Redox titrations [compatibility mode]

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Redox titrations [compatibility mode]

  1. 1. Redox titrations Pharma.analytical chemistry IIDr.Jehad M Diab, faculty of pharmacy Damascus University
  2. 2. REDOX titrationsA volumetric method of analysis whichrelies on oxidation or reduction of theanalyte using redox indicators orpotentiometry.This unit covers: changes in solution potential during atitrationbasic calculationsmethods of sample preparation andcommon titrantsExample applications Dr.Jehad diab
  3. 3. Dr.Jehad diab
  4. 4. Determination of equivalence point potential:Fe2+ Ce4+ = Fe3+ Ce3+E0Fe3+= 0.771 v , E0Ce4+=1.70 vEeq = E0Fe3+ - 0.0592/1 log [Fe2+] / [Fe3+] (1)Eeq = E0Ce4+ - 0.0592/1 log [Ce3+] / [Ce4+] (2)1+2:2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1 log [Fe2+][Ce3+] / [Fe3+] [Ce4+]At the equivalence point:[Fe3+] =[Ce3+ ][Fe2+]=[Ce4+ ]Rearrangement:2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1 log [Ce4+][Ce3+] / [Ce3+] [Ce4+]Eeq = (E0Ce4+ + E0Fe3+) /2= 1.70+0.771/2= 1.24 volt Dr.Jehad diab
  5. 5. draft2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1 log [Fe2/ [Fe3+]-0.0592/1log [Ce3+] / [Ce4+]2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1(log [Fe2]/ [Fe3+]+ log[Ce3+] / [Ce4+]2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1( (log [Fe2]-log [Fe3+]+log[Ce3+]-log[Ce4+]2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1(log [Fe2]+log[Ce3+]-log[Fe3+]-log[Ce4+]2Eeq = E0Ce4+ + E0Fe3+ - 0.0592/1 log ([Fe2+][Ce3+] / [Fe3+] [Ce4+]) Dr.Jehad diab
  6. 6. E eq =(n1e0ox+n2e0red) / n1+n2 Dr.Jehad diab
  7. 7. Dr.Jehad diab A more complex example Determine of Eeq for the following reaction: Mno4- +5 Fe2+ 8H+ = Mn2+ + 5Fe3++ 4H2O e eq= 0.77+0.059/1 × log [Fe3+] / [Fe2+] (1) e eq=1.51+0.059/5 × log [Mno4-][H+]8/[Mn2+] (2) × 5 1+2:6eeq= 0.77+5 × 1.51+0.059 × log [Fe3+][MnO4- ][H+]8/[Fe2+][Mn2+] [Fe2+]=5[MnO4-] , [Fe3+]=5[Mn2+] , [Fe3+]/[Fe2+]= [Mn2+] / [MnO4-]6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4-][Mn2+]
  8. 8. 6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4-][Mn2+]6eeq=0.77+5×1.51+0.059 log[H+]8eeq=(0.77+5×1.51)/6 + 0.059/6 log[H+]8eeq=(0.77+5×1.51)/6 - 0.079 pHIf [H+]=1M (pH= 0):eeq=(0.77+5×1.51)/6 + 0.059/6 log[1]8 =1.39 vSo, the potential at equivalence point is dependenton [H+]. If [H+]=1M we can calculate from thefollowing equation: eeq=(n1e0ox + n2e0Red ) / n1+n2 Dr.J.Diab
  9. 9. Dr.Jehad diabE,volt E,volt Volume of reagent,ml
  10. 10. Complete titration curve Dr.Jehad diab (or volume of titrant)
  11. 11.  100 ml 0.1 M Fe2+ with 0.1 M Ce4+ solution Ce4+ + Fe2+ Ce3++ Fe3+ Dr.Jehad diab
  12. 12. Dr.Jehad diab
  13. 13. Ce4+ + Fe2+ Ce3++ Fe3+ Dr.Jehad diab
  14. 14. Ce4+ (0.1 M) + Fe2+ (100 ml 0.1 M) Ce3++ Fe3+ Addition of 20 ml of Ce4+: 100 0.1  20  0.1 0.059 100  20 EFe  0.771 log  0.735v 1 20  0.1 100  20  Addition of 50 ml of Ce4+: 100  0.1  50  0.1 0.059 100  50 EFe  0.771 log  0.771v 1 50  0.1 100  50 Dr.Jehad diab
  15. 15. Dr.Jehad diab
  16. 16. Dr.Jehad diab
  17. 17. Fe2+ Ce4+ = Fe3+ Ce3+ , e0Fe3+/Fe2+= 0.771 v , e0Ce4+/Ce3+=1.70 v E eq =n1e0ox+n2e0red / n1+n2 Dr.Jehad diab
  18. 18. Dr.Jehad diab
  19. 19. Ce4+ + Fe2+ Ce3++ Fe3+ Dr.Jehad diab
  20. 20. Dr.Jehad diab
  21. 21. Ce4+ + Fe2+ Ce3++ Fe3+ 200 x0.05= 210 x[Ce3+] 10ml x0.1M=210ml x [Ce4+] Dr.Jehad diab
  22. 22. Ce4+ + Fe2+ Ce3++ Fe3+ E = e0 – 0.059 log [Ce3+]/ [Ce4+] (200 x0.05/ 210) 1.70 - (10ml x0.1M/210ml) 1.64 v voltOr E =1.70-0.059log 100%/10%=1.64 v 1.66 V Dr.Jehad diab
  23. 23. HomeworkYou are titrating 50 ml 0.1M of Co2+ solutionwuth0.1 M Ce4+ titrant.E0Co=0.85 v ,E0Ce=1.70.What is the potential for the titrationsystem after addition of:a. 0 ml of Ce4+b. 25 ml of Ce4+c. 50 ml of Ce4+d. 75 ml of Ce4+ Dr.Jehad diab
  24. 24. Ce4+ + Co2+ Ce3++ Co3+Equivalent point volume:CCe4+ VCe4+ = CCo2+ V Co2+0.1 x VCe4+ = 50 x0.1 ==> VCe4+ = 50 mla) Addition of 0 ml of Ce4+ ,absence of Co3+ ,E can not be calculatedb) Addition of 25 ml of Ce4+ before the equivalent point; excess of Co2+E= 0.85+0.059/1xlog([25x0.1)/75/(50 x0.1-25x0.1)/75=0.85vc)Addition of 50 ml of Ce4+;at the equivalent point:E= (e0Co+ e0Ce)/2= (0.85 +1.70)/2= 1.275 v Dr.Jehad diab
  25. 25. c)Addition of 75 ml of Ce4+;after the equivalent pointexcess of Ce4+E= e0Ce+0.059/1xlog[Ce4+]/[Ce3+] E =1.70+0.059/1xlog(75x0.1-50x0.1)/125 /(50x0.1)/125= 1.682 v Dr.Jehad diab
  26. 26. Dr.Jehad diab (True)Self indicators: KMnO4 (purple) → Mn2+ (colorless) I2 (yellow) → 2I- (colorless) Ce4+(yellow) → Ce3+(colorless)
  27. 27. Redox indicatorsSpecific indicators:Starch:Starch +I2 <--> blue complexIt is an easy to detect and rapid indicator. Thisexplains why iodine is a common titrant eventhough it is a weak oxidant.KSCN:Fe3++ SCN- ( Indicator) → FeSCN2+(red complex)Determination of Fe3+ with Ti3+ in presence ofSCN-Fe3+ with Ti3+ --> Fe2+ with Ti4+when [Fe3+] decreases then color of red complexdisappears which indicates the end point.
  28. 28. Dr.Jehad diab
  29. 29. ____ 1E=E0ind - 0.059/n (color of reducing form)E=E0ind + 0.059/n (color of oxidizing form) Dr.Jehad diab
  30. 30. Dr.Jehad diab
  31. 31. general (Ferroin) 1.150v Dr.Jehad diab
  32. 32. Dr.Jehad diab
  33. 33. Dr.Jehad diab OxPotential range for color change is:e0-0.059/n - e0+0.59/n:From 0.80-0.59/1 to 0.80 +0.59/1=(0.741→ 0.859 v)
  34. 34. √√√√ Dr.Jehad diab
  35. 35. Potential IndicatorEnd point is determined by measuring the potential ofindicator electrode against reference electrode and plottingthe potential against the volume of titrant. (Pt for E( mv) and glass electrode for pH) Dr.Jehad diab
  36. 36. Potentiometric titration ‫اﻟﻣﻌﺎﯾرة اﻟﻛﻣوﻧﯾﺔ‬It is possible to monitorthe course of a titrationusing potentiometricmeasurements. The Ptelectrode, for example,is appropriate formonitoring an redoxtitration and determiningan end point in lieuof an indicator. Theprocedure has beencalled a potentiometric titration. The end pointoccurs when the easuredpotential undergoes asharp change—when allthe oxd. or red. in thetitration vessel is reacted Dr.Jehad diab
  37. 37. The potential scale is calibrated in pH units (59.16 mV/pH at 25o C). A temperature adjustment feature changes the slope by 2.303RT/F. Fig. 13.10. Typical pH meter.©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)
  38. 38. Dr.Jehad diab
  39. 39. Dr.Jehad diab
  40. 40. ‫اﻟﻌوﻣل اﻟﻣؤﻛﺳدة واﻟﻣرﺟﻌﺔ اﻟﻣﺳﺎﻋدة‬ ‫‪Dr.Jehad diab‬‬
  41. 41. Dr.Jehad diab
  42. 42. ‫اﻟﻌواﻣل اﻟﻣرﺟﻌﺔ اﻟﻣﺳﺎﻋدة‬ ‫‪Dr.Jehad diab‬‬
  43. 43. Dr.Jehad diab Reducing column
  44. 44. Dr.Jehad diabAuxiliary reducing agent Fe2++ Fe3+ (reductor ) → Fe2+ , Cr3+ +e →Cr2+ Fe2++ Fe3+ (reductor ) → Fe2+ , Cr3+ not reduced Than that of Walden n
  45. 45. Dr.Jehad diab
  46. 46. Auxiliary oxidizing agents Sodium bismuthate-NaBiO3 ‫ﺑﯾزﻣوﺗﺎت اﻟﺻودﯾوم‬ Dr.Jehad diab
  47. 47. ammonium persulfate, ‫ﻓوق ﻛﺑرﯾﺗﺎت اﻷﻣوﻧﯾوم‬Ammonium peroxydisulphate- (NH4)2S2O8 Dr.Jehad diab
  48. 48. H2O2,Na2O2Oxidizes Co2+, Fe2+, Mn2+ Dr.Jehad diab
  49. 49. Some common oxidants Dr.Jehad diab indicator‫٭‬Oxidizing Reduction stability agent product eo volt standardization ‫٭٭‬ AS2O3 , KMnO4 Mn2+ 1.51 Na2C2O4 Mno4- (b) Potassium permanganate , Fe KBrO3Potassium Br- 1.44 KBrO3 (1) (a)bromate 1.44 Na2C2O4 (H2SO4) Ce4+ Ce3+ , AS2O3 , (2) (a) 1.70 Cerium(iv) Fe (HClO4)
  50. 50. Some common oxidantsK2Cr2O7 Cr3+ 1.33 K2Cr2O7 (3) (a)Potassiumdichromate H5IO6 IO3- 1.60 AS2O3 starch (b) Periodic acid KIO3 ICl2- 1.24 KIO3 (4) (a)Potassium iodate BaS2O3. I2 I- 0.536 H2O , starch (c)Iodine AS2O3 Dr.Jehad diab
  51. 51. *(1) α- Naphthoflavone; (2) Ferroin;(3) diphenyl amine sulfonicacid;(4) disapperance of I2 from chlorofom. **(a) stable ;(b) moderately stable ;(c) unstableDr.Jehad diab
  52. 52. Dr.Jehad diab E0 =1.33 vCr2O7-2 +14H+ +6e- 2Cr3++ 7H2O
  53. 53. Dr.Jehad diabPrimary applications of Cr2O7-2 - Determination Fe2+ - Indirect determination of oxidizing agents;A known excess of fe2+ is added to thesample which is oxidant such as MnO4- andthe excess of fe2+ is back titrated withCr2O7-2- Ethanol (C2H5OH)Reactions:6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O3C2H5OH + 2Cr2O72- + 16H+ → 4Cr3+ + 3CH3COOH + 11H2O
  54. 54. of titrantMoles of Fe2+ = 6 moles of Cr2O7-2W Fe/ Mw Fe = 6 CCr2O7-2 x VCr2O7-2 (L)0.2464/56 = 6 CCr2O7-2 x 39.31/1000CCr2O7-2 = 0.01871 M Dr.Jehad diab
  55. 55. Dr.Jehad diab
  56. 56. Dr.Jehad diab
  57. 57. MnO4- at pH < 1(H2SO4 is used ,HCl can not be usedbecause MnO4- oxidize 2Cl- to Cl2)MnO4- + 8H+ + 5 e- → Mn2+ + 4 H2O (E0 = 1.51 V strong acidic med)MnO4- + 2H2O +3e- →MnO2 +4OH- (E0=0.59 v weak basic med)MnO4- + 4H+ +3e- →MnO2 +2H2O (E0=1.69 v weak acidic to neutral )MnO4- + e- →MnO4-2 (Eo=0.56 v strong basic med) Standardization of MnO4- Dr.Jehad diab
  58. 58. Application of KMnO4 in Redox - Titrations MnO4- + 8 H+ + 5 e-  Mn2+ + 4 H2OMo3+ Mo3+ +4H2O →MoO42- + 8H+ +3eH2O2 H2O2 → O2 +2H+ + 2e-Ti3+ Ti3+ +H2O→ TiO2 +2H+ + eH3AsO3 H3AsO3 + H2O →H3AsO4+ 2H++ 2e Dr.Jehad diab
  59. 59. Common titrantsOxidizing titrants Cerium (IV) (Ce4+),E0=1.44 v and 1.70 v in 1M H2SO4 and 1M HClO4 respectively - Commonly used in place of KMnO4 - Works best in acidic solution - Can be used in most applications in previous table - Used to analyze some organic compounds - Color change not distinct to be its own indicator ,Ferroin indicator is used Yellow colorless Dr.Jehad diab
  60. 60. Common titrants Dr.Jehad diabOxidizing titrantsI2 (Iodimetry)I2 + 2e- →2I-I2 + I- → I3- (I- added to increase solubility of I2)I3- + 2 e- ------> 3 I- , E0 =0.536 VWeek oxidizing agentsUnstable Needed to be re-standardized 2- →starchI2 + 2S2O3 S4O62- + 2I-( end point: disappearance of blue color) Less popular than Ce4+, MnO4- ,Cr2O72- used for the determination of strong reductants
  61. 61. Applications of Iodine in Redox Titrations(Iodimetry) ‫اﻟﻣﻘﯾﺎس اﻟﯾودي‬ I3- + 2 e- → 3 I- , I2 + 2e- →2I-Iodimetric titrations carried out in weak acidic to weak basicmedium because in strong basic I2 + OH- → OI-+H+And 3OI-disproportionate to IO3- +2I- this resulted in error bydisturbing the stoichiometry of the reaction. In strong acidicmedium starch decomposes.H2S + I2 → S + 2I- + 2H+H3AsO3 +I2+ H2O →H3AsO4 +2I- +2H++ 2eSO32- + I2+H2O →SO42- + 2I- + 2H+Sn2+ + I2 → Sn4+ + 2 I-AsO33- + I2 + H2O →AsO43- +2 I- +3H+N2H4 + 2I2 → N2 +4H+ +4I- Dr.Jehad diab
  62. 62. Iodimetric determination of hydroquinone by back titrationExcess of standard II2 is added and excess is back 2titrated with Na2S2O3 I2 + 2S2O32- → S4O62- + 2I- Dr.Jehad diab
  63. 63. Dr.Jehad diab
  64. 64. → +2H+ + 2eC6H8O6 + I3-  C6H6O6 + 3I- + 2H+ Dr.Jehad diab
  65. 65. Oxidizing titrantsPotassium iodate:KIO3 ‫ﯾودات اﻟﺑوﺗﺎﺳﯾوم‬IO3- + 5I- (excess) + 6H+(0.1-1M) → 3I2 + 3H2OIO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2OIO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2OIn intermediate step iodate is converted to I2;I2 In presence of CHCl3 resulted in violetcolor. in final step iodine converted to ICl2-and violet color disappear which match theend point. Dr.Jehad diab
  66. 66. Application: Determination of iodine(I2) and iodide(I-) in an aqueous mixturethe iodine in an aliquot of a mixture is determinedby standard sodium thiosulfate solution inpresence of starch as indicator . the concentrationof iodine plus iodide is then determined withstandard potassium iodate solution in strong HClin presence of chloroform as indicator.I2 + 2S2O32- → S4O62- + 2I-IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2OIO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O Dr.Jehad diab
  67. 67. BromometryBrO3- (standard soln.) + 5Br- (excess) + 6H+ →3Br2 + 3H2O (BrO3 ≡3Br2 ≡6e-)Determination of phenol (back and substitutetitration):C6H5OH + 3Br2 (Excess) →C6H2Br3OH3Br2 + 6I-→ 3I2 +6Br- , 3 I2 + 6S2O32- → 3S4O62- + 6I- 3Br2 ≡ 3I2 ≡ 6S2O32- C6H5OH ≡3Br2 ≡6e- Dr.Jehad diab
  68. 68. (Iodometry) ‫اﻟﻣﻘﯾﺎس اﻟﯾودوي‬ I2O2 Dr.Jehad diab
  69. 69. Iodometry Dr.Jehad diab
  70. 70. S4O62- S4O62- is 4 Dr.Jehad diab
  71. 71. Application of Iodide in Redox Titrations that Produce I3- ,triiodide (Iodometry)Iodometric titrations carried out in strong acidic medium2Cu2+ + 4I- → 2CuI + I22Ce4+ + 2I- →2 Ce3+ +I22MnO4- + 10I- + 2H+ →5 I2+ 2 Mn2+ 8H2OH2O2 +2I- +2H+ → I2 +2H2OIO3- + 5I- + 6H+ → 3I2 +3H2OHNO2 +2I- →I2 + 2NO+H2O Fe3+ + 2I- → Fe2++ I2I2 + 2- starchS O 2- 2S2O3 → 4 6 + 2I- Dr.Jehad diab
  72. 72. Reducing titrants Na2S2O3 Dr.Jehad diab
  73. 73. Reactions:2S2O32- + I3- → S4O62- + 3I- Dr.Jehad diab
  74. 74. 0 . 121 214 1 C  0 . 04164 6C  0 . 08147 M Dr.Jehad diab
  75. 75. iron(II) ammonium sulfate (Mohrs salt)Application :determination of strong oxidizingagents as MnO4- ,Cr2O7 2+, Ce4+ …. Dr.Jehad diab
  76. 76. I2 + SO2 + H2O → 2HI + SO3C5H5N.I2 + C5H5N.SO2 + C5NSN+H2O → H5 O2C5H5N.HI +C5H5N.SO3 Dr.Jehad diab
  77. 77. The titration’s end point is signaled when the solutionchanges from the yellow color of the products to thebrown color of the Karl Fisher Reagent. Dr.Jehad diab
  78. 78. Example: A 1.0120 g mineral sample was crushedand dissolved in acid solution. This sample waspassed through a Jones reactor (Fe3+ convertedto Fe2+). Titration of the Fe(II) required 23.29 mlof 0.01992 M KMnO4. What is the %Fe in thesample? wFe 2  2 moles of Fe 56 5    moles of MnO4 0.02329  0.01992 1 2+ wFe2  0.12990 g 2+ 0.12990 100 % Fe2   12.8320% 1.0120 Dr.Jehad diab
  79. 79. Example:A 0.1165 g of primary standard Cu wasdissolved and then treated with an excess ofkI. Reaction: 2Cu2+ + 4I-  CU2I2(s) +I2Calculate the normality of Na2S2O3 soln. if36.24 ml were needed titrate the librated I2I2 + 2S2O32- → S4O62- + 2I- , 2Cu2+ ≡ I ≡2e- 2 0.1165  N  0.03624 63.54 N  0.10125 Dr.Jehad diab
  80. 80. Dr.Jehad diab Example:The amount of ascorbic acid, C6H8O6, in orange juice wasdetermined by oxidizing the ascorbic acid todehydroascorbic acid, C6H6O6, with a known excess of I3–,and back titrating the excess I3– with Na2S2O3. A 5.00-mLsample of filtered orange juice was treated with 50.00 mLof excess 0.01023 M I3–. After the oxidation was complete,13.82 mL of 0.07203 M Na2S2O3 was needed to reach thestarch indicator end point. Report the concentration ofascorbic acid in milligrams per 100 mL. C6H8O6 + I3-  C6H6O6 + 3I- + 2H+ (I3- =I2) I3- + 2S2O32-  3I- + S4O62-
  81. 81. C6H8O6 + I3- +  C6H6O6 + 3I- + 2H+ I3- + 2S2O32-  3I- + S4O62-Moles (I3-)total= moles (I3-)ascorbic acid +(mole I3-)backtitrationMV(I3-) = WmgC6H8O6/Fw + 0.5(MV)S2O32-50 ×0.01023= W/176.13+ 0.5 × 13.82 × 0.07203W= (50*0.01023-0.5 × 13.82 × 0.07203) × 176.13=2.43mgC6H8O6 in 5 ml sample or (2.43 ×100) /5 =48.60 mg/100 ml orange juice.Or:50 × 0.01023=5×M + 0.5 ×13.82 × 0.07203M=0.0028C(g/l)= 0.0028 × 176.13=0.493C(mg/100ml)= (0.486 × 1000)/10=48.60 mg/100 ml Dr.Jehad diab
  82. 82. ‫ﺗﻣرﯾن:ﺗوزن 02 ﻣﺿﻐوطﺔ وﺗﺳﺣق ،ﺛم ﯾﺣل ﻣن اﻟﻣﺳﺣوق ﻣﺎ ﯾﻛﺎﻓﺊ 051‬‫ﻣﻎ ﻣن ﺣﻣض اﻷﺳﻛورﺑﻲ ﻓﻲ ﻣزﯾﺞ ﻣؤﻟف ﻣن ﺣﻣض اﻟﻛﺑرﯾت واﻟﻣﺎء‬‫،ﯾﻌﺎﯾر اﻟﻣﺣﻠول ﺑﻣﺣﻠول ‪ 0.1 M‬ﻣن اﻟﺳﯾرﯾوم اﻟرﺑﺎﻋﻲ،ﻓﻛﺎن اﻟﻣﺻروف‬‫71 ﻣلﻋﻠﻣﺎ ً أن ﻣﺣﺗوى اﻟﻣﺿﻐوطﺔ اﻟواﺣدة ﻣن ﺣﻣض اﻷﺳﻛورﺑﻲ ھو‬ ‫.‬ ‫005 ﻣﻎ،و اﻟوزن اﻟﺟزﯾﺋﻲ ﻟﺣﻣض اﻷﺳﻛورﺑﻲ 31.671 غ.‬‫ﻣﺎ ھﻲ اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ ﻟﺣﻣض اﻷﺳﻛورﺑﻲ ، وﻣﺎ ھو اﻟﻣﺣﺗوى اﻟﻔﻌﻠﻲ ﻣن‬ ‫ﺣﻣض اﻷﺳﻛورﺑﻲ ﻓﻲ اﻟﻣﺻﻐوطﺔ اﻟواﺣدة .‬‫+‪C6H8O6 + 2Ce4+→ C6H6O6 + 2Ce3+ + 2H‬‬ ‫1 31 . 671 / 6 ‪moles C 6 H 8 O 6 W C 6 H 8 O‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪moles Ce 4‬‬ ‫‪MV Ce 4‬‬ ‫2‬ ‫6 ‪2  W C 6 H 8O‬‬ ‫1 . 0 ‪ 17 ‬‬ ‫31 . 671‬ ‫31 . 671 ‪17  0 . 1 ‬‬ ‫‪W ‬‬ ‫‪ 149 . 7 mg‬‬ ‫‪Dr.Jehad diab‬‬ ‫2‬
  83. 83. ‫اﻟﻧﺳﺑﺔ اﻟﻣﺋوﯾﺔ:‬ ‫001 ‪149 .7 ‬‬‫‪% Ascorbicac id ‬‬ ‫08. 99 ‪‬‬ ‫051‬ ‫ﻣﺣﺗوى اﻟﻣﺿﻐوطﺔ اﻟﻔﻌﻠﻲ ﻣن ﺣﻣض اﻷﺳﻛورﺑﻲ:‬‫08 . 99 ‪500 ‬‬ ‫‪ 499 mg‬‬ ‫001‬ ‫‪Dr.Jehad diab‬‬
  84. 84. Problem: Calculate the normality of the solutionproduced by dissolving 2.064 g of primarystandard K2Cr2O7 in sufficient water to give 500ml.N=2.064 g / (294/6)=0.421g/0.500L=0.0842 NProblem: calculate the molarity of the I2 solutionthat is 0.04N with respect to the following reaction:I2 + H2S →2I- +2H++SM=0.04/2=0.02 MProblem: calculate the molarity of theKIO3solution that is 0.04 N with respect to thefollowing reaction:IO3-+2I-+6H++6Cl- →3ICl2-+3H2OM=0.04/4=0.01 M Dr.Jehad diab
  85. 85. The End

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