Redox Reaction and Equivalent Concept Solved Examples

REDOX REACTION AND EQUIVALENT CONCEPT SOLVED EXAMPLES Ex. 1 Sol. The weight of sodium bromate required to prepare 55.5 mL of 0.672 N solution for cell reaction, BrO3– + 6H+ + 6e–  Br– + 3H2O, is (A) 1.56 g (B) 0.9386 g (C) 1.23 g (D) 1.32 g Meq. of NaBrO3 = 55.5 × 0.672 = 37.296 Let weight of NaBrO3 = W  W × 6 × 1000 = 37.296 (equivalent weight = M/6) of n-factor = 6 M NaBrO3 M × 6 × 1000 = 37.296 151  W = 0.9386 g Hence, (B) is the correct answer.  Ex. 2 NaIO3 reacts with NaHSO3 according to equation IO3– + 3HSO3–  I – + 3H+ + 3SO42– The weight of NaHSO3 required to react with 100 mL of solution containing 0.68 g of NaIO3 is (A) 5.2 g (B) 0.2143 g (C) 2.3 g (D) none of the above Sol. Meq. of NaHSO3 = Meq. of NaIO3 = N × V =  WNaHSO3 M NaHSO3 × 2 × 1000 = 0.68 × 6 × 1000 (I 5+ + 6e–  I –) 198 0.68 × 6 × 100 198 0.68 × 6 × 100 × 104 = 0.2143 198 × 1000 Hence (B) is the correct answer. WNaHSO 3 = Ex. 3 Sol. If 0.5 moles of BaCl2 is mixed with 0.1 moles of Na3PO4, the maximum amount of Ba3(PO4)2 that can be formed is (A) 0.7 mol (B) 0.5 mol (C) 0.2 mol (D) 0.05 mol Let us first solve this problem by writing the complete balanced reaction. 3BaCl2 + 2Na3PO4  Ba3(PO4)2  + 6NaCl 3 times the moles of Na 3PO 4. Therefore, to react with 2 3 0.1 mol of Na3PO4, the moles of BaCl2 required would be 0.1 × = 0.15. Since BaCl2 is 0.5 mol, we can conclude that 2 3 Na3PO4 is the limiting reagent. Therefore, moles of Ba3(PO4)2 formed is 0.1 × = 0.05 mol. 2 Hence, (D) is the correct answer. We can see that the moles of BaCl 2 used are Ex. 4 A 0.1097 g sample of As2O3 required 36.10 mL of KMnO4 solution for its titration. The molarity of KMnO4 solution is. (A) 0.02 Sol. (B) 0.04 n-factor = 5 As2O3 + MnO4– 2AsO4 3– + Mn2+ n-factor = 4 (C) 0.0122 (D) 0.3 Let, molarity of KMnO4 solution be M  Eq. of As2O3 = Eq. of KMnO4 solution 0.1097 36.10 × M × 5 198 ×4= (Equivalent weight As2O3 = ) 198 1000 4 Molarity = 0.0122 M Hence, (C) is the correct answer. Ex. 5 In basic medium, CrO42– oxidize S2O32– to form SO 42– and itself changes to Cr(OH)4– . How many mL of 0.154 M CrO42– are required to react with 40 mL of 0.246 M S2O32– ? (A) 200 mL Sol. (B) 156.4 mL (C) 170.4 mL 2– 3 (D) 190.4 mL 2– 4 40 × 0.246 × 8 = V × 0.154 × 3 (Meq. of S2O = Meq. of CrO )  V = 170.4 mL Hence, (C) is the correct answer. Ex. 6 10 mL of 0.4 M Al2(SO4)3 is mixed with 20 mL of 0.6 M BaCl2. Concentration of Al3+ ion in the solution will be. (A) 0.266 M (B) 10.3 M Sol. Al2(SO4)3 Initial Meq. (C) 0.1 M + 10 × 0.4 × 6 BaCl2  BaSO4  + AlCl3 20 × 0.6 × 2 0 = 24 = 24 0 0 Final Meq. (D) 0.25 M 24 0 24 24 = 0.266 M 30 × 3 Hence (A) is the correct answer. [Al3+] = Ex. 7 0.52 g of a dibasic acid required 100 mL of 0.2 N NaOH for complete neutralization. The equivalent weight of acid is (A) 26 Sol. (B) 52 (C) 104 (D) 156 Meq. of Acid = Meq. of NaOH 0.52 × 1000 = 100 × 0.2 E  E = 26 Hence (A) is the correct answer. Ex. 8 34 g hydrogen peroxide is present in 1120 mL of Solution : This solution is called (A) 10 volume Sol. Wt. of H2O2 in 1 mL = (B) 20 volume (C) 30 volume 34 g 1120 1 O2 2 34 g of H2O2 gives 11200 mL of O2 at STP H 2O 2  H 2O +  34 11200 34 g of H2O2  × = 10 mL of O2 at STP.. 1120 34 1120 Hence, (A) is the correct answer. (D) 32 volume REDOX REACTION AND EQUIVALENT CONCEPT Ex. 9 The number of moles of KMnO4 that will be required to react with 2 mol of ferrous oxalate is 6 5 7+ Mn + 5 e–  Mn2+ ] × 3 (A) Sol. (B) 2 5 (C) 4 5 (D) 1    Fe2+  Fe3+ + e– C2O42–  2CO2 + 2e– 3 moles of KMnO4 = 5 moles of FeC2O4 6 mole of KMnO4 5 Hence, (A) is the correct answer.  2 mol of ferrous oxalate  Ex. 10 The minimum quantity of H2S needed to precipitate 63.5 g of Cu2+ will be nearly. (A) 63.5 g Sol. (B) 31.75 g (C) 34 g (D) 2.0 g 2+ Meq. of H2S = Meq. of Cu  WH 2S 34 / 2 × 1000 = 63.5 × 1000 63.5 / 2  WH 2 S  34 g Hence, (C) is the correct answer. Ex. 11 Which of the following is / are correct? (A) g mole wt. = mol. wt. in g = wt. of 6.02 × 1023 molecules (B) mole = N molecule = 6.02 × 1023 molecules (C) mole = g molecules (D) none of the above Sol. (A), (B) and (C) Ex. 12 8 g of O2 has the same number of molecules as (A) 7 g of CO Sol. (B) 14 g of CO (C) 28 g of CO (D) 11 g of CO2 (A) and (D) Ex. 13 What volume of 0.1 M KMnO4 is needed to oxidize 100 mg of FeC2O4 in acid solution ? (A) 4.1 mL Sol. (B) 8.2 mL (C) 10.2 mL Meq. of KMnO4 = Meq. of FeC2O4 n-factor = 2 2×3 MnO4– + Fe C2 O4 + H+ n-factor = 5 n-factor = 1  V = 4.1 mL Hence, (A) is the correct answer. 2×4 Mn2++ Fe3++ 2 C O2 + H2O (D) 4.6 mL Ex. 14 What volume of 6 M HNO3 is needed to oxidize 8 g of Fe2+ to Fe3+, HNO3 gets converted to NO? Sol. (A) 8 mL (B) 7.936 mL Meq. of HNO3 = Meq. of Fe 2+ or 6 × 3 × V = (C) 32 mL (D) 64 mL 8 × 1000 56 V = 7.936 mL n-factor = 3 +5 (NO3 2+ – NO) Hence, (B) is the correct answer. Ex. 15 0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N KOH. The percentage of free SO3 in the sample is (A) 30.6% Sol. (B) 40.6% (C) 20.6% (D) 50% Meq. of H2SO4 + Meq.of SO3 = Meq. of KOH  (0.5 – x) x × 1000 + × 1000 = 26.7 × 0.4 98 / 2 80 / 2  x = 0.103 0.103 × 100 = 20.6% 0.5 Hence, (C) is the correct answer.  % of SO3 = Ex. 16 The eq. wt. of a substance is the weight which either combines or displaces. (A) 8 part of O Sol. (B) 2 part of H (C) 35.5 part of Cl (D) none of the above (A) and (C) Ex. 17 'A' g of a metal displaces V mL of H2 at NTP. Eq. wt of metal E is / are : (A) E = A × Eq. wt. of H Wt. of H 2 displaced A ×1.008 Volume of H 2 displaced × 0.0000897 (A), (B) and (C) (C) E = Sol. (B) E = A ×1.008 × 22400 × Eq. wt. of H Volume of H 2 displaced × 2 (D) none of the above Ex. 18 Which of the following is/are redox reaction (s)? Sol. (A) BaO2 + H2SO4  BaSO4 + H2O2 (B) 2BaO + O2  2BaO2 (C) 2KClO3  2KCl + 3O2 (D) SO2 + 2H2S  2H2O + 3S (B), (C) and (D) Ex. 19 Which of the following statements is/are true if 1 mol of H3POx is completely neutralized by 40 g of NaOH? Sol. (A) x = 2 and acid is monobasic (B) x = 3 and acid is dibasic (C) x = 4 and acid is tribasic (D) x = 2 and acid does not form acid salt (A) and (D). REDOX REACTION AND EQUIVALENT CONCEPT Ex. 20 In the reaction, 3Br2 + 6CO32– + 3H2O  5Br– + BrO3– + 6HCO3– Sol. (A) bromine is oxidized and carbonate is reduced (B) bromine is oxidized (C) bromine is reduced (D) it is disproportionation reaction or autoredox change (B), (C) and (D) Ex. 21 A 20 g sample of only CuS and Cu2S was treated with 100 mL of 1.25 M K2Cr2O7. The products obtained were Cr3+, Cu2+ and SO2. The excess oxidant was reacted with 50 mL of Fe2+ Solution : 25 mL of the same Fe2+ solution required 0.875 M KMnO4 under acidic condition, the volume of KMnO4 used was 20 mL. Find the % of CuS and Cu2S in the sample. Sol. 1.25× 6 ×100 = 0.75 1000 0.875 × 5 × 5 × 20 Equivalents of Fe2+ in 25 mL = = 0.0875 1000 Equivalents of Fe2+ in 50 mL = 0.0875 × 2 = 0.175 Equivalents of dichromate initially = Equivalents of excess dichromate = 0.175  Equivalents of dichromate consumed by (CuS and Cu2S) = 0.75 – 0.175 = 0.575 If x g is the mass of CuS, the mass of Cu2S is (10 – x) g x (10 – x) × 6+ × 8 = 0.575 159 95.5  x = 5.74 g % CuS = 5.74 × 100 = 28.7 % 20 % Cu2S = 71.3 % Ex. 22 2.249 g of a sample of pure BaCO3 and impure CaCO3 containing some CaO was treated with dil. HCl and it evolved 168 mL of CO2 at N.T.P. From this solution, BaCrO4 was precipitated filtered and washed. The precipitate was dissolved in dil. H2SO4 and diluted to 100 mL. 10 mL of this solution when treated with KI solution liberated iodine which required exactly 20 mL of 0.05 N Na2S2O3. Calculate the percentage of CaO in the sample. Sol. Let, weight of BaCO3, CaCO3 and CaO are x,y and z respectively.  x + y + z = 2.249 BaCO3  BaCrO4 Redox change Cr6+ + 3e–  Cr3+ 2I–  I2 + 2e– Meq. of BaCO3 = Meq. of BaCrO4 = Meq. of I2  x 100 × 1000 = 20 × 0.05 × 10 197 / 3  x = 0.657 g ....(1) The equivalent weight of BaCrO4 is M/3, therefore for BaCO3, it should be M/3 also because mole ratio of BaCO3 and BaCrO4 is 1 : 1. Applying POAC for C atom, Moles of C in BaCO3 + Moles of C in CaCO3 = Moles of C in CO2  x y 168 + = 22400 197 100  200 x + 294 y = 295.5 ....(2) From equation (1) and (2) y = 0.416 g  0.657 + 0.416 + z = 2.249 z = 1.176 % of CaO = 1.176 = 52.29 % 2.249 Ex. 23 For estimating ozone in the air, a certain volume of air is passed through an alkaline KI solution when O 2 is evolved and iodide is oxidized to iodine. When such a solution is acidified, free iodine is evolved which can be titrated with standard Na2S2O3 Solution : In an experiment, 10 L of air at 1 atm and 27°C were passed through an alkaline KI solution, and at the end, the iodine was entrapped in a solution which on titration as above required 1.5 mL of 0.02 N Na2S2O3 solution. Calculate volume percentage of ozone in the sample. Sol. The chemical reaction is, H2O + KI + O3  I2 + O2 + KOH Milliequivalents of iodine = Milliequivalents of KI = Milliequivalents of O3 reacted Milliequivalentsof Na2S2O3 = 1.5 × 0.02 = 3 × 10–2 3 × 1 0 –2 = 1.5 × 10–2 2 Millimoles of ozone = 1.5 × 10–2 Millimoles of iodine = [n-factor for iodine = 2] nRT 1.5 × 10 –5 × 0.0821 × 300 Volume of ozone = = = 36.945 × 10–5 litre P 1 36.945 × 10 –5 Volume percent of ozone = × 100 = 3.6945 × 10–3 10 Ex. 24 What volume of 0.40 M H2SO4 is required to produce 34.0 g of H2S by the reaction, 8 KI + 5 H2SO4  4 K2SO4 + 4 I2 + H2S + 4 H2O ? Sol. 1 mole of H2S  5 moles of H2SO4 34 g of H2S  5 moles of H2SO4 0.40 × VH 2SO 4 = 5  VH 2 SO 4 = 5 = 12.5 litre 0.40 Ex. 25 To 50 L of 0.2 N NaOH, 2.5 L of 2N HCl and 15 L of 0.1 N FeCl3 solutions are added. What weight of Fe2O3 can be obtained from the precipitate? Also report the normality of NaOH left in resultant solution : Sol. Eq. of NaOH = 50 × 0.2 = 10 Eq. of HCl = 2.5 × 2 = 5 Eq. of NaOH left after reaction with HCl = 10 – 5 = 5  FeCl3 + NaOH  Fe (OH)3   Fe2O3 FeCl3 reacts with NaOH to give Fe(OH)3 which on ignition gives Fe2O3  Eq. of NaOH used for FeCl3 = Eq. of Fe(OH)3 = Eq. of Fe2O3 = 15 × 0.1 = 3.5 REDOX REACTION AND EQUIVALENT CONCEPT  Eq. of NaOH left finally = 5 – 1.5 = 3.5 Normality of NaOH in the resultant solution = WFe 2O3 M Fe 2O3 × 6 = 1.5 WFe 2 O 3 = 3.5 = 0.05 70 (n-factor for Fe2O3 = 6) 1.5 × 160 = 40g 6 Ex. 26 One litre of a mixture of O2 and O3 at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 30 mL of M/10 Na2S2O3 solution for titration. What is the weight percent of ozone in the mixture ? Ultraviolet radiation of wavelength 300 mm can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture? Sol. O3 + 2 KI + H2O  2KOH + I2 + O2 2Na2S2O3 + I2  Na2S4O6 + 2Nal Millimoles of O3 = Millimoles of I2 = ½ × Millimoles of Na2S2O3 = ½ × 30 × 1/10 = 1.5 millimoles = 0.0015 moles Total moles of O3 and O2 in the mixture PV = n RT 1 × 1 = n × 0.0821 × 273  n = 0.044 Moles of O2 = 0.044 – 0.0015 = 0.0425 WO 2 = 0.0425 × 32 g = 1.36 g WO 3 = 0.0015 × 48 g = 0.072 g  % of O3 = 0.072 × 100 = 5.028 % 1.432 No. of photons required for decomposition of Ozone molecules = 0.0015 × 6.023 × 10 23 = 0.90 × 1021 Ex. 27 A 2.18 g sample contains a mixture of XO and X2O3. It reacts with 0.015 moles of K2Cr2O7 to oxidize the sample completely to form XO4– and Cr3+. If 0.0187 mole of XO4– is formed, what is the atomic mass of X ? Sol. XO + K2Cr2O7  Cr3+ + XO4– X2O3 + K2Cr2O7  Cr3+ + XO4– Let, wt. of XO in the mixture be x g Equivalent of K2Cr2O7 consumed by the mixture= 0.015 × 6 Equivalents of XO = x ×5 x + 16 Equivalents of X2O3 =  2.18 – x 8 2x  48 2.18 – x x × 8 = 0.015 × 6 ×5 + 2x – 148 x + 16 Since 1 mole of XO gives 1 mole XO4– and 1 mole of X2O3 gives 2 moles of XO4–,  x 2x(2.18 – x) + = 0.0187 x + 16 2x + 48 On solving, x = 99 Ex. 28 An aqueous solution containing 0.10 g KIO3 (formula wt. = 214.0) was treated with an excess of KI Solution : The solution was acidified with HCl. The liberated I2 consumed 55 mL of thiosulphate solution to decolourize the blue starch-iodine complex. Calculate molarity of the sodium thiosulphate solution : Sol. Moles of KIO3 = 0.1 = 0.00047 214  Moles of I2 liberated from KIO3 = 0.00047 = 0.000235 2 Moles of KI reacting = 0.00047 × 5 = 0.00235 Moles of I2 produced from KI = ( n-factor for KIO3 and KI are 5 and 1 respectively) 0.00235 = 0.001175 2 Total moles of I2 produced and reacted = 0.000235 + 0.001175 = 0.00141 equivalents of I2 reacted = 0.00141 × 2 = 0.00282 = equivalents of thiosulphate Solution : Molarity = 0.00282 × 1000 =0.0512 M (For thiosulphate n-factor = 1) 55 Ex. 29 Mg can reduce NO3– to NH3 in basic medium. NO3– + Mg (s) + H2O  Mg (OH)2 (s) + OH– (aq.) + NH3(g) A 25.0 mL sample of NO3– solution was treated with Mg. The NH3 (g) was passed into 100 mL of 0.15 N HCl. The excess of HCl required 32.10 mL of 0.10 N NaOH for neutralization. What was the molarity of NO3– ions in the original sample ? Sol. Meq. of NH3 formed = Meq. of HCl used = 100 × 0.15 – 32.10 × 0.10 = 11.79 Here, n-factor of NH3 is 1 (acid - base reaction) For redox change, 5  3 8 e NO 3   NH 3 (n - factor = 8)  Meq. of NH3 for n-factor 8 = 8 × 11.79  Normality of NO3– = Molarity of NO3– = 94.32 = 3.77 25 3.77 = 0.47125 8 REDOX REACTION AND EQUIVALENT CONCEPT Ex. 30 Chile salt peter a source of NaNO3 also contains NaIO3. The NaIO3 can be used as a source of iodine produced in the following reactions : Step 1 : IO3– + 3HSO3–  I – + 3H+ + 3SO42– Step 2 : 5I – + IO3– + 6H+  3I 2 + 3H4O One litre of chile salt peter solution containing 6.80 g NaIO3, is treated with stoichiometric quantity of NaHSO3. Now additional amount of same solution is added to the reaction mixture to bring about the second reaction. How many grams of NaHSO3 are required in step 1 and what additional volume of chile salt peter must be added in step 2 to bring in complete conversion of I– to I2 ? Sol. 6.8 × 6 × 1000 198 (For KIO3, n-factor = 6) Meq. of NaHSO3 = Meq. of NaIO3 = N × V = I5+ + 6e–  I –  Meq. of NaHSO3 = 206.06 WNaHSO3 M/2 × 1000 = 206.06 206.06 × 104 = 10.715 g 2 × 1000 Also, Meq. of I – formed using n-factor 6 = 206.06  WNaHSO 3 = In step 2, n-factor I – is 1 and n-factor for IO3– is 5 206.06 6 206.06 Meq. of NaIO3 used in 2nd step = 6 Meq. of I – formed using n-factor 1 =  6.8 206.06 × 5×V= 198 6  VNaIO = 199.99 mL 3 Ex. 31 1 g sample of AgNO3 is dissolved in 50 mL of water. It is titrated with 50 mL of KI Solution : The AgI precipitated is filtered off. Excess of KI in filtrate is titrated with M/10 KIO3 in presence of 6 M HCl till all I – converted into ICI. It requires 50 ml of M/10 KIO 3 Solution : 20 mL of the same stock solution of KI require 30 mL of M/10 KIO3 under similar conditions. Calculate % of AgNO3 in the sample. The reaction is : KIO3 + 2KI + 6HCl 3ICI + 3KCl + 3H2O. Sol. AgNO3 + KI AgI  + KNO3 As present in AgNO3 is removed as AgI by adding 50 mL of KI solution of which 20 mL requires 30 mL of M/10 KIO3 Meq. of KI in 20 mL = Meq. of KIO3 = 30 × 1 ×4 10 For KIO3, I5+ + 4e–  I+ I –  I1+ + 2e– Eq. wt. of KI =  Meq. of KI in 50 mL added to AgNO3 Meq. of KI left unused by AgNO3 = 50 × = 1 × 4 = 20 10 30 × 4 × 50 = 30 10 × 20 M 2  Meq. of KI used for AgNO3 = 30 – 20 = 10  Meq. of AgNO3 = 10 WAgNO3 170 × 2 × 1000 = 10  WAgNO 3 = 0.85 g = 85% Mole of AgNO 3 1 = Mole of KI 1  If equivalent weight of KI = M 2 Equivalent weight of AgNO3 = M 2 Ex. 32 A sample of Mg was burnt in air to give a mixture of MgO and Mg3N2. The ash was dissolved in 60 Meq. of HCl and the resulting solution was back titration with NaOH. 12 Meq. of NaOH were required to reach the end point. An excess of NaOH was then added and the solution was distilled. The ammonia released was then trapped in 10 Meq. of acid. Back titration of this solution required 6 Meq. of base. Calculate the % of Mg burnt to the nitride. Sol. Let, total moles of Mg used for MgO and Mg3N2 be a and b respectively 2Mg + O2  2MgO a 0 0 a 3Mg + N2  Mg3N2 b 0 0 b/3 b moles respectively of MgO and Mg3N2 are present in the mixture. 3 MgO + 2HCl  MgCl2 + H2O Now a, Mg3N2 + 8HCl  3MgCl2 + 2NH4Cl or solution contains (A) moles of MgCl2 from MgO and (B) moles of MgCl2 from Mg3N2 and Now, moles of HCl or Meq. of HCl (monobasic) = 30 – 12 = 48 8b = 48 ..........(1) 3 Moles of NH4Cl formed = Moles of NH3 liberated = Moles of HCl used for absorbing NH3. 2a + 2b = (10 – 6) = 4 3 8× 6 From (1), 2a + = 48 or a = 16 3  % of Mg used for Mg3N2 = 6 × 100 = 27.27% (6 +16) ..........(2) 2b moles of NH4Cl. 3 REDOX REACTION AND EQUIVALENT CONCEPT Ex. 33 30 mL of a solution containing 9.15 g/litre of an oxalate K xH y (C 2O 4) z .nH 2O are required for titrating 27 mL of 0.12 N NaOH and 36 mL of 0.12 N KMnO4 separately. Calculate x,y,z and n. Assume all H atoms (except H2O) are replaceable and x,y,z are in the simple ratio of g atoms. Sol. Let molecular weight of oxalate salt is M (i) n - factor in acid - base reaction = 2 (ii) n - factor in redox titration = 2 × z (C2O42– 2CO2 + 2e)  Meq. of acid in 30 mL = Meq. of NaOH used 9.15 × y = 27 × 0.12 M ...(1) 9.15 × (2z) = 36 × 0.12 M ...(2) 30 × Also, 30 × y 27 y 3 =  = 2z 36 z 2 Also, total cationic charge = total anionic charge From equations (1) and (2) ...(3)  x + y = 2z ...(4) By equations (3) and (4) x : y : z :: 1 : 3 : 2 These are in simplest ratio and molecular formula is KH3 (C2O4)2.nH2O Molecular weight of salt = 39 + 3 + 176 + 18n = 218 + 18n From equation (1), M = 30 × 9.15 × 3 = 254.16 27 × 0.12  218 + 18n = 254.15 n=2  Oxalate salt is KH3(C2O4)2.2H2O Ex. 34 Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO4 (20 mL) acidified with dilute H2SO4. The same volume of the KMnO4 solution is just decolourized by 10 mL of MnSO4 in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO2. The brown precipitate is dissolved in 20 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H2SO4. Write the balanced equations involved in the reactions and calculate the volume strength of H2O2. Sol. Meq. of MnO2 = Meq. of Na2C2O4 = 20 × 0.2 × 2 = 8 (C2O42–  2CO2, n-factor = 2) Millimoles of MnO2 = 8 4 2 n-factor = 3 – +2 MnO4 + Mn H 2O MnO2 brown (ppt.) n-factor = 2 or 2MnO4– + 3Mn2+  5MnO2 (Mole ratio is reciprocal of n-factor ratio) – 4 Millimoles of MnO 2 = Millimoles of MnO 2 5 2 × Millimoles of MnO2 5 2 8 = × 4= 5 5 2KMnO4 + 5H2O2 + 3H2SO4  2 MnSO4 + K2SO4 + 8H2O + 5O2 Millimoles MnO4– = Millimoles of H 2O 2 5 = – Millimoles of MnO 4 2  Millimoles of H2O2 = 5 8 × =4 2 5 N H 2 O 2 × 20 = 4 × 2 (n-factor for H2O2 = 2)  N H 2 O 2 = 0.4 Volume strength of H2O2 = 5.6 × N H 2 O 2 = 5.6 × 0.4 = 2.24 Ex. 35 1.0 g of moist sample of mixture of potassium chlorate (KClO3) and potassium chloride (KCl) was dissolved in water and solution made upto 250 mL. This solution was treated with SO2 to reduce all ClO3– to Cl– and excess of SO2 was removed by boiling. The total chloride was precipitated as silver chloride. The weight of precipitate was found to be 1.435 g. In another experiment, 25 mL of the original solution was heated with 30 ml 0.2 N FeSO4 and unused FeSO4 required 37.5 mL of 0.08 N KMnO4 solutions. Calculate the molar ratio of the ClO3– to the given mixture. Given that, ClO3– + 6Fe2+ + 6H+  Cl– + 6Fe3+ + 3H2O 3SO2 + ClO3– + 3H2O  Cl– + 3SO42– + 6H+ Sol. ClO3– is reduced to Cl– by SO2 and ClO3– is also reduced to Cl– by Fe2+, hence AgCl is formed due to total Cl– Meq. of Fe2+ initially taken = 30 × 0.2 = 6 Meq. of Fe2+ unused = 37.5 × 0.08 = 3  Meq. of Fe2+ = 6.0 – 3.0 = 3.0 Thus, Meq. of ClO3– in 25 mL = 3.0 3.0 = 0.0005 1000  6 +5 –1 ClO3–  Cl– Moles of ClO3– in 25 mL = O.N. 5 (n-factor = 6) –1 Thus, moles of ClO3– in 25 mL solution = 0.0005 ClO3– is also reduced to Cl– by SO2 in first experiment and precipitated as AgCl. Thus, Cl– formed from ClO3– = AgCl from ClO3– = 1.435 g Total AgCl formed both from actual and Cl– from ClO3– = 1.435 g 1.435 = 0.01 mol 143.5 Thus, AgCl formed due to Cl– only = 0.01 – 0.0005 = 0.0095 mol Thus, ClO3– and Cl– are in molar ratio = 1 : 19 =