (e) Equiv. wt. of any oxidant and reductant is = Equiv. wt. of oxidant = Mol. wt. of oxidant xfactor Mol. wt. of reduc tant xfactor Do, remember the given formula not at all applicable in disproportionation reaction. (f) Equiv. wt. of acid or base in a acid-base reaction (not the redox reaction) is Equiv. wt of reductant = Equiv. wt of acid = Mol. wt. of acid Basicity of the acid The equiv. wt. of acid = The equiv. wt of base = 158 . 3 Mol. wt. of acid x Mol. wt. of base acidity of base Mol. wt. of base x The basicity or x for acid in a acid base reaction is equal to the total number of replacable Hatoms present in the acid. Replacable H-atoms The equiv. wt. of base = x = 7 – 4 = 3. equiv. wt. of KMnO4 = Thus, it is evident that the equivalent weight of the substances may vary depending on the "x" factor. EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS are such type of H-atoms which are attached with the atoms of group VI or group VII elements, viz, O, S, Se, Te and F, Cl, Br, I. The acidity or x for base in a acid-base reaction is equal to the number of replacable OH groups present in the formula unit of base. Thus, acidity of Al (OH)3 = 3. The acidity of NH4OH = 1. 3. DETERMINATION OF EQUIVALENT WEIGHT OF ELEMENT, ACIDIC AND BASIC OXIDES AND SALT (a) The oxidation state of the elements in the pure state = 0. The equivalent wt of the element E = Atomic wt of the element x (b) The equivalent wt of any compound = sum of the equivalent wt of component elements or ions. By using this relation we can calculate the equivalent wt. of any acidic and basic oxide. It is already known to us equiv wt of O = 8 Equiv. wt. of the other element = Atomic wt of the element x or valency of that element Now you can also calculate the equiv. wt. of R2O3 = 46.34+8 = 54.34 Again you may calculate the equivalent wt. of R in otherway by applying the given concept as the valency of R = 3  56 + 8 = 26.5 3 For example, = equiv. wt of As2O3 = 75 + 8 = 33 3 (c) Let us illustrate another problem : The molecular wt. of R2O3 is 326. What is the equivalent wt of R ? Let. the at. wt. of R is Ar.  2 Ar + 48 = 326 or Ar = 326  48 = 139 2 E = = 139 gm of Ar = 46.34 gm of Ar 3 the equivalent wt of R = 46.34 Ans.  formula wt of Ca3 (PO 4 ) 2 310 = 6 32 = 51.6 Ans. (e) Equivalent wt of an ion formula wt (or At. wt.) of ion its valency 4. LAW OF EQUIVALENTS AND ITS APPLICATION IN THE DETERMINATION OF EQUIVALENT WEIGHTS OF ELEMENTS 4.1Law of Equivalents (a) It is in accord with the law of equivalent proportion which states, "Substances (elements or compounds) combine in the ratio of their gram equivalent weights or equivalents". (b) Let g1 and g2 be the weights of the two reacting substances and E1 and E2 their equiv wts. From the law of equivalent we can write, wt of the Ist substan ce in grams Equiv. wt of the Ist subs tance in grams = 2  139 8 gm O combines with × 8 gm of Ar 3  16 = formula wt of the salt total valence of designated ion For example, the equivalent wt E of Ca3(PO4)2  3 × 16 gm O combines with 2×139 gm of Ar  FG 326  48 IJ H 2 K 139 = 46.34 3 But this method is not convenient one to calculate the equivalent wt. of the metal in mixed oxides. There we are to use oxidation state instead of valency. (d) The equivalent weight of salt E, which is not behave as oxidizing or reducing agent is given by, = atomic wt of Fe + 8 3 1 3 = For example, Equiv. wt. of Fe2O3 = The equiv. wt of R = or wt of the IInd substance in grams Equiv. wt of the IInd subs tance in grams g2 g1 = E2 E1 or g1 E1 = g2 E2 EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS (c) In general we can write for the two reacting substances X and Y as, wt. of X Equiv wt. of X = wt. of Y Equiv wt. of Y 4.2 Determination of equiv. wt. of metals by hydrogen displacement method (a) The equiv wt of metals viz, K, Na, Ca, Fe, Zn, Mg, Al and Sn, which can displace hydrogen from any acid or from H containing compound, can only be determined by this method. (b) For example, w1 gm. of any metal displaces w2 gm of H from an dilute acid. The, w 1 Equiv. wt. of metal [the equiv. wt. of H = 1] w2 = 1  w1 Equiv. wt. of metal = w 2 (c) 0.224 gm of a metal when dissolved in dilute acid liberates 285 c.c. of hydrogen, measured at 17 ºC and 78 cm of mercury pressure. Find the equiv. wt. of the metal. w1 = 0.224 wt. of hydrogen, w2 = 78 PVM 0.285  2 = × 76 0.082  290 RT = 0.0246 gm.  Equiv wt of metal = w1 0.224 = = 9.105 w2 0.0246 4.3 Determination of equivalent wt. of element from the oxide formation method (a) From the formation of oxide the equiv. wt of the element can be easily determined. (b) Let w1 gm of element X, combines with w2 gm of O.  w2 gm. O combines with w1 gm of X.  w1 8 gm. O combines with w × 8 gm of X 2 [Equiv. wt. of O = 8]  equiv. wt of X = (c) From 3.36 gm is obtained. What In iron oxide, the In iron oxide, the = 1.44 gm. w1 × 8. w2 of Fe, 4.80 gm of iron oxide is the equivalent wt of Fe ? amount of Fe = 3.36 gm amount of O = 4.8 – 3.36 1  1.44 gm oxygen combines with 3.36 gm of Fe 3.36  8 gm oxygen combines with × 8 gm of 144 . Fe = 18.67 gm of Fe  equiv. wt of Fe in iron oxide = 18.67 (d) Some times the information about the composition of the oxides of the element is given indirectly. To find the equiv. wt of the element we are to proceed logically. (e) When 31.6 gm of pure copper oxide are completely reduced by dry hydrogen, 7.2 gm of water are produced. Calculate the equiv. wt of copper. wt of oxygen, w2 present in given copper oxide = 7.2 × 16 gm = 6.4 gm 18  wt of copper w1 = 31.6 – 6.4 = 25.2 gm Hence, equiv wt of copper w1 25.2 = w × 8 = × 8 = 31.5 6.4 2 4.4 Determination of equivalent wt. of the elements from their chloride formation (a) The equiv. wt of the element is that weight which combines with 35.5 gm of Cl. (b) Let, w1 gm of the element 'x' combines with w2 gm of Cl.  w2 gm Cl combines with w1 gm of X  w1 35.5 gm Cl combines with w × 35.5 gm of 2 X  w1 The equiv. wt. of X = w × 35.5 2 (c) Sometimes composition of the chlorides are given indirectly. (d) The equivalent wt. of a metal is 30. What volume of chlorine would be liberated at 27 ºC and 750 mm pressure by 0.6 gm of the metal ? (Aq. tension at 27 ºC = 26.6 mm) Here weight of metal = W1 = 0.6gm Here, wt of chlorine = w2 gm.  0.6 30 w 2 = 35.5 or w2 = 0.6  35.5 = 0.71 gm. 30 The volume of chlorine liberated = FG w IJ HMK 2 RT P EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS 0.082  300 (750  26.6) 0.71 = × 760 71 = 0.2584 lit. 4.5 Determination of equivalent wt. of metal from the displacement/substitution reactions (a) When one metal displaces another from a solution of one of its compounds, the equivalent weight of one metal displaces the equivalent weight of the other, so that of the other can be found. (b) For instance, if metallic zinc or iron be added to a solution of silver nitrate or copper sulphate, finely divided silver or copper is precipitated : so, we can write = wt. of zn / or / Fe wt of Ag Equiv. wt. of zn / or / Equiv. wt. of Fe = Equiv. wt of Ag (c) 0.13 gm of copper displaces 0.432 gm of silver from silver nitrate solution. 0.13 gm of aluminium displaces 0.47 gm of copper from copper sulphate solution. While 0.13 gm of hydrogen is displaced by 1.17 gm of aluminium calculate from this data the equivalent weight of silver. wt of cu Equiv. wt. of Cu = ............... (1) wt of Ag Equiv . wt. of Ag wt of Al Equiv. wt. of Al = ............... (2) wt of Cu Equiv. wt. of Cu wt of Al = wt of H Equiv. wt. of Al Equiv. wt. of H ............... (3) Now, introducing the informations in relation (3) wt of Al 117 . Equiv. wt. of Al = × 1 = = 9 wt of H 0.13 Now, introducing equiv. wt of Al = 9 in relation (2) we get Equiv. wt of Cu = = wt of Cu × Equiv. wt of Al wt of Al 0.47 × 9 = 32.54 0.13 1 Now putting the equiv. wt. of Cu = 32.54 in relation (1) wt of Ag Equiv. wt of Ag = × Equiv. wt. of Cu wt of Cu = 0.432 × 32.54 = 108.1 0.13 4.6 Determination of equivalent weight of metal by double decomposition method (a) Double decomposition method is based on the law of equivalent. As per this law, substances react in the proportion of their equivalent amount. This procedure is also applicable for nonmetallic reactants. (b) Let us consider a reaction as PmQn + ROSP  products Let, amount of Pm Qn reacted = w1 gm Let, amount of RoSp reacted = w2 gm  Equivalent wt of Pm Q n w1 = Equivalent wt of R o Sp w2 w1  w 2 = Equivalent wt of P  Equivalent wt of Q Equivalent wt of R  Equivalent wt of S If w1, w2 and Equiv wt of Q, R and S are known to us, we can calculate the equiv wt of P. (c) Chloride of metal 'M' contains 47.23% of the metal. 1 gm of this metal displaced from a compound 0.88 gm of another metal 'N'. Find the equivalent weight of 'M' and 'N' respectively. Let, the equiv wts of M and N are EM and EN respectively. 35.5  47.23 52.77  EM 47.23 = 35.5 52.77 or EM = 31.77  EM 1 EN = 0.88 or EN = 0.88 EM = 0.88 × 31.77 = 27.96 or EM = EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS SOLVED PROBLEMS Ex.1 In the reaction Br2 + Na2CO 3  NaBr + NaBrO3 + CO2 The equiv. wt. of NaBrO3 is (A) Mol. wt 1 (B) Mol. wt Mol. wt Mol. wt (C) (D) 10 5 4 Sol. Per mole of the formation NaBrO3 the x factor = + 5  equiv. wt. of NaBrO3 Mol. wt = 5 (Ans. C) Ex.2 The equiv. wt. of hypo in the reaction Na2 S2 O 3 + Cl2 + H2O  Na2SO 4 + H2SO4 + HCl is – (A) Mol. wt 2 Sol. Na 2S2 O 3 + Cl2 + H2O  (B) Mol. wt Mol. wt Mol. wt (C) (D) 4 1 8 2 6 62  5 equiv. of P4 31 = 10 equiv. of P4  62 gm P4 = The equiv. wt. of HNO3 = Mol. wt 63 = 1 1  the wt. of HNO3 required = 10 × 63 = 630 gm (Ans. B) Ex.6 The equiv. wt. of an element is 9. If it forms volatile chloride of vapour density 58.5. What is the approximate at wt. of the element ? 6 Na 2 S O 4 + H2 S O 4 + HCl  x factor for Na2S2O3 = 2|(2 – 6)| = 8  equiv. wt. of Na2S2O3 = Ex.5 What weight of HNO3 is needed to convert 62 gm of P4 in H3PO4 in the reaction ? P4 + HNO3  H3PO4 + NO2 + H2O (A) 63 gm (B) 630 gm (C) 315 gm (D) 126 gm 31  4 31 Sol. The equiv. wt. of P4 = = 54 5 Mol. wt (Ans.D) 8 Ex.3 In acting as a reducing agent, a piece of metal M weighing 16 grams gives up 2.25 × 1023 electrons, what is the equiv.alent weight of the metal (A) 42.83 (B) 21.33 (C) 83.32 (D) 32 Sol. NA no of electron will be removed by 6.023  10 23 × 16 gm of metal M 2.25  10 23 = 42.83 gm of metal M  equiv. wt. of metal is 42.83 (Ans.A) Ex.4 The equiv. wt. of the salt KHC2O4, H2C2O4, 4H2O (to be used as a reducing agent) is Mol. wt Mol. wt Mol. wt Mol. wt (B) (C) (D) 1 2 4 3 Sol. Per moles of this salt moles of C2O42– = 2  no of electrons transferred per moles of this salt is 2C2O42–  2 × 2e = 2CO2 × 2  'x' factor (as a reducing) of the given salt = 4 (A) Mol. wt  equiv. wt. of the given salt = 4 (Ans. D) (A) 9 (B) 18 (C) 27 (D) 54 Sol. Let, the molecular formula of the chloride is MClx and at wt. of the element is a  9x + x × 35.5 = 58.5 × 2 58.5  2 x = = 2.63 44.5 The nearest whole no. of 2.63 = 3  approximate at wt. of the element = 9 × 3 = 27 (Ans. C) Ex.7 6.90 gm of a metal carbonate were dissolved in 60 ml of 2(N) HCl. The excess acid was neutralized by 20 ml of 1(N) NaOH. What is the equiv.alent wt. of metal ? (A) 40 (B) 20 (C) 19 (D) 39 Sol. Equiv. of HCl taken = 60 × 2 × 10–3 Equiv. of HCl present after the reaction = 20 × 1 × 10–3  Equiv. of HCl utilized = 100 × 10–3  100 × 10–3 equiv. of metal carbonate = 6.90 gm  1 equiv. of metal carbonate 6.90 = = 69 gm 10 1  equiv. wt. of metal = 69 – 30 = 39 [becuase equiv. wt. of carbonate = 30] (Ans. D) Ex.8 10 ml of 75 ml of FG NIJ H 5K FG NIJ H 2K HCl, 30 ml of FG N IJ H 10 K HNO3 and HNO3 are mixed, the normality of H+ in the resulting solution is(A) 0.2 (B) 0.1 (C) 0.5 (D) 0.25 EQUIVALENT CONCEPT AND VOLUMENTRIC ANALYSIS Sol. The equiv. of H+ in 10 ml of = 10 × 10–3 2 The equiv. of H+ in 30 ml of = 30 × 10–3 10 The equiv. of H+ in 75 ml of = FG N IJ H 10 K FG NIJ H 5K FG NIJ H 2K HCl HNO3 HNO3 = 250 × 10–3 equiv.. 100  wt. of H2C2O4, 2H2O required = 250 × 10–3 × 63 gm = 0.1575 gm 100 (Ans. B) Ex.11 Which of the following has the highest Hence, total equiv. of H+ = (5 + 3 + 15) × 10–3 = 23 × 10–3 total volume of solution = 115 ml Hence, normality of H+ in the resulting mixture 23  10 3  10 3 = (N) 115 = 126 = 63 2 250 ml of seminormal solution 75 × 10–3 5 FG NIJ H 5K  equiv. wt. = = 0.2 (N) (Ans. A) normality ? (consider each of the acid is 100% ionised.) (A) 1 (M) H2SO4 (B) 1 (M) H3PO3 (C) 1 (M) H3PO4 (D) 1 (M) HNO3 Sol. The normality of 1(M) H2SO4 = 2(N) The normality of 1(M) H3PO3 = 2(N) The normality of 1(M) H3PO4 = 3(N) The normality of 1(M) HNO3 = 1(N) (Ans. C) Ex.9 500 ml of 0.2 (M) H2SO4 are mixed with 250 ml of 0.1 (M) Ba(OH)2, the normality of Ex.12 0.45 gm of an acid of mol. wt. 90 was resulting solution is - neutralised by 20 ml of 0.5 (N) caustic potash. (A) 0.2 (B) 0.1 (C) 0.5 (D) 0.25 Sol. 500 ml of 0.2 (M) H2SO4 = 500 × 0.2 × 10–3 moles of H2SO4 The basicity of acid is(A) 1 (B) 2 Sol. 20 ml of 0.5 (N) caustic potash = 20 × 0.5 × 10–3 equiv. of caustic potash = 0.2 equiv. of H2SO4  20 × 0.5 × 10–3 equiv. of acid = 0.45 gm 0.45  1 equiv. of acid = = 45 gm 10  10 3 0.1 (M) Ba(OH)2 = 250 × 0.1 × 10–3 moles of Ba (OH)2 = 250 × 0.1 × 10–3 × 2  x for acid = = 0.05 equiv. of Ba (OH)2 Excess H2SO4 = 0.2 – 0.05 = 0.15 equiv. Hence normality of H2SO4 in resulting solution 0.15 = × 1000 = 0.2 (N) (Ans. A) 750 Ex.10 What weight of H2C2O4, 2H2O (mol. wt. = 126) should be dissolved to prepare 250 ml of centinormal to be used as a reducing agent ? (A) (D) 4 = 0.5 × 0.2 × 2 equiv. of H2SO4 250 ml of  (C) 3 0.63 gm (C) 0.126 gm (B) 0.1575 gm (D) 0.875 gm Sol. Per mole of given acid C2O42– present = 1 mole  C 2O4 2–  CO2 + 2e  x factor for H2C2O4, 2H2O as a reducing agent = 2 90 = 2 45 Hence, basicity of acid = 2 (Ans. B) Ex.13 The equivalent wt. of a metal is double that of oxygen. How many times is weight of its oxide greater than the wt. of metal ? (A) 2 (B) 3 (C) 1.5 (D) 0.25 Sol. The equiv. wt. of the metal = 16 The equiv. wt. of the metal oxide = 16 + 8 = 24 24  oxide is 16 = 1.5 times greater than the wt. of metal (Ans. C)