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°
HNO3HNO3Ag
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
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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
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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
pó
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
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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
vé
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
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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
++ excexc
= 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
-1-11,
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
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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
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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
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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
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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
pó
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
-2-2mol /L
mol /L
(V
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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
33AsO
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
22O
O
33--
----
----
---2
-2 H
H
33AsO
AsO
33197,8g
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
22O
O
332-2-↔
↔
2Fe
2Fe
2+2++ S
+ S
44O
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
22O
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
44O
O
66(s) + 6H
(s) + 6H
22O
O
↔↔4H
4H
33AsO
AsO
333Ag
3Ag
+++ AsO
+ AsO
3 333 --↔ ↔Ag3AsO
Ag3AsO
33AsO
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
22O
O
3322--
+
+ 4H
4H
++↔
↔
AsO
AsO
4433--
+ I
+ I
33--
+ S
+ S
44O
O
6622--
+2H
+2H
22O
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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