Esercizi svolti della parte di teoria di Chimica analitica 1

MI

15 =

= . ,

.

COz--

zn En Cob

+

S -

0 0

1 15

- .

.

f 05

0 0

05

- . . 3

mmal

[Cos] -

05

0 2 M

no

= = .

. 25 ML

zitCosaal k1k2

Xcoz

ZnCoz(s) 398

0

= = .

2

[H ]

] ke[H +

+ knka

[Cg2]) +

+

S (S a coz2-

+ · 11

[zn2[co32] s($[0]).

kps -

2.

=

= =

* [zn2]

10

Kps 51 unico in

2 M

S presente Sz

a

=

= .

= .

[COz2] Scoza-

. Ent]

Ezn25 ? ?

all'ex

b) libero

+ot S

ZnCa

mL M

10 01

0 .

+ invariato

0 comme

01

Nazloz pH

15 ML lo

M - =

.

+ Znlt

H LOG Kf 3

e =

↳ pkan 8

HL 5

Per = , I

+

Xzn2 0103

0

= = ,

En2taa) 032 kf Cy)

1

=

zn(os(s) #

+

+ . Ec

- =

I

I [CO32])

(S

S <Coz2-

Xzn2 + 103

+ .

. k+ 969

kf 969

11 0 =

= .

=

- . 8 5

][CO32] -

[CO32]) .

[zn2 (5 THath 969

Xco

+

Kps Xzn2

+. XL

S : = 0

-

.

= = =

.

= : .

1085

+

100

S [en2]

KPS 46

2 M Tot

=

= .

= ,

l I

+.

Xzn2 Xcoz-

2 .

. [zn(t]

]

+

[zn2 +

=

100 -

Jea

Ezn2 + 2 46

S += 0103

[zna 53

2

0

M 10

.

=

.

= .

. , ,

.

Verifica ApproX :

10-3 12 3

[C0329 -

2 no

· . . .

↳ Trascurabile

Ent

E

* C +

= 0

22

]

C' -

+ =

(zn( 1

0

0

G 1

=

= -

- .

. 10-0 8 10-6

[zn2]ex

J

[zn( Ezn2 En2t)

+

]

+ S 46 2

2 10- 43

53

2

Tot =

= =

-

= . .

- .

-

, ,

.

S ? 8

pH

AgCN

2) = 16

AgCN -

Per Kps 2 no

= .

9

Per pka

HCN 4

= . KaHen

CN-

+

AgCNAg

a) ↓ 0383

0

CN-

+ = = .

3

[H

S ka

+

S XCN- IN

+ +

.

Kps [Agt][CN] S S XCN -

=

= . =

. M Ag

=

EPS 32

S 7 .

= = . -

pH P

b) Sagan 2 M

10

= . CN-

+

AgCNAg +

S S XCN-

.

J [CN]

[Ag S

+

Kps XCN-

S

= .

= .

7)2

10-16 (2 10- XCN-

.

2 = . .

16 10-3

- 5

XCN 2 10

= .

. =

10-1

6 . ka

XCN - = 3

[H + wa

+ 3

10-9 4

103 -

[H 92

7 pH

M 7

10 1

5 . = =

=>

= .

. , ,

10-9 4

[H ]

+ ,

+

Il

&

Esercizi : Agt Tit cl-0

100

1) mL con 11

. 1

-

Kps

10 8 10

1

C

mL

per = = .

,

UtoT

inizio

a 100 ML

=

+ -AgC

Ag C

-

+ Aginizio

mmal c 1M

0

mmal 10 ml Immo

= =

.

= .

pea

[Ag] mma pAg

1 01M 2

0

= =

= .

wo mL

b) Vc 5 Vot 105

mL ML

= =

5

mmac mL 0

0 1r 5 mmal

=

= . . .

=

mmaAg 100 01M

0

ML immo

=

. .

Ag

+ C -AgC

-

+ 10

[Agt] Mmal

.

0 5

i 4 M

76

- PAg

OS 2 32

1 = . =

= ,

& ,

d ML

los

5 0 5

0

F - .

.

Va Mom)

c) Utot oped

mc

10 =

=

mmac- += 1 mmal

mmol Ag

=

+

Ag C AgC

- -

+

i -

1

M

f t

- 1

Ag((s) + Cal

Ag cox) + S

S 10-

[Agt] S pAg 87

Kps .

4

1 34 M

= = .

= =

,

d) va Vot 112mL

12 mL =

= mma

+=

mmd Ag 1 1r

0

mma ML mmol

2

12 1

- . =

= . .

Age

+ -

a

Ag +

i S

2

1

1 ,

f 1b

0

- 2

. 103M

[C] mmal

.

0 2 1 78

= .

= ,

112mL Agcoal C19Q)

Age(s) + 13

S S 78

1

+ .

,

3)

S($

[Agt][C]

Kps 1 78 -

1

+

=

= .

,

78.13

10-0 5 1

8

,

1 S

= .

. , pAy

= ]

[Ag

+ 7

S pag

-

0

1 10 = =

.

= . -

ma e e

ot B

32

2 8

, 87

4

2 ⑤

.

7 O &

⑤ I D

I id ML

12

CrOn20

+

PD2 (peal

con8mL

2) 5

3

pH

TIT 1 M

25 ML =

. ,

16

2

KPSpp2 -

10

.

+ =

HeCro pkar 6

0

phan 52

75

X = =

= . ,

OML 25 ML

a Von- Viot

= =

Grou2-

+

PD2 PbCrOu

>

+ Pb2

+ 0

Mmal

Grou

Mmol 1 M

8 mL 0 mmal

inizio 8

= =

= . .

.

per

]

[Pb2 + 0

mma PPD 49

8 032

0 1

M

= =

=

. ,

.

23 mL

b) Vo- 29m

UTOT

4 ML =

= 0 8

mmdpb2 += mmol

.

co mmal

0 4

MI 0

4 er

mmal .

= = ,

.

Grou2-

+

PD2 PbCrOu

>

+

i08 0 4 S

,

I 4 d

0

F 0

4 - .

.

[PD2 ]

+ ppb 1

0 mmol 86

0138M

4 0

= =

=

. . ,

29mL

Vo 33

8 ML UTOT

c) -

ML pea

= =

mmal goc-

+= 0

mmapb2 8 mmol

= .

Grou2-

+

PD2 PbCrOu

>

+

i 08 0 S

8

,

S t

S

& 8

F 0 . Gonna

Pb2caal

PDCOn(s) - + Xow-

S

S . um coniugata di acido diprotico

Base 10-4

k1k2 9

cro- 52

X .

= = ,

2 ]

j

[H kn[H k1k2

+ + +

+ * ]

[Pb2

S 6

KpS pPb

10 +

58 34

4 M =

=

.

= = , ,

Xcr2- Vot

d) Vo 8 33

8 m

8

mL

= =

. ,

mmal += mmol

pb2 8

0 . I Pb

Go

Gu

mma 0

8 88M

1m P

mi

0

8 mc .

= = .

, .

Croc-

+

PD2 49e

1

Pacron

>

-

+ ,

S

i 86

4

80 .

1

88

0 ·

.

. 34

6

0

08 8 2

f 0

- ·

,

, 8 8 &

05

10

OlTRe-D

10 % I

.

il

[con] 10-3 pea

mmal

08

0 2 37 M

= = .

. ,

33 8 mL ↑

v

ppb

,

Pb2caa) GOnlaal

PbGOn (s) + 103)

(S

S Non

37

2

+ ·

-

, ⑨

--

10-3)

S[(X don]

Kps 37

2

+ .

= .

,

16 10-1

- 36

8

S 2 l

.

= = .

.

10-3 10-

9 52

2 37 I

.

. . .

, &

]

[Pb2 +

= &

pPb 10 05

= I

! 8 D

, 8

,

AgNOz

30

3) ML 3

[NH3] - M

2 10

= - 0

11 ma 01M

- .

10-10

Kps =

+

Ag 3

LGBr 3

/NH3 = ,

LG Br 2

7

= , + 1

Ag" Br

Ag(NHy)

NH3 = =

+ AgINHalà

* Ba

Agt =K

2NH3 a

+

Va - OML 30mL

a) UTOT =

= 0

Aginiziali 01 11

0 mmol

mmalci per M

mmol 11 ml =

= = . ,

.

3 m ]

] [Ag(N +

CAg - [Ag(N

+

[Ag

mmal 3)2

+

0 3) +

11 3 67 +

10 +

+

= . =

=

. ,

30mL 3 105m

[Agt] <Ag XAg - 5 36

67

3 0146

l

CAg 10 0

+

= · = . . .

=

= .

.

,

B2[NH3]2

Br[NH3]

1 +

+

27

pAg 4

= ,

b) 35

5 mL

5

Va UTOT 5 mL

= =

. ,

mmal Agt mmd

0 11

= . 055

0

Mmolc - mmel

= .

Ag

+ AgC

a - >

-

+

0 0055

- 11 -

. & /

055 0 055

-0 ,

,

b 3

10

055

.

0

CAg 1

mmal 55 M

= = .

,

35 5 ML

,

10.3

[NH3] 10 M

30 ml

2 69

= 1

. . = .

,

5

35 mi

. 103 10-3M

]

[Ag l

+ CAg pAg

3 50

12

55 4

Tag 1

= = =

. .

· =

. .

,

Ba[NH3]2

Br[NH3]

1 +

+

2) UTOT 0

41 m1

Uc per

ML

11 o

=

= ,

Agt

mmd mma 21

0 mmol

- = = .

Ag + AgC

C -

-

+

i0 0 11

1 S

.

, S 114

S

F 0

.

Agtaal

Ag((s) claal

+ 10-3. 10-3M

[NH3]

S

S 30m)

2 46

XAg 1

+ = .

.

. ,

=

41mL

A

(S

Kps XAgt) S & 0264

0

Agt

. .

= = = ,

2

BaINH5]

Br[NH3]

1 +

+

M

= = A

6

S 15 .

= . 0

[Agt] 79

S

<Ag 5

Xg+ pAg

<Ag+ 62 - M

10

1 =

= =

= .

- . ,

,

d) Vot 1

G2

Vc 12 1 mi

ML =

= ,

, 121

0

01

1

Mmdc- 12 mmol

0 =

.

= .

.

,

Ag 1

1

Mma 0

=

Ag

+ Aga

-

C

+

i I

0110 121

.

↑ / 0116

0 0

011 ,

, "M

[4] 011 10

0 61

2

= . .

= ,

42 1

, 3

3

[NH3] 2 10

-

10 30 41

1

= . .

. = ,

.

42 1

Ag( +

Ag a -

, + (aQ

col 10-4

S 61

S

+

XAy 2

+ .

. , 0282

0

XAgt = ,

10-4)

↓ps (S

(S 61

<Ag ) 2

+ +

= . .

. 10"

10-10 5 (S 61

0 0282 2

+

= . .

. ,

. *

5 M [Agt] 10

S CAg 63

CAg

29 bAg

-

1 3

(Ag M

1 44

+ 6

= = = =

=

+

. . ,

, ,

I pAg

mcc pAga

·

O 27

4 a

, 50

4

5

5 o

.

, -

519

11 &

I

6 44 &

1

.

OlTRe-D 12

10 % ,

il pea & D

mic-

10/11

Esercizi 10-2

+

Ag pH

EDTA E

M

1) mL

10 + =

Cl-10-3 ]

[Ag

+

20 m) M

TIT con Agy3-

y4

+ - -

iniziale Ag

Sz

al +

: 2

13 S

i -

2 10

. 10-3 3

-

f 2.

8 l

.

1 0

l 0123

XAg =

= Ci .

1048 3

ki -

1 10

1 +

+ .

. 10

* kzk3ku

Xy" 1 .

5 02 .

= =

<

]" kn

]" J

[H

[H [H + /A*]knkzk3

+ Knkak3ka

+ knkz +

+ + +

10 - 4

Kr kf xy - 1

= = .

. 153 10-3M

2