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Pdf report-3 compress - Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM

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Pdf report-3 compress - Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM

111 56 lượt tải Tải xuống





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

4.4.

Then the standardized KMnOThen the standardized KMnO

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

solution: calculate the weight of KMnOsolution: calculate the weight of KMnO

required to prepare of a 0.05 Nrequired to prepare of a 0.05 N

KMnOKMnO

solution. After weighing the required amount KMnOsolution. After weighing the required amount KMnO

, 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

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

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

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

SOSO

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

and titrate the hot solutionsand titrate the hot solutions

against the KMnOagainst the KMnO

solution.solution.

4.4. Determination of unknown concentration HDetermination of unknown concentration H

solution: pipet separate 10 mL ofsolution: pipet separate 10 mL of

unknown concentration solution of Hunknown concentration solution of H

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

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

SOLUTION WITH STANDARD HSOLUTION WITH STANDARD H

SOLUTIONSOLUTION

Normality of the standard HNormality of the standard H

solution,solution, NN

H2C2O4H2C2O4

= 0.05 N= 0.05 N

Volume of the standard HVolume of the standard H

solution used,solution used, VV

H2C2O4H2C2O4

= 10 mL= 10 mL

Trial Trial # # Burette Burette reading reading (mL) (mL) Volume Volume of of KMnOKMnO

(mL) (mL) Normality Normality of of KMnOKMnO

(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

KMnO4KMnO4

x Nx N

KMnO4KMnO4

= V= V

H2C2O4H2C2O4

x Nx N

H2C2O4H2C2O4

The Normality of KMnOThe Normality of KMnO

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

== (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

SOLUTION WITHSOLUTION WITH

STANDARD KMnOSTANDARD KMnO

SOLUTIONSOLUTION

Normality of the standard KMnONormality of the standard KMnO

solution,solution, NN(KMnO(KMnO

) = 0.05 N) = 0.05 N

Volume of the unknown HVolume of the unknown H

solution used,solution used, VV(H(H

) = 10 mL) = 10 mL

Trial Trial # # Burette Burette reading reading (mL) (mL) Volume Volume of of KMnOKMnO

(mL)(mL)

Normality ofNormality of

(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

KMnO4KMnO4

x Nx N

KMnO4KMnO4

= V= V

H2C2O4H2C2O4

x Nx N

H2C2O4H2C2O4

The Normality of KMnOThe Normality of KMnO

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

== (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

SOLUTION WITHSOLUTION WITH

STANDARD KMnOSTANDARD KMnO

SOLUTIONSOLUTION

Normality of the standard KMnONormality of the standard KMnO

solution,solution, NN(KMnO(KMnO

) = 0.05 N) = 0.05 N

Volume of the unknown HVolume of the unknown H

solution used,solution used, VV(FeSO(FeSO

) = 10 mL) = 10 mL

Trial Trial # # Burette Burette reading reading (mL) (mL) Volume Volume of of KMnOKMnO

(mL) (mL) Normality Normality ofof

FeSOFeSO

(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



KMnO4KMnO4

x Nx N

KMnO4KMnO4

= V= V

FeSO4FeSO4

x Nx N

FeSO4FeSO4

The Normality of KMnOThe Normality of KMnO

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

== (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