Download - Redox titrations [compatibility mode]
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Redox titrations
Dr.Jehad M Diab, faculty of pharmacy Damascus University
Pharma.analytical chemistry II
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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 a
titrationbasic calculationsmethods of sample preparation andcommon titrantsExample applications
Dr.Jehad diab
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Dr.Jehad diab
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Determination of equivalence point potential:Fe2+ Ce4+ = Fe3+ Ce3+
E0Fe3+= 0.771 v , E0
Ce4+=1.70 v
Eeq = E0Fe3+ - 0.0592/1 log [Fe2+] / [Fe3+] (1)
Eeq = E0Ce4+ - 0.0592/1 log [Ce3+] / [Ce4+] (2)
1+2:2Eeq = E0
Ce4+ + E0Fe3+ - 0.0592/1 log [Fe2+][Ce3+] / [Fe3+] [Ce4+]
At the equivalence point:[Fe3+] =[Ce3+ ][Fe2+]=[Ce4+ ]Rearrangement:2Eeq = E0
Ce4+ + E0Fe3+ - 0.0592/1 log [Ce4+][Ce3+] / [Ce3+] [Ce4+]
Eeq = (E0Ce4+ + E0
Fe3+) /2= 1.70+0.771/2= 1.24 volt
Dr.Jehad diab
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2Eeq = E0Ce4+ + E0
Fe3+ - 0.0592/1 log [Fe2/ [Fe3+]-0.0592/1 log [Ce3+] / [Ce4+] 2Eeq = E0
Ce4+ + E0Fe3+ - 0.0592/1(log [Fe2]/ [Fe3+]+ log
[Ce3+] / [Ce4+]2Eeq = E0
Ce4+ + E0Fe3+ - 0.0592/1( (log [Fe2]-log [Fe3+]+
log[Ce3+]-log[Ce4+]2Eeq = E0
Ce4+ + E0Fe3+ - 0.0592/1(log [Fe2]+log[Ce3+]-log
[Fe3+]-log[Ce4+]2Eeq = E0
Ce4+ + E0Fe3+ - 0.0592/1 log ([Fe2+][Ce3+] / [Fe3+] [Ce4+])
draft
Dr.Jehad diab
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Dr.Jehad diab
E eq =(n1e0ox+n2e0
red) / n1+n2
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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+]
Determine of Eeq for the following reaction:
Mno4- +5 Fe2+ 8H+ = Mn2+ + 5Fe3++ 4H2O
A more complex example Dr.Jehad diab
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6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4
-][Mn2+]
6eeq=0.77+5×1.51+0.059 log[H+]8
eeq=(0.77+5×1.51)/6 + 0.059/6 log[H+]8
eeq=(0.77+5×1.51)/6 - 0.079 pH
If [H+]=1M (pH= 0):
eeq=(0.77+5×1.51)/6 + 0.059/6 log[1]8 =1.39 v
So, the potential at equivalence point is dependent on [H+]. If [H+]=1M we can calculate from the following equation:
Dr.J.Diabeeq=(n1e0ox + n2e0
Red ) / n1+n2
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Volume of reagent,ml
E,vo
lt
E,vo
ltDr.Jehad diab
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Complete titration curve
(or volume of titrant) Dr.Jehad diab
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Ce4+ + Fe2+ Ce3++ Fe3+ 100 ml 0.1 M Fe2+ with 0.1 M Ce4+ solution
Dr.Jehad diab
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Dr.Jehad diab
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Ce4+ + Fe2+ Ce3++ Fe3+
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Addition of 50 ml of Ce4+:
vEFe 771.0
501001.05050100
1.0501.0100
log1059.0771.0
Addition of 20 ml of Ce4+:
vEFe 735.0
201001.02020100
1.0201.0100
log1059.0771.0
Ce4+ (0.1 M) + Fe2+ (100 ml 0.1 M) Ce3++ Fe3+
Dr.Jehad diab
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Dr.Jehad diab
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Dr.Jehad diab
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E eq =n1e0ox+n2e0
red / n1+n2
Fe2+ Ce4+ = Fe3+ Ce3+ , e0Fe3+/Fe2+= 0.771 v , e0
Ce4+/Ce3+=1.70 v
Dr.Jehad diab
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Dr.Jehad diab
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Ce4+ + Fe2+ Ce3++ Fe3+
Dr.Jehad diab
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Dr.Jehad diab
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10ml x0.1M=210ml x [Ce4+]200 x0.05= 210 x[Ce3+]
Ce4+ + Fe2+ Ce3++ Fe3+
Dr.Jehad diab
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E = e0 – 0.059 log [Ce3+]/ [Ce4+]
volt
Ce4+ + Fe2+ Ce3++ Fe3+
1.70 -1.64 v
1.66 V
Or E =1.70-0.059log 100%/10%=1.64 v
Dr.Jehad diab
(200 x0.05/ 210)
(10ml x0.1M/210ml)
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Homework
You are titrating 50 ml 0.1M of Co2+ solution wuth0.1 M Ce4+ titrant.E0
Co=0.85 v ,E0Ce=1.70.
What is the potential for the titration system 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
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Ce4+ + Co2+ Ce3++ Co3+
Equivalent point volume: CCe4+ VCe4+ = CCo2+ V Co2+
0.1 x VCe4+ = 50 x0.1 ==> VCe4+ = 50 ml a) 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= (e0
Co+ e0Ce)/2= (0.85 +1.70)/2= 1.275 v
Dr.Jehad diab
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c)Addition of 75 ml of Ce4+;after the equivalent point excess 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
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Self indicators: KMnO4 (purple) → Mn2+ (colorless) I2 (yellow) → 2I- (colorless)Ce4+(yellow) → Ce3+(colorless)
Dr.Jehad diab
(True)
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Redox indicatorsSpecific indicators:Starch:Starch +I2 <--> blue complexIt is an easy to detect and rapid indicator. This explains why iodine is a common titrant even though it is a weak oxidant.KSCN:Fe3++ SCN- ( Indicator) → FeSCN2+(red complex)Determination of Fe3+ with Ti3+ in presence of SCN-
Fe3+ with Ti3+ --> Fe2+ with Ti4+
when [Fe3+] decreases then color of red complex disappears which indicates the end point.
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Dr.Jehad diab
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____
1
E=E0ind - 0.059/n (color of reducing form)
E=E0ind + 0.059/n (color of oxidizing form) Dr.Jehad diab
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Dr.Jehad diab
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general
(Ferroin)
1.150v
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Dr.Jehad diab
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Ox
Potential 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)
Dr.Jehad diab
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Dr.Jehad diab
√
√√
√
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Potential IndicatorEnd point is determined by measuring the potential ofindicator electrode against reference electrode and plottingthe potential against the volume of titrant.
Dr.Jehad diab
(Pt for E( mv) and glass electrode for pH)
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Potentiometric titrationIt is possible to monitor the course of a titration using potentiometric measurements. The Pt electrode, for example, is appropriate for monitoring an redox titration and determining an end point in lieuof an indicator. The procedure has been called a potentiometrictitration. The end point occurs when the easuredpotential undergoes a sharp change—when all the oxd. or red. in the titration vessel is reacted
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المعایرة الكمونیة
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Fig. 13.10. Typical pH meter.
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.
©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)
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Dr.Jehad diab
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Dr.Jehad diab
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العومل المؤكسدة والمرجعة المساعدة
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Dr.Jehad diab
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Dr.Jehad diab
العوامل المرجعة المساعدة
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Dr.Jehad diab Reducing column
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Auxiliary reducing agent
nThan that of Walden
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Fe2++ Fe3+ (reductor ) → Fe2+ , Cr3+ +e →Cr2+
Fe2++ Fe3+ (reductor ) → Fe2+ , Cr3+ not reduced
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Dr.Jehad diab
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Auxiliary oxidizing agents
Sodium bismuthate-NaBiO3 بیزموتات الصودیوم
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Ammonium peroxydisulphate- (NH4)2S2O8
,ammonium persulfateفوق كبریتات األمونیوم
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Oxidizes Co2+, Fe2+, Mn2+
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H2O2,Na2O2
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stability standardizationeo voltReduction٭indicator٭٭
productOxidizing
agent
(b)Mno4-
AS2O3 , Na2C2O4
, Fe1.51Mn2+KMnO4
(a)(1)KBrO31.44Br-KBrO3
Potassium
bromate
(a)(2)Na2C2O4, AS2O3 ,
Fe
1.44 (H2SO4)
1.70(HClO4)
Ce3+Ce4+
Some common oxidants Dr.Jehad diab
Potassium permanganate
Cerium(iv)
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(a)(3)K2Cr2O71.33Cr3+K2Cr2O7
(b)starchAS2O31.60IO3-H5IO6
(a)(4)KIO31.24ICl2-KIO3
(c)starchBaS2O3.H2O , AS2O3
0.536I-I2
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Some common oxidants
Potassium dichromate
Periodic acid
Potassium iodate
Iodine
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*(1) α- Naphthoflavone; (2) Ferroin;(3) diphenyl amine sulfonicacid;(4) disapperance of I2 from chlorofom.
**(a) stable ;(b) moderately stable ;(c) unstable
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E0 =1.33 v
Cr2O7-2 +14H+ +6e-2Cr3++ 7H2O
Dr.Jehad diab
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Primary applications of Cr2O72-
- Determination Fe2+
- Indirect determination of oxidizing agents;A known excess of fe2+ is added to the sample which is oxidant such as MnO4
- and the excess of fe2+ is back titrated with Cr2O7
2 -
- Ethanol (C2H5OH) Reactions:6Fe2+ + Cr2O7
2- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
3C2H5OH + 2Cr2O72- + 16H+ → 4Cr3+ + 3CH3COOH + 11H2O
Dr.Jehad diab
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Moles of Fe2+ = 6 moles of Cr2O7-2
W Fe/ Mw Fe = 6 CCr2O7-2 x VCr2O7-2 (L)0.2464/56 = 6 CCr2O7-2 x 39.31/1000CCr2O7-2 = 0.01871 M
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of titrant
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Dr.Jehad diab
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Dr.Jehad diab
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MnO4- at pH < 1(H2SO4 is used ,HCl can not be used
because 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-
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-Application of KMnO4 in Redox Titrations
MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O
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H2O2 → O2 +2H+ + 2e-H2O2
Ti3+ Ti3+ +H2O→ TiO2 +2H+ + e
H3AsO3 H3AsO3 + H2O →H3AsO4+ 2H++ 2e
Mo3+ Mo3+ +4H2O →MoO42- + 8H+ +3e
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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 usedYellow colorless
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Common titrantsOxidizing titrantsI2 (Iodimetry)I2 + 2e- →2I-I2 + I- → I3- (I- added to increase solubility of I2)I3- + 2 e- ------> 3 I- , E0 =0.536 VWeek oxidizing agentsUnstable Needed to be re-standardizedI2 + 2S2O3
2- → S4O62- + 2I-
( end point: disappearance of blue color)
Less popular than Ce4+, MnO4- ,Cr2O7
2-
used for the determination of strong reductants
starch
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Applications of Iodine in Redox Titrations(Iodimetry)
I3- + 2 e- → 3 I- , I2 + 2e- →2I-Iodimetric titrations carried out in weak acidic to weak basic medium because in strong basic I2 + OH- → OI-+H+
And 3OI-disproportionate to IO3- +2I- this resulted in error by
disturbing the stoichiometry of the reaction. In strong acidic medium starch decomposes.H2S + I2 → S + 2I- + 2H+
H3AsO3 +I2+ H2O →H3AsO4 +2I- +2H++ 2eSO3
2- + I2+H2O →SO42- + 2I- + 2H+
Sn2+ + I2 → Sn4+ + 2 I-AsO3
3- + I2 + H2O →AsO43- +2 I- +3H+
N2H4 + 2I2 → N2 +4H+ +4I- Dr.Jehad diab
المقیاس الیودي
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Iodimetric determination of hydroquinone by back titration
I2Excess of standard I2 is added and excess is back titrated with Na2S2O3
I2 + 2S2O32- → S4O6
2- + 2I-
Dr.Jehad diab
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Dr.Jehad diab
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C6H8O6 + I3- C6H6O6 + 3I- + 2H+
→ +2H+ + 2e
Dr.Jehad diab
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Oxidizing titrants
Potassium iodate:KIO3
IO3- + 5I- (excess) + 6H+(0.1-1M) → 3I2 + 3H2O
IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2O
IO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O
In intermediate step iodate is converted to I2;I2 In presence of CHCl3 resulted in violet color. in final step iodine converted to ICl2-
and violet color disappear which match the end point.
Dr.Jehad diab
یودات البوتاسیوم
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Application:
Determination of iodine(I2) and iodide(I-) in an aqueous mixture
the 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 + 2S2O3
2- → S4O62- + 2I-
IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2O
IO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O
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BrO3- (standard soln.) + 5Br- (excess) + 6H+ →3Br2 + 3H2O (BrO3 ≡3Br2 ≡6e-)
Bromometry
Determination of phenol (back and substitute titration):
C6H5OH + 3Br2 (Excess) →C6H2Br3OH 3Br2 + 6I-→ 3I2 +6Br- , 3 I2 + 6S2O3
2- → 3S4O62- + 6I-
3Br2 ≡ 3I2 ≡ 6S2O32-
C6H5OH ≡3Br2 ≡6e- Dr.Jehad diab
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(Iodometry)
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O2
I2
المقیاس الیودوي
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Dr.Jehad diab
Iodometry
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4S4O62- is
S4O62-
Dr.Jehad diab
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2Cu2+ + 4I- → 2CuI + I22Ce4+ + 2I- →2 Ce3+ +I22MnO4
- + 10I- + 2H+ →5 I2+ 2 Mn2+ 8H2OH2O2 +2I- +2H+ → I2 +2H2OIO3
- + 5I- + 6H+ → 3I2 +3H2OHNO2 +2I- →I2 + 2NO+H2O Fe3+ + 2I- → Fe2++ I2
Application of Iodide in Redox Titrations that Produce I3- ,triiodide (Iodometry)
I2 + 2S2O32- → S4O6
2- + 2I-
Iodometric titrations carried out in strong acidic medium
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starch
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Reducing titrants Na2S2O3Dr.Jehad diab
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Reactions:
2S2O32- + I3- → S4O6
2- + 3I-
Dr.Jehad diab
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MCC
08147.061
04164.0214121.0
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Application :determination of strong oxidizing agents as MnO4
- ,Cr2O7 2+, Ce4+ ….
iron(II) ammonium sulfate (Mohr's salt)
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NSO
I2 + SO2 + H2O → 2HI + SO3
C5H5N.I2 + C5H5N.SO2 + C5H5N+H2O →
2C5H5N.HI +C5H5N.SO3Dr.Jehad diab
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Dr.Jehad diab
The titration’s end point is signaled when the solution changes from the yellow color of the products to the brown color of the Karl Fisher Reagent.
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Example: A 1.0120 g mineral sample was crushed and dissolved in acid solution. This sample was passed through a Jones reactor (Fe3+ converted to Fe2+). Titration of the Fe(II) required 23.29 ml of 0.01992 M KMnO4. What is the %Fe in the sample?
%8320.120120.1
10012990.02%
12990.0215
01992.002329.0562
4
2
Fe
gwFe
wFe
MnOof molesFe of moles
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2+
2+
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Example:A 0.1165 g of primary standard Cu was dissolved and then treated with an excess of kI. Reaction:
2Cu2+ + 4I- CU2I2(s) +I2Calculate the normality of Na2S2O3 soln. if 36.24 ml were needed titrate the librated I2I2 + 2S2O3
2- → S4O62- + 2I- ,
10125.0
03624.054.63
1165.0
N
N
2Cu2+ ≡ I2 ≡2e-
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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.
Example:
C6H8O6 + I3- C6H6O6 + 3I- + 2H+
I3- + 2S2O32- 3I- + S4O6
2-(I3- =I2)
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C6H8O6 + I3- + C6H6O6 + 3I- + 2H+
I3- + 2S2O32- 3I- + S4O6
2-
Moles (I3-)total= moles (I3-)ascorbic acid +(mole I3-)back titrationMV(I3-) = WmgC6H8O6/Fw + 0.5(MV)S2O3
2-
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
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150 یكافئ ما المسحوق من یحل ،ثم وتسحق مضغوطة 20 توزن:تمرین والماء الكبریت حمض من مؤلف مزیج في األسكوربي حمض من مغ
0.1بمحلول المحلول ،یعایر M المصروف الرباعي،فكان السیریوم من ھو األسكوربي حمض من الواحدة المضغوطة محتوى أن علماً .مل 17
.غ 176.13 األسكوربي لحمض الجزیئي الوزن ،ومغ 500 من الفعلي المحتوى ھو وما ، األسكوربي لحمض المئویة النسبة ھي ما
. الواحدة المصغوطة في األسكوربي حمضC6H8O6 + 2Ce4+→ C6H6O6 + 2Ce3+ + 2H+
mgW
W OHC
7.1492
13.1761.017
1.01713.176
2 686
2113.176/
44
686686 Ce
OHC
Ce
OHC
MVW
molesmoles
Dr.Jehad diab
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: النسبة المئویة
80.99150
1007.149%
idAscorbicac
:محتوى المضغوطة الفعلي من حمض األسكوربي
mg499100
80.99500
Dr.Jehad diab
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Problem: Calculate the normality of the solution produced by dissolving 2.064 g of primary standard K2Cr2O7 in sufficient water to give 500 ml.N=2.064 g / (294/6)=0.421g/0.500L=0.0842 NProblem: calculate the molarity of the I2 solution that is 0.04N with respect to the following reaction:I2 + H2S →2I- +2H++SM=0.04/2=0.02 MProblem: calculate the molarity of the KIO3solution that is 0.04 N with respect to the following reaction:IO3
-+2I-+6H++6Cl- →3ICl2-+3H2OM=0.04/4=0.01 M Dr.Jehad diab
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The End