Chapter 20

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

Chapter 20

20-1

(a)  2 Mn

2+

! 2 O 8

+

2−

(b)  #a$iO' &  s % + 2Ce

(c)  H 2O 2

 +

("

+

(d)  *&OH % "

−

(e)  2 MnO"

(f)  ClO'

20-2

−

+

−

−

+

+

20-3

+

6H

+

2 MnO "

−

+

16H

+

+

−

+  

+

2H  

+

+

+

2H  

 

 →

*O

2+

!O2 + 2Mn 2

'+ 2 + Cl

  →

→

−

+

+

+

A)Cl& s % + 'H 2O

+

8H 2O

'H 2O

+

+

O "

2−

+

"H

+

+ ',n& s % + 12H + →  ',n 2+ + 2MoO'+ + 8H 2 O !H#O2 −

2−

−

+

!$r 

(e)  2 HAsO '   (f)  2 +

2+

 

(b)  2 H 2 MoO"

(d)  $rO'

(O   2

!H 2 O2 + 6H

−

2−

+ "H + → $iO  + + 2Ce "+ + 2H 2 O + #a +

O 2 & g % + 2 H 2 O →  2 Fe 2

+

(c)  2MnO"

'+

A)& s % + Cl

+

6+

+

(a)  2 Fe '+

+

→

8H 2 O → 10  O "

+

+

+

H

+

→  

C6 H ! #H 2

2Mn

+

2+

'H 

+  

+

! #O'

→

−

+

'H 2O

$r 'C6 H ! #H 2

+

−

'$r 

+

'H 2 O

+ O 2 &  g % →  2 HAsO " 2−  

2 H#O 2

+

2H

+

+

→  2 +

2 #O + 2 H 2 O

Only in the presence of Cl- ion is A) a sufficiently )ood reducin) a)ent to e /ery useful for prereductions. +n the presence of Cl- the half-reaction occurrin) in the alden reductor is A)& s % + Cl

−

− → A)Cl   & % +e

 

s

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

he e3cess HCl increases the tendency of this reaction to occur y the common ion effect.

20-4

Amal)amation of the 4inc pre/ents loss of rea)ent y reaction of the 4inc 5ith hydronium ions. 2+

+ 2A)& % + "H +  +  2 Cl −

→ ←

+ 2A)Cl& % + H 2 O

(O 2

20-6

2iO

20-7

tandard solutions of reductants find some5hat limited use ecause of their susceptiility

2+

 s

+

,n& s %

+

" H 

+

  → ←

2i

'+

+

(

"+

20-5

,n

2+

 s

+

2H 2 O

to air o3idation.

20-8

tandard MnO" solutions are seldom used to titrate solutions containin) HCl ecause of the tendency of MnO"- to o3idi4e Cl- to Cl2 thus causin) o/erconsumption of MnO "-.

20-9

Cerium &+*% precipitates as a asic o3ide in al7aline solution.

20-10 2 MnO "

−

+

'Mn 2

+

+

 

2 H  2 O →   !MnO 2 & s % + " H

+

20-11 Freshly prepared solutions of perman)anate are ine/italy contaminated 5ith small

amounts of solid man)anese dio3ide 5hich cataly4es the further decompositions of  perman)anate ion. $y remo/in) the dio3ide at the outset a much more stale standard rea)ent is produced.

20-12 tandard perman)anate and thiosulfate solutions are )enerally stored in the dar7 ecause

their decomposition reactions are cataly4ed y li)ht.

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

20-13 " MnO "

−

Chapter 20

+ 2H 2 O → " MnO    2 & s % + 'O 2 & g % + "OH −  ro5n

20-14 olutions of  2Cr 2O are used e3tensi/ely for ac7-titratin) solutions of Fe 29 5hen the

latter is ein) used as a standard reductant for the determination of o3idi4in) a)ents.

20-15 +odine is not sufficiently sufficiently solule in 5ater to produce a useful standard rea)ent. +t is uite

solule in solutions containin) e3cess + - ecause of formation of triiodide. 20-16 he solution concentration of + '- ecomes stron)er ecause of air o3idation of the e3cess

+-. he reaction is

6+

−

 

+

O 2 & g % + " H  

+

20-17  2 O '

2−

+

→

2+ '

+

H →     HO'

−

+

−

+

2H 2 O

& s %

20-18 hen a measured /olume of a standard solution of +O' is introduced into an acidic

solution containin) an e3cess of iodide ion a 7no5n amount of iodine is produced as a conseuence of a reaction. −

−

+

+O '

 

+

!+

+

 

→

6H

+

'+ 2

'H 2 O

20-19

$rO' +2

+

−

+

6+

−

e3cess

2 2 O '

2−

+

6H

→

+

2+

−

−

→

$r 

+

+

"O6

2−

'+ 2

+

'H 2O

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

20-20

Cr 2 O 

2−

+

6+

20-21 2+ 2

+

e3cess 2−

+

+2

−

 # 2 H " 

−

→

2 2 O ' +

1" H 2+

→

+

→

2Cr '

+

+

'+ 2

+

H 2O

2−

+

"O6

 

 # 2 &  g % + " H

+

+

"+

−

20-22 tarch is decomposed in the presence of hi)h concentrations of iodine to ) )i/e i/e products

that do not eha/e satisfactorily satisfactorily as indicators. his reaction is pre/ented y delayin) the addition of the starch until the iodine concentration c oncentration is /ery small.

20-23

0.22!6 ) sample ×

(a) 

(b) 

(c) 

1000 mmol Fe2

".0';61 mmol Fe

2+

"+

× 1 mmol Ce2+ =  0.11"2 M Ce "+

'!.' m<

mmol Fe

".0';61 mmol Fe 2+ '! .' m<

'!.' m<

  =

!!.8" )

".0';61 mmol Fe 2 +

".0';61 mmol Fe 2+

+

2−

× 1 mmol Cr 2 O2+ =  0.01;0" M Cr 2 O  2−

×

6 mmol Fe

1 mmol MnO " ! mmol Fe

2+

−

=  0.0228" M MnO " −

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

(d) 

(e) 

".0';61 mmol Fe 2+ '!.' m<

".0';61 mmol Fe 2

+

Chapter 20

×

1 mmol * &OH % "

×

1 mmol +O '

'! .' m<

mmol Fe

" mmol Fe

2+

−

2

+

=  0.11"2 M *&OH % " +

=  0.028!! M +O ' −

+

20-24

0.02!00 mmol  2 Cr 2 O 

× !00.0 m< × 2;".18! )  2 Cr 2 O  = '.6 )  2 Cr 2 O  1000 mmol

m<

=issol/e '.6 )  2Cr 2O in 5ater and dilute to !00.0 m MnO 2

20-34 'C& #H 2 % 2

+

"$rO'

−

+

 

 

'H 2O → 'CO& #H   2 %2

+

'O "

2−

+

−

" $r 

+

6H

+

" mmol $rO'- ≡ ' mmol C&#H2%2

 0.008'' mmol $rO ' ' mmol C& #H 2 % 2 6.122 ) C& #H 2 % 2      × 1".1 m< $rO ' × ×   " mmol $rO 1000 mmol m< '     0.01! ) sample ×100 > = ;.'8> C& #H 2 % 2

20-35 MnO "

−

+

!Fe 2

+

+

8H   →   Mn 2 +

 

+

+

!Fe '

+

+

"H 2 O

1 mmol MnO"- ≡ ! mmol Fe29 ≡ !?2 mmol Fe2O'

mmol MnO "

=

0.02086 MnO " m<

 

× ';.21  m< MnO " = 0.81; mmol MnO "

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

(a)

  ! mmol Fe 2+ 1 mmol Fe !!.8" ) Fe   0.81; mmo< MnO   2+ " mmol MnO mmol Fe 1000 mmol "    × × ×  ×100>   0.120 ) sample

= '2.08> Fe (b)

     0.81; mmo< MnO " × ! mmol Fe 2 O ' × 1!;.6;2 ) Fe 2 O '   2 mmol MnO " 1000 mmol      × 100> = "!.86> Fe 0.120 ) sample

'+ 20-36 'n

+ Cr 2 O  2− + 1" H + 

→ 'n " + ←

+ 2Cr '+ +  H 2 O

1 mmol Cr 2O2- ≡ ' mmol n'9 ≡ ' mmol nO2

2

O'

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

(a)

mmol n

=

0.01'! mmol  2 Cr 2 O 

' mmol n '+

× 2;. m<  2 Cr 2 O  ×

m<

mmol  2 Cr 2 O 

×

1 mmol n '

mmol n

= 1.!";!' mmol n

+

  118.1 ) n     1.!";!' mmol n × 1000 mmol    ×100> = "2.2> n   0."'!2 ) sample

(b)

mmol nO 2

=

0.01'! mmol  2 Cr 2 O  m<

× 2;. m<  2 Cr 2 O  ×

' mmol nO 2 mmol  2 Cr 2 O 

  1!0.1 ) nO 2   1.!";!' mmol nO 2 ×   1000 mmol      ×100> = !'.66> nO 0."'!2 ) sample

20-37

2 H 2 #OH Cr 2 O 

2−

+

+

" Fe '

6 Fe 2

+

 # 2 O & ) % + " Fe 2

+

→ ←

1" H

+

+

→ ←

2Cr '

+

+

+

+

"H

6 Fe '

+

+

+

+

H 2O

H 2O

2

= 1.!";!' mmol nO 2

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

 #OH 1 mmol Cr 2O2-≡ 6 mmol Fe'9 ≡ ' mmol H2 #OH

 0.0'2! mmol  2 Cr 2 O     × 1;.8' m<  2 Cr 2 O  × ' mmol H 2 #OH   mmol   Cr  O   m< 2 2      = 0.0'8 M H 2 #OH !0.00 m< sample 20-38

,n 2

+

+

H 2C 2O "

2 MnO "

−

+

→

,nC 2 O " & s % + 2 H

!H 2 C 2 O "

+

6H

+

→

+

2 Mn 2

+

+

10CO 2 & g % + 8 H 2 O

2 mmol MnO"- ≡ ! mmol #a 2C2O" ≡ ! mmol ,nO

 0.01!08 mmol MnO " ! mmol ,nO 81.1'; ) ,nO      × '.81 m< MnO " × × m< 2 mmol MnO " 1000 mmol      ×100> 0.;280 ) sample

= 12."6> ,nO

20-39 &#ote: +n the first printin) of the te3t the ans5er in the ac7 of the oo7 5as in error.%

ClO'

−

+

6Fe

2+

+

6H

+

 

  Cl

→

−

+

'H 2 O + 6Fe

'+

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

mmol Fe 2+

=

2+

0.08;'0 mmol Fe 2+ m<

mmol Fe titrated  y Ce 1 mmol Fe mmol Ce

2+

"+

"+

=

Chapter 20

× !0.00 m< Fe 2+ = "."6!0 mmol Fe 2+

0.08'610 mmol Ce "+ m<

×1".;' m< Ce "+ ×

= 1.2"8' mmol Fe 2+

mmol Fe 2+ reacted 5ith ClO '

= "."6!0 − 1.2"8' = '.216 mmol Fe 2+

  122.!"; ) ClO '   '    '.216 mmo< Fe 2+ × 1 mmol ClO ×   6 mmol Fe 2+ 1000 mmol  × 100> = !1.'> ClO   0.12; ) sample

20-40

+ + 2 → @&C 2 H ! % ' + + C 2 H ! + 2− 2− + 2 + 2 2 O ' → 2+ − +  " O 6 @&C 2 H ! % "

1 mmol +2 ≡ 2 mmol 2O'2- ≡ 1 mmol @&C2H!%"

'

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

mmol + 2

=

0.020;! mmol + 2 m<

mmol + 2 titrated  y  2 O ' 1 mmol + 2 2 mmol  #a 2 2 O '

2−

×1!.00 m< + 2 = 0.'1"2!0 mmol + 2

=

0.0'"6! mmol  #a 2 2 O '

20-41 H ' AsO '

m<

× 6.0; m<  #a 2 2 O ' ×

= 0.10!!1 mmol + 2

mmol + 2 reacted 5ith @ & C 2 H ! % "

0.208" mmol + 2

Chapter 20

= 0.'1"2!0 − 0.10!!1 = 0.208" mmol + 2

'2' ." m) @ & C 2 H ! % " mmol < 2! .00 m< sample × 1000 m<

×

1 mmol @ & C 2 H ! % " mmol + 2

×

=

2.0 × 10 ' m) @ & C 2 H ! % " < sample

  H '  AsO " + 2+ − + 2 H + + + 2 + H 2O →

1 mmol +2 ≡ 1 mmol H'AsO' ≡  mmol As2O'

 0.01;8! mmol + 2 1 mmol As 2 O ' 1; .8"1 ) As 2 O '      × 2".!6 m< + 2 × ×   2 mmol + 1000 mmol m< 2  ×100 >   ."1 ) sample

= 0.6!1 > As

O 2

'

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

20-42 Cr 2 O 

2−

+

'( "

+

+

+

 

2H   → '(O 2

2+

+

Chapter 20

2Cr '

+

+

H 2O

1 mmol Cr 2O2- ≡ ' mmol ("9 ≡ 1 mmol #aCl

   0.100 mmol  2 Cr 2 O   ×1;.; m<  2 Cr 2 O  × 1 mmol  #aCl × !8.""2 )  #aCl   m< mmol  2 Cr 2 O  1000 mmol    ×100>   2!.0 m< 0.800 ) sample × !0.0 m<

= 2;.1>  #aCl

20-43 2C 2 H !H

mmol + 2

+ + 2 → C 2  H !C  2 H ! + 2+ − + 2 H +

=

0.012;' mmol + 2 m<

mmol + 2 titrated  y  2 O ' 0.01"2! mmol  #a 2 2 O ' m<

2−

× !0.0 m< + 2 = 0.6"6! mmol + 2

=

× 1!.2 m<  #a 2 2 O ' ×

mmol + 2 reacted 5ith C 2 H !H

1 mmol + 2 2 mmol  #a 2 2 O '

= 0.1120 mmol + 2

= 0.6"6! − 0.1120 = 0.!'"! mmol + 2

     0.!'"! mmo< + 2 × 2 mmol C 2 H !H × 62.1" ) C 2 H !H   mmol + 2 1000 mmol      × 100> = ".''> C 1.!'" ) sample

2

H !H

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

20-44

'eO2

2−

Cr  O 2

+

Cr 2 O 

+

6Fe 2

2−

+

+

+

8H

+

1"H

+

→

'H 2 eO"

→

2Cr '

+

+

+

2Cr '

+

6Fe'

+

+

+

H 2O

H O



2

mmol  2 Cr 2 O 

=

0.0'11" mmol  2 Cr 2 O  m<

mmol  2 Cr 2 O  titrated  y Fe 2+ 0.11'! mmol Fe 2+

× !0.00 m<  2 Cr 2 O  = 1.!!0 mmol  2 Cr 2 O 

=

×10.0! m< Fe 2+ ×

m< mmol  2 Cr 2 O  reacted 5ith eO 2

1 mmol  2 Cr 2 O  2+

= 0.1;01 mmol  2 Cr 2 O 

6 mmol Fe = 1.!!0 − 0.1;01 = 1.'668 mmol  2 Cr 2 O 

    1.'668 mmo<  2 Cr 2 O  × ' mmol eO 2 × 1!;.6 ) eO 2   mmol  2 Cr 2 O  1000 mmol      × 100> = 1'.16> eO ".;1 ) sample

20-45

2+

−

+O' +2

+ −

+

$r 2

+

!+

→ −

22 O '

+ 2−

+2

+

6H →

+

−

2 $r  →

2+

−

'+ 2 +

+

'H 2 O

"O 6

2−

2

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

1 mmol + ≡ 1 mmol +O'- ≡ ' mmol +2 ≡ 6 mmol 2O'2-

 0.0!;82 mmol  #a 2 2 O ' 1 mmol + 166 .00 ) +   2 2 ' m<    ×1;.;6 m<  #a  O × 6 mmol  #a 2 2 O ' × 1000 mmol    ×100> 1.'0; ) sample

= 2.!2"> +

20-46 MnO "

−

  Mn 2 + + !Fe '+ + " H 2 O + !Fe 2+ + 8 H +   → ←

mmol Fe

=

0.01;20 mmol MnO " m<

×1'.2 m< MnO " ×

! mmol Fe mmol MnO "

×

!00.0 m< !0.00 m<

= 1'.112 mmol Fe mmol Fe and Cr  =

0.01;20 mmol MnO " m<

× '6."' m< MnO " ×

= 1."86" mmol Fe and Cr  mmol Cr  = (1."86" − 1'.112 ) = ".'1!2   mmol Cr 

(1'.112 mmol Fe ) ×

!!.8" ) Fe

1000 mmol 1.06! ) sample

( ".'1!2 mmol Cr ) ×

×100> = 6;.0> Fe

!1.;;6 ) Cr  1000 mmol

1.06! ) sample

×100> = 21.0> Cr 

! mmol Fe mmol MnO "

× !00.0 m< 100.0 m<

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

20-47 +n the alden reductor

* &OH % "

+

+ 2 H + +  e −

→ ←

*O 2 +

+ 'H 2 O

* 2+

+ "H 2O

+n the Bones reductor

* & OH % "

+

+ " H + + '  e −

→ ←

+n the first titration

Ce "+

+ Fe 2+ → Fe '+ + Ce '+ + Ce "+ + *O 2+ + 'H 2 O → *&OH % " + Ce '+ + 2 H +

mmol Fe and *

= mmol Ce

"+

=

0.1000 mmol Ce "+     m<

×1." m< Ce "+ = 1."0 mmol Fe and *

+n the second titration

Ce "+

+ Fe 2+ → Fe '+ + Ce '+ + 'Ce "+ + * 2+ + " H 2 O → *&OH % " + 'Ce '+ + " H +

mmol Fe and ' × *

= mmol Ce "+ =

0.1000 mmol Ce " + m<

× "".6 m< Ce "+ = "."60 mmol Fe and ' × *

utractin) the first euation from the second euation eu ation )i/es

 

Fundamentals of Analytical Chemistry: 8th e  ed d. "."60 −1."0

=

2.6;'0

=

Chapter 20

2 × mmol *

2.6;'0 = 1.'"6! mmol * 2 1.'"6! mmol * mmol *2 O ! = = 0 .6'2! mmol * O 2 ! 2 mmol Fe = 1."0 −1.'"6! = 0."2! mmol Fe mmol *

=

mmol Fe O 2

' =

0."2! mmol Fe 2

( 0.6'2! mmol *2 O ! ) ×

=

0.21'! mmol Fe 2 O '

181 .88 ) *2 O ! 1000 mmol

×100> = ".8!> *2 O !

2.!!; ) sample × !0.00 m< !00 .0 m< 1!; .6; ) Fe 2 O ' 1000 mmol !0.00 m< 2.!!; ) sample × !00 .0 m<

( 0.21'! mmol Fe 2 O ' ) ×

× 100> = 1'.'"> Fe 2 O '

20-48

2l

+

+

CrO "

2−

→

l 2 CrO " &s%

2l 2 CrO " &s% + 2 H Cr 2 O 

2−

+

6Fe 2

+

+

+

→

1" H

"l +

+

→

+

Cr 2 O 

2Cr '

+

+

2−

+

H 2O

6Fe '

+

+

H 2O

6 mmol Fe29 ≡ 1 mmol Cr 2O2- ≡ " mmol l9

0.10"" mmol Fe 2+ m<

× ';.!2 m< Fe

2+

" mmol l + ×   6 mmol Fe 2 +

×

20".'8 ) l + 1000 mmol

= 0.!622 ) l +

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

20-49

O2 &  g % + 2OH −

+O

+

−

2H O

'

2

→ +

O'

2Cl

−

'

2−

→

+

H 2O −

+Cl 2

+

2−

O

+

2H

+

"

1 mmol +O'- ≡ 2 mmol H2O' ≡ 2 mmol O2

+n

2.!0 <

× 6".00 min = 160 < of  sample there are

min 0.00'12! mmol +O ' m<

× ".;8 m< +O ' ×

2 mmol O 2 mmol +O '

     1.;;"02 ×10 −' ) O   6 2  ×10  ppm = 10."  ppm O 2  1 . 20 )    160 < ×   <     20-50

+ 2 O ! & s % + !CO& g %  → !CO 2 & g % + + 2 & g % +2

+ 2 2 O '2− → 2+ − +  " O 6 2−

×

6".06! ) O 2 1000 mmol

= 1.;;"02 ×10 −' ) O 2

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

1 mmol +2 ≡ ! mmol CO ≡ 2 mmol  2O'2-

0.00221 mmol  2 O ' m<

2−

× .6 m<  2 O ' 2− ×

! mmol CO 2 mmol  2 O '

     1.200; ×10 −' ) CO   6   ×10  ppm = "0.!  ppm CO  2". < × 1.2 )       <   20-51

 2−

+ + 2  → & s % + 2+ − 2− 2− + 2 + 2 2 O ' → 2+ − +  " O 6

1 mmol +2 ≡ 1 mmol H2 ≡ 2 mmol 2O'2-

2−

× 28.01 ) CO = 1.200; ×10 −' ) CO 1000 mmol

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

=

mmol + 2 added

0.0100 mmol + 2 m<

=

mmol + 2 in e3cess

Chapter 20

× 10.00 m< + 2 = 0.100 mmol + 2

0.01'"" mmol  2 O ' m<

2−

×12.8! m<  2 O ' 2− ×

1 mmol + 2 2 mmol  2 O '

= 0.086'! mmol + 2 = mmol H 2 = 0.100 − 0.086'! = 0.0206! mmol H 2

mmol + 2 reacted

 0.0206! mmol H  × '".082 ) H 2      2 1000 mmol    × 10 6 ppm = 1;.!  ppm H 2 1 .2 )    '0.00 < ×    <     20 20-52 -52 (a)

A)+ + 22 O ' '$r 2 !+

−

+

+

+

−

+O'

+ −

2−

→

A)&2 O ' % 2

'H 2 O → +O'

+

6H

+

→

2−

+

2 +

2 O 2

'

'+ 2

−

+

+

2+

O "

−

6$r 

2−

+

+

'H 2 O

−

→

'−

6

+

−

+

6H

+

2−

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

Chapter 20

(b) 1 mmol +O'- ≡ ' mmol +2 ≡ 1 mmol A)+ ≡ 6 mmol 2O'2-

0.0'!2 mmol  2 O '

2−

×1'. m<  2 O ' 2− ×

m<

1 mmol A)+ 6 mmol  2 O '

2−

×

2'" . m) A)+ mmol

".00 cm 2

=

".2 m) A)+ cm 2

20-53

O2

+

" Mn&OH % 2 & s %

" Mn& OH % ' & s %

+

12 H

+

+

0.00;"2 mmol  2 O ' m<

2H 2O +

2−

"+

−

→ ←

→ ←

" Mn& OH % ' & s % " Mn 2

+

+

2+ 2

+

6H 2 O

× 1'.6 m< 2 O ' 2− ×

1.0' m) O 2

     2! m< sample × 1!0 m<   1!" m<      

=

0.0"2' m) O 2 m< sample

1 mmol O 2 " mmol  2 O '

2−

×

'2.0 m) O 2 mmol

= 1.0' m) O 2

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

A

Chapter 20

B

C

D

E

F

G

H

25.00 0 mL of 0.025 0.025 M SnCl2 it! 0.050 M F"Cl# 1   20-54 (a) Titration of 25.0 2

Reaction: Sn2+ + 2Fe3+  →  Sn4+  + 2Fe2+

#

For Fe3+/Fe2+, Eo

4 5

For Sn /Sn , E

0.154Equivalence point will e at 25.00 ! 0.025 ! 2/0.050"25.00 #$

%nitial concentration Sn 2+

0.025

$

&oncentration Fe3+

%

'olu#e Sn&l2 (o  (olution

&

Equivalence point volu#e

'

4+

2+

0.771

o

0.05 25.00 25.00 #

"r"nta*"+

,ol. F"  mL

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10

10

2.50

0.0023

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12

30

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1#

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14

50

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G22(7B7#267B74)# E2# 2#( (7B 7B7 7$6 $6B2 B2##-7B 7B7 756 567B 7B7% 7%662) 2)( (7B 7B7% 7%B B2# 2#)) F2#(7 F2#(7B75 B7567B 67B7%6 7%62) 2)(7B (7B7% 7%B2# B2#))

C10( (7 7B7 B7$6 $6B1 B10 02 2) )(7 (7B7 B7% %B1 B10 0) 2'   C10 D10(7 (7B75 B7567B 67B7%7%-7B7 7B7$6B $6B10. 10.2) 2)(7B (7B7% 7%B10 B10)) #0   D10 #1   G107B74-(0.05'22)6L8G10(D10C10) #2 ## #4 #5 #$ #% #& #' 40 41 42 4# 44 45

 

G2#7B7#-0.05'26L8G10(F2#E2#)

 

Fundamentals of Analytical Chemistry: 8th e  ed d.

A

B

Chapter 20

C

D

E

F

G

H

-

off 0.0&4$% M :a2S28# it! 0.102#5 M ;2 (;# ) 1   20-54 (9) Titration of 25.00 mL o 2

Reaction: 2S232- + %3-  →  S4)2-  + 3%-

#

For S232-/S4)2-, Eo

4

For %3-/%-, Eo

5

%nitial concentration S232-

0.0*

Equivalence point will e at 25.00 ! 0.0*4)7 / 2 / 0.10235"10.34 0.53)#$

$

&oncentration %3

%

'olu#e a2S23 (  (o olution

&

Equivalence point volu#e

'

0.0*4)7

-

0.10235 25.00 10.34 #

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