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Môn: Chemistry Laboratory (CH012IU)

Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

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Tác giả:

Profile Mai Nguyệt
CHEMISTRY LABORATORYCHEMISTRY LABORATORY
REPORTREPORT
Experiment 3: REDOXExperiment 3: REDOX
TITRATIONTITRATION
ToTo:: InstructorInstructor: PhD Hoang Le Son: PhD Hoang Le Son
FromFrom: Do Truong Anh Thu -: Do Truong Anh Thu - Student ID:Student ID:BTBCIU14046BTBCIU14046
Subject:Subject:General General Chemistry LaboratoryChemistry Laboratory
DateDate
: October 12: October 12
thth
, 2015, 2015
I. INTRODUCTIONI. INTRODUCTION
This experiment shows us the technique to determine the unknown solution by the knownThis experiment shows us the technique to determine the unknown solution by the known
solution of the oxidation-reduction reaction - TITRATION. The mainly titrant used insolution of the oxidation-reduction reaction - TITRATION. The mainly titrant used in
this experiment is potassium manganate with N=0.05 to titrate against a standard solutionthis experiment is potassium manganate with N=0.05 to titrate against a standard solution
of Hof H
22
OO
22
OO
4.4.
Then the standardized KMnOThen the standardized KMnO
44
solution will be used to determine thesolution will be used to determine the
concentration of unknown oxalic acid and unknown Feconcentration of unknown oxalic acid and unknown Fe
3+3+
solution. Furthermore, thesolution. Furthermore, the
experiment provides the overview concept about the nature of redox reaction and theexperiment provides the overview concept about the nature of redox reaction and the
relationship between the GEW (gram equivalent weight), the volume and relationship between the GEW (gram equivalent weight), the volume and with thewith the
normality among the oxidizing/reducing agentsnormality among the oxidizing/reducing agents
II. PROCEDUREII. PROCEDURE
A.A.
Instruments:Instruments:
One 50 mL buretOne 50 mL buret
One 250 mL volumetric flaskOne 250 mL volumetric flask
One 100 mL graduated cylinderOne 100 mL graduated cylinder
Three 250 mL beakersThree 250 mL beakers
One 10 mL volumetric pipetOne 10 mL volumetric pipet
There 250 mL Erlenmeyer flasksThere 250 mL Erlenmeyer flasks
One glass watchOne glass watch
One funnel (small size)One funnel (small size)
One stirring rodOne stirring rod
One medicine dropperOne medicine dropper
Water bathWater bath
B. Experimental Procedure:B. Experimental Procedure:
1.1. Prepare KMnOPrepare KMnO
44
solution: calculate the weight of KMnOsolution: calculate the weight of KMnO
44
required to prepare of a 0.05 Nrequired to prepare of a 0.05 N
KMnOKMnO
44
solution. After weighing the required amount KMnOsolution. After weighing the required amount KMnO
44
, transfer it to a 250 mL, transfer it to a 250 mL
beaker with 250 mL of distilled water. Mix the solution thoroughly by vigorous swirbeaker with 250 mL of distilled water. Mix the solution thoroughly by vigorous swirling.ling.
Then transfer it to a dark brown bottle, dThen transfer it to a dark brown bottle, discard undissolved solid.iscard undissolved solid.
2.2. Clean the buret with distilled water and then rinse it Clean the buret with distilled water and then rinse it three times with 5 mL portionsthree times with 5 mL portions
prepared KMnOprepared KMnO
44
solution, allowing the rinse solution to drain though solution, allowing the rinse solution to drain though the tip of the buretthe tip of the buret
each time. Discard the rinse solution. Fill the buret with KMnOeach time. Discard the rinse solution. Fill the buret with KMnO
44
solution and allow it tosolution and allow it to
drain tough the buret tip until no drain tough the buret tip until no air bubbles remain in the tip. Record air bubbles remain in the tip. Record the buret readingthe buret reading
before beginning the titration.before beginning the titration.
3.3. Standardization of prepared KMnOStandardization of prepared KMnO
44
solution: pipet separate 10 mL of standard oxalicsolution: pipet separate 10 mL of standard oxalic
acid solution into three 250 mL Erlenmeyer flasks. Add acid solution into three 250 mL Erlenmeyer flasks. Add approximately 40 mL of distilledapproximately 40 mL of distilled
water to each flask. In the fume hood, cautiously add 20 mL of 6 N Hwater to each flask. In the fume hood, cautiously add 20 mL of 6 N H
22
SOSO
44
solution tosolution to
each flask. Warm the flasks in the water bath to each flask. Warm the flasks in the water bath to 8585
––
9090
00
and titrate the hot solutionsand titrate the hot solutions
against the KMnOagainst the KMnO
44
solution.solution.
4.4. Determination of unknown concentration HDetermination of unknown concentration H
22
CC
22
OO
44
solution: pipet separate 10 mL ofsolution: pipet separate 10 mL of
unknown concentration solution of Hunknown concentration solution of H
22
CC
22
OO
44
into three 250 mL Erlenmeyer flasks andinto three 250 mL Erlenmeyer flasks and
proceed as directed in the standardization process. After finishing the titraproceed as directed in the standardization process. After finishing the titration, calculatetion, calculate
the normality of the unknown concentration Hthe normality of the unknown concentration H
22
CC
22
OO
44
solution; determine the average andsolution; determine the average and
the standard deviation.the standard deviation.
III. DATA AND DISCUSSIONIII. DATA AND DISCUSSION
1. TITRATION OF KMnO1. TITRATION OF KMnO
44
SOLUTION WITH STANDARD HSOLUTION WITH STANDARD H
22
CC
22
OO
44
SOLUTIONSOLUTION
Normality of the standard HNormality of the standard H
22
CC
22
OO
44
solution,solution, NN
H2C2O4H2C2O4
= 0.05 N= 0.05 N
Volume of the standard HVolume of the standard H
22
CC
22
OO
44
solution used,solution used, VV
H2C2O4H2C2O4
= 10 mL= 10 mL
Trial Trial # # Burette Burette reading reading (mL) (mL) Volume Volume of of KMnOKMnO
44
(mL) (mL) Normality Normality of of KMnOKMnO
44
(N)(N)
1 1 0.00-10.00 0.00-10.00 10.00 10.00 0.050.05
2 2 10.00-19.80 10.00-19.80 9.80 9.80 0.0510.051
3 3 19.80-29.60 19.80-29.60 9.80 9.80 0.0510.051
Data Calculation:Data Calculation:
Normality of the solution is calculated by the relationshipNormality of the solution is calculated by the relationship
VV
KMnO4KMnO4
x Nx N
KMnO4KMnO4
= V= V
H2C2O4H2C2O4
x Nx N
H2C2O4H2C2O4
The Normality of KMnOThe Normality of KMnO
44
is: Nis: N
KMnO4KMnO4
= (V= (V
H2C2O4H2C2O4
x Nx N
H2C2O4H2C2O4
)/V)/V
KMnO4KMnO4
For the Trial 1: NFor the Trial 1: N
KMnO4KMnO4
= 0.05x10/10 = 0.05= 0.05x10/10 = 0.05
For the Trial 2: NFor the Trial 2: N
KMnO4KMnO4
= 0.05x10/9.8 = 0.051= 0.05x10/9.8 = 0.051
For the Trial 3: NFor the Trial 3: N
KMnO4KMnO4
= 0.05x10/9.8 = 0.051= 0.05x10/9.8 = 0.051
Average Normality of KMnOAverage Normality of KMnO
44
== (0.05 + 0.051 + 0.051)/3 = 0.0507 (N)(0.05 + 0.051 + 0.051)/3 = 0.0507 (N)
2. TITRATION OF UNKNOWN CONCENTRATION H2. TITRATION OF UNKNOWN CONCENTRATION H
22
CC
22
OO
44
SOLUTION WITHSOLUTION WITH
STANDARD KMnOSTANDARD KMnO
44
SOLUTIONSOLUTION
Normality of the standard KMnONormality of the standard KMnO
44
solution,solution, NN(KMnO(KMnO
44
) = 0.05 N) = 0.05 N
Volume of the unknown HVolume of the unknown H
22
CC
22
OO
44
solution used,solution used, VV(H(H
22
CC
22
OO
44
) = 10 mL) = 10 mL
Trial Trial # # Burette Burette reading reading (mL) (mL) Volume Volume of of KMnOKMnO
44
(mL)(mL)
Normality ofNormality of
HH
22
CC
22
OO
44
(N)(N)
1 1 0.00-7.90 0.00-7.90 7.90 7.90 0.03950.0395
2 2 10.00-18.10 10.00-18.10 8.10 8.10 0.04050.0405
3 3 10.00-17.80 10.00-17.80 7.80 7.80 0.0390.039
Data Calculation:Data Calculation:
Normality of the solution is calculated by the relationshipNormality of the solution is calculated by the relationship
VV
KMnO4KMnO4
x Nx N
KMnO4KMnO4
= V= V
H2C2O4H2C2O4
x Nx N
H2C2O4H2C2O4
The Normality of KMnOThe Normality of KMnO
44
is: Nis: N
H2C2O4H2C2O4
= (V= (V
KMnO4KMnO4
x Nx N
KMnO4KMnO4
)/V)/V
H2C2O4H2C2O4
For the Trial 1: NFor the Trial 1: N
H2C2O4H2C2O4
= 0.05x7.9/10 = 0.0395= 0.05x7.9/10 = 0.0395
For the Trial 2: NFor the Trial 2: N
H2C2O4H2C2O4
= 0.05x8.1/10 = 0.0405= 0.05x8.1/10 = 0.0405
For the Trial 3: NFor the Trial 3: N
H2C2O4H2C2O4
= 0.05x7.8/10 = 0.039= 0.05x7.8/10 = 0.039
Average Normality of KMnOAverage Normality of KMnO
44
== (0.0395+0.0405+0.039)/3 = 0.0397 (N)(0.0395+0.0405+0.039)/3 = 0.0397 (N)
3. TITRATION OF UNKNOWN CONCENTRATION FeSO3. TITRATION OF UNKNOWN CONCENTRATION FeSO
44
SOLUTION WITHSOLUTION WITH
STANDARD KMnOSTANDARD KMnO
44
SOLUTIONSOLUTION
Normality of the standard KMnONormality of the standard KMnO
44
solution,solution, NN(KMnO(KMnO
44
) = 0.05 N) = 0.05 N
Volume of the unknown HVolume of the unknown H
22
CC
22
OO
44
solution used,solution used, VV(FeSO(FeSO
44
) = 10 mL) = 10 mL
Trial Trial # # Burette Burette reading reading (mL) (mL) Volume Volume of of KMnOKMnO
44
(mL) (mL) Normality Normality ofof
FeSOFeSO
44
(N)(N)
1 1 10.00-15.80 10.00-15.80 5.80 5.80 0.0290.029
2 2 10.00-15.90 10.00-15.90 5.90 5.90 0.02950.0295
3 3 10.00-16.00 10.00-16.00 6.00 6.00 0.030.03
Data Calculation:Data Calculation:
Normality of the solution is calculated by the relationshipNormality of the solution is calculated by the relationship
VV
KMnO4KMnO4
x Nx N
KMnO4KMnO4
= V= V
FeSO4FeSO4
x Nx N
FeSO4FeSO4
The Normality of KMnOThe Normality of KMnO
44
is: Nis: N
FeSO4FeSO4
= (V= (V
KMnO4KMnO4
x Nx N
KMnO4KMnO4
)/V)/V
FeSO4FeSO4
For the Trial 1: NFor the Trial 1: N
FeSO4FeSO4
= 0.05x5.8/10 = 0.029= 0.05x5.8/10 = 0.029
For the Trial 2: NFor the Trial 2: N
FeSO4FeSO4
= 0.05x5.9/10 = 0.0295= 0.05x5.9/10 = 0.0295
For the Trial 3: NFor the Trial 3: N
FeSO4FeSO4
= 0.05x6/10 = 0.03= 0.05x6/10 = 0.03
Average Normality of FeSOAverage Normality of FeSO
44
== (0.029+0.0295+0.03)/3 = 0.0295 (N)(0.029+0.0295+0.03)/3 = 0.0295 (N)
IV. CONCLUSIONIV. CONCLUSION
With the method Titration in this experiment, we can calculate the unknown concentrationWith the method Titration in this experiment, we can calculate the unknown concentration
solution by adding the known volume of the standardized solution until the reaction betweensolution by adding the known volume of the standardized solution until the reaction between
them reaches neutralization through the relationship Vthem reaches neutralization through the relationship V
oxidoxid
x Nx N
oxidoxid
= V= V
redred
x Nx N
redred
. At the end of. At the end of
titration, three of four variables will be known and the unknown variable can be determined.titration, three of four variables will be known and the unknown variable can be determined.
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  CHE C M HE I M ST I R ST Y R Y LABO LAB R O AT R O AT R O Y R REPORT Experimen Experime t n t 3 : 3 : REDOX R TITR TI AT TR I AT O I N To T : Instructor: : Ph D Ph D H o H a o n a g n g L e L e So n So From: : D o T r T u r o u n o g n g A n A h n h T h T u h u - St S u t d u e d n e t n t ID: I BTBCIU BCI 1 U 4 1 0 4 4 0 6 4 Su S b u j b e j c e t c : t G  e G n e e n r e a r l a l Che Ch m e i m s i t s r t y r y L a L b a o b r o a r t a o t r o y r Date: O c O t c o t b o e b r e r 1 2th t , ,2 0 2 1 0 5 1   I. INTRODUCTION Th T i h s i s e x e p x e p r e i r m i e m n e t n t s h s o h w o s w s u s u s t h t e h e t e t c e h c n h i n q i u q e u e t o o d e d t e e t r e m r i m n i e n e t h t e h e u n u k n n k o n w o n w n s o s l o u l t u i t o i n o n b y b y t h t e h e k n k o n w o n w so s l o u l t u i t o i n o n o f o f t h t e h e o x o i x d i a d t a i t o i n o - n r - e r d e u d c u t c i t o i n o n r e r a e c a t c i t o i n o n - - T I T T I RA T T RA I T O I N O . N . T h T e h e m a m i a n i l n y l y t i t tr t a r n a t n t u s u ed e d i n i th t i h s i s e x e p x e p r e i r m i e m n e t n t i s i p o p t o a t s a s s iu i m u m m a m n a g n a g n a a n t a e t e w i w t i h t h N = N 0 = . 0 0 . 5 0 5 t o t o t i t t i ra r t a e t e a g a a g i a n i s n t s t a a s ta t n a d n a d r a d d s ol o u l t u i t o i n o of o f H 2O2O4. 4 Th T e h n e n t h t e h e s tanda d r a d r i d z i e z d e KMn K O Mn 4s  o s l o u l t u i t o i n o n w i w l i l l be b e u sed t o t o d eter e m r i m n i e n e t he co c n o c n e c n e t n r t a r t a i t o i n o n o f o f u n u k n n k o n w o n w n o x o a x l a i l c i c a c a i c d i d a n a d n d u n u k n n k o n w o n w n Fe 3+s  o s l o u l t u i t o i n o . n Furthermore, , t he ex e p x e p r e im i e m n e t n t p r p o r v o i v d i e d s e s t h t e h e o v o e v r e v r i v e i w e w c o c n o c n e c p e t p t a b a o b u o t u t t h t e h e n a n t a u t r u e r e o f o f r e r d e o d x o x r e r a e c a t c i t o i n o n a n a d n d t h t e h re r l e a l t a i t o i n o s n h s i h p i p b e b t e w t e w e e n e n t h t e h e G E G W E W ( g ( r g a r m a m e q e u q i u va v l a e l n e t n w e w i e g i h g t h ) t , ) , t he h e v o v l o u l m u e m e a n a d n d with the h normality a mong t he o xidizing/reduc u ing a gents II. PR OCEDURE A. A Inst In r st u r m u ents: t On O e n e 5 0 5 0 m L m L b u b r u e r t e On O e n e 2 5 2 0 5 0 m L m L v o v l o u l m u e m t e r t i r c i c f l f a l s a k s On O e n e 1 0 1 0 0 0 m L m L g r g a r d a u d a u t a e t d e d c y c l y i l n i d n e d r e Th T r h e r e e e 2 5 2 0 5 0 m L m L b e b a e k a e k r e s r On O e n e 1 0 1 0 m L m L v o v l o u l m u e m t e r t i r c i c p i p p i e p t e Th T e h r e e r e 2 5 2 0 5 0 m L m L E r E l r e l n e m n e m y e e y r e r f l f a l s a k s s k On O e n e g l g a l s a s s s w a w t a c t h c On O e n e f u f n u n n e n l e l ( s ( m s a m l a l l l s iz i e z ) e On O e n e s t s ir i r r i r n i g n g r o r d o On O e n e m e m d e i d c i i c n i e n e d r d o r p o p p e p r e Wa W t a e t r e r b a b t a h t    B. B. Exp Ex e p r e i r m i e m n e t n a t l a l Pro Pr c o e c d e u d r u e r : e 1. 1  Pre Pr p e a p r a e r e KMn O KMn 4s  o s l o u l t u i t o i n o : n c a c lculate the e we w i e g i h g t h t o f o f KMn O KMn 4r  e r q e u q i u r i e r d e d t o prepa p r a e e o f o f a a 0 . 0 05 5 N KMnO4s  o s l o u l t u i t o i n o . n Af ter wei we g i h g i h ng g t he r e r q e u q i u red a mount KMn O KMn 4, , t r t a r n a s n f s e f r e r i t i t t o t o a a 2 5 2 0 5 0 m L m beak a er with 250
5 mL of distilled water. Mix the e solution t horough g ly by y vigorous swirling. The Th n e n t r t a r n a s n f s e f r e r i t i t t o t o a a d a d r a k r k b r b o r wn o wn b o b t o t t l t e l , e , d is i card u n u dissol o v l e v d e d s o s l o i l d. 2. 2  Cle Cl a e n a n t he b u b r u e r t e t wi with d i d s i t s illed e d w a water a n a d n t h t e h n e n r inse it three t im i e m s e s wi t wi h t h 5 5 m L mL po p r o t r ions p  r p e r p e a p r a e r d e d KMn O KMn 4s  o s l o u l t u i t o i n o , n a l a l l o l wi o n wi g n g t he r i r nse s olution t o drain t hough the t ip o f the h e buret ea e c a h c h t i t m i e m . e . Di s Di c s a c r a d r d t h t e h e r i r n i s n e s e s o s l o u l t u i t o i n o . n . F i F l i l l l t h t e h e b u b r u e r t e t wi t wi h t h KMn O KMn 4s  o s l o u l t u i t o i n o n a n a d n d a l a l l o l w o i w t t o t dr d a r i a n i n t ough g h t he h e b u b r u e r t e t t i t p i p u n u t n i t l i l n o n o ai a r i r b u b b u bl b es rema m i a n i n i n t h t e h e t i t p. Re c Re o c r o d r d th t e h e b uret r e r a e d a i d n i g n b  e b f e o f r o e r e b e b g e i g n i n n i n n i g n g t h t e h e t i t t i r t a r t a i t o i n o . n 3. 3  Sta St n a d n a d r a dizat a i t o i n o n o f o f p r p e r p e a p red d KMn O KMn 4solut u i t o i n o : n p i p pe p t e t s ep e a p r a a r t a e t e 1 0 1 0 m L m o L f o f s t s a t n a d n a d r a d r d o x o a x l a i l c i ac a i c d i d s o s l o u l t u i t o i n o n i n i t n o t o t h t r h e r e e e 2 5 2 0 5 0 m L m Er L l Er en e m n e m y e e y r e r f l f a l s a k s s k . s . Ad d Ad d ap a p p r p oxi x m i a m t a e t l e y l y 4 0 4 mL o L f distilled wat wa e t r e r t o t o e a e c a h c h f l f a l s a k s . k . I n I n t h t e h e f u f m u e m e h o h o o d o , d , c a c u a t u i t o i u o s u l s y l y a d a d d d 2 0 2 0 m L m o L f o f 6 6 N H N 2SO4s  o s l o u l t u i t o i n o n t o t – – 0 ea e c a h c h f l f a l s a k s . k . W a W r a m r m t h t e h e f l f a l s a k s s k s i n i n t h t e h e wa t wa e t r e r b a b t a h t h t o t o 85 8 9  0 9 a  n a d n d t i t tr t a r t a e t e t h t e h e h o h t o t s ol o u l t u i t o i n o s n ag a ainst the KMn O4s  o s l o u l t u i t o i n o . n 4. 4  Det De e t r e m r i m n i a n t a i t o i n o n o f o f u n u k n no n wn o wn c on o c n e c n e t n r t a r t a i t o i n o n H 2C2O4s  o s l o u l t u i t o i n o : n p i p pe p t e t s ep e a p r a a r t a e e 1 0 1 0 m L m o L f o unkno n wn conce
c ntration solution of H2C2O4 in i t n o t o t hr h e r e e e 2 5 2 0 5 0 m L m L Erl Er e l n e m n e m y e e y r e f l f a l s a k s s k s a n a d n p  r p o r c o e c e e d e d a s a s d i d r i e r c e t c e t d e d i n i n t h t e h e s t s a t n a d n a d r a d r i d z i a z t a i t o i n o n p r p o r c o e c s e s s . s . Af t Af e t r e r f i f n i i n s i hi h n i g n g t h t e h e t i t t i r t a r tion, , c a c l a cul u a l t a e t th t e h e n o n r o ma m l a i l t i y t y o f o f t h t e h e u n u k n n k o n wn o wn c o c n o c n e c n e t n r t a r t a i t o i n o n H 2C2O4s  o s l o u l t u i t o i n o ; n d e d termine the e a verage g e a n a d n th t e h e s t s a t n a d n a d r a d d d e d v e i v a i t a i t o i n o . n III I . . DATA T AND DIS I CUS U SION O 1. . TITRA TR TION ON OF KMn KM O4S  O S LU OLUTION TI W I W T I H S TH TA T N A D N A D RD R D H 2C2O4S  OL SO UTI LU O TI N ON No  r No m r a m l a i l t i y t y o f o f t h t e h e s t s a t n a d n a d r a d r d H 2C2O4s  o s l o u l t u i t o i n o , n NH2 H C 2 2 C O4 O = 0 = . 0 0 . 5 0 5 N Vol Vo u l m u e m e o f o f t h t e h e s t s a t n a d n a d r a d r d H 2C2O4 so s l o u l t u i t o i n o n u sed, VH2C2O4=  1 = 0 m L m Tria i l # # Bure Bur t e t t e t e re r a e d a i d n i g n g (mL) L Volu l me of o f KMn KM O4(mL) m L) Norm r alit i y of o KMn KM O4(N) 1 1 0. 0 0 . 0 0 - 0 1 - 0 1 . 0 0 . 0 0 0 10 1 . 0 0 . 0 0 0 0. 0 0 . 5 0  2 2 10 1 . 0 0 . 0 0 - 0 1 - 9 1 . 9 8 . 0 8 0 9.80 8 0 0. 0 0 . 5 0 1 5 3 3 19 1 . 9 8 . 0 8 - 0 2 - 9 2 . 9 6 . 0 6 0 9.80 8 0 0. 0 0 . 5 0 1 5 Da D t a a t a Ca C l a c l u c l u a l t a i t o i n: Normality of the e solution is cal a culated b y the e relationship VKMnO4x  x N KMnO4=  = V H2C2O4 H2C2O4x N H2C2O H2C 4 2O The Th e No r No m r a m l a i l t i y t y o f KMn O KMn 4 is: : NKMnO4 = ( = V ( H2 H C 2 2 C O 2 4x  x N H2 H C 2 2 C O 2 4)/VKMn M O n 4 O For Fo r t h t e h e Tr i Tr al a l 1 : 1 : N KMn K O Mn 4 = 0 = . 0 0 . 5 0 x 5 10/10 1 0 = 0 = . 0 05 For Fo r t h t e h e Tr i Tr al a l 2 : 2 : N KMn K O Mn 4 = 0.05x 5 1 x 0 1 /9. 9 8 . 8 = 0. 0 0 . 5 0 1 5 For Fo r t h t e h e Tr i Tr al a l 3 : 3 : N KMn K O Mn 4 = 0.05x 5 1 x 0 1 /9. 9 8 . 8 = 0. 0 0 . 5 0 1 5 Aver e age e Norma m lity t of KMn KM O4 = (0 ( . 0 0 . 5 0 5 + 0.05 0 1 5 1 + + 0 . 0 0 . 5 0 1 5 )/3 3 = 0 = . 0 0 . 5 0 0 5 7 0 7 ( N)  2. . TITRA TR TION ON OF UNKN N OWN OW CONC ON EN E TRA TR TION ON H2C2O4 SOLU SOLUTION TION WI W TH I STAN TA DARD D KMn KM O4S  O S LUTI OLU O TI N ON No  r No m r a m l a i l t i y t y o f o f t h t e h e s t s a t n a d n a d r a d r d KMn O KMn 4 so s l o u l t u i t o i n o , n N(KMn ( O KMn 4) = 0.05 0 5 N Vol Vo u l m u e m e o f o f t h t e h e u n u k n n k o n wn o wn H 2C2O4 so s l o u l t u i t o i n o n u s u e s d e , d V(H2C2O4) = 1 = 0 mL Tria i l # # Bure Bur t e t t e t e re r a e d a i d n i g n g (mL) L Volu l me of o f KMn KM O4 Norm r alit i y of (m ( L m ) L H2C2O4(  N ( ) N 1 1 0. 0 0 . 0 0 - 0 7 - . 7 9 . 0 9 0 7.90 0 0. 0 0 . 3 0 9 3 5 9 2 2 10 1 . 0 0 . 0 0 - 0 1 - 8 1 . 8 1 . 0 1 0 8.10 0 0. 0 0 . 4 0 0 4 5 0 3 3 10 1 . 0 0 . 0 0 - 0 1 - 7 1 . 7 8 . 0 8 0 7.80 0 0.039 3 Da D t a a t a Ca C l a c l u c l u a l t a i t o i n: Normality of the e solution is cal a culated b y the e relationship VKMnO4x  x N KMnO4=  = V H2C2O4 H2C2O4x N H2C2O H2C 4 2O The Th e No r No m r a m l a i l t i y t y o f KMn O KMn 4 is: : NH2C2 C O 2 4 O = ( = V ( KMnO4x NKMnO4)/VH2C2O 2 4 O  For Fo r t h t e h e Tr i Tr al a l 1 : 1 : N H2 H C 2 2 C O4 = 0.05x 5 7 x . 7 9/10 = 0. 0 0 . 3 0 9 3 5 For Fo r t h t e h e Tr i Tr al a l 2 : 2 : N H2 H C 2 2 C O4 = 0 = . 0 0 . 5 0 x 5 8.1/10 = 0 = . 0 0405 For Fo r t h t e h e Tr i Tr al a l 3 : 3 : N H2 H C 2 2 C O4 = 0 = . 0 0 . 5 0 x 5 7. 7 8 . / 8 10 = 0 = . 0 03 0 9 3 Aver e age e Norma m lity t of KMn KM O4 = (0 ( . 0 0 . 3 0 9 3 5 9 +0. +0 0 . 4 0 05 0 +0 5 . +0 0 . 3 0 9 3 )/3 = = 0. 0 0 . 3 0 9 3 7 9 (N) 3. . TITR TI ATI A ON ON OF O UNKN N OWN OW CON O CENTR T ATION TI Fe SO S 4S  OL SO UTI LUTION ON W I W TH TH STAN TA DARD D KMn KM O4S  O S LUTI OLU O TI N ON No  r No m r a m l a i l t i y t y o f o f t h t e h e s t s a t n a d n a d r a d r d KMn O KMn 4 so s l o u l t u i t o i n o , n N(KMn ( O KMn 4) ) = 0 = . 0 0 . 5 0 5 N Vol Vo u l m u e m e o f o f t h t e h e u n u k n n k o n wn o wn H 2C2O4 so s l o u l t u i t o i n o n u s u e s d e , d V(Fe ( SO Fe 4) = 10 0 m L m Tria i l # # Bure Bur t e t t e t e re r a e d a i d n i g n g (mL) L Volu l me of o f KMn KM O4 (mL) L) No N r o m r a m l a i l t i y of FeSO S 4(N ( ) N 1 1 10 1 . 0 0 . 0 0 - 0 1 - 5 1 . 5 8 . 0 8 0 5.80 0 0.0 . 2 0 9 2 2 2 10 1 . 0 0 . 0 0 - 0 1 - 5 1 . 5 9 . 0 9 0 5.90 0 0.0295 9 3 3 10 1 . 0 0 . 0 0 - 0 1 - 6 1 . 6 0 . 0 0 0 6.00 0 0. 0 0 . 3 0 Da D t a a t a Ca C l a c l u c l u a l t a i t o i n: Normality of the e solution is cal a culated b y the e relationship  VKMnO4x  x N KMnO4=  = V FeSO4 FeSO4x  x N FeSO4 FeSO4 The Th e No r No m r a m l a i l t i y t y o f KMn O KMn 4 is: : NFeSO4 = ( = V ( KMn MnO4x x N KMnO4 )/VFeSO4 For Fo r t h t e h e Tr i Tr al a l 1 : 1 : N FeS F O eS 4 O = 0 = . 0 0 . 5 0 x 5 5. 5 8 . / 8 10 1 0 = 0.029 2 For Fo r t h t e h e Tr i Tr al a l 2 : 2 : N FeS F O eS 4 O = 0 = . 0 0 . 5 0 x5. 5 9 . / 9 1 / 0 1 0 = 0.029 2 5 9 For Fo r t h t e h e Tr i Tr al a l 3 : 3 : N FeS F O eS 4 O = 0 = . 0 0 . 5 0 x 5 6 x / 6 10 1 0 = 0 = . 0 0 . 3 0 Aver e a r ge e N o N r o m r a m l a i l t i y t y o f o f Fe S Fe O S 4 = (0 ( . 0 0 . 2 0 9 2 +0 9 . +0 0 . 2 0 9 2 5 9 +0 5 . +0 0 . 3 0 ) 3 / ) 3 / 3 = = 0 . 0 0 . 2 0 9 2 5 9 5 ( N) ( IV I . V . C ON C C ON L C US LU I S O I N ON Wi W t i h t h t h t e h e m e m t e h t o h d o d T i T t i r t a r t a i t o i n o n i n i n t h t i h s i s e x e p x e p r e im i e m n e t n , t , we we c a c n a n c a c l a c l u c l u at a e t e t h t e h e u n u k n n k o n wn o wn c o c n o c n e c n e t n r t a r t a i t o i n o so s l o u l t u i t o i n o n b y b y a d a d d i d ng n g t h t e h e k n k o n wn o wn v o v l o u l m u e m e o f o f t h t e h e s t s a t n a d n a d r a d r i d z i e z d e d s o s l o u l t u i t o i n o n u n u t n il i l t h t e h e r e r a e c a t c i t o i n o n b e b t e we t e we n e th t e h m e m r ea e c a h c e h s e s n e n u e t u r t a r l a i l z i a z t a i t o i n o n t h t r h o r u o g u h g h t h t e h e r e r l e a l t a i t o i n o s n h s i h p i p V ox o idx Noxid i = Vred r x x N red. At the h e e nd n d o f ti t t i r t a r t a i t o i n o , n , t h t r h e r e e e o f o f f o f u o r u r v a v r a i r a i b a l b e l s e s wi l wi l l l b e b e k n k o n wn o wn a n a d n d t he h e u n u k n n k o n wn o wn v a v r a i r ab a l b e l e c a c n a n b e b e d e d t e e t r e m r i m n i e n d e . d