Phu Luc 5 - Tai Lieu Giang Day Hoa Ly 2(1)

7/29/2019 Phu Luc 5 - Tai Lieu Giang Day Hoa Ly 2(1)

1/30

Ph lc 5TRNG I HC TR VINHKHOA HA HC NG DNG

B MN DCDLH&VLNN

TI LIU GING DY

MN THC TP HA L 2

GV bin son: NGUYN TH ANH TH

Tr Vinh, nm 2013

Lu hnh ni b


2/30

MC LC

Ni dung Trang

Ti liu ging dy Mn Thc tp Ha L 2


3/30

BI 1

XC NH DN IN NG LNG CC I CACHT IN LY MNH V CHT IN LY YU

I. C s l thuyt

dn in ng lng cc i ca cht in li mnh CH 3COONa, NaCl, HCl

c xc nh t dn in cc nng khc nhau nh nh lut Conraus

(1)

o dn in ring ca cc chui dung dch ca cc cht trn cc nng

bit trc. dn in ng lng c tnh t dn in ring o c theo biu

thc

(2)

l dn in ring ca nc dng pha dung dch. Nu


4/30

III. Thc nghim:

1. Chun b cho th nghim:

Ra sch tt c m dng c thy tinh, trng bng nc ct.

Ra sch in cc nhiu ln bng nc ct, cho n khi dn ca nc ra

xung di 2 cm-1.

Ghi nhn dn ca nc ct dng pha dung dch, lu nc ct phi c

dn di 2 cm-1.

Lu : ch dng nc ct cho th nghim.

2. Hiu chnh my

Ly 10ml dung dch KCl 0.1M cho vo bnh nh mc 100ml. Thm nc n

vch nh mc c dung dch chun KCl 0.01M.

a nhit quy chiu ca my o dn v 25 0C.

a h s nhit ca dn v 1.91%.

Ly khong 40ml dung dch KCl 0.01M va pha vo cc 50ml. Trng in cc

nhiu ln.

b dung dch . Ly li 50ml dung dch chun KCl 0.01M, o v ghi dn

. Ghi nhn hng s in cc K1.

Tnh hng s in cc mi K2 theo biu thc:

= = K1.

Chnh hng s in cc v gi tr K2 va tnh c.

Chnh h s nhit v 0 dn in hin th l dn nhit th nghim.

3. Xc nh dn in ng lng cc i ca CH3COOH

Ti liu ging dy Mn Thc tp Ha L 2 2


5/30

Pha 100ml dung dch vi nhng nng sau t dung dch CH3COONa 0.1N.

5.00 x 10-2N 3.00 x 10-2N 1.50 x 10-2N

1.00 x 10-2

N 6.00 x 10-3

N 3.00 x 10-3

N

Lu : Th tch dung dch ly pha long khng c di 10ml m bo

chnh xc cn thit. Dng buret c chnh xc 0.05ml ly dung dch. Cn la

chn cc dung dch dng pha long sao cho t phi ra, trng dng c nht. Ghi s

liu dng pha dung dch vo bng.

STT

Dung dch trc pha long Dung dch sau pha long

Nng Th tch

pha longNng

Th tch sau

pha long

dung dch pha vo cc 50ml sch. Nh trng cc bng chnh dung dch sng.

Cho cc cc ng dung dch va pha vo bnh n nhit. Ch cc cc cng

ngp su trong nc cng tt. Ch cho cc dung dch n nh v nhit phng.

Trong lc ch i ra sch in cc bng nc ct cho n khi dn ca nc ra

xung di 2cm-1. Trong lc ch n nh nhit hy pha cc chui dung dch k tip

tit kim thi gian.

o dn ca tng dung dch, bt u t dung dch long nhtnh sau. t

nh giy lau pha bn ngoi cc l ca in cc thm kh in cc. Tuyt i

khng a giy lau vo trong in cc. a in cc vo dung dch sao cho dung dch

ngp l trn thnh in cc. Trng in cc vi ln bng cch nng ln v h in cc

xung. Ch n khi n nh mi ghi nhn gi tr dn v nhit vo bng.

Lu : Nhit ca cc dung dch khng c chnh lch qu 0.1oC.



6/30

o ln lt t dung dch long n dung dch c.

Cht in li:

dn in ring ca nc ct:

STT CN T(0C)

1. Lp li bc 1 n bc 3 cho NaCl v HCl.

2. Tnh cc i lng cn li trong bng cho mi chui dung dch. Lu n v

ca cc i lng.

3. Dng th cho mi chui dung dch. Dng ng hi quy

tuyn tnh. Tnh phng trnh hi quy. Xc nh dn in ng lng cc i.

4. Ngoi suy ng hi quy v tung gc xc nh dn in ng lng

cc i. So snh gi tr ly t th v gi tr t phng trnh hi quy.


Cu hi (bi tp) cng c:1. Tnh dn in ng lng cc i ca CH3COOH.

2. So snh dn in ng lng cc i thc nghim ca CH3COONa,

NaCl, HCl v CH3COOH vi cc gi tr tra cu t s tay ha l.

3. Hy tnh dn in ng lng ca AgIO3 298oK, bit dn in

ng ca NaIO3, CH3COONa, CH3COOAg 298oK ln lt l 9,11; 9,10;

10,28cm2

/om. lg.


7/30

BI 2

PHNG PHP CHUN pH

NH LNG HN HP ACID H2SO4 V H3PO4

I. C s l thuyt

Khi trung ha mt acid (n hay a acid) bng baz mnh, pH tng dn trong qu

trnh trung ha. ng pH = f(V) (vi V l th tch dung dch NaOH thm vo) c

nhng dng khc nhau th theo acid c trung ha l acid mnh hay acid yu. Vi

acid a chc, nu cc chc ca acid c pKa khc nhau qu 4 n v, ta c th trung

ha tng chc mt. T gi tr th tch NaOH mi im tng ng, ta suy ra nng

tng lng ca acid.

Trong bi thc hnh ny, chng ta s tin hnh chun hn hp hai acid H2SO4

v H3PO4bng dung dch NaOH chun. T s liu thu c, v ng pH = f(V),

ng cong ny c hai im un ti hai bc nhy tng ng vi hai im tng

ng. im tng ng th nht: chun H 2SO4 v chc th 1 ca H3PO4. im

tng ng th 2: chun chc th 2 ca H3PO4.

vic xc nh Vt chnh xc, ta c th da vo:


Mc tiu hc tp: Sau khi hc xong bi ny, ngi hc c th:- nh lng hn hp acid H2SO4 v H3PO4 bng phng php chun pH


8/30

- th pH/V theo Vtb

- Tnh 2pH/(V)2

II. Ha cht v dng c

Buret My o pH NaOH 0.1NPipete My khuy t Acid oxalic

Erlen a khuy Hn hp H2SO4 v H3PO4

Bnh nh mc 100 ml C t Phenolphtalein

Becher cc c

III. Thc nghim

1. Xc nh nng dung dch chun NaOH

Chun li dung dch NaOH 0.1N bng H 2C2O4 0.1N vi ch th phenolphtalein.

Np dung dch NaOH vo buret 25ml, dng pipet ly 10ml dung dch acid oxalic 0.1N

vo erlen 250ml, thm vo 3 git phenolphtalein. Chun bng dung dch NaOH

n khi dung dch xut hin mu hng nht bn trong 30 giy, ghi th tch NaOH tiu

tn. Lp li th nghim t nht 3 ln, ly kt qu trung bnh. T tnh ra nng

chnh xc ca NaOH.

2. Chun dung dch hn hp H2SO4 v H3PO4

a. Chun th

Chun my o pH vi dung dch m l 4; 7; 10

Dng pipet ht 10 ml hn hp phn tch cho vo beaker 100 ml, thm nc ct n

ngp in cc (khong n vch 60 ml). Cho c t vo khuy trn u dung dch

(trnh ng c t chm in cc), ghi gi tr pH trn my khi s hin ln n nh.

Sau , mi ln thm vo 1ml dung dch NaOH (thm khong 20 ln), ghi gi tr pH

ng vi th tch NaOH thm vo. T xc nh Vt1vVt2 gn ng (ng vith tch 1ml NaOH thm vo pH tng cao nht).

b. Chun tinh

Ht 10ml dung dch hn hp cho vo beaker 100ml, thm nc ct n ngp in

cc (khong n vch 60ml). Lp li qu trnh nh chun th n khi cch Vt1

gn ng 2ml th mi ln thm vo 0.2ml NaOH. Khi cch Vt1 gn ng 1ml th mi

ln thm 0.1ml NaOH. Cho n khi qua Vt1 gn ng 1ml, tng th tch NaOH mi



9/30

ln thm l 0.2ml. Khi qua Vt1 gn ng 2ml, thm vo 1ml NaOH. Lm tng t nh

trn i vi im tng ng th hai.

Ngng chun khi qua im tng ng th hai khong 3ml.

Ra sch in cc bng nc ct v ngm in cc trong dung dch KCl c nng thch hp vi in cc.

\

Bi 3

XC NH TH IN CC OXY HA KH V HNG S

CN BNG CA PHN NG OXY HA KH BNG

PHNG PHP IN TH K

I. C s l thuyt

Th chun ca h Fe+3/Fe+2 c xc nh bng phng php chun dung dch

Fe+2 bng dung dch KMnO4 chun. y l h oxy ha kh nn in cc ch th phi l

in cc Pt. Trong h lun c cn bng ca phn ng oxy ha kh

MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O

Nn th ca hai h Fe+3/Fe+2 v MnO4-,H+/Mn2+ trong dung dch lun bng nhau.

Tuy nhin trc tng ng dung dch gm ion Fe+2 d, ion Fe3+ v Mn2+ hnh thnh,

nng ion MnO4- rt nh nn cn bng c quyt nh bi nng v th chun

ca h Fe

+3

/Fe

+2

, v c tnh bi phng trnhFe3+ + e Fe2+


Cu hi (bi tp) cng c:1. V ng biu din pH = f(V).

2. Tnh Vt1, Vt2, CH2SO4 , CH3PO4

3. Ti sao phi tin hnh chnh m pH trc khi o pH hoc chun pH?

4. Thit lp cng thc tnh nng H2SO4 v H3PO4

Mc tiu hc tp: Sau khi hc xong bi ny, ngi hc c th:- p dng phng php chun in th xc nh th chun ca in cc

oxy ha kh


10/30

[ ][ ]+

+

+=++ 2

3

,

0 lg303.2

23

Fe

Fe

F

RTEE

FeFe

Sau tng ng h MnO4-,H+/Mn2+ d nn th ca in cc c quyt nh bi

t l nng v th chun ca h MnO4-/Mn2+

MnO4- + 8H+Mn2+ + 4 H2O

[ ][ ][ ]+

+

+=++ 2

8

4

,,lg

5

303.22

4 Mn

HMnO

F

RTEE

o

HMnMnO

Ni chung trc im tng ng th ca in cc l th ca h kho st nm

di cc, sau tng ng l th ca h cht chun nm trn buret. Ti im tng

ng th thay i t ngt.

im tng ng VE c xc nh t ng chun tch phn hay vi phn.

E/mV

V/mLE/V

VE V

Hnh 1: ng chun tch phn (trn) v vi phn (di)

Cho ti im tng ng nng Fe 3+ t l vi th tch dung dch chun thm

vo V. Nng ion Fe2+ lc ban u t l vi th tch tch dung dch chun ti im

tng ng Vt, v trc tng ng th t l vi Vt V. T biu thc tnh th

in cc trc tng ng c dng:

VV

V

F

RT

EEt

o

FeFe+=

++ lg

303.223 ,



11/30

Nh vy E s bng Eo ti im V = 1/2Vt. Eo l tung gc ca ng tuyn tnh

biu din ph thuc gia E v lgVV

V

t . S in t trao i c tnh t h s gc

ca ng ny.II. Ha cht v dng c

FeSO4 0.1M My o in th ng ong

KMnO4 0.025M My khuy t Phu thy tinh

H2SO4 m c Cc 100 ml

KCl 0.1M Buret 25 ml

Giy lc pipet 10 ml

III. Thc nghim

Do Fe2+ khng bn, d b oxy ha thnh Fe3+ nn khng th pha trc dung dch

Fe2+ khng cha Fe3+ . dung dch FeSO4 khng cha Fe3+ th phi thm Fe v

H2SO4 vo dung dch. Fe s a Fe3+tr v Fe2+. 2Fe3+ + Fe = 3Fe2+

H2SO4 c tc dng trnh s kt ta ca ion Fe 3+, lm chuyn dch cn v pha to

Fe2+. Nng Fe2+ trong dung dch s thay i theo thi gian. V vy trc mi th

nghim sinh vin phi xc nh li nng Fe

2+

.1. Lc ly khong 50 ml dung dch FeSO4 ca phng th nghim. Ly 2 ml cho

vo bnh nn. Thm 10 ml nc v 1 ml H2SO4 m c. Chun bng dung dch

KMnO4 0.025N cho n khi dung dch chuyn sang mu hng. Tnh nng FeSO 4.

T dung dch ny pha 50 ml FeSO4 0.025N.

2. Ly 10 ml dung dch mi pha v chun li tng t nh trong bc. Ghi nhn

th tch tng ng.

Lu : H2SO4 m c rt nguy him. Cn thn khi ly H 2SO4 m c.

Khng mang H2SO4 m c v ch lm vic.

3. Ly 10 ml dung dch FeSO4 0.025N cho vo beaker 100 ml c c t. Thm

30 ml nc ct v 1 ml H2SO4 m c. cc ln my khuy t. Rp h thng o

nh hnh 2.

Ti liu ging dy Mn Thc tp Ha L 2 9 Pt

Volt

k kkl

v

-

Ag/AgCl


12/30

Nu dng in cc ORP tch hp th khng cn in cc so snh na.4. Nh 0.5 ml dung dch KMnO4 0.025N vo dung dch FeSO4. Khuy u. Tt

my khuy. Ch th n nh mi ghi nhn gi tr th hin th. Tip tc chun cho ti

gn im tng ng Gim lng KMnO4 thm vo xung cn 0.1 ml. Lp li cho

n khi th tng t ngt. Chun tip vi lng KMnO4 mi ln thm l 0.5 ml. Dng

khi tng th tch KMnO4 thm vo gp khong 1.5 ln th tch tng ng.. Ghi s

liu vo bng.

5. Dng th Eo v th theo VKMnO4. Xc nh im tng ng t hai

th trn.

Th tch dung dch FeSO4 (ml):

Nng dung dch KMnO4:

Nhit th nghim:Ln 1 Ln 2 Ln 3

VKMnO4 (ml)Eo (mV)

VE /

6. Tnh lgVV

V

t , ++ 23 ,FeFeE ,

o

FeFeE ++ 23

, . Ghi s liu vo bng. Tnho

FeFeE ++ 23

,

trung bnh. (Ch cn tnh Eo cho khong 4 im trc v 4 im sau gn vi im V=

Vt).

++ 23,FeFe

E

= Eo + Eso snh

Ly th ca in cc so snh cho trong phn l thuyt

VV

V

F

RTEE

t

o

FeFeFeFe

=++++

lg303.2

2323 ,,

Th tch KMnO4 ti im tng ng (ml):

Nng dung dch KMnO4:VKMnO4 (ml) Eo (mV) lg

VV

V

t

++ 23,FeFe

E oFeFe

E ++ 23,


FeSO4 0.025N +1 ml H2SO4


13/30

Eo trung bnh:

7. Dng th++ 23

,FeFeE

theo lg

VV

V

t

. Xc nh s in t trao i n v

o

FeFeE ++ 23

, t th.

8. So snh gi tr th in cc, s in t trao i thu c v gi tr l thuyt, t

a ra nhn xt v cc phng php o v tnh p dng trong bi thc nghim.

BI 4

XC NH SUT IN NG V TH IN CC PINNNG

I. C s l thuyt

Thit lp pin Daniel Jacobi vi cc nng dung dch khc nhau. o S.

(-) Zn| ZnSO4 || Cu SO4| Cu (+)

Do Zn hot ng mnh hn Cu nn khi ni mch nguyn t trn phn ng xy ral

Zn + Cu2+ = Zn2+ + Cu

Ti in cc Zn xy ra phn ng oxy ha nn in cc Zn l anod. Trn in cc

Cu ion Cu2+ b kh v Cu nn in cc Cu l catod. o S phi ni in cc Cu

vi cc dng, cn in cc Zn th ni vi cc m ca my o. Khi o S hin th

c gi tr dng. Nu rp ngc li S hin th s c gi tr m.


Mc tiu hc tp: Sau khi hc xong bi ny, ngi hc c th:- Thit lp pin Galvani v o sut in ng ca pin- To pin t cc vt liu thng thng

Cu hi (bi tp) cng c:1. Vit cc phn ng xy ra in cc v vit s pin to thnh trong phn

ng oxi ha kh trn

2. Tnh sut in ng P = 1,013.105

N.m2

, T = 298o

K v vit s pin.Trong xy ra phn ng thun nghch:

2FeCl2 + H3AsO4 + 2HCl 2FeCl3 + HAsO2 + 2H2O

Bit: Hot cc cht trong pin:

02.03 =+Fea , 005.02 =+Fea , 2.043 =AsOHa , 1.02 =HAsOa , 01.0=+Ha , 12 =OHa


14/30

S ca pin Daniel Jacobi c tnh bi phng trnh Nernst

(1)

Th ca cc in cc c o nh s tr gip ca in cc so snh Cl -/AgCl, Ag.

in cc kho st ng vai tr catod c ni vi cc dng ca my o, cn in

cc so snh ng vai tr anod c ni vi cc m ca my o.

(2)

Trong Ein cc kho st = + (3)

T th o c SV cn nhn xt v mi tng quan gia hot tnh ca kim loi v

th in cc.

Th ca in cc kim loi c tnh nh biu thc

++

+=zz MMM

azF

RTEE

ln

0

/ (4)Trong pin nng khng ti t hai in cc kim loi in cc nm trong dung dch

c nng thp s l anod (a1 < a2). o S.

(-) Cu | CuSO4 a1 || CuSO4 a2 | Cu (+)

(-) Zn | ZnSO4 a1 || ZnSO4 a2 | Zn (+)

S ca pin nng khng ti c tnh nh sau:

1

2

ln a

a

zF

RT

E = (5)

So snh cc gi tr o c vi cc gi tr l thuyt tnh theo phng trnh Nerst.

Hot ca c tnh nh php gn ng th nht Debye-Huykel.

ai=fi x Ci (6)

lg fi = -0.509 |zi2| (7)

(8)



15/30

Thit lp pin t dy km, dy ng, dy nhm v cc loi qu c c nh c

chua, cam, chanh. o S.

I. DNG C V HA CHT

CuSO4 rn My o in th

ZnSO4 rn Cc 100 ml

KCl 0.1M Pipete 10 ml

Dy Zn, Cu, Fe, Al in cc Cu: 2

C chua, cam, chanh in cc Zn: 2

Giy nhm nc Bnh nh mc 100 ml

II. THC NGHIM

1. Chun b thc nghim

a. Dng giy nhm nc nh sch cc in cc, v cc dy kim loi

b. Pha dung dch CuSO4 v dung dch ZnSO4 0.1M. Pha long 10 ln c cc

dung dch nh cho trong bng. ra cc 50 ml.

CuSO4 CuSO4 CuSO4 ZnSO4 ZnSO4 ZnSO40.1 0.01 0.001 0.1 0.01 0.001

c. Ly 50 ml dung dch KCl 0.1 M vo cc 100 ml. a in cc so snh clorua

bc vo cc. Ch kim tra dung dch trong in cc so snh. Nu dung dch cn th

cn chm thm dung dch KCl bo ha. Lu th ca in cc so snh ph thuc vo

nng dung dch KCl trong in cc ch khng ph thuc vo nng dung dch

KCl trong cc.

2. Xc nh S ca pin Daniel-Jacobi

a. Thit lp nguyn t Daniel-Jacobi vi nng dung dch thay i nh cho

trong bng

ST

T

CuSO4

(M)

ZnSO4

(M)

Eo

(V)

JCuSO4 JZnSO4 fCu2+ fZn2+ aCu2+ aZn2+ Etnh(V)

Sai s

(%)



16/30

1 0.01 0.1

2 0.01 0.01

3 0.1 0.1

4 0.1 0.01

b. Rp nguyn t in ha nh hnh bn vi in cc Cu (+) nm trong dung dch

CuSO4, in cc Zn (-) nm trong dung dch ZnSO4. Ni hai dung dch bng cu mui

aga.

c. o S. Ghi kt qu vo bng.

d. Tnh lc ion cho cc dung dch.

e. Tnh h s hot v hot ca cc ion.

f. Tnh S t gi tr hot .

g. So snh gi tr o v gi tr tnh.

3. Xc nh th in cc

a. Thit lp cc nguyn t trong anod l in cc so snh v catod l in cc

Cu hay in cc Zn. Ni dung dch ca in cc kim loi v dung dch KCl bng cu

mui aga.

b. o S. Ghi kt qu vo bng.

c. Tnh th in cc t gi tr o c.

d. Nhn xt v mi lin h gia hot tnh kim loi v gi tr th in cc.

e. Tnh lc ion cho cc dung dch.

f. Tnh h s hot v hot ca cc ion.

g. Tnh th in cc t gi tr hot .

h. So snh gi tr o v gi tr tnh.



17/30

STT Eo

(V)

Ec

(V)

JM2+ fM2+ aM2+ Etnh

(V)

Sai s

(%)1 CuSO4

(M)

0.1

2 CuSO4

(M)

0.01

3 ZnSO4

(M)

0.1

4 ZnSO4

(M)

0.01

4. Xc nh S ca pin nng khng ti

a. Thit lp cc pin nng khng ti sau. Ni hai dung dch bng cu mui aga.

(-) Cu | CuSO4 0.001 M || CuSO4 0.1 M | Cu (+)

(-) Zn | ZnSO4 0.001 M || ZnSO4 0.1 M | Zn (+)

b. o S. Ghi kt qu vo bng.

c. Tnh lc ion cho cc dung dch.

d. Tnh h s hot v hot ca cc ion.

e. Tnh S t gi tr hot .

f. So snh gi tr o v gi tr tnh.

5. Pin tri cy

a. Cm tng cp dy hay ng xu kim loi vo qu chanh, cam hay c chuaCh

khng hai kim loi chm nhau.

b. Ni hai u kim loi vi cc ca ng h in t. Ch du ca cc cc. Theo

di ng h c chy khng. Ghi kt qu vo bng.

c. o S. Ghi kt qu vo bng.



18/30

Tri cy Anod Catod ng h chy E/VSTT Chanh Fe Cu1 Cam Zn Cu2 Chui Zn Fe

3 C Al Cu4 .. .. ..

d. Rp ni tip v song song cc pin. o in th.


Cu hi (bi tp) cng c:

1. Vit cc phng trnh phn ng xy ra cc in cc2. Hy vit phn ng xy ra trong pin v tnh sut in ng ca pin sau iu

kin chun v 25oC (-) Cd|Cd2+|| Cu2+| Cu (+)

3. Nhn xt v kh nng to in th v to dng ca cc pin tri cy trn.


19/30

BI 5

XC NHG,H,S CA MT PHN NG IN HA

I. C s l thuyt:

Nhit ng hc c lin quan n s thay i nng lng din ra trong qu trnh

ha hc v vt l. Nhng nguyn l c bn cho s thay i nng lng ny gm:

Nguyn l th nht nhit ng hc (trng thi bo ton nng lng): Nng lng

khng t nhin sinh ra cng khng t nhin mt i m ch chuyn ha t dng nysang dng khc, nhng nng lng ton phn ca h khng i. Trong cc phn ng

ha hc, din ra s sp xp li cc nguyn t v cc electron dn n s hnh thnh

cht mi. ng thi cc cht cng tng tc vi mi trng xung quanh v chnh iu

ny cng lm thay i in th ha hc ca h. Lng nhit ta ra hoc cn cung cp

bng bin thin entalpi ca h:

H = Hcui Hu

Nguyn l th hai nhit ng hc lin quan n qu trnh t din ra ca mt h ha

hc. Nguyn l ny c thit lp trn c s quan st nhng qu trnh din ra mt

chiu v khng din ra theo chiu ngc li. V d: bn lm ri mt quyn sch xung

sn nh quyn sch ny s khng t tr li tay bn. iu ny c p dng vo trong

ha hc hoc vt l. Mt qu trnh din ra khng lin quan n xo trn hay mt

trt t ca mt h. Nhit ng hc minh ha n mt cch nh lng bng entropi

S. S thay i entropi tun theo phng trnh:

S = Scui Su

Nng lng t do: mt qu trnh t din ra mt nhit v p sut khng i th

s bin i nng lng t do c minh ha bng phng trnh:

G =H -S (1)DuG lin quan n mt phn ng t din bin:

Phn ng din ra theo chiu thun, chiu t din bin l chiu lm gim nng

lng t do iu kin ng nhit, ng p .


Mc tiu hc tp: Sau khi hc xong bi ny, ngi hc c th:- Xc nh nhng c tnh nhit ng hc bng k thut in ha


20/30

Phn ng t trng thi cn bng .

Phn ng khng xy ra, trn nguyn tc, n xy ra theo chiu ngc li

Gi tr G, H, S ch ph thuc vo trng thi u v trng thi cui ca h,

ch khng ph thuc vo cch bin i ca qu trnh. Do G, H, S l cc hm strng thi. Vic nghin cu nhng phn ng in ha, cho ta nhng bin php n

gin v trc tip cho vic xc nh cc gi tr nhit ng hc bi v nng lng in

ha c th sinh ra hoc tiu th cho mt phn ng in ha c th o c rt chnh

xc.

S thay i nng lng t do ca mt phn ng in ha lin quan n in th E

ca h in ha:G = -nFE (2)

n: s mol electron trao i cho 1 mol cht phn ng

F: hng s Faraday = 96500 coulombs (in tch ca 1 mol in t)

Nu in th c o nhng nhit khc nhau, s thay i entropi p sut

khng i c biu din bng phng trnh:

( )

dT

GdS

= (3)

Th (2) vo (3):

dT

dEnFS = hay

nF

S

dT

dF = (4)

T (4) ta thydT

dEl h s gc ca th s ph thuc in th E vo nhit .

Trong th nghim ny, chng ta kho st mt h in ha gm hai in cc Zn 2+ /

Zn v Cu2+ /Cu tip xc nhau qua mt cu mui KCl bo ha .

eZn 2 +2Zn (Qu trnh oxi ha xy ra anod, cc m ca pin)

Cu2+ +2e Cu (Qu trnh kh ang xy ra catod, cc dng ca pin)

Cu2+ + Zn Cu + Zn2+

S ph thuc ca E vo nng , nhit ,... c xc nh bng phng trnh

Nernst:


( )

dT

nFEdS

=


21/30

Epu = E0putc

sp

a

a

nF

RTln (5)

Trong hu ht cc trng hp, hot ng c thay th bi nng . Phng trnh

(5) c vit:Epu = E0pu -

[ ][ ]tcsp

nlog

0591,0(6)

Khi nng Zn2+ v Cu2+ trong dung dch nh nhau th log[ ]tc

sp= 0. Trong iu

kin ny, th E ca pin bng vi E0. T E da vo (2) ta tnh c G.

Ngoi ra, trong th nghim ny, gi tr th cng c o nhng nhit khc

nhau.

II. Ha Cht v dng c:

Cu mui agar-agar

2 ming kim loi Zn v Cu CuSO4 0,5M

Chu thy tinh KNO3

Volt k ZnSO4 0,5M

Bnh iu nhit

III. Thc nghim1. Chun b cu mui

Ha tan 100mg agar-agar (agar,aga) v 500mg KNO3 vo 5 mL H2O, un nh cho

hn hp chy lng ra. Dng ng ht Pasteur cho dung dch cn nng vo mt ng

thy tinh hnh ch U. Dung dch s to gel trong khong 10 n 15 pht (khi ngui).

2. Chun b dung dch v in cc

Chun b 7 mL mi dung dch CuSO4 0,5M v ZnSO4 0,5M.

Ct 2 ming kim loi Zn v Cu c dng hnh ch nht 5x40mm (c sn, sinh vin

khng cn ct li).

Lu : Kt qu th nghim khng ph thuc vo chnh xc ca nng dung

dch.

3. To h in cc phn ng

Thit lp pin Gavalni. Ni hai dung dch bng cu mui aga.

(-) Zn | ZnSO41 M || CuSO4 1M | Cu (+)



22/30

4. Kho st nh hng ca in th vo nhit

- nhit nc lnh, ngm h trong chu nc (ly nc vi) cho n khi h

cn bng nhit (khong 10 pht), ghi nhit dung dch mui ca pin v gi tr sc

in ng c trn volt k .- Lm lnh h trong chu nc , khi h cn bng nhit (khong 10 pht), ghi

nhit dung dch mui trong pin v gi tr c trn volt k .

- Lm nng h n 500C (bng bnh iu nhit hoc bng cch ngm h pin trong

mi trng nc nng pha 520C, h pin trong mi trng ny i khong 10 pht

nhit dung dch ca h pin v mi trng u bng 50 0C, lc ny c t s sc

in ng trn volt k) cng i khi h cn bng nhit (15 pht), ghi nhit

phng v gi tr trn volt k .

- C th c thm vi nhit khc, lu ch c khi h tr nn cn bng

nhit .

Ch :

- Sau khi lm xong th nghim nh em cu mui hnh ch U trong nc nng

agar chy lng v ra sch ng nghim hnh ch U ny.

- Sau khi th nghim sinh vin nh vn nt volt k.

IV. Tnh ton kt qu

- Dng excel v th s ph thuc ca E vo nhit . H s gc ca th ny l

tr gi cadT

dE, thay vo (4) ta c S . Hoc v E theo T (dng giy k ly v),

xc nh c dc hay h s gc ca ng thng (xc nh c dE/dT).

- Tnh G t (2)

- Tnh c t H t S v G t (1)



23/30


Cu hi (bi tp) cng c:1. Trong nguyn t Galvani :

- Ti anol xy ra qu trnh g ?

- Ti catod xy ra qu trnh g ?

- Dng electron sinh ra nh th no ?

2. Mt pin Cu-Pb cho kt qu th nghim nh sau :

T0(0C) 26,5 2,5 80

E(V) 0,465 0,454 0,493

Tnh H,S v G 26,50C

3. T cc kt qu th nghim, phn ng no t din bin? Gii thch .


24/30

BI 6

KHO ST MT S H KEO

I. C s l thuyt

Ho hc cht keo - hay chnh xc hn, ho hc cc h phn tn - nghin cu cc h

bao gm mt pha phn tn dng chia nh thnh ht, phn b trong mt mi trng

phn tn lin tc. Cc ht c kch thc quy c t 1 m 100 m (107 105 cm),

ngha l ln hn kch thc phn t, nn h l d th nhng cha ln c th pht

hin bng knh hin vi nn h l siu vi d th.Ho hc cht keo cn nghin cu cc h cha cc ht c kch thc ln hn 100

m, c gi l cc h phn tn vi d th hoc cc h phn tn th. Cc h th c

gi l huyn ph nu cc ht l pha rn, c gi l nh tng nu cc ht l pha lng

v gi l bt nu ht l kh c phn b trong cc mi trng lng.

Do cc ht keo c kch thc nh nn chng c b mt ring rt ln. V vy, h

keo c nng lng b mt rt ln v khng bn vng nhit ng hc. Cc ht c xu

hng lin kt li vi nhau thnh nhng ht ln hn (lm cho b mt ring nh i v

nng lng b mt gim) ri lng xung. Qu trnh c gi l s keo t. gi

cho cc h keo c bn vng mt mc no ngi ta phi thm vo h cc

cht n nh (hay cht lm bn). C ch ca s lm bn trong i a s trng hp l

do ht keo hp ph mt loi ion ca cht lm bn, lm tng in tch ca ht keo,

khin cho lc y tnh in gia cc ht keo tng ln, ngn cn s lin kt ca cc ht.

C hai cch ch to cc h phn tn. Cch th nht l chia nh vt cht cho n khi

thu c cc ht c kch thc c ht keo, cc phng php theo cch ny gi l

phng php phn tn.V d, ngi ta xay lu hunh ri phn b vo nc, lu hunh

l pha phn tn v nc l mi trng phn tn. Cch th hai l tp hp cc phn t,

nguyn t hay ion c sn trong mi trng li thnh cc ht c kch thc c ht keo.

Cc phng php theo cch ny gi l phng php ngng t. V d, keo AgI c

ch to bng cch cho AgNO3 phn ng vi KI. Cc ion Ag+ v I trong mi trng

s tp hp li thnh cc ht keo AgI.


Mc tiu hc tp: Sau khi hc xong bi ny, ngi hc c th:- iu ch v tnh ngng keo t ca mt s h keo.


25/30

Mt iu kin thit yu cho c hai phng php trn l pha phn tn phi thc t

khng tan trong mi trng phn tn hay ni cch khc pha phn tn khng c tng

tc vi mi trng. Chnh v th m cc h keo ch to bng hai cch trn c gi l

h keo k lu. Ngoi ra cn c cc h keo a lu, in hnh l dung dch cc cht caophn t, trong cc phn t c tng tc vi mi trng v to thnh dung dch tht.

V vy, ch to chng ngi ta ch cn cho cht cao phn t vo mi trng thch

hp, cht ny s t ho tan vo mi trng . Song nu mi trng l mt

dung mi khng tt i vi cht cao phn t th ngi ta vn c th ch to dung

dch keo ca cht cao phn t bng phng php phn tn hay ngng t v thu c

cc h keo k lu.

Ht keo ch to trong mi trng lng thng c cu to phc tp, tu thuc vo

iu kin ch to n. Ngy nay, ngi ta quan nim ht keo l mt mixen trung ho

in bao gm mt nhn, thng c cu to tinh th, gia v mt lp in kp bao

quanh n. Lp kp gm hai bn. Bn trong gm cc ion cng mt du in nm trn

b mt ht keo, to cho ht c mt in tch no nn c gi l cc ion quyt nh

th hiu. Bn ngoi gm cc ion in tch ngc du va trung ho in tch ca

ht keo - l cc ion nghch. Cc ion nghch chu tc dng ca lc ht tnh in, lchp ph vo b mt ht v lc khuch tn do chuyn ng nhit gy ra, nn c phn

b thnh hai lp. Lp trong gm mt s ion nm st b mt ht keo gi l lp hp ph,

lp ngoi bao gm s ion nghch cn li, l lp khuch tn. lp ny nng cc ion

nghch gim dn t trong ra ngoi v khi lp kp kt thc th nng bng nng

cc ion nghch trong dung dch. Ngi ta thng biu din cu to ht keo bng s

hay bng cng thc. V d ht keo AgI hnh thnh do phn ng:

vi gi thit KI ly d lm cht lm bn, c s v cng thc c ch ra

hnh 1



26/30

Hnh 1: S v cng thc cu to ca mixen ca son AgI trong dung dch KI

long

Tu thuc vo nng cc ion trong dung dch m lp kp c th dy hay mng.

Khi lp kp c chiu dy ln th ht keo bn vng, khi lp kp mng th ht keo tinst li gn nhau khong cch m lc ht tc dng mnh khin cho cc ht lin kt li

vi nhau thnh nhng ht ln - l s keo t. Chnh v vy m khi thm cht in li

vo h keo thng xy ra s keo t. S keo t bt u xy ra mt nng cht in

li ti thiu gi l ngng keo t, c tnh bng s mili mol hoc mili ng lng

cht in li trn mt lt dung dch keo.

S keo t bi cht in li tun theo quy tc Schultze Hardy. Tc dng keo t ca

mt cht in li l do ion ngc du vi du in ca ht keo v tc dng tng ln

rt nhanh theo s tng ho tr ca ion . L thuyt cng nh thc nghim ch ra rng

ngng keo t t l nghch vi ho tr Z lu tha 6.

S c mt ca mt cht cao phn t thch hp trong mt h keo k lu c th lm

tng bn vng ca h keo . Tc dng c gi l tc dng bo v ca chtcao phn t i vi h keo k lu. C ch ca tc dng bo v l do cht cao phn t

hp ph ln b mt ht keo, lm thnh mt mng bo v ngn cn s lin kt ca cc

ht khi chng n gn nhau.

Nh trn, ta bit ht keo thng c in tch nht nh, tu thuc vo bn cht

ion quyt nh th hiu. Do , khi t h keo vo mt in trng th cc ht keo tch

in s di chuyn v mt cc, l hin tng in di (phn ht di chuyn bao gm:

nhn cng lp ion quyt nh th hiu v lp hp ph ca ion nghch); mi trng vi



27/30

phn ion nghch cn li l lp khuch tn s di chuyn v cc khc, l hin tng

in thm.

Nhng h keo k lu sau khi ch to thng cha cc cht ln, thng l cht in

li. Khi nng cht in li ln, chiu dy lp kp gim lm cho bn vng ca hgim i. Mun cho h keo bn vng ngi ta phi loi bt cht in li bng phng

php thm tch, ngha l cho h keo vo mt ti mng bn thm, cht in li lt qua

mng thot dn khi h, cn cc ht keo c kch thc ln nn khng lt qua mng v

gi li trong h. Cng v kch thc ht ln m cc h keo cn c c im l khuch

tn rt chm so vi dung dch thc.

ch to cc nh tng ngi ta dng phng php phn tn v thng phi

thm cht lm bn gi l cht nh ho. C hai loi nh tng:

a. Nh tng loi I hay nh tng thun, c pha phn tn l cht lng khng phn

cc, mi trng phn tn l nc.

b. Nh tng loi II hay nh tng nghch, c pha phn tn l nc, mi trng

phn tn l cht lng khng phn cc (thng gi l du).

Nh tng loi I c k hiu l D/N (du trong nc), nh tng loi II c k hiu l

N/D (nc trong du). Cht nh ho thng l cht hot ng b mt, th d kali oleat,

canxi oleat, nhm stearat v.v...

II. Dng c v ha cht

Bp in FeCl3 1N

Bnh tam gic 250ml KCl

ng ong 250ml K 2SO4

ng nghim Tinh bt

Cc 250ml Trng gBong bng heo

III. Thc nghim

1.`iu ch h keo Fe(OH)3

Dng ng ong ly 200ml nc ct vo erlenmeyer (bnh nn) 250ml, un si

nc trn bp in cch li amian. Tt bp, thm tng git dung dch FeCl 3 1N vo

erlenmeyer n khi ht 5 ml. Ta c ta keo dng Fe(OH)3 mu nu sm, trc khi

th dung dch FeCl3 c mu vng).



28/30

Khi thm ht 5ml th ly bnh keo ra khi bp, ngui trong khng kh .

2. Keo t keo Fe(OH)3 bng cht in li

t 10 ng nghim vo gi ng ng nghim. Cho vo cc ng nghim th tch

cc dung dch ha cht (mL) theo bng sau:ng nghim 1 2 3 4 5 6 7 8 9 10Keo Fe(OH) (mL) 5 5 5 5 5 5 5 5 5 5

Nc ct (mL) 4,5 4 3,5 3 2,5 2 1,5 1 0,5 0Cht in li (mL) 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5

Nhn xtCht in li l dung dch KCl 3N. Lp li th nghim vi dung dch cht in li

K2SO4 0,01N. Sau khi thm cht in li, mi ng u c lc mnh ri yn 15

pht. Sau quan st ri nh u cng + nhng ng b keo t (ng b c), du tr- nhng ng khng b keo t (ng trong).

Tnh ngng keo t cho mi cht in li, tc nng ca cht in li ti thiu

trong ng gy nn keo t:

V

Nv 310..=

v v N l th tch (mL) v nng ban u (lg/L) ca cht in li ng u tin

b keo t.V: l tng th tch ha cht trong ng nghim (10mL).

Khi c n v l mmol/L.

3. Keo a lng

a. iu ch

- Keo tinh bt: cn 0,5 gam tinh bt cho vo cc (becher) 250 mL cha sn 100

mL nc ct.

un nh trn bp in, va un va khuy u n khi tinh bt tan ht. Nhc

xung ngui.

- Keo albumin: ly mt lng trng trng g cho vo cc 250 mL ri thm 100 mL

nc ct, khuy u l c keo albumin. (Sinh vin ly trng g t lnh ca

phng th nghim).

Hai dung dch keo ny c dng tip cho phn b v c di y.

b. Thm tch keoLy 50 mL keo tinh bt cho vo ti thm tch (bng mng bong bng heo).



29/30

(Sinh vin ly bong bng heo t trong t lnh ca phng th nghim. Nu bong

bng b ng cng th sinh vin phi ngm bong bng trong nc x ng). Thm

tip 10mL dung dch K2SO4 0,1N vo bong bng, dng dy cao su ct kn ming bong

bng ri nhng vo erlen 1000mL cha sn 500mL nc ct, khuy lin tc cc ncbn ngoi ti khon 10 pht.

Ly vo hai ng nghim, mi ng khong 10mL nc trong bnh thm tch

trn (nc ngoi bong bng). Mt ng th nhn bit ion SO4 2- bng dung dch Ba2+

(dung dch BaCl2), mt ng th xem c tinh bt khng bng dung dch I 2 (nhm bit

ion SO42-, tinh bt c c thm tch ra khi mng bong bng heo hay khng).

Nhn xt kt qu v gii thch.

c. Bt

Ly vo ba ng ong 100 mL ln lt 5, 10 v 15 mL keo albumin. Thm vo cc

ng ln lt 15, 10 v 5 mL nc ct. y nt (hoc bt tay), lc k mi ng trong 5

pht. Ghi th tch hn hp c bt trong mi ng.

Th bn ca bt trong mi ng bng cch nhng mt vng dy cao su vo mi

ng. Nhc t t vng dy c keo ra ngoi. Ghi thi gian mng keo b t.

Nhn xt kt qu.

Sau khi th nghim, sinh vin b bong bng heo v phn trng g cn li vo thng

rc, t ngoi phng th nghim.

TI LIU THAM KHO

TI LIU THAM KHO BIN SON NI DUNG MN HC:

1. PGS. TS. V Ngc Ban - Gio Trnh Thc Tp Ha L - NXB i Hc Quc

Gia H Ni - 2007


Cu hi (bi tp) cng c:1. Vit phng trnh phn ng iu ch keo Fe(OH)3

2. Vit cng thc cu to ca h keo Fe(OH)3

3. Tnh ngng keo t theo n v mN/L

LmNVV

VC

k

/1000..

+=

V: s mililit dung dch cht in li cho vo dung dch keo.

C: Nng dung dch cht in li cho vo.

Vk: S mililit dun d ch keo dn keo t .


30/30

2. TS.Trn Th Ngc Lan - in Ha Hc - i Hc Khoa Hc T Nhin

TI LIU THAM KHO NGH CHO HC VIN:

1. T chuyn ngnh Ha L - Gio Trnh Thc Tp Ha L - Trng i hc CnTh - 2011

Documents

Phu Luc 5 - Tai Lieu Giang Day Hoa Ly 2(1)