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phương pháp chuẩn độ oxi hóa khử
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CHNG VIII: PHNG PHP CHUN OXI HA KH
CHNG VIII: PHNG PHP CHUN OXI HA KH
Bi 1.
Chun dung dch FeSO4 0,02M bng dung dch K2Cr2O7 0,03M pH = 0. Tnh th ca in cc Pt trong dung dch cn chun nu sai s chun q = - 0,1%.
Li gii
Phn ng chun :
Khi q = - 0,1% = 10-3 => Trc im tng ng.
Bi 89.
Chun 100ml dung dch FeSO4 0,02M v Fe2(SO4)3 0,03M bng dung dch Ce(SO4)2 0,04M pH = 0. Tnh th in cc Pt trong dung dch cn chun sau khi thm: a. 49ml; b. 50ml; c. 51ml dung dch Ce(SO4)2 .
Cho v 300C.
Li gii
Phn ng chun ( 300C):
Theo QTL:
a.V = 49ml => Trc T
Vy:
b. V = 50ml => Ti T
c. V = 51ml. Sau T.
Bi 2
. Chun 50ml SnCl2 0,01M bng FeCl3 trong dung dch HCl.
1. Tnh th ti cc thi im chun sau khi thm: 9ml; 10ml; 10,5ml FeCl3.
2. Tnh sai s chun nu kt thc chun ti E = 0,35V.
Li gii
Phn ng chun :
1.Theo QTL:
a. V = 9ml. Trc T
b. V = 10ml. Ti T.
c. V = 10,5ml. Sau T.
2.Khi dng chun ti E = 0,35 > ET = 0,32=> ti thi im sau tng ng.
Bi 3
Chun 100ml dung dch Fe2+ 0,15N bng KMnO4 0,15N pH = 0.
Tnh th ti cc thi im chun sau khi thm: 99,6ml v 100,4ml dung dch KMnO4.
Li gii
Chun :
1. Phn ng chun :
Theo QTL:
a.V = 99,6ml. Trc T.
b. V = 100,4ml. Ti sau T.
Bi 4
. Chun 25ml Fe2(SO4)3 0,02M bng TiCl3 0,02M pH=0.
1. Tnh bc nhy chun nu coi sai s .
2. Tnh [Fe3+] ti T.
3. Tnh th tch TiCl3 dng lm i mu r cht ch th metilen xanh ( E = 0,33V) v tnh sai s chun . Cho
Li gii
Phn ng chun :
1.Tnh bc nhy chun .
* Ti q = - 0,1% = -10-3. im trc tng ng.
* Ti q = 0,1% = 10-3. Sau T.
2. Phn ng C:
C K>> nn xy ra phn ng nghch.
Ti T, theo QTL ta c: 2C0.V0 = C.V => V= 50ml
C0
C[ ]
=>
Gi s ti T:
=> Ti T nng [Fe3+]
3. Ti T
Ta c:
Ecm< ET => Trc T
Bi 5
Tnh th ti T ca php chun dung dch FeSO4 0,02M bng Br2 0,02M pH = 0.
Li gii
Phn ng C:
K = 106,525
Ti T:
Thnh phn dung dch ti T:
Gi s xy ra cn bng nghch.
n0 C0.V0 2.C.V
n[ ]
Gi s ti T, coi:
( vi C0.V0 = 2.C.V theo QTL)
Thay vo (*) ta c:
Vy:
Bi 6
Feroin c dng lm ch th trong php chun FeSO4 bng dung dch Ce(SO4)2. Phng trnh phn ng nh sau:
Xanh nht
Tnh sai s chun nu chun dung dch FeSO4 0,02M bng dung dch Ce(SO4)2 0,04M n chuyn mu cht ch th t sang xanh r.
300C.
Li gii
Phn ng chun :
Ti T:
Phn ng chuyn mu:
Xanh nht
Ecm > ET => Tnh sai s q ti im sau tng ng.
ES = Ecm = ECe + 0,06.lgq
=> 1,09 = 1,44 + 0,06.lgq
=> q = 1,17.10-6 = 1,17.10-4%
Bi 7:
Chun 20,00 ml dung dch urani (IV) 0,1000 M bng dung dch Ce(SO4)2 0,1000 M pH = 0. Phng trnh chun xy ra:
1) Tnh th ti im tng ng ca php chun .
2) Tnh nng ion U4+ ti im tng ng.
3) Tnh th tch dung dch Ce(SO4)2 phi cho vo dung dch cn chun sao cho th in cc plantin trong dung dch l 0,378V.
4) Tnh th in cc plantin trong dung dch cn chun sau khi thm 50,00 ml Ce(SO4)2 .
5) Tnh sut in ng ca pin:
Hg | Hg2Cl2,KCl bo ho dung dch cn chunti im tng ng | Pt.
Cho .
Li gii
Theo quy tc ng lng : S ng lng U4+ = S ng ca Ce4+
20.0,1=0,1.v
v= 20 ml
1).Tnh th ti im tng ng:
Ta c:
2). Tnh [U4+] ti T:
V k>> nn ta c cn bng ti im tng ng:
k = 10-10,169
nbd c.v c0.v0
[] 2x x
Ta c ti T:
Vy [U4+]=2,91.10-7 (M)
3). th in cc E=0,378(V).Trc im tng ng:
4).Sau khi thm 50 ml Ce(SO4)2 Sau T:
5).Tnh S ca pin ti T :
EMBED Equation.DSMT4
EMBED Equation.DSMT4
EMBED Equation.DSMT4
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