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Exercícios Resolvidos Envolvendo Volumetria de Precipitacao

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U

Um

ma

a m

mo

oe

ed

da

a d

de

e p

prra

atta

a p

pe

essa

an

nd

do

o 0

0,,5

50

00

00

0g

g é

é d

diisssso

ollv

viid

da

a cco

om

m a

acciid

do

o

n

nííttrriicco

o e

e a

a p

prra

atta

a é

é ttiittu

ulla

ad

da

a cco

om

m 3

38

8,,2

22

2m

mll d

de

e u

um

ma

a sso

ollu

uççã

ão

o d

de

e K

KS

SC

CN

N

q

qu

ue

e cco

on

ntté

ém

m 4

4,,5

50

00

00

0 g

g//LL d

de

e sso

ollu

uççã

ão

o.. C

Ca

allccu

ulla

arr a

a p

po

orrcce

en

ntta

ag

ge

em

m d

de

e

p

prra

atta

a na

na mo

moed

eda

a an

anal

alis

isad

ada.

a.

Ag°

Ag°

HNO3HNO3

Ag

Ag

++

(aq)

(aq)

+ SCN

+ SCN

--(aq)(aq)

AgSCN

AgSCN

(aq)(aq)

KS

KS

CN

CN

---

---

-

-

4,

4,

5

5

g/L

g/L

97 ---1,0L ---1,0mol

97 ---1,0L ---1,0mol

4,5000----1,0L ---X

4,5000----1,0L ---X

X= 0,046 mol/L

X= 0,046 mol/L

Nº d

Nº de m

e mmol

molss de

de Ag

Ag

++

= Nº

= Nº de m

de mmol

molss de

de SCN

SCN

-

-(aq) (aq)

= V(ml)x[SCN

= V(ml)x[SCN

--

]

]

= 38,22 x 0,046mol/L

= 38,22 x 0,046mol/L

Nº d

Nº de m

e mmol

molss de

de Ag

Ag

++

=

= 1

1,,7

75

58

8 m

mm

mo

ollss

1

1,,0

0 m

mm

mo

oll d

de

e Ag

Ag

++

----108/1000

----108/1000

1,758 ---y

1,758 ---y

y = 0,1869 g de prata

y = 0,1869 g de prata

0,1869g---K

0,1869g---K

0,5000g---100%

0,5000g---100%

K=38% de prata

K=38% de prata

(2)

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Q

Qu

ua

all d

de

ev

ve

e sse

err a

a m

mo

olla

arriid

da

ad

de

e d

de

e u

um

ma

a sso

ollu

uççã

ão

o d

de

e A

Ag

gN

NO

O

3

3

p

pa

arra

a q

qu

ue

e

sse

en

nd

do

o u

ussa

ad

da

a n

na

a d

de

ette

errm

miin

na

aççã

ão

o d

do

o g

grra

au

u d

de

e p

pu

urre

ezza

a d

de

e a

am

mo

ossttrra

ass d

de

e

N

Na

aC

Cll,, tto

orrn

na

an

nd

do-

o-se

se m

mas

asssas

as d

de

e a

am

mo

ossttrra

ass d

de

e 0

0,,5

50

00

00

0g,

g, cca

ad

da

a 1

1,,0

00

0m

mll d

de

e

sso

ollu

uçção

ão cco

orrrre

essp

po

on

nd

da

a a

a 1

1,,0

0%

% d

de

e N

Na

aC

Cll n

na

a a

amo

mosstr

tra

a an

anal

aliissad

ada?

a?

1,0% ---X

1,0% ---X

100%----0,5

100%----0,5

X=0,005g de NaCl

X=0,005g de NaCl

1

1

mm

mm

ol

ol

de

de

NaC

NaC

l

l

---

---

---

---

58,

58,

45

45

/10

/10

00

00

Y---0,005g

Y---0,005g

Y=0,08

Y=0,08

55 mm

55 mm

ols

ols

de Na

de Na

Cl

Cl

Nº de

Nº de mmols

mmols de AgNO

de AgNO

33

= Nº de mols de

= Nº de mols de NaCl

NaCl

V(

V(mL

mL) x[AgNO

) x[AgNO

33

] = 0,0855

] = 0,0855

1,0mLx[AgNO

1,0mLx[AgNO

33

] = 0,0855mol/L

] = 0,0855mol/L

[AgNO

(5)

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A

Am

mo

ossttrra

a d

de

e 2

20

0 tta

ab

blle

ette

ess d

de

e ssa

acca

ari

rin

na

a sso

ollú

úv

ve

ell ffo

oii ttrra

atta

ad

da

a cco

om

m 2

20

0,,0

0mL

mL

d

de

e sso

ollu

uççã

ão

o 0

0,,0

08

81

18

81

1m

mo

oll//LL d

de

e A

Ag

gN

NO

O

3

3

.. A

Ap

óss rre

em

mo

oççã

ão

o d

do

o ssó

ólliid

do

o,, a

a

ttiittu

ulla

aççã

ão

o d

do

o ffiillttrra

ad

do

o n

ne

ecce

essssiitto

ou

u d

de

e 2

2,,8

81

1 m

mLL d

de

e sso

ollu

uççã

ão

o

0

0,,0

04

41

12

24

4m

mo

oll//LL d

de

e K

KS

SC

CN

N.. D

De

ette

errm

miin

ne

e a

a m

ma

assssa

a e

em

m m

mg

g d

de

e ssa

acca

arriin

na

a e

em

m

ccad

ada

a ttab

able

lette.

e. Se

Segu

gun

ndo

do a

a R

Re

eaç

ação

ão::

NaC

NaC

6

6

H

H

6

6

CONSO

CONSO

2(aq)

2(aq)

+ Ag+

+ Ag+

(aq)

(aq)

AgC

AgC

6

6

H

H

6

6

CONSO

CONSO

2(s)

2(s)

+ Na+

+ Na+

(aq)

(aq)

Nº de mm

Nº de mm

ols

ols

de Sa

de Sa

car

car

ina = Nº de m

ina = Nº de m

mol

mol

s

s

de Ag+

de Ag+

(adc)(adc)

 –

 –

Nº de

Nº de

mmo

mmo

ls

ls

de Ag

de Ag

+

+

(exc)(exc)

=

=

20

20

ml

ml

x

x

0,

0,

08

08

181

181

mo

mo

l/

l/

L

L

-

-

0,

0,

115

115

9

9

=1,6362

=1,6362

 –

 –

0,1159

0,1159

Nº de mmol

(8)

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U

Um

ma

a sso

ollu

uççã

ão

o d

de

e FFe

eC

Cll

3

3

.. 6H

6H

2

2

O cco

O

on

ntte

em

m e

em

m cca

ad

da

a m

mLL,, u

um

ma

a m

ma

assssa

a d

de

e

ffe

errrro

o q

qu

ue

e p

prro

od

du

uzziirriia

a a

attrra

av

éss d

de

e u

um

m p

prro

occe

esssso

o q

qu

uíím

miicco

o a

ad

de

eq

qu

ua

ad

do

o

0

0,,3

30

00

0 m

mg

g d

de

e FFe

e

2

2

O

O

3

3

.. Q

Qu

ua

an

ntto

oss m

mLL d

de

e A

Ag

gN

NO

O

3

3

0,

0,05

050

0 N

N se

serrão

ão n

nec

eces

essá

sári

rios

os

p

pa

arra

a ttiittu

ulla

arr 5

50

0,,0

0 m

mLL d

da

a sso

ollu

uççã

ão

o d

de

e FFe

eC

Cll

3

3

.. 6H

6H

2

2

O?

O?

C = 1,015 g/L de

C = 1,015 g/L de

FeCl

FeCl

3

3

. 6H

. 6H

2

2

O

O

N

N

=

=

m

m

..

=

=

1,015

1,015

..

N = 0,0112 N

N = 0,0112 N

eq

eq g

g .

. V(L)

V(L) 90,06

90,06 .

. 1,0

1,0 LL

PF

PF

3

3

FeCl

FeCl

3

3

. 6H

. 6H

2

2

O

O

Fe

Fe

2

2

O

O

3

3

540,40g

159,70g

540,40g

159,70g

X

X

0,300

0,300 g

g

X= 1,0

X= 1,015m

15mg ou 1,0

g ou 1,015 .

15 . 10

10

-3-3

= 1,015g/L

= 1,015g/L

Nº de

(11)

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O

O cloreto

cloreto contido

contido em

em uma

uma amostra

amostra de

de salmoura

salmoura foi

foi determinado

determinado

pelo

pelo Método

Método de

de Volhard

Volhard.. Uma

Uma alíquota

alíquota de

de 10

10,,00

00 mL

mL da

da salmoura

salmoura

foi

foi tratada

tratada com

com 15

15,,00

00 mL

mL de

de solução

solução 0

0,,1182

1182 mol/L

mol/L de

de AgNO

AgNO

3

3

padronizada

padronizada.. O

O excesso

excesso de

de prata

prata foi

foi titulada

titulada com

com solução

solução 0

0,,101

101

mol/L

mol/L de

de KSCN

KSCN padronizada,

padronizada, requerendo

requerendo 2

2,,35

35 mL

mL até

até a

a formação

formação

do

do complexo

complexo vermelho

vermelho Fe(SCN)

Fe(SCN)

2

2+

+

.. Calcule

Calcule a

a concentração

concentração de

de NaCl

NaCl

na

na salmoura,

salmoura, expressando

expressando o

o resultado

resultado em

em g/L

g/L..

Nº d

Nº d

e mm

e mm

ol

ol

s

s

de N

de N

aC

aC

l

l

= Nº d

= Nº d

e mm

e mm

ol

ol

s

s

de A

de A

g

g

++

adic

adic

 –

 –

Nº de

Nº de

mmo

mmo

ls

ls

de A

de A

g

g

++ exc exc 

= 15,00 x 0,1182 molL

= 15,00 x 0,1182 molL

-1-1

-

-

0,2374 mmols

0,2374 mmols

= 1,773

= 1,773

 –

 –

0,2374

0,2374

Nº de

Nº de

mmo

mmo

ls

ls

de Na

de Na

Cl

Cl

= 1,5

= 1,5

356 m

356 m

mol

mol

s

s

Nº de

Nº de

mmo

mmo

ls

ls

de A

de A

g

g

++ exc 

exc 

= Nº de m

= Nº de m

mol

mol

s

s

de KS

de KS

CN

CN

= 2,3

= 2,3

5 mL

5 mL

x 0,1

x 0,1

01 mo

01 mo

l L

l L

-1-1

1,

1,

0 mm

0 mm

ol

ol

s

s

Na

Na

Cl

Cl

---

---

---

---

58

58

,4

,4

5 / 1

5 / 1

000

000

1,5356mmol NaCl----W

1,5356mmol NaCl----W

(14)

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O

O m

mééttooddoo ddee F

Faajjaannss uussaa uum

m iinnddiiccaaddoorr fflluuoorreesscceennttee,, qquuee ssee aaddssoorrvvee

eelleettrroossttaattiiccaam

meennttee ssoobbrree aa ssuuppeerrffíícciiee ddee uum

m pprreecciippiittaaddoo ccoollooiiddaall

im

imed

edia

iata

tame

ment

ntee ap

após

ós oo po

pont

ntoo es

este

tequ

quio

iomé

métr

tric

ico.

o. T

Tai

aiss in

indi

dica

cado

dorres

es po

poss

ssue

uem

m co

corres

es

ddiiffeerreenntteess nnooss eessttaaddooss lliivvrree ((eem

m ssoolluuççããoo)) ee aaddssoorrvviiddoo.. O

O iinnddiiccaaddoorr m

maaiiss

ccoom

muum

meennttee uussaaddoo éé aa eeoossiinnaa ((tteettrraabbrroom

moofflluuoorreesscceeíínnaa)).. O

O ddiirreettoorr ddee uum

m

llaabboorraattóórriioo ddee aannáálliisseess qquuíím

miiccaass ffeecchhaa aaccoorrddoo ccoom

m uum

m cclliieennttee ppaarraa

ddeetteerrm

miinnaaççããoo ddee bbrroom

meettoo nnuum

m ggrraannddee nnúúm

meerroo ddee aam

moossttrraass.. P

Paarraa eevviittaarr

ppeerrddaa ddee tteem

mppoo ccaallccuullaannddoo oo rreessuullttaaddoo ffiinnaall,, oo qquuíím

miiccoo rreessppoonnssáávveell ppeellaass

aannáálliisseess qquueerr qquuee oo vvoolluum

mee ddee A

AggN

NO

O

33

uussaaddoo nnaa ttiittuullaaççããoo ddee F

Faajjaannss sseejjaa

iigguuaall,, nnuum

meerriiccaam

meennttee,, àà ppeerrcceennttaaggeem

m ddee bbrroom

moo nnaa aam

moossttrraa.. SSee ddee ccaaddaa

aam

moossttrraa sseerrããoo ppeessaaddooss 550000,,00 m

mgg,, qquuaall aa ccoonncceennttrraaççããoo m

moollaarr ddee A

AggN

NO

O

33

que

que

de

deve

verá

rá se

serr us

usad

ada?

a?

V

V

AgNO3AgNO3

= % Br

= % Br

 – –

1,0

1,0

mL

mL

=

=

1%

1%

de

de

Br

Br

-

-Nº de m

Nº de m

mol

mol

s

s

de B

de B

r

r

--

= N

= N

º de

º de

mm

mm

ol

ol

s

s

de

de

Ag

Ag

++

0,062257 mmol

(17)

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SSee 2255,,00 m

mL

L ddee uum

maa ssoolluuççããoo ddee N

NaaC

Cll ffooii rreeqquueerriiddoo ppaarraa

pprreeccip

ipiita

tarr qquuaannttiittaattiivvaam

meennttee aa pprraattaa ddee uum

maa ssoolluuççããoo oobbttiiddaa ppeellaa

ssoolluubbiilliizzaaççããoo ddee 00,,22336655 gg ddee pprraattaa m

meettáálliiccaa ccoom

m ppuurreezzaa ddee 9988%

%..

Q

Quuaall aa ccoonncceennttrraaççããoo ddaa ssoolluuççããoo ddee N

NaaC

Cll eem

m gg//m

mL

L ee

normalidade.

normalidade.

Pure

(20)

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U

Um

maa ssoolluuççããoo ddee A

AggN

NO

O

33

00,,11000000 m

mooll L

L--11 ffooii ppaaddrroonniizzaaddaa ccoom

m oo

ssaall K

KC

Cll ddee 9999,,6600%

% ddee ppuurreezzaa.. U

Um

maa aam

moossttrraa ddee 00,,33225544gg ddoo ssaall éé

ttiittuullaaddaa ddee aaccoorrddoo ccoom

m oo m

mééttooddoo M

Moohhrr ggaassttaannddoo 4444,,2200m

mL

L ddaa

ssoolluuççããoo ddee A

AggN

NO

O33.. C

Caallccuullee oo ffaattoorr ddee ccoorrrreeççããoo ddaa ssoolluuççããoo ddee

AgNO

(22)

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U

Um

ma

a a

am

mo

ossttrra

a iim

mp

pu

urra

a d

de

e K

KII p

pe

essa

an

nd

do

o 2

2,,1

14

45

5g

g ffo

oii d

diisssso

ollv

viid

da

a e

em

m

a

ag

gu

ua

a,, a

acciid

diiffiicca

ad

da

a e

e ttrra

atta

ad

da

a cco

om

m 5

50

0,,0

0 m

mLL d

de

e A

Ag

gN

NO

O

3

3

0,0243mol/L.

0,0243mol/L.

A

Ap

óss a

a p

prre

ecciip

piitta

aççã

ão

o d

do

o A

Ag

gII,, a

a m

miissttu

urra

a rre

essu

ulltta

an

ntte

e ffo

oii ttiittu

ulla

ad

da

a cco

om

m

K

KC

CS

SN

N 0

0,,1

12

21

1 m

mo

oll//LL ,, o

ob

bsse

erv

rva

an

nd

do

o--sse

e u

um

m cco

on

nssu

um

mo

o d

de

e 3

3,,3

32

2 m

mLL p

pa

arra

a

ttiittu

ulla

arr o

o e

exxcce

esssso

o d

de

e A

Ag

gN

NO

O

3

3

.. C

Ca

allccu

ulla

arr a

a p

po

orrcce

en

ntta

ag

ge

em

m d

de

e K

KII n

na

a

amostra.

amostra.

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N

Noo m

mééttooddoo ddee L

Liieebbiigg,, oo cciiaanneettoo éé ddeetteerrm

miinnaaddoo ppoorr ttiittuullaaççããoo ccoom

m

AgNO

AgNO

33

pa

padr

drão

ão,, qu

quan

ando

do as

as se

segu

guin

inte

tess rrea

eaçõ

ções

es ac

acon

onte

tece

cem:

m:

2CN

2CN

--

+

+ Ag

Ag

+

+

[Ag(CN)

[Ag(CN)

22

]]

--[Ag(CN)

[Ag(CN)

22

]]

--

+

+ Ag

Ag

+

+

Ag[Ag(CN)

Ag[Ag(CN)

22

]]

A

A sseegguunnddaa rreeaaççããoo iinnddiiccaa oo ppoonnttoo ffiinnaall ddaa ttiittuullaaççããoo.. U

Um

maa

aam

moossttrraa ddee N

NaaC

CN

N,, ppeessaannddoo 00,,44002299 gg,, éé ddiissssoollvviiddaa eem

m áágguuaa ee

ttiittuullaaddaa ccoom

m 4400 2255 m

mL

L ddee A

AggN

NO

O 00 11001122 m

mooll L

L

-1

-1

C

Caallccuullee oo ggrraauu

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E

Em

m uum

maa ttiittuullaaççããoo ddee 5500,,00 m

mL

L ddee N

NaaC

Cll 00,,11000000 m

mooll L

L

-1

-1

ccoom

m A

AggN

NO

O

33

00,,11000000 m

mooll L

L

-1

-1

,, ddeetteerrm

miinnaarr aass ccoonnce

cennttrraaççõõeess ddooss ííoonnss cclo

lorreeto

to ee ííoonnss

pprraattaa nnaa ssoolluuççããoo,, eem

m ccaaddaa uum

maa ddaass sseegguuiinntteess ssiittuuaaççõõeess::

Kps

Kps

AgCl

AgCl

=

= 11,5

,566xx1100

-10

-10

Após a adiçã

Após a adição de 20,0 mL

o de 20,0 mL de

de AgN

AgNO

O

33

;;

[Cl

[Cl

--

] = (V

] = (V

Cl-Cl-

x C

x C

Cl-Cl-

)

)

 – 

 – 

(V

(V

Ag+Ag+

x C

x C

Ag+Ag+

) =

) =

(0,05 x 0,1000)

(0,05 x 0,1000)

 – 

 – 

(0,02 x 0,1000) = 4,28 x 10

(0,02 x 0,1000) = 4,28 x 10

-2-2

mol /L

mol /L

(V

(28)

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U

Um

maa aam

moossttrraa ddee N

NaaC

Cll ee N

NaaII ppeessaannddoo 00,,44000000 gg pprroodduuzziiuu aattrraavvééss ddee

uum

m ttrraattaam

meennttoo qquuíím

miiccoo aaddeeqquuaaddoo uum

m pprreecciippiittaaddoo ddee A

AggC

Cll ee A

AggII

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U

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25,00 mL

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É

É ssaabbiiddoo qquuee oo m

mééttooddoo ddee V

Voollhhaarrdd ppooddee sseerr eem

mpprreeggaaddoo ppaarraa aa

ddeetteerrm

miinnaaççããoo ddee hhaalleettooss.. C

Coonnssiiddeerree uum

maa aam

moossttrraa ccuujjaa m

maassssaa eem

m

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Um químico deseja determinar o grau de

Um químico deseja determinar o grau de pureza de um lote de

pureza de um lote de ferrocianeto

ferrocianeto

de potássio, cujo r

de potássio, cujo rótulo diz estar

ótulo diz estar trihidratado

trihidratado {K4[Fe(CN)6] . 3H2O}

{K4[Fe(CN)6] . 3H2O}. Ele

. Ele

pesa, então, 1,6801 g

pesa, então, 1,6801 g do sal como

do sal como amostra, transfere para um erlenmeyer

amostra, transfere para um erlenmeyer,,

dissolve em água e titula com uma solução de Ce(IV) em

dissolve em água e titula com uma solução de Ce(IV) em H2SO4, gastando

H2SO4, gastando

39,63 mL. Esta mesma solu

39,63 mL. Esta mesma solução gastou na padr

ção gastou na padronização 49,06 mL

onização 49,06 mL para titular

para titular

50,00 mL

50,00 mL de uma solução contendo 4,848 g/L

de uma solução contendo 4,848 g/L de

de As2O3. Qual é o grau de

As2O3. Qual é o grau de

pureza do ferrocianeto de potássio?

pureza do ferrocianeto de potássio?

As

As

22

O

O

33

+ 6OH

+ 6OH

--

2AsO

2AsO

33

3-

3-

+ 3H

+ 3H

22

O

O

AsO

AsO

33

3-

3-

+ 3H

+ 3H

+

+

H

H

33

AsO

AsO

33

H

H

33

AsO

AsO

33

+ 2Ce

+ 2Ce

4+

4+

+ H

+ H

22

O

O ↔

HAsO

HAsO

44

2-

2-

+ 2Ce

+ 2Ce

3+

3+

+ 4H

+ 4H

+

+

Ce

Ce

4+

4+

+ Fe

+ Fe

2+

2+

Ce

Ce

3+

3+

+ Fe

+ Fe

3+

3+

Nº d

Nº de m

e mmol

molss de

de

Ce

Ce

4+4+

= (nº de

= (n

º de mmo

mmols

ls deH

deH

33

AsO

AsO

33

)x 2

)x 2

49,06[Ce

49,06[Ce

4+4+

]

] = (

= (50ml

50ml x

x 0,0489)x2

0,0489)x2

[Ce

[Ce

4+4+

] =(50 x 0,0489)x2 = 0,0996mol/L

] =(50 x 0,0489)x2 = 0,0996mol/L

As

As

22

O

O

33

--

----

----

---2

-2 H

H

33

AsO

AsO

33

197,8g

197,8g ---251,84

---251,84gg

4,848---W

4,848---W

W= 6,170 g

W= 6,170 g

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U

Um

maa aam

moossttrraa ddee m

miinnéérriioo ddee ffeerrrroo,, ppeessaannddoo 775500,,00 m

mgg,, ffooii ddiissssoollvviiddaa eem

m áácciiddoo ee

ttrraattaaddaa ppaarraa ooxxiiddaarr ttooddoo oo ffeerrrroo aaoo ííoonn fféérrrriiccoo.. A

Appóóss eelliim

miinnaarr ttooddoo oo eexxcceessssoo ddee

aaggeennttee ooxxiiddaannttee,, uum

m eexxcceessssoo ddee K

KII ffooii aaddiicciioonnaaddoo.. O

O II22 lliibbeerraaddoo rreeqquueerreeuu 1188,,5500

m

mL

L ddee N

Naa22SS22O

O33 00,,00775500 m

mooll L

L--11 ppaarraa ttiittuullaaççããoo.. Q

Quuaall éé aa ppeerrcceennttaaggeem

m ddee ffeerrrroo

na

na am

amos

ostr

tra?

a?

Re

Reaç

açõe

õess qu

quím

ímic

icas

as:: 2F

2Fee

3+

3+

+

+ 22II-- ↔

2Fe

2Fe

2+

2+

+

+ II

22

II

22

+

+ 2S

2S

22

O

O

33

2-

2-

2I

2I

--

+

+ SS

44

O

O

66

2-

2-2Fe

2Fe

3+3+

+ 2S

+ 2S

22

O

O

332-2-

2Fe

2Fe

2+2+

+ S

+ S

44

O

O

662-

2-Nº de

Nº de mm

mmol

olss de

de

Fe

Fe

3+3+

= n

= nº d

º de m

e mmo

mo,s

,s de

de SS

22

O

O

332-

2-= 18,50ml x 0,0750 mol/L

= 18,50ml x 0,0750 mol/L

Nº de

Nº de mm

mmol

olss de

de

Fe

Fe

3+3+

=1,3875 mmols

=1,3875 mmols

1,

1,0 mm

0 mmol

ol de

de

Fe

Fe

3+3+

---0,05585 g

---0,05585 g

1,3875mmols---X

1,3875mmols---X

X= 0,07749g

X= 0,07749g

750mg

750mg ---

--- 0,75g

0,75g ---100%

---100%

0,0774---Y

0,0774---Y

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A

A prata contida em 24,00 mL

prata contida em 24,00 mL de uma solução 0,1000 mol L-1 é precipitada

de uma solução 0,1000 mol L-1 é precipitada

completamente como

completamente como Ag3AsO4. O precipitado isolado é

Ag3AsO4. O precipitado isolado é dissolvido em ácido

dissolvido em ácido

forte e o

forte e o arsen

arseniato

iato deter

determinad

minado iodomet

o iodometrica

ricament

mentee com Na2S2O

com Na2S2O3 requ

3 requere

erendo

ndo

20,00 mL

20,00 mL desta solução. Qual

desta solução. Qual é a concentração da s

é a concentração da solução de tiossulfato

olução de tiossulfato em mol

em mol

L-1? Resposta: 0,0800 mol L-1.

L-1? Resposta: 0,0800 mol L-1.

As

As

44

O

O

66

(s) + 6H

(s) + 6H

22

O

O

↔↔

4H

4H

33

AsO

AsO

33

3Ag

3Ag

++

+ AsO

+ AsO

3 333

--↔ ↔

Ag3AsO

Ag3AsO

33

AsO

AsO

33

33

--

+ I

+ I

33

--

+ H

+ H

22

O

O ↔

AsO

AsO

44

3-

3-

+ 3I

+ 3I

--

+ 2H

+ 2H

+

+

IO

IO

33

--

+ 8I

+ 8I

--

+ 6H

+ 6H

+

+

3I

3I

33

--

+ 3H

+ 3H

22

O

O

2S

2S

22

O

O

33

22

--

+ I

+ I

33

--

SS

44

O

O

66

22

--

+ 3I

+ 3I

--AsO

AsO

3333

--

+

+ IO

IO

33

--

+ 2I

+ 2I

--

+

+ 2S

2S

22

O

O

3322

--

+

+ 4H

4H

++

AsO

AsO

4433

--

+ I

+ I

33

--

+ S

+ S

44

O

O

6622

--

+2H

+2H

22

O

O

Nº d

Nº de mm

e mmol

olss de

de

Ag

Ag

++

= n

= nº de

º de mm

mmols

ols de

de

AsO

AsO

443-

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Uma amostra de minério de ferro, pesando 750,0 mg, foi dissolvida

Uma amostra de minério de ferro, pesando 750,0 mg, foi dissolvida

em ácido e tratada para oxidar todo

em ácido e tratada para oxidar todo o ferro ao íon

o ferro ao íon férrico.

férrico. Após

Após

eliminar todo o excesso de agente oxidante, um excesso de KI foi

eliminar todo o excesso de agente oxidante, um excesso de KI foi

adicionado. O I

adicionado. O I

2

2

libera

liberado re

do requereu

quereu 18,50 m

18,50 mL

L de Na

de Na

2

2

S

S

2

2

O

O

3

3

0,0750

0,0750

mol L

mol L

-1

-1

 para titulação. Qua

 para titulação. Qual é a percentagem

l é a percentagem de ferro na amostra?

de ferro na amostra?

Reações químicas:

Reações químicas:

2Fe

2Fe

3+

3+

+ 2I

+ 2I

--

2Fe

2Fe

2+

2+

+ I

+ I

22

II

22

+ 2S

+ 2S

22

O

O

33

2-

2-

22

I-

I-

+ S

+ S

44

O

O

66

2-

2-m=

m= 750mg=

750mg= 0,75g

0,75g

f=

f= 0,0750x10=

0,0750x10= 0,75

0,75

Fe

Fe

3+3+

%

%

=

=

1

1

0

0

0

0

x

x

0

0

,

,

7

7

5

5

x

x

0

0

,

,

0

0

0

0

5

5

5

5

8

8

5

5

x

x

1

1

8

8

,

,

5

5

F

F

e

e

3+3+

%

%

=

=

1

1

,

,

7

7

4

4

9

9

2

2

F

F

e

e

3+3+

% = 10,33

% = 10,33

0,75

0,75

0,75

0,75

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