gt-hoa-keo

B GIO DC V O TO TRNG I HC NNG NGHIP I

TS. PHAN XUN VN (Ch bin) TS. NGUYN TIN QU

GIO TRNH

HO KEO (Dng cho ngnh Nng Lm Ng nghip)

H NI 2006

LI NI U

Trng i hc Nng nghip H Ni Gio trnh Ho keo 1

http://www.ebook.edu.vn 1

Ho keo l mt mn hc trng qu trnh o to giai on 2 cho cc ngnh sinh hc

ca trng i hc Nng nghip I H Ni. Nh trng chng ta thc hin quy trnh t nm 1996 nhng cho n nay vn

cha xut bn ring mt gio trnh ca mn hc HO KEO. Da vo mc tiu o to, ni dung mn hc v kinh nghim ging dy, chng ti vit

cun gio trnh HO KEO ny. Cun sch gm 7 chng km theo cu hi v bi tp cui mi chng, tng ng vi 30 tit v Ho keo ang c ging trong trng.

Cun gio trnh ny phc v sinh vin hc tp v c th lm ti liu tham kho cho cc bn c thuc ngnh khoa hc lin quan.

Chng ti chn thnh cm n v hoan nghnh nhng kin ng gp ca cc bn s dng, lm cho gio trnh khng ngng hon thin.

H Ni, thng 02 nm 2006 T.M. CC TC GI Nguyn Tin Qu



CHNG I KHI NIM V CC H KEO

H keo l mt h phn tn, nhng cht phn tn phn b dng cc ht nh c kch

thc ln hn nhng phn t v ion n gin, gi l cc ht keo. Tuy nhin, cc ht keo vn khng b giy lc gi li, chng ch b gi li bi cc mng t bo sinh vt.

Do cht phn tn dng cc ht keo nn h keo c nhng c im khc vi cc h phn tn khc.

I. Cch phn loi cc h phn tn 1. Theo kch thc ht phn tn Da vo kch thc hoc ng knh ca ht phn tn, cc h phn tn c chia lm

3 loi chnh sau: H phn tn phn t: Trong h, cht phn tn dng nhng phn t rt nh, kch thc nh hn 10-7cm,

chng l nhng phn t v ion n gin. Cc h phn tn phn t c gi l dung dch tht hay dung dch thuc loi h ng th v c nghin cu nhiu. V d: cc dung dch phn t v in ly.

H phn tn keo Gm cc ht phn tn c kch thc 10-7 n 10-4cm, gi l cc ht keo1. H phn tn

keo thng c gi l h keo hoc son (sol). V d: keo AgI, keo Protit.. trong nc. Trong cc dung dch long, mi phn t protit cng nh phn t polyme khc x s

nh 1 ht c kch thc ht keo. Mi ht keo khc ni chung gm hng nghn n hng trm phn t, ion n gin to thnh.

So vi phn t, ion n gin th ht keo c kch thc ln hn, nhng chng ta khng nhn thy bng mt thng. quan st c cc ht keo c bit l cc ht c kch thc khong 10-7cm ngi ta dng knh siu hin vi in t. Vy h keo l h phn tn siu vi d th, trong ht phn tn c kch thc khong t 10-7 n 10-4cm. Cc h keo l i tng nghin cu ca ho keo.

H phn tn th Gm cc ht c kch thc ln hn 10-4cm, thng gi l h th. Ni chung h th l h vi d th khng bn vng. Chng hn, trong mi trng lng c

ht phn tn rn kch thc ln hn 10-4cm, th ht c th s nhanh chng lng xung hoc ni nn trn b mt lng (tu theo khi lng ring ca ht v ca mi trng) ngha l tch khi mi trng ca h.

Trong h th c 2 loi quan h quan trng l huyn ph v nh tng. Huyn ph l h th gm cc ht rn phn b trong mi trng lng nh: nc ph

sa Nh tng l h th gm cc ht hoc git lng phn b trong mi trng lng nh: cc ht du m trong nc. Trong nhiu trng hp phi thm cht lm bn vo huyn ph v nh tng cc h phn tn bn vng.

Cc huyn ph v nh tng dng trong thc t l nhng h vi d th tng i bn. Cc h c bn cht ca h keo nn c th coi l cc h keo khi nghin cu v s dng.

Ho keo cng nghin cu cc h vi d th c tnh bn. Trong gio trnh ny chng ta coi h th c tnh bn v h keo u thuc loi h vi d th.

1 Mt s ngi s dng khong 10-7 n 10-5cm, nhng hin ti khng c quy nh cht ch no.



2. Theo trng thi tp hp pha ca h Phng php n gin cho cch phn loi ny l da vo pha mi trng ca h

phn loi cc h vi d th. Mi trng phn tn kh. Gi chung l son kh (aeorosol) gm cc h: H L/K (cc git lng phn b trong pha

kh) nh: my, sng m H R/L (cc ht rn phn b trong pha kh) nh: khi, bi (H K/K l h phn tn phn t).

Mi trng phn tn lng Gm cc h: H K/L (Cc bt kh phn b trong pha lng) nh: bt x phng trong

nc H L/L (cc git lng phn b trong pha lng) nh: huyn ph, keo v c trong nc.

Trng hp h gm cc ht phn tn rn, lng hoc kh, c kch thc ca ht keo, trong mi trng lng th gi chung l son lng (lyosol), trong mi trng nc, ru th tng ng c cc h hydro sol, alcol sol

i tng nghin cu ch yu ca chng ta l cc h keo gm nhng ht phn tn rn trong mi trng nc.

Mi trng phn tn rn. Gm cc h: H K/R (cc ht kh phn b trong pha rn) nh: bt kh trong thu tinh,

cc vt liu xp.. H L/R (cc git lng phn b trong mi trng rn) nh nhng git lng trong m ng, thc vt H R/R (cc ht phn tn rn trong pha rn) nh: thu tinh mu, hp kim

Khi cc ht phn tn rn, lng hoc kh, c kch thc ht keo, trong pha rn th gi l h son rn (xerosol).

3. Theo cng tng tc gi ht phn tn v mi trng ca h Cc h vi d th trong mi trng lng c chia lm 2 loi l cc h keo ght lu v

h keo a lu. H keo ght lu. H gm cc ht phn tn hu nh khng lin kt vi mi trng th c gi l h keo

ght lu hoc h keo ght dung mi(lyophobe), nu mi trng nc th gi l h keo ght nc (hydrophobe). H keo ght lu thng gp l cc h keo v c trong nc. V d: cc keo AgI, As2S3, keo kim loi, keo oxt kim loi trong nc.

Cc h keo in hnh hu ht l cc h ght lu, do trong h c b mt phn cch pha r rng gia ht phn tn v mi trng ca h. H keo ght lu thuc loi h d th, nhiu tnh cht b mt nh tnh hp ph, tnh cht in biu hin rt r rt.

H keo a lu. H gm cc ht phn tn lin kt cht ch vi mi trng ca h c gi l h keo a

lu hay h keo a dung mi (lyophile), nu mi trng nc th gi l h keo a nc (hydrophile).

Mi ht keo a lu c bao bc bi lp sonvat ho gm cc phn t mi trng, nn h keo a lu thuc loi h ng th v thng c gi l dung dch. H keo a lu thng gp l dung dch cao phn t. V d: cc dung dch nc ca protit, gluxit

H keo a lu cng c tnh cht ca dung dch tht nh: s thm thu v l h ng th, nhng cng c nhng tnh cht ca h keo ght lu v ht keo c kch thc ln hn so vi phn t n gin.

Tuy nhin, khng c ranh gii tuyt i gia 2 loi h keo nu trn. V d: h keo gm cc ht keo c to thnh t cc phn t cht bn keo (nh x phng C17H35COONa.) gi



l h bn keo, c tnh cht b mt trung gian gia h keo a lu v h keo ght lu nhng h bn keo rt gn vi h keo ght lu.

II. Nhng c im ca h phn tn keo 1. B mt d th B mt phn chia cc pha ca h d th gi l b mt d th ca h. i vi mt h

phn tn d th, th b mt d th ca h tnh bng tng din tch b mt cc ht phn tn. Kch thc ht cng nh th b mt d th S ca h cng ln. V d:

Phn chia 1cm3 mt cht rn thnh cc ht hnh lp phng cnh l. Nu l = 1cm, th ch c 1 ht, din tch b mt ca n l 6cm2. Nu l = 10-4cm, th s c 1012 ht, tng din tch b mt cc ht l S=6.104cm2. Nu l = 10-7cm, th s c 1021 ht, tng din tch b mt cc ht l S=6.107cm2.

R rng l cng vi mt lng cht phn tn dng ht th kch thc ht cng nh, s ht cng nhiu, tng din tch b mt cc ht cng ln. Khi kch thc ht bng 10-7cm th b mt d th ca h rt ln xem bng I.1.

Bng I.1: S bin thin din tch b mt ca mt h ng vi 1cm3 lp phng, cht

phn tn, khi chia thnh cc ht hnh lp phng c kch thc gim dn.

Kch thc ca ht hnh lp phng

cnh l(cm) S ht n Th tch 1 ht (cm3)

Din tch b mt 1 ht s

(cm2)

Tng din tch b mt cc ht

S= ns (cm2) 1 10-1 10-2 10-3 10-4 10-5 10-6 10-7

1 103 106 109 1012 1015 1018 1021

1 10-3 10-6 10-9 10-12 10-15 10-18 10-21

6 6. 10-2 6. 10-4 6. 10-6 6. 10-8 6. 10-10 6. 10-12 6. 10-14

6 6. 10 6. 102 6. 103 6. 104 6. 105 6. 106 6. 107

Nu phn chia cht phn tn thnh nhng phn t rt nh, kch thc khong 10-8cm

th S=0. Nhng phn t l cc phn t v ion n gin, chng khng c b mt ngn cch vi mi trng ca h.

2. B mt ring v phn tn B mt ring ca h phn tn l tng din tch b mt ca cc ht, ng vi 1 n v th

tch cht phn tn nghin nh:

(1.1) VSSr =

S: tng din tch b mt ca cc ht V: th tch cht phn tn nghin nh Sr: b mt ring

n gin cho tnh ton ngi ta cho ht c hai dng l hnh lp phng v hnh cu, chng hn:

H gm n ht hnh lp phng cnh l th



(1.1a) 66 32

lnlnl

VSSr ===

nhng ht hnh cu bn knh r th

(1.1b) r3

34

rn43

2

===rnV

SSr

i vi cht phn tn nghin th vic xc nh khi lng n gin hn so vi vic xc nh th tch, nn b mt ring c tnh bng tng din tch b mt ca cc ht, ng vi 1 n v khi lng cht phn tn nghin nh:

(1.2) 'mSS r =

m: khi lng cht phn tn nghin Sr: b mt ring tnh theo khi lng (th nguyn l cm2g-1, m2g-1)

Thay th m= .V, vi l khi lng ring ca ht vo cng thc I.2 v tnh ton tng t nh trn s c cc cng thc tnh Sr khi ht dng hnh lp phng.

(1.2a) l

6' =rS v khi ht dng hnh cu

(1.2b) 3'r

S r = Hnh cu l dng ph bin ca ht keo, nn cng thc 1.2b thng c ng dng V d: Nghin SiO2 thnh cc ht hnh cu bn knh r = 10-5cm. Tnh b mt ring ca

SiO2? Bit khi lng ring ca SiO2 l = 2,7g.cm-3 Gii: p dung cng thc 1.2b:

121255 1,1110.11,110.7,2

3 === gmgcmSr

T cng thc I.1a v I.1b suy ra

(1.3) dkSr =

v tng t, qua cc cng thc 1.2a v 1.2b chng ta c:

(1.4) 'd

kS r = k: hng s ph thuc hnh dng ht

d: kch thc ca ht, d = l nu ht hnh lp phng cnh l v d = r nu ht hnh cu bn knh r.

Vy b mt ring t l nghch vi kch thc ht phn tn. H keo gm cc phn tn kch thc nh (10-7 10-4cm) l h c b mt ring cc i hoc c b mt ring rt pht trin. y l c im c bn ca h keo. so snh b mt d th ca cc h ngi ta da vo b mt ring. Nu cng lng cht phn tn th h keo l h c b mt ring rt pht trin, do c b mt d th rt ln. Theo quan im ca nhit ng hc th s c mt ca mt b mt phn cch ln gn lin vi s c mt ca mt nng lng b mt ng k iu nh hng rt nhiu n cc tnh cht ha keo ca h nh tnh hp ph, tnh cht in, tnh bn, tnh ng t.



Cc vt liu xp c mt h mao qun. H thng mao qun c b mt ring thng gi l b mt trong (tng t b mt ring ca h phn tn keo) c ng dng rt ph bin trong thc t.

i lng tnh bng nghch o ca kch thc ht phn tn gi l phn tn ca h. Kch thc ht cng nh th phn tn ca h cng cao. V d: cc h keo c phn tn rt cao khong t 104cm-1 n 107cm-1.

B mt ring v phn tn l nhng i lng c trng cho mc phn tn ca ht. B mt d th rt pht trin v phn tn rt cao l nhng c im ca cc h keo.

III. Khi nim v h a phn tn Trong nhiu trng hp cc ht phn tn khng ch khc nhau v kch thc m c

hnh dng. Mt h phn tn, nu ch gm cc ht cng dng th gi l h n dng, nu cc ht

khc nhau v hnh dng th gi l h phn tn a dng, nu ch gm cc ht c cng kch thc th gi l h n phn tn, nu cc ht c kch thc khc nhau th gi l h a phn tn.

H a phn tn gm nhiu cp ht.

1. Cp ht Cp ht l mt tp hp nhiu ht c bn knh trong khong t ri n rk no Chng hn: H gm ht bn knh t 10-2 n 5(1) c th phn chia thnh mt s cp

ht nh sau: cp 1 gm cc ht c bn knh r t 10-2 n 5.10-2, cp 2 gm cc ht c t 5.10-2 n 0,1, cp 3 gm nhng ht c r t 0,1 n 0,5.

Mi cp ht c mt bn knh trung bnh ca cc ht. Do c th ni: cp ht l mt tp hp nhiu ht c bn knh trung bnh r no .

i vi h a phn tn gm n cp ht, ht c dng hnh cu th tnh b mt ring theo cng thc:

= (I.5) %3'i

ir r

as

ai%: thnh phn phn trm khi lng ca cp ht i so vi tng khi lng ca cc cp ht

ir : bn knh trung bnh ca ht cp i V d: Mt h keo gm 3 cp ht hnh cu: cp 1 c cmri

510= chim 45%, cp ht 2 c cmri

610.5,2 = chim 35% v cp ht 3 c cmri 710.2 = chim 20% khi lng ring ca SiO2 chim. Tnh b mt ring ca h ? Bit khi lng ring ca SiO2 l =2,65g.cm-3

Gii: p dng cng thc I.5:

12765

321

.14,13410.220,0

10.5,235,0

1045,0

65,23%%%3' =

++

++= gm

rc

rb

raS r

Vic phn tch cc cp ht c tin hnh bng nhiu phng php. i vi cc ht

th, thng dng phng php ry. Ngi ta dng cc ry c kch thc bit tch mt h thnh nhiu cp theo kch thc ca mt ry, sng. i vi cc h c phn tn tng i cao th phng php phn tch sa lng c dng ph bin.



A

B

h

Hnh I.1: S sa lng ca ht phn tn

)(34

03 = rm

Nguyn tc phng php phn tch sa lng. Hin tng ri t do ca ht trong mi trng ca h do tc dng ca trng lc, gi l

s sa lng. V khi lng ht t l vi lp phng kch thc ht, nn ht c kch thc tng i

ln s sa lng. Lc cn tr s sa lng l lc ma st ca ht vi mi trng. Khi lc ma st (f) bng trng lc ca ht (P) th ht sa lng vi tc khng i (v).

V f = B.v v P=m.g nn: Bv = mg

B: h s ma st m: khi lng hiu dng ca ht g: gia tc trng trng

i vi ht hnh cu bn knh r chuyn ng trong mi trng c nht th B=6r, nu khi lng ring ca ht l v ca mi trng l 0 th T suy ra:

v phng trnh tnh tc sa lng nh sau:

V d: Tnh tc sa lng ca ht SiO2 hnh cu bn knh r=10-3cm v khi lng

ring =2,7g.cm-3 trong nc? Gi s

Gii: p dng cng thc I.6:

Nu < o ht s ni ln (hin tng sa ni), nu > o th ht s ri xung (hin

tng sa lng). Nguyn tc phng php phn tch sa lng l: da vo phng tnh tnh tc sa lng xc nh kch thc ht phn tn.

(1.6) )(92 20 grv

=

grrv )(346 0

3 =

poacmg OHOH 0115,0 .1 223 == v

1223 .10.219,3980.)10(0115,09

)0,17,2(2 == scmv



Gi s mt ht sa lng t A n B xem hnh I.1, trong thi gian t(s), cao AB = h(cm), th tc sa lng ca ht l

Kt hp cng thc I.7 vi phng trnh I.6 suy ra kch thc ca ht:

hoc

vi

i vi h phn tn c th mt nhit xc nh, th k l mt hng s nn vic xc nh kch thc ht cn li l vic o cao h m ht sa lng trong thi gian t.

Trong cc h n phn tn, tc sa lng cc ht bng nhau, s phn lp s xy ra sau mt thi gian xc nh. Cui cng trong h ch c lp mi trng trong sut pha trn v lp cc ht sa lng pha di.

Trong h a phn tn, tc sa lng cc ht c kch thc khc nhau, khng bng nhau, nn bin gii phn cch 2 lp nh trn khng r rt. Sau mt thi gian nht nh, nhng cao khc nhau chng ta rt c cc cp ht khc nhau ra khi h.

Cn lu rng, phng trnh I.8 ch cho php xc nh kch thc ca ht sa lng hnh cu hoc dng hnh cu, khng b sonvat ho v ht sa lng l ht n hay ht c thn.

Phng php phn tch sa lng xc nh kch thc ht phn tn ch p dng vi cc h huyn ph. i vi h c phn tn cao nh h keo, do tc sa lng ca ht rt nh nn phi s dng my ly tm hay siu ly tm sa lng ht.

Cch phn chia cp ht, tu thuc vo yu cu nghin cu v kh nng cho php ca phng php phn cp. Cn nh rng, mi cp ht l mt h a phn tn hp. Mt h a phn tn hp cng c th coi l h n phn tn, bn knh ca ht l r . V d:

H phn tn gm 4 loi ht: loi 1 c r = 10-6cm chim 10%, loi 2 c r = 2.10-6cm chim 25%, loi 3 c r = 3.10-6cm chim 35% v loi 4 c r = 4.10-6cm chim 30% khi lng ca tt c cc cp ht, nu coi l h n phn tn th bn knh ht l cmr 610.26,2 =

2. Mc a phn tn c mt s phng php biu th mc a phn tn ca cc h keo, sau y l

phng php biu th bng a phn tn ca h. Cc ht keo c coi l nhng phn t ln tng t cc phn t cht cao phn t. Do

chng ta phn bit khi lng trung bnh s nM v khi lng trung bnh khi wM ca ht. Khi lng trung bnh s hoc khi lng trung bnh theo s lng ht thng gi tt

l khi lng trung bnh ca ht, tnh theo cng thc:

(I.7) thv =

(I.8) )(2

9

0 thr

=

(I.9) thkr =

gk

)(29

0

=



ni : s ht i trong h Mi: khi lng 1 ht i

Tr s nM tnh c t cc phng php cho php xc nh nng cht phn tn. Khi lng trung bnh khi hoc khi lng trung bnh tnh theo khi lng ca ht

tnh theo cng thc.

wi: khi lng ca tt c cc ht i Tr s wM c suy ra t cc phng php cho php xc nh kch thc ht. Lun thy wM > nM , nu l cc phn t n gin th nM = wM a phn tn ca h tnh bng t s gia khi lng trung bnh khi v khi lng

trung bnh s ca cc ht.

: a phn tn Nu = 1 hoc nM = wM th h l n phn tn, thng thy h gm nhng phn

t n gin. Nu >1 hoc wM > nM th h l a phn tn. Khi >>1 th mc a phn tn ca

h rt rng, l h gm cc ht rt khc nhau v kch thc hoc khi lng. V d: C 2 h a phn tn A v B gm cc ht c khi lng nh sau (quy c 1 n

v khi lng y bng 103vC cho ph hp vi cc ht, khi lng ht ny l 100n v): H A gm 100 ht, khi lng mi ht l 1 n v v 1 ht khi lng 100 n v. H B gm 100 ht, khi lng ca mi ht l 1 n v v 100 ht khi lng mi ht

l 100 n v. Hy tnh a phn tn ca mi h? Gii: p dng cc cng thc I.11; I.10 v I.12 i vi h A:

v i vi h B:

(I.10) =

i

iin n

MnM

(I.11) 2

=

i

iiw w

MnM

(I.12) n

w

MM=

37,2599,1

5,50 =

99)100100()1001(

)100100()1100( 22 =++=wM n v

5,50100100

)100100()1100( =++=nM n v

5,50)1001()1100(

)1001()1100( 22 =++=wM n v

99,11100

)1001()1100( =++=nM n v



Nh vy cc h A v B u l a phn tn, nhng mc a phn tn ca h A ln

hn h B hng chc ln. Ngha l s sai khc ca cc ht (v khi lng hoc kch thc) trong A ln hn trong B nhiu ln.

IV. iu ch v tinh ch cc h keo 1. iu ch C 2 phng php chnh iu ch l phng php phn tn v phng php ngng t. Phng php phn tn: bao gm cc bin php chia nh cc ht phn tn c kch

thc ln thnh cc ht c kch thc nh, thch hp. V d: nghin, xay, gi, dng h quang, siu m

Phng php ngng t: th ngc li vi phng php phn tn bao gm cc bin php tp hp cc phn t nh thnh cc ht c kch thc thch hp. V d: s thay i tnh cht mi trng (nhit , pH, dung mi) u c th lm cho cc phn t ca cht tan ngng kt li thnh cc ht, cng c th thc hin phn ng ho hc (oxy ho - kh, trao i, thu phn) to ra cc phn t ca cht kh tan chng tp hp li thnh cc ht.

Vn kh nht trong vic iu ch khng phi l tm c bin php phn tn hay ngng t m phi tm c bin php khng ch kch thc ht keo. Ngha l s chia nh khng lm cho ht qu ln. Ngoi cc yu t nhit , pH ngi ta thng ch n vic thm vo h mt cht khc hoc 1 cht lm bn thch hp, n va c tc dng khng ch kch thc ht va c tc dng chng li s ng vn hoc s kt dnh gia cc ht, trong qu trnh iu ch.

2. Tinh ch keo Trong qu trnh iu ch, do nguyn liu dng, do phi thm cht lm bn nn

dung dch keo thu c thng khng sch. Trong s cc cht lm bn th cht in ly l cht nh hng ln n tnh cht ca h keo. Do vic tinh ch keo, trc ht nhm tch cc cht in ly ra khi h bng phng php thm tch, cch tin hnh nh sau:

96,15,50

99 =

mng thm tch

Hnh I.2 S tinh ch h keo bng phng php thm tch

nc vo

nc ra

h keo



Cho h keo vo mt bnh thm tch, nhng pha di bnh c bt bng mt mng thm tch. c im ca mng thm tch l ch cho cc phn t v ion n gin i qua, cc ht keo khng i qua c. C bnh trn c t trong mt chu nc sch c dng chy xem hnh I.2. Cc ion ca cht in ly khuch tn qua mng thm tch t h keo vo nc v b nc cun i. Cui cng trong bnh thm tch ch cn li l h keo. tng tc qu trnh v hiu qu tinh ch, ngi ta t bnh thm tch trong in trng ca dng in mt chiu. l nguyn tc ca phng php in thm tch tinh ch keo.

Ngoi phng php thm tch cn c th dng phng php siu lc. Thc cht l s lc dung dch keo qua cc mng lc c bit, mng lc c cc l nh

vi kch thc xc nh. Cc ion v phn t nh lt qua mng lc, cn cc ht keo k c phn t cht polyme b gi li trn phu lc. Bng cch chn cc mng c l thch hp, phng php siu lc chng nhng cho php tnh ch cc h keo m cn tch ring c cc ht keo theo kch thc ca chng.



CU HI V BI TP CHNG 1 1. Cch phn loi cc h phn tn? Phn bit cc h: huyn ph v nh tng, keo ght lu v keo a lu? Cho v d. 2. Hy chng t rng kch thc ht phn tn cng nh th b mt d th cng ln. 3. B mt ring v phn tn? Cng thc tnh b mt ring theo kch thc ht? 4. c im ca h phn tn keo? 5. Phn bit h n phn tn vi h a phn tn? Cp ht v nguyn tc phng php phn tch sa lng? 6. a phn tn ca h? Phng php tnh? 7. Nguyn tc cc phng php iu ch v tinh ch cc h keo? 8. Mt dung dch protit X. B mt ring ca protit l 8,24.104 m2g-1, khi lng ring ca ht l = 1,1616g cm-3. Tnh bn knh trung bnh r ca cc ht protit X trong dung dch?

Tr li: 3,13.10-5cm 9. Mt loi t st c khi lng ring l =2,68g.cm-3 c nghin thnh 3 cp ht: cp 1 c cmr 61 10.5,2

= chim 45%, cp 2 c cmr 52 10.6,4 = chim 28%, cp 3 c cmr 43 10.8,1

= chim 27% khi lng tt c cc cp ht. Tnh b mt ring ca loi t st nghin trn.

Tr li: 20,91m2.g-1 10. Trong th tch ca mt h keo Ag c 0,105g Ag. Gi s ht dng hnh lp phng c di cnh l 2.10-6cm. Tnh

a/ S ht keo v nng mol ht ca h b/ B mt d th ca ht Khi lng ring ca Ag l 1,05g.cm-3

Tr li: 1,25.1015ht, 2,083.10-9mol ht.l-1; 3m2 11. Tnh thi gian cn thit ht SiO2 bn knh 5.10-4cm lng trong nc ct 250C, nht 0,01poa, c 50cm? Bit khi lng ring ca SiO2 l 2,6g.cm-3 v ca nc l 0,982g.cm-3.

Tr li: 94,69 pht 12. Mt hn hp gm 0,5mol cht A khi lng phn t A l 100.000 v 0,5 mol cht B khi lng phn t ca B l 200.000. Tnh khi lng trung bnh s v khi lng trung bnh khi ca phn t.

Tr li: 150.000 v 167.000



CHNG II

TNH CHT NG HC PHN T V S KHUCH TN NH SNG CA CC H KEO

Sau nhng nghin cu v l thuyt v thc nghim, ngi ta khng nh rng

thuyt ng hc phn t c th p dng c cho tt c cc h c ht tng i nh c th tham gia vo chuyn ng nhit. l nhng h cha cc ht c kch thc ca nhng ht keo. Nh vy cc h keo c tnh cht ging cc dung dch tht nh: s khuych tn, s thm thu, tnh nht S khuch tn nh sng ca h keo cng l mt bng chng cho thy tnh ng hc phn t ca h keo.

I. Tnh ng hc phn t 1. Chuyn ng Brao (Brown) Chuyn ng hn lon ca cc ht phn tn keo c gi l chuyn ng Brao. Theo thuyt ng hc phn t, chuyn ng Brao c gii thch nh sau: Chuyn ng hn lon ca ht keo ngoi nguyn nhn do chnh bn thn chuyn ng

nhit ca n gy ra, th s va chm x y hn lon ca cc phn t mi trng (vn l chuyn ng nhit) ng vai tr ch yu. Ht phn tn c kch thc nh, nn s va chm xy ra theo cc phn khc nhau rt khng u nhau. Kt qu l ht b x y v pha ny v pha n, nn chiu chuyn ng ca ht b thay i rt nhanh.

Vy, bn cht chuyn ng Brao l chuyn ng nhit, l chuyn ng ng hc ca cc phn t. Rt kh quan st ng i thc ca ht v khng th o c tc chuyn ng ca ht. Hnh II.1. V s qu o tng trng ca mt ht khi quan st chuyn ng Brao ca n t A n B theo phng x, chiu chuyn di ca ht t v tr ny n v tr khc c quy c biu din bng mt on thng.

Hnh II.1. S tng trng chuyn ng Brao ca mt ht t A n B.

Nu gi l ri trung bnh ca ht th tnh theo cng thc: (II.1)

... 22221

nn+++=

22

21, . l cc bnh phng hnh chiu ca cc on AM, MN trn trc x, m ht

di chuyn c trong cng khong thi gian t

B

A N

x

M



din t st hn chuyn ng Brao trong khng gian ngi ta dng gi tr di trung bnh bnh phng 2 trong thi gian t

Da vo thuyt ng hc phn t Anhstanh (Einstein) thit lp c cng thc.

2 =2 Dt (II.2) D: h s khuch tn ca h

2. S khuch tn Nu trong h c hin tng khng ng u v mt hay nng ht phn tn th c

s di chuyn cc ht t vng nng cao ti vng nng thp hn ( san bng nng ), chng ta ni l c s khuych tn.

nh lut Fich I (Fick1) v s khuych tn phn t dng cng thc nh sau:

m: lng cht khuch tn D: h s khuch tn ca h : gradien nng S: b mt thng m ht khuch tn qua t: thi gian

S khuch tn xy ra theo chiu nng gim ( 00. Trng hp khng thay i theo thi gian chng ta c:

Nu 1,1,1 === tSdxdC th m = D. Vy, h s khuch tn l lng cht chuyn

qua mt n v thit din thng trong mt n v thi gian khi grdien nng bng -1. Trong h CGS th th nguyn ca D l cm2.s-1 Theo Anhstanh:

B: h s ma st ca ht trong mi trng phn tn. Cc k hiu R, T v N0 xem phn ph lc cui cun sch

i vi ht phn tn cu bn knh r, mi trng ca h c nht th theo Stc (Stock): B = 6r. Do :

Cng thc II.6 cho thy: - mt nhit khng i, i vi h phn tn xc nh, h s khuch tn c trng

cho kh nng chuyn ng ca ht v ch ph thuc kch thc ca ht. So vi cc phn t

(II.3) SdtdxdCDdm =

dxdC

dxdC

StdxdCDm = (II.4)

.0 BN

RTD = (II.5)

61.

0NRTD = (II.6)



n gin th ht keo c kch thc ln hn, di trung bnh bnh phng rt nh, nn kh nng khuch tn ca h keo nh hn nhiu so vi cc dung dch phn t.

- Bit h s khuch tn D s tnh c bn knh r ca ht, l cch xc nh kch thc ht phn tn bng phng php o h s khuch tn.

Trong th nghim, da vo cc phng php xc nh nng , s tnh c D theo cng thc (II.4)

Bit khi lng ring v bn knh r ca ht, tnh c khi lng M ca 1mol ht phn tn.

Do c th da vo h s khuch tn ca dung dch cao phn t tm khi lng

phn t cht polyme, nu chp nhn rng phn t polyme dng ht hnh cu trong dung dch.

3. p sut thm thu Tnh ng hc phn t l nguyn nhn gy ra hin tng thm thu v p sut thm

thu ca cc dung dch v cc h keo p sut thm thu l p sut thy tnh nn ln mng bn thm c tc dng lm ngng

s thm thu gia dung mi nguyn cht v dung dch p sut thm thu cc dung dch phn t long, theo Vanhp(Vant Hoff) c biu

th bng phng trnh:

C: nng cht tan tnh theo mol. l-1 m: khi lng cht tan (tnh theo g) trong mt lt dung dch M: khi lng mt mol ca cht tan

Phng trnh p dng c cho dung dch keo nh sau:

: nng mol ht tnh theo s mol ht, 1 mol ht gm No ht. m : khi lng cht phn tn (tnh theo g)c trong 1 lt h keo. M: khi lng 1 mol ht keo.

Khi m= m, nhng M > M th < C p mt thm thu ca h keo rt nh so vi dung dch tht c cng khi lng cht

phn tn, do M

Cc h keo ght lu u km bn vng, do ht c kch thc ln th c khuynh hng sa lng, cn cc ht nh th c khuynh hng lin kt li (do tc dng ca lc tng tc phn t) thnh s t ht ln hn lm cho s ht keo (ht n) gim, nn p mt thm thu ca h khng n nh.

Tri li, trong dung dch cao phn t long, cc phn t polyme trng thi phn tn, s ht keo n nh, nn p sut thm thu ca cc dung dch cho d rt nh, nhng xc nh.V vy, c th xc nh c khi lng phn t cht cao phn t da vo phng php o p sut thm thu ca dung dch, m c s l phng trnh II.8. c kt qu chnh xc hn nn s dng phng trnh:

MmRTRTCP == (II.8)

'

'

,'MmRTRTP == (II.9)

(II.7) 0

3

34 NrM =



Na+ Cl- cl-

M

Na+ Cl-

Hnh II.2: S thm tch ca 2 dung dch tht NaCl tip xc vi nhau qua mng M

B: hng s ph thuc bn cht cht cao phn t v dung mi.

Xy dng th biu din s ph thuc ca vo m (dng ph thuc y=ax+b) v

ngoi suy ti m = 0, s tnh c M theo cng thc i vi cht cao phn t in ly, nh prott hoc cc ht keo mang in khc th s

phn b khng u nhau ca cht in ly hai bn mng cng nh hng n kt qu o p sut thm thu. Lc p mt thm thu ca dung dch c quan h n hiu ng nnan (xem di y)

p sut thm thu ca h keo sau khi loi tr hiu ng nnan gi l p sut thm thu keo.

Phng php o p sut thm thu ca dung dch xc nh khi lng phn t cht cao phn t theo phng trnh II.8, c p dng i vi dung dch long cht cao phn t khng in ly hoc trung ho v in, nu khi lng phn t ca n khng qu ln

V d: 250C, p sut thm thu ca dng dch cha 10g.l-1 cht cao phn t trung ho in

tch, o c l 1,22210-2 atm. Tm khi lng phn t cht cao phn t y? Gii: p dng cng thc II.8: 4. Cn bng mng nnan (Donnan) Cn bng mng nnan l mt c trng rt quan trng v tnh ng hc phn t ca

h keo. Chng ta nhn thy iu trong vic kho st cc th nghim sau:

MRT

mP

m=0lim (II.11)

mP

997.1901222,0

10.298.082,0 ===PmRTM

BmMRT

mP += (II.10)



i vi dung dch tht: Gi s c 2 dung dch tht NaCl nng l C1 v C2 tip xc vi nhau qua mng thm

tch M - xem hnh II.2 . Nng ban u C1 C2.Qu trnh thm tch din ra theo xu hng san bng nng . Sau mt thi gian th nng 2 bn mng bng nhau, khng xut hin cn bng nnan

i vi h keo: Xt mt h, gm h keo NanR nng mol ht l C1 tip xc vi dung dch tht NaCl

nng mol l C2 qua mng thm tch - xem hnh II.3. Mt cch tng qut chng ta biu din cng thc ht keo mang in l NanR tng t

cng thc ho hc ca mt cht, trong Rn- l ion keo. Ht keo in ly nh sau: NanR n Na+ + Rn- V chu nh hng ca Rn- (kh nng khuch tn nh, nhiu in tch m) cc ion Na+

trong h keo khng linh ng bng cc ion Na+ trong dung dch tht

Hnh II.3: S phn b cc ion 2 bn mng thm tch (3), h keo (1) v dung dch tht (2): a. trng thi u b. trng thi sau khi thm tch

Lc u nng cc ion thm tch (Na+,Cl -) v ion khng thm tch (Rn-) phn b

2 bn mng nh trnh by hnh II.3a. Sau c s khuch tn cc ion thm tch qua mng (3). Do nh hng ca ion keo, lng ion Na+v Cl- khuc tn d dng t h (2)sang h (1).

Gi x l lng ion (tnh theo nng ) Na+ hoc Cl- chuyn t dung dch tht vo dung dch keo th nng ion trong mi h c biu din hnh II.3b.

Theo quan im ng hc th tc khuch tn cc ion thm tch ca h t l thun vi nng ca chng trong h .

Tc khuch tn cc ion Na+ v Cl- t h (1) sang h (2) l :

v tc khuch tn cc ion Na+ v Cl- t h (2) sang h (1) l :

)1(12 ).( = + uNa CCkv

)).( )2(21 = + uNa CCkv

n Na+ Rn-

(nc1) (c1)

(1) (3) (2) (1) (2)

Na+ Cl-

(c2) (c2)

Rn - Na+ Cl-

(c1) (nc1+x) (x)

Na+ Cl-

(c2-x) (c2-x)

a b

(3)



vi k l hng s tc .

Khi th hnh thnh cn bng mng nnan, l trng thi cn bng ca s khuch tn cc ion thm tch gia h keo v dung dch tht tip xc vi nhau qua mng thm tch. Khi :

Thay nng cc ion Na+v Cl- bit hnh II.3b vo cng thc II.10 s tm c: T tnh c tng nng cc ion thm tch h keo:

v trong dung dch tht:

Suy ra: Qua cc cng thc II.12 v II.14 chng ta c nhn xt sau: - Cn bng mng c thit lp khi tch s cc nng ion thm tch hai bn mng

bng nhau. - Trng thi cn bng mng c thit lp ngay khi nng cc ion thm tch hai

bn mng khc nhau nhng tng nng cc ion h keo ln hn. l c im quan trng nht ca cn bng mng trong cc h keo. iu c

ngha rt ln i vi cc qu trnh sinh l: C th sng c cu to di nhiu nhiu hnh thc khc nhau ca trng thi keo. S

trao i cht gia c th sng v mi trng lun ti trng thi cn bng mng gia h keo ca c th v mi trng cha cc cht in ly khc nhau, nhng vn bo m cho h keo lun c p mt thm thu ln hn- mt yu t khng th thiu cho nc v cc cht dinh dng c vn chuyn t mi trng vo cc b phn ca c th.

Hin tng chnh lch nng cht in ly 2 bn mng gi l hiu ng nnan. Hin tng lm pht sinh mt in th b mt tip xc gia h keo vi dung dch tht c gi l th nnan, cn tr s khuch tn cc ion thm tch t dung dch in ly vo h keo.

5. nht nht l mt i lng c trng cho lc ma st ni trong s chy ca mt cht

lng gy ra do cc lp cht lng chy vi tc khc nhau v trt ln nhau. Nh lc ht phn t lp chy nhanh li ko lp chy chm, cn lp chy chm km hm lp chy nhanh.

2112 vv =

21

22

CnCCx += II.13

21

22

21122

1 222).

CnCCCnCCnxxnC +

++=++

)2()1( ).().( ++ +>+ ClNaClNa CCCC (II.14)

)2()1( ).().( 11 ++ += CnNaCnNa CCkCCkII.12 )2()1( ).().( 11 ++ = CnNaCnNa CCCC

21

2221

22 222)()(CnC

CCnCxCxC ++=+



Trc ht, chng ta xem xt v nht ca mt cht lng nguyn cht. Gi thit c mt cht lng chy trong ng hnh tr bn knh r, theo hng trc y - xem

hnh II.4. Tc chy u l mt hm s ca x, (x: l khong cch tnh t trc y v phng gc vi y) u = u max khi x = 0 v u = 0 khi x = r khi s di chuyn cc lp tr nn u hay

dxdu l mt hng s, th theo nh lut Niutn (Newton) v nht chng ta c:

f: lc ma st . S: din tch tip xc gia 2 lp lng

dxdu : gradien tc gia cc lp theo hng x thng gc vi trc y

: h s t l c gi l nht Trong h CGS th nguyn ca l yn.cm-2s, c tn l poa, cn trong h SI th th

nguyn cat l N.m-2.s, 1N.m-2s = 10 poa,. nht ca cc h keo S chy ca son khc vi cht lng nguyn cht v dung dch tht ch, trong mi

trng ca h c nhng ht keo m kch thc ln hn rt nhiu so vi cc phn t n gin. Cc ht keo chon khng gian ca cht lng, lm cho cc phn t trong s chy lc i v lm tng gradien tc trung bnh gia cc lp, do nht ca h keo cao hn nht ca mi trng.

Theo Anhstanh th nht ca h keo ph thuc vo th tch v hnh dng ht keo, nu l ht keo rn dng hnh cu th:

= o(1 + 2,5 ) (II.16)

o : nht ca mi trng phn tn : nng th tch ca pha phn tn trong 1ml, l tng th tch cc ht phn

tn rn c trong 1 ml ca h. Thc t c mt s trng hp m phng trnh Anhstanh nu trn khng nghim.V

d:

II.15 dxduSf =

y

r x

x

Hnh II4. gii thch nh lut Niutn v nht



- Khi ht c dng hnh que hay hnh tm th nht ca h lun ln hn so vi kt qu tnh theo cng thc II.16. . Nguyn nhn: cht lng nm trong vng th tch quay ca ht (vn c chuyn ng Brao quay) gn lin vi ht dn n s tng biu kin th tch ca ht. Do , dng tng qut phng trnh Anhstanh v nht ca h keo c th nh sau

= o(1 + . ) (II.17)

: tha s ph thuc hnh dng ht.

- Nng ht cng ln th nht ca h cng ln, do nht ca h keo tng theo s gim kch thc ht.

- Khi gia cc ht c lc tng tc in hay cc lc tng tc khc. Nu cc ht keo tch in cng du th nht ca h tng, do b mt ht xut hin lp in kp. S xut hin lp v sonvt ht keo cng lm tng nht ca h, do n lm gim linh ng ca cc phn t dung mi, lm tng biu kin nng th tch ca pha phn tn nht ca dung dch cao phn t

C nhiu yu t nh hng n nht ca dung dch cao phn t nh: kch thc v hnh dng phn t polyme, dung mi, nng , cu to v khi lng phn t polyme Khi lng phn t cht cao phn t rt ln nhng khng tp trung thnh mt khi m tri ra theo mch , nn lc lng tng tc gia cc phn t vi nhau v gia chng vi mi trng rt mnh.

So vi dung dch phn t nh th cc dung dch cao phn t c nht ln hn nhiu l do tnh thu ng hc ca cc cao phn t mm do v ko di, do t nhiu c xut hin cc lin hp phn t trong dung dch.

Nhiu tc gi cho rng: hnh dng v khi lng phn t ca cht cao phn t c th xc nh bng phng php o nht.

Trong php o nht ngi ta quy c: nht ring:

trong : l nht ca dung dch, 0 l nht ca dung mi.

nht rt gn: l nht ring quy v n v nng . nht c trng hay nht rt gn:

i vi dung dch cao phn t long th r tnh theo phng trnh: r=KMC (II.18)

K: hng s, c trng cho mt dy ng ng cao phn t trong mt dung mi xc nh.

M: khi lng phn t ca cao phn t C: nng , tnh theo s mol c bn trong 1lt ( l s mol mt xch polyme c

trong 1 lt dung dch, v d: 1mol c bn ca polybutadien bng 54g) Da vo phng trnh II.18 tnh c M, nhng ch p dng khi mch ngn hay di

nhng phi l cao phn t cng, ngha l gi c dng ko di hoc dng hnh que hi cong. Cng theo phng trnh II.18 th nht rt gn r/C=KM khng ph thuc nng . Nhng

0

0

=r

Cr

[ ]

= Cr

C

0

lim



thc t nht rt gn ca dung dch cao phn t ph thuc nng theo phng trnh bc nht:

a: hng s S ph thuc phn nh tng tc ln nhau gia cc phn t cht cao phn t trong

dung dch. xc nh M ngi ta thng s dng phng trnh sau:

[] =K.M (II.20) K: hng s i vi mt dy ng ng cao phn t trong mt dung mi xc nh : hng s, c trng cho tnh mm do ca mch.

Nu =1 th phng trnh II.20 tr thnh phng trnh II.18, nu =0,5 th mch cao phn t mm do. Nhng nu =0 th cao phn t dng ht hnh cu, nht ca dung dch khng ph thuc kch thc hoc M ca cht cao phn t. Thc t thng thy rng: 0



S khuch tn nh sng c th nghim thy trong th nghim ca Tinan (Tyndahl)nh sau:

HnhII.7: S th nghim ca Tindan. Chiu chm sng t ngun sng S qua mt knh hi t H vo h keo cha trong mt

bnh c 2 thnh bnh song song, xem hnh II.7. ng bn quan st, chng ta thy trong bnh xut hin mt di sng hnh nn ct, l hnh nn Tinan, cn gi l hiu ng Tinan.

C hin tng v mi ht keo sau khi nhn c nh sng ti tr thnh mt im sng th cp, khuch tn nh sng v mi pha. Hin tng Tinan l mt biu hin c trng v tnh cht quang hc ca h keo, n gip chng ta nhn bit c cc h keo

Cng nh sng khuch tn ca h keo long c biu din bng cng thc sau

I : cng nh sng khuch tn k : hng s ph thuc cng tia ti, chit xut ca mi trng v ca pha phn

tn. n: s ht keo trong mt n v th ca h keo : th tch mi ht phn tn

Tia Tia l

cc tia phn x

Tia ti

>

Hnh II.5: hin tng phn x nh sng Hnh II.6: hin tng

khuch tn nh sng

4

2

nkI = II.21

di sng Tinan

S

h keo



: bc sng tia ti. Cng thc II.21 cho thy: - Cng nh sng khuch tn t l thun vi bnh phng ca th tch ht phn tn.

iu c ngha l i vi cc dung dch tht (kch thc phn t cht tan rt nh), th hin tng khuch tn nh sng rt yu, thm ch khng ng k. Tri li, khi kch thc ht tng i ln (ht keo) th I tng nhanh, cn i vi ht ln (ht th) th I khuch tn tr thnh I phn x.

- Cng nh sng khuch tn t l thun vi nng ht. V vy, i vi mt h keo xc nh, mi nng ng vi mt cng nh sng khuch tn xc nh v ngc li. o c cng nh sng khuch tn s suy ra nng ca h keo, l nguyn tc xc nh nng keo theo phng php o cng nh sng khuch tn bng Tindanmet v cc my o c (Nephelomet).

V nguyn tc, Nephelomet c cu to ging t sc k hoc my so mu. iu khc c bn gia chng ch: trong my so mu tia ti i thng n mt chng ta qua dung dch, cn trong Nephelomet th tia ti c chiu vung gc vi mt chng ta qua h keo (v chng ta o cng nh sng khuch tn ) - xem hnh II.8.

Cch xc nh nng keo bng Nephelomet nh sau. iu ch mt dung dch keo chun nng C0, cha trong cuvet A c b dy ct dung

dch l h0. Cho dung dch keo (cng loi v cng kch thc ht vi keo chun) nng cha bit Cx vo cuvet B c b dy ct dung dch l h. Chiu vo 2 cuvet trn bng mt ngun sng t theo phng vung gc thnh cuvet - xem hnh II.8, v iu chnh b dy ct dung dch trong cuvt B n hx, sao cho nhn c 2 tia l c cng bng nhau. Khi chng ta c:

Hnh II.8: S v tr ngun sng ti (1) trong: a. t sc k. b. Nephelomet v v tr quan st (2) Phng trnh II.21 khng p dng cho h keo kim loi (v ht dn in, hp th v

phn x rt mnh nh ng) v h c mu (v nh sng b hp th).

xx h

hCC 00= II.22

tia l

dung dch tht h keo

a

tia ti

b

tia t

i tia

l

(1)

(1)

(2) (2)



Cc dng c quang hc hin i hn, chng nhng ch gip chng ta xc nh c nng cc h keo theo phng php trn, m cn gip chng ta m c cc im sng trong mt n v th tch h keo long (khi dng siu hin vi), xc nh c nng ht chnh xc hn. Vic chp nh cc im sng (bng knh hin vi electron) cho php xc nh c hnh dng v kch thc ca ht, suy ra khi lng ca ht. Knh hin vi electron c s dng trong vic phn tch t v xc nh khi lng phn t cht cao phn t trong t

Kho st s khuch tn nh sng lm sng t chuyn ng Brao ca ht phn tn v c im siu v th ca cc h keo.

Hin tng hp th v khuch tn nh sng ca cc h phn tn gip chng ta gii thch mu sc cc h keo, cc hin tng mu sc trong t nhin, trong thin vn, hng hi, hng khng . Do c vn dng rt nhiu trong cc lnh vc k trn (tn hiu sng, chp nh, d bo thi tit)



CU HI V BI TP CHNG 2 1 . Chuyn ng Brown? c im v tnh ng hc phn t ca h keo? 2 . H s khuch tn v cc yu t nh hng? Xc nh kch thc ht phn tn bng phng php o h s khuch tn 3. c im v p mt thm thu ca cc h keo? So snh dung dch phn t vi h keo v p sut thm thu? 4. Xc nh khi lng phn t polyme bng phng php o p sut thm thu ca dung dch? 5. Cn bng mng nnan? c im v ngha. 6. nht ca cht lng? Cng thc tnh nht ca h keo v ng dng ? 7. Hin tng khuch tn nh sng ca h keo ? Th nghim ca Tyndahl? ng dng ca vic o cng nh sng khuch tn? 8. Da vo nht ca dung dch xc nh M ca cht cao phn t nh th no? 9. Dung dch cha 2g protit A trong 100ml c p sut thm thu 7,083.10-3 atm b qua hiu ng onnan, 250C. Tnh khi lng phn t protit A?

Tr li: 69.000 10. Huyt tng mu cha 40g albumin M=69.000 v 20g globulin M= 160.000 trong 1lt dung dch. Tnh p sut thm thu keo ca dung dch 370C ?

Tr li:0,0179 atm 11. Tnh h s khuch tn trong nc 250C nht l 0,01 poa mt phn t hnh cu X khi lng l 105 ? Bit khi lng ring ca X l 1,37g cm-3.

Tr li :7,7 . 10-7 cm2 .s-1 12. H s khuch tn ca glubilin trong 1 dung dch nc l 4.10-7 cm2 .s-1 250C. Gi s rng phn t c dng hnh cu, tnh khi lng phn t ca n ? Bit rng 250C nht ca nc l 0,01005 poa v khi lng ring ca glubilin l 1,33 g. cm-3

Tr li : 51.000



CHNG III

NNG LNG B MT. S HP PH S phn chia cc cht rn, lng, thnh nhng ht nh to cho lp b mt ca cc ht

c tnh cht khc vi tnh cht cc lp trong lng ht, l tnh cht b mt biu hin bng s hp ph

Tnh cht b mt l tnh cht c trng ca cc h phn tn d th, c bit l cc h keo.

I. Nng lng b mt v sc cng b mt Do khng cn bng v lc tng tc phn t, nn cc phn t b mt cht (rn,lng)

chu tc dng ca lc ht vo trong lng cht. Nh vy mun a mt phn t cht t trong lng ln b mt phi tn mt nng lng

chng li lc ht . Ni cch khc l mi phn t b mt c mt nng lng ln hn so vi cc phn t trong lng hoc trong th tch ca cht.

Nng lng d ca tt c cc phn t b mt so vi nng lng trung bnh ca cc phn t trong th tch ca cht c gi l nng lng t do b mt, gi tt l nng lng b mt thng k hiu F.

Nng lng b mt quy v mt n v din tch b mt c gi l sc cng b mt, k hiu l . V mt nhit ng hc th sc cng b mt ng bng tr s ca cng to ra 1 n v din tch b mt trong qu trnh thun nghch ng nhit.

Trong h CGS, th nguyn ca l erg.cm-2 hoc yn.cm-1 nhng trong h SI, thi th nguyn ca l J.m-2 hoc N.m-1 . T c nh ngha v sc cng b mt nh sau: Sc cng b mt l lc tc dng trn mt n v chiu di b mt phn chia 2 pha chng li s ko cng b mt.

V d: 200C, H2O = 0,07275J.m-2 hoc 0,07275N.m-1 Sc cng b mt l kt qu ca lc tng tc phn t, nn n ph thuc bn cht ca

cht v nhit . cng mt nhit th: hp cht ion > hp cht phn cc > hp cht khng phn cc. cht rn > cht lng.

Bng III.1: Sc cng b mt ca mt s cht lng v cht rn trn gii hn vi khng kh

Tn cht t(0C) 103 (N.m-1) Tn cht t(0C) 103 (N.m-1) Hydro(lng) Oxy(lng) te tylic(lng) Hecxan(lng) tylic(lng) Clorofooc(lng)

-252 -198

20 20 20 20

2 17 17

18,5 21,627,1

Bezen (lng) Nc (lng) Thu ngn (lng) Thic (lng) PbF2 (rn) Ba504 (rn)

20 20 20

920 920 25

28,972,75

485 510 900

1250 B mt cht lng l ng nht, nn nng lng b mt ca cht mi im nh nhau.

i vi cht rn, mt phn t cnh, gc v trn b mt khng bng nhau, nn nng lng b mt cc im l khc nhau. Ni cch khc, b mt cht rn khng ng nht v nng lng, nn sc cng b mt ca cht ch c gi tr trung bnh.



Khi tng nhit th lc tng tc phn t gim, do sc cng b mt ca cht gim. Do , khi nhit tng sc cng b mt ca cht lng gim, nhit ti hn khng cn b mt phn chia, sc cng b mt ca cht bng 0.

Bng III.2: Sc cng b mt ca nc nhit khc nhau.

Nhit (0C) 103 (N.m-1)

0 10 15 20 25 40 60 80

75,64 74,22 73,49 72,75 71,97 69,58 66,18 62,61

Quan h gia nng lng b mt v sc cng b mt theo cng thc

F = .S (III.1) S: din tch b mt

i vi h phn tn th S l b mt d th ca h Do c b mt d th rt ln, nn nng lng b mt ca h rt ln, v vy theo nhit

ng hc th h keo l h khng bn vng, s xy ra cc qu trnh lm gim , lm gim S gim F ca h.

II. Khuynh hng gim din tch b mt ca h. Khuynh hng to kin trc khi cu. c im ny thng xuyn xy ra cc git cht lng , v vi cng mt th tch th

dng hnh cu l dng khi c din tch b mt cc tiu. Cc ht rt nh nh nguyn t, phn t v c ht keo c coi l cc ht dng hnh cu l c c s thc t, c tnh ph bin.

S hnh thnh ht kp. Vi cng lng cht phn tn dng ht (ht n) kch thc ht cng nh th s ht

cng nhiu, b mt d th ca h cng ln. Theo nhit ng hc th h nh th rt khng bn, cc ht n s tp hp hoc kt dnh li thnh cc ht i, ht ba gi chung l cc ht kp, gim b mt d th, nhm gim nng lng b mt ca h.

S hnh thnh cc ht kp d xy ra khi mt ht ln, c th dn n s sa lng ca ht. V vy, trong thc t s lng cc ht keo trong cc h keo khng ln hay nng mol ht ca cc h rt nh. V d: mt h keo vng bn, c ht 10-7cm, nng mol ht khng vt qu 1,67.10-6 mol ht. l-1 .

S tan ca cc ht nh to ra s t ht ln hn. Hin tng xy ra khi lm kt tinh cht kh tan trong dung dch in ly, khi ngng

t my m thnh macng l nhng hin tng gim b mt d th (S) gim nng lng b mt (F) ca h.

i vi mt s h do c im ca n, din tch b mt ca h khng thay i, vic gim F ch c th gim sc cng b mt b mt. l hin tng hp ph, trong nhiu trng hp s gim S v gim xy ra ng thi .



III. S hp ph. 1. Mt s khi nim c bn S hp ph. l hin tng b mt nhm thu ht cht b hp ph ln b mt cht hp ph, lm

gim sc cng b mt ca cht hp ph. Ngc vi s hp ph , qu trnh i ra ca cht hp ph khi b mt cht hp ph gi

l s gii hp hoc phn hp. Khi tc hp ph bng tc phn hp th s hp ph trng thi cn bng.

Hp ph vt l v hp ph ho hc. Da vo bn cht ca lc hp ph, ngi ta phn bit hp ph vt l v hp ph ho

hc. Hp ph vt l gy ra bi lc vt l (lc tng tc phn t ), cn hp ph ho hc gy ra bi lc ho hc(lc ca lin kt ho hc).

V d: s hp ph ca than hot tnh i vi cc phn t kh hoc hi CO2, C2H5OH, H2O(gi l hp ph phn t) l hp ph vt l, nhng s hp ph ca cht rn AgI i vi Ag+ trong dung dch l hp ph ho hc v n kin ton cu trc b mt mng tinh th hp cht ho hc AgI.

Do lc hp ph yu, nn hp ph vt l c tnh thun nghch. Khi nhit tng lc tng tc phn t gim nn hp ph vt l gim. V vy hp ph vt l thng tin hnh nhit thp (thp hn nhit si ca cht b hp ph ). Hp ph vt l c th l hp ph n lp hoc n phn t, cng c th l a lp hoc a phn t. Hp ph ho hc c bn cht ca mt phn ng ho hc, nn hp ph ho hc c tnh bt thun nghch (rt kh thc hin s phn hp). Khi nhit cng tng, tc phn ng ho hc tng, nn hp ph ho hc tng. Do hp ph ho hc thng xy ra nhit cao.

Nhit hp ph. S hp ph pht nhit, nhit hp ph vt l rt nh, nhit hp ph ho hc ln; tng

ng hiu ng nhit ca phn ng ho hc. Tuy nhin ranh gii gia hp ph vt l v hp ph ho hc ch l tng i. c bit

khi cht hp ph l nhng ion hu c th hp ph vt l v hp ph ho hc thng xy ra ng thi. Ion hu c d b hp ph bi cc b mt phn cc, b mt c in tch ngc du, cng c trng hp b mt mang in tch cng du th c th do lc tng tc phn t ln hn lc y tnh in.

Tnh chn lc nh hng. S hp ph c tnh chn lc v nh hng theo quy tc nh sau: Nhng cht c tnh cht tng t nhau d hp ph vo nhau. Nhng phn c tnh phn

cc nh nhau hoc gn ging nhau s hng vo nhau. Nu l hp ph ho hc i vi cc ion, th ion b hp trc phi l ion c trong thnh

phn cu to tinh th b mt, sau l nhng ion tng t c kh nng hon thnh cu to mng li tinh th b mt. Th t hp ph i vi cc ion y theo s u tin cho ion c in trng ln hn (in tch ln, sonvat ho t hn).

V d: cht rn AgI hp ph mnh i vi I- khi trong dung dch c KI, nhng cng c kh nng hp ph ion Cl- (hoc SCN-) khi trong dung dch c NaCl hoc NaSCN

2. hp ph v ph hp ph ca mt cht hp ph l lng cht b hp ph(mol, milimol) b hp

ph trn mt n v din tch hoc trn mt n v khi lng cht hp ph khi c trng thi cn bng hp ph, mt nhit xc nh. Thng k hiu hp ph l G:

SnG = (III.2)



hoc

n: s mol hay s milimol cht b hp ph S: din tch b mt (m2 , cm2 ) m: khi lng cht hp ph (g)

nhit khng i, hp ph ca mt cp cht hp ph v cht b hp ph ph thuc vo bn cht cp cht y, b mt cht hp ph, nng (hoc p sut) ca cht b hp ph.

Khi tng nng cht b hp ph th hp ph tng n cc i v khng i cho d nng cht b hp ph tip tc tng xem hnh III.1.

Vy : hp ph cc i Gmax l hp ph ti a ca cht hp ph ng vi nng

cn bng xc nh ca cht b hp ph, mt nhit xc nh. ph hoc ph b mt l t s gia din tch b mt cht hp ph hp ph v

tng din tch b mt ca cht hp ph , thng tnh theo phn trm:

Sr: b mt ring. Sp: din tch b mt ca cht hp ph hp ph : ph.

Khong xc nh ca ph: Khi 0



Din tch b mt hp ph c lin h vi hp ph: Sp = G.No.So, y So gi l ph c bn hoc ph nguyn t, l din tch ti thiu ca mt phn t hoc mt ion chim ch b mt cht hp ph.

V d: 10g mt loi than ci c b mt ring l Sr =300m2.g-1 hp ph c 0,46g ru C2H5OH, 250C. Tnh ph ca loi than ny i vi C2H5OH? Bit rng So(C2H5OH)= 21,62 ,1=10-8cm.

Gii hp ph ca than: Din tch b mt ca 1g than hp ph: ph b mt ca than: Trng hp ny xy ra hp ph n lp khng hon ton. 3. S hp ph trn b mt rn. Cc cht rn c sc cng b mt rt ln so vi cht lng v cht kh ho lng. Do

b mt cht rn d hp ph cc phn t ca cht lng v cht kh. Sau y l mt s phng trnh hp ph ng nhit cho php xc nh hp ph. Phng trnh hp ph ng nhit Lang mua (Langmur). Tc gi gi thit rng: hp ph vt l trn b mt rn l hp ph n lp, b mt

cht hp ph ng nht v nng lng, do nhit hp ph mi im l nh nhau. Trong mt h kn ng nhit, mt cht rn hp ph tip xc vi mt cht kh c nng

C. Nu tng s im c kh nng hp ph trn mt cm2 b mt cht hp ph l x, trong y im hp ph nng cht b hp ph l C, s im cn li cha hp ph l: x-y th tc hp ph (Vh) t l thun vi s x-y v nng C:

Vh= kh(x-y)C. Tc phn hp (Vp) t l thun vi y: Vp= kp.y. Khi c cn bng hp ph th Vh=Vp , nn:

kh=(x-y)C=kp.y (III.6) hoc

khv kp: cc hng s ca tc hp ph v tc phn hp. Do t s :

(III.8) max CaCGG +=

Phng trnh (III.8) c gi l phng trnh hp ph ng nhit Lang mua. Phng trnh III.8 cho bit quan h gia hp ph v nng cn bng ca cht b hp ph , trong

1.001,01046

46,0 == gmolx

G

1241623 .10.13010.10.02,6001,0 == gcmxS p

maxGG

xy = a

kk

p

h =v t nn c phng trnh:

%33,4310.30010.130

4

4

==

Ckk

Cxy

p

h +=

(III.7)



Gmax v a l cc hng s ph thuc nhit i vi mt cp cht hp ph v cht b hp ph xc nh, G v Gmax tnh theo mol. g-1 v C tnh theo mol. l-1

Phng trnh hp ph ng nhit Lang mua ch ph hp vi thc nghim trong hai trng hp sau:

Khi nng cht b hp ph rt nh th C a, nn G = Gmax, ng ng nhit hp ph l ng song song so vi trc honh.

ng vi khong nng trung bnh (t C1 n C2 ) cc ng ng nhit hp ph Lang mua khng ph hp vi ng cong thc nghim - xem hnh III.2. Trn hnh III.2 trnh by cc ng ng nhit hp ph Lang mua, ng cong thc nghim MN ch ph hp vi phng trnh III.9. Mt khc, nhit thp c th c trng hp hp ph n gii hn Gmax, nhng s hp ph vn tip tc khi tng nng cht b hp ph.

Nhng hn ch bt ngun t nhng gi thit gn ng ban u ca tc gi. Thc t b mt cht hp ph khng hon ton ng nht v hp ph vt l khng ch n lp m c th a lp.

Phng trnh hp ph ng nhit Lang mua cho s hp ph n lp khng ch p dng i vi cht kh m c dung dch phn t long.

Phng trnh hp ph ng nhit Frendlich (Freundlich). Trng hp nng cht b hp ph c gi tr trung bnh ngi ta s dung phng

trnh hp ph ng nhit Frendlich:

G: hp ph ( tnh theo mol.g-1) x: s mol cht b hp ph m: khi lng cht hp ph (g) C: nng cn bng hp ph (mol.l-1) k v n: cc hng s i vi mt cp cht hp ph v cht b hp ph, n>1.

nCkmxG

1

.. == (III.9)

CkCa

GG ..max == (III.9 )

Hnh III.2: Cc ng ng nhit hp ph Lang mua:(1) v (2)

G = Gmax G = Gmax

C

M

(1)

N (2) (2) G

C1 C2



Hng s k chnh l hp ph ng vi nng C = 1 Cc hng s k v n d xc nh bng thc nghim da vo cng thc:

Lp th v s ph thuc ca lgG vo lgC, dng y = ax+b, s tnh cn1 v k.

Phng trnh III.9, khng p dng khi nng cht b hp ph ln, v n ch l b xung cho phng trnh hp ph ng nhit Lang mua.

Thc t thy rng s hp ph ca than hot tnh i vi cht hu c trong nc ph hp vi phng trnh III.9. Trong dung dch nc than hot tnh (b mt khng phn cc) hp ph rt mnh nhiu cht hu c v chng km phn cc hn nc.

Phng trnh BET l phng trnh hp ph ng nhit ca 3 tc gi l Brunaoe, Emt v Tenle, c

dng nh sau:

vm: th tch kh b hp ph khi ton b b mt cht hp ph b ph 1 lp n phn

t c kht. v: th tch tng cng ca kh b hp ph p mt cn bng p po: p sut hi bo ho ca cht b hp ph cng nhit vi nhit xy ra hp

ph c: hng s ph thuc nhit

Bit vm c th tnh c din tch b mt cht hp ph. Phng trnh BET l phng trnh hp ph vt l a lp ca cht hp ph rn i vi

cht kh. Phng trnh c p dng xc nh b mt ring ca cht xc tc v cht hp ph rn.

V d: che ph 1g silicagel (SiO2.H2O) bng mt lp n phn t cn th tch kh nit l vm =129ml ( 1atm v 00C). Tnh din tch b mt ca silicagel, nu ph c bn ca nit l 16,22.

Gii:

Phng trnh hp ph trong dung dch. Trong thc nghim, vic xc nh hp ph ca cht hp ph rn trong dung dch

phn t v dung dch in ly nhit khng i, da vo phng trnh:

G: hp ph (mol.g-1) C0 v C: nng u v nng cn bng (mol.l-1) ca dung dch cht b hp

ph. V: th tch (l) dung dch trong xy ra hp ph. m: khi lng (g) cht hp ph.

(III.10) )1(.1

)( 00 cpvpc

cvppvp

m

+=

1228233

.7,561)10.(2,16.10.02,6.4,22

10.129 == gmSr

(III.11) )( 0m

VCCG =

kCn

G lglg1lg +=



V d: Cho 5g bt TiO2vo 100ml dung dch Natri Dodecylsunfat vit tt l NaDS (C12H25SO4Na) nng 0,002M. Sau mt thi gian, nng NaDS l 0,0014M. Tnh ph ca TiO2? Bit b mt ring ca TiO2 l 7,8m2g-1 v So,NaDS=27,32.

Gii: hp ph ca TiO2:

Din tch b mt hp ph:

ph ca TiO2:

%28,25%100.8,7

972,1 == Ion b hp ph y l ion C12H25SO-4, chng t b mt ht TiO2 mang in tch

dng. 4. S hp ph trn b mt lng Nc l cht lng c sc cng b mt ln. Cc cht ho tan c th b hp ph ln b

mt lng lm gim sc cng b mt ca nc. Quan h gia lng cht b hp ph (G) trong lp b mt, nng cht tan trong dung

dch (C) v sc cng b mt () ca dung dch theo phng trnh: (III.11) .

dCd

RTCG =

G: hp ph (mol.cm-2)

dCd : bin thin sc cng b mt ca dung dch.

Phng trnh III.11 gi l phng trnh hp ph Gip (Gibbs). Da vo phng trnh chng ta phn bit 2 trng hp sau:

Khi dCd >0 th G



i vi nc, cht in ly v c nh axt, kim, mui l nhng cht khng hot ng b mt. Dung dch cc cht c sc cng b mt ln hn cht t so vi nc.

Khi dCd 0, l s hp ph dng, nng cht tan trong lp b mt ln

hn trong th tch dung dch. Nhng cht ho tan lm gim sc cng b mt ca dung mi gi l nhng cht hot ng b mt. i vi nc, cht hot ng b mt thng l cc hp cht hu c phn cc nh ru, axt bo, mui ca axit bo (thng gi l x phng), cc ankyl sunfat (nh NaDS)

Cu to phn t cht hu c phn cc gm 2 phn: phn khng phn cc l gc hydrocacbon v phn phn cc l cc nhm chc xem hnh III.3.

Trong dung dch nc cht hot ng b mt b hp ph sp xp nh hng nh sau:

phn phn cc hng vo nc (mi trng phn cc), phn khng phn cc hng vo khng kh (mi trng khng phn cc). Tu theo nng dung dch m lp b mt c th cha bo ho - xem hnh III.4a, hoc bo ho - xem hnh III.4b. Tng t, cc phn t cht hot ng b mt sp xp nh hng b mt 2 pha lng c phn cc rt khc nhau l H2O v C6H6 xem hnh III.4c v III.4d. (Ch : H2O c sc cng b mt ln hn so vi C6H6).

lp hp ph bo ho cc phn t cht hot ng b mt xp kht nhau to thnh mng b mt che ph b mt dung dch.

Mch cc bon cng di, tnh ght nc ca cc phn t cht hot ng b mt cng tng, chng cng c khuynh hng tp trung trn b mt, sc cng b mt ca dung dch cng gim. l c s l thuyt ca quy tc Trao b (Traube): Kh nng b hp ph ca mt dy ng ng cc hp cht hu c phn cc trn b mt nc tng vi s tng di mch cc bon. V vy phng trnh Gp rt ph hp vi cc cht hot ng b mt m phn t c mch cc bon ln (cc cht tng i t tan).

Ch : ngi ta gi nhng cht hot ng b mt nh: C17H35COONa, C12H25SO4Na, C16H33N(CH3)3NBr, l cc cht hot ng b mt ion ho, nhng ion C17H35COO-, C12H25SO4- v C16H33C(CH3)3N+ l nhng anion v cation hot ng b mt.

nc

dc

benzen

Hnh III.4: S v s sp xp nh hng cc phn t cht hot ng b mt: a v b: ranh gii khng kh v pha lng c v d: ranh gii gia 2 pha lng

dung dch nc

a b

khng kh



IV. S thm t v s ngng t mao qun. 1. S thm t. Nh git cht lng ln b mt mt cht rn nu cht lng lan trn b mt cht rn th

l s thm t. Cht lng to vi b mt rn mt gc - gi l gc thm t xem hnh III.5. Khi 0 th gi cht lng l cht lng thm t v b mt rn l b mt a lng, khi = 00 hay cos = 1 th ni rng cht lng thm t hon ton b mt rn.

Cht lng thm t b mt rn l do lc tng tc gia cc phn t cht lng yu hn lc tng tc gia cc phn t lng v rn. T suy ra:

- Cht lng c sc cng b mt cng nh cng thm t tt cc b mt rn. Nc c

sc cng b mt ln ch thm t hon ton mt s cht nh thch anh, thu tinh silict. Thu ngn lng c sc cng b mt rt ln, ch thm t mt s t b mt rn. Nhng cht lng khng phn cc (v d: hydrocacbon) c sc cng b mt rt nh khong 20.10-3 n 30.10-3 N.m-1, thc t thm t mi b mt rn.

- tng tnh thm t ca nc cn thm vo cht hot ng b mt. S thm t l mt hin tng b mt tng t s hp ph v do sc cng b mt gia

cc cht quyt nh. iu kin cn bng ca s thm t l c s cn bng v sc cng b mt gia cc cp pha tng ng xem hnh III.5.

(III.12) cos. LKRLRK += RK, RL, LK ln lt l sc cng b mt ca 2 pha R K, ca 2 pha R L v ca 2

pha L - K. Phng trnh III.12 gi l phng trnh Iung (Young). Do RK v LK l cc hng s, nn o c th tnh c RL . l cch xc nh

sc cng b mt theo phng php thm t. S thm t c tnh chn lc: b mt s thm t cht lng no m hiu s sc cng

b mt vi pha rn l b nht. V d: Nc v mt hydrocacbon no , th gia 2 cht lng s c s thm t chn lc. Nu b mt a nc th n s thm t chn lc nc.

2. S ngng t mao qun. S ngng t mao qun l qu trnh chuyn th hi sang th lng trong mao qun

iu kin ng nhit. Ch c cht lng thm t thnh mao qun th hi ca n mi d dng ngng t trong mao qun.

Nh chng ta bit, khi nhng mao qun vo cht lng thm t th cht lng dng ln trong mao qun thnh ct lng c mt khum lm xem hnh III.6. Nguyn nhn ca hin tng nh sau:

Do c s hp ph a phn t hoc a lp nhng phn t hi trn b mt mao qun, tng hp ph a phn t gn cht vi thnh mao qun chuyn cc lp hp ph thnh cht

RK

LK

T

K

L R RL

Hnh III.5. S git thm t vi gc : cht lng L, cht rn R v khng kh K



lng c mt khum lm. l s ngng t mao qun. Do p sut hi bo ho mt khum lm (P) thp hn p sut hi bo ho trn b mt thong phng (P0), nn cht lng b ht vo trong mao qun. t ti p sut hi bo ho l iu kin cn thit cho s ngng t mao qun.

C th ni rng thc cht ca s ngng t mao qun l s hp ph vt l, Tm xn

Ken vanh (Thompson Kelvin) thit lp quan h gia p sut hi bo ho trong mao qun vi sc cng b mt ca cht lng () v bn knh mao qun (r) nh sau:

(III.13) 2ln0 RTr

VPP =

V: th tch mol ca cht lng khi ho hi Theo cng thc III.13 th sc cng b mt ca cht lng cng ln hoc bn knh mao

qun cng nh th p sut hi bo ho trong mao qun cng nh, ngha l hi cng d ngng t thnh lng. Nc l cht lng c sc cng b mt ln, cc h mao qun trong t, trong cc c cu sinh vt rt nhiu, do hi nc d ngng t thnh lng .

S ngng t mao qun gn lin vi s thm t ca cht lng, da vo s ngng t mao qun xc nh sc cng b mt ca cht lng gi l phng php mao qun. Ni dung nh sau:

Nhng mt mao qun vo cht lng thm t th cht lng dng ln trong mao qun n cao h xem hnh III.6, do c cn bng gia trng lc P1 v lc y P2. Trng lc do ct cht lng gy ra l P1=r2hg. Lc y gy ra bi sc cng b mt ca cht lng l P2=2cos = 2 nu ~00. T xc nh c sc cng b mt ca cht lng theo cng thc:

r: bn knh mao qun : khi lng ring ca cht lng h: cao ct cht lng trong mao qun g: gia tc trng trng

h

p

P0

Hnh III.6: S dng ca cht lng trong mao qun: : gc thm t P v P0: p sut hi bo ha trong mao qun v trn b mt thong phng

(III.14) ...21 ghr =



Trong thc t, ngi ta dng phng php t i loi b cc i lng khng xc nh (nh r, g). xc nh sc cng b mt ca cht lng kho st, cch tin hnh nh sau: Nhng mao qun vo mt cht lng bit sc cng b mt o v khi lng ring o, o c cao ho. Sau li nhng mao qun vo cht lng kho st cng nhit , o c cao h. p dng cng thc III.14 cho mi trng hp chng ta c:

..21 hr = v 000 .2

1 hr = T c cng thc:

(III.15) .

000 h

h =

: khi lng ring ca cht lng kho st V d: 200C, khi nhng mt ng mao qun thy tinh vo benzen th ct cht lng dng ln n hx =5cm, cn khi nhng vo nc th ct nc dng ln n h = 11,3cm. Tnh sc cng b mt ca benzen nhit trn?

Bit: 3301

0 .899,0,.997,0,.75,72 6622 === cmgcmgcmdyn HCHH

Gii: p dng cng thc III.14

131 .10.02,29.02,293,11997,0

5899,075,72 === mNcmdyn



CU HI V BI TP CHNG 3

1. Nng lng b mt v sc cng b mt? Cc yu t nh hng?. 2. Khuynh hng gim nng lng b mt ca h phn tn d th? 3. S hp ph. Phn bit hp ph vt l v hp ph ho hc? 4. Tnh chn lc v nh hng ca s hp ph? V d? 5. hp ph? Cc yu t nh hng. ph v khong xc nh ca ph? 6. Phng trnh hp ph ng nhit Langmur v hn ch ca phng trnh ? ng dng ca phng trnh Fruendlich? 7. Phng trnh hp ph Gibbs? Quy tc Traube? Gii thch. 8. Cht hot ng b mt? C ch hp ph cht hot ng b mt trn b mt lng. 9. S thm t? iu kin cn bng ca s thm t. S ngng t mao qun? 10. Xc nh sc cng b mt bng phng php mao qun? 11. nhit 250C thy rng 1g than ci hp ph c ti a 46mg hi nc. Tnh b mt ring hot ng ca than ci ny i vi nc? Bit rng: khi lng ring ca nc lng 250C l = 0,982g.cm-3, coi phn t nc c dng hnh lp phng, hp ph ca than l n lp.

Tr li: 132m2.g-1 12. Cho 10g t st, khi lng ring =2,7g.cm-3 vo 1 lt dung dch CaCl2 nng 0,4mol.l-1 Lc cho qu trnh hp ph xy ra. Sau mt thi gian thy nng CaCl2 ch cn 0,2mol.l-1. Tnh:

a/ hp ph ca t st i vi Ca2+.nH2O? b/ ph b mt ca t st ? Bit: b mt ring ca t st ny l 3m2.g-1, ph c bn ca ion Ca2+.nH2O l

10,22 Tr li: 0,02milimol.g-1; 40,9%.

13. Cho 5g bt mt loi TiO2 vo 100ml dung dch NaDS nng 5.10-3M. Sau mt thi gian nng cn bng hp ph NaDS ca dung dch l 2.10-3M. Tnh ph? Gii thch.

Bit rng: b mt ring ca TiO2ny l 7,8m2g-1, ph c bn ca NaDS l 27,32. Tr li: 126,41%; hp ph a lp.

14. Cho 2 g mt loi than hot tnh vo 40ml dung dch CH3C00H 0,22M. Lc cho xy ra s hp ph. Khi c cn bng hp ph th nng CH3COOH cn li l bao nhiu?

Bit rng s hp ph ca loi than hot tnh trn i vi axt CH3COOH tun theo phng trnh Frendlich vi k = 3.10-3 v n = 2.

Tr li: 0,16M 15. Tnh din tch b mt ca 1g cht xc tc, bit rng to ra 1 lp n phn t n hp ph 103ml kh nit (tnh 760mml Hg v 00C). Hp ph c o -1950C, din tch hiu dng do 1 phn t nit chim trn b mt xc tc nhit l 16,22.

Tr li 449m2g-1 16. C dung dch axit palmitic trong bezen cha 4,24g.l-1. Khi cho dung dch ln b mt nc th benzen b bay hi, cn axit palmitic to ra lp mng n phn t c kht. Nu cn ph mt b mt din tch 500cm2 bng lp ny th cn ly th tch dung dch axit trn l bao nhiu? B mt do mi phn t axit palmitic chim l 21 2. 17. 200C sc cng b mt ca toluen l = 28,4.10-3N.m-1, khi lng ring ca toluen bng 0,866g.cm-3. Tnh kch thc ln nht ca mao qun ct cht lng dng ln c 2cm?

Tr li: 0,0335cm.



CHNG IV TNH CHT IN CA CC H KEO

Tnh cht in ca h phn tn cng l tnh cht b mt ca h vi d th nhng l tnh

cht c trng ca h keo, n quyt nh tnh in ng, tnh phn tn, tnh hp ph trao i ion ca h l c s ca cc phng php phn tch tch nhng vi ht, nhng cht in ly cao phn t lm bn hoc n nh h v nhiu phng php k thut cng ngh khc.

I. Hin tng in ng. in di v in thm: Khi t mt h keo trong mt in trng th cc ht phn tn mang in chuyn dch

v in cc tri du vi chng; c th lm cho mi trng y vn c, cn mi trng phn tn cha cc ion n gin hydrat ho i v in cc i lp vi in cc trn.

Di tc dng ca in trng cc phn t mi trng phn tn chy v mt in cc gi l in thm, cn cc ht phn tn chy v mt in cc khc gi l in di

in di, in thm, gi chung l hin tng in ng ca h phn tn. Hin tng gip chng ta xc nh c du in tch ca ht phn tn, l c s cho phng php phn tch in di (c s dng rng ri trong ho sinh, th nhng ) v phng php lm kh cc vt liu xp

Hin tng in ng ca h phn tn bt ngun t cu to ca ht phn tn. C th coi ht phn tn nh mt phn t ln MnR , c kh nng in ly ra cc ion n gin M v ion R , nhng ion R c kch thc tng i ln hn cc ion n gin, nn c b mt phn cch vi mi trng.

II. Cu to ht keo ght lu. Hai thnh phn ch yu ca ht keo l nhn keo v lp in kp.

Hnh IV:1: S cu to ht keo dng AgI

mAgI

mAgI

+ + + +

+ + + +

+ + + +

+

+ + +

+ + +

+ +

+

+

+ +

tng khuch tn

tng hp ph

ion quyt nh th

nhn keo



Nhn keo Nhn keo do rt nhiu phn t, nguyn t hoc ion n gin tp hp li, cng c

trng hp do s chia nh ca ht ln hn. Nhn keo c th c cu to tinh th hoc v nh hnh, nhng l phn vt cht n nh, hu nh khng c bin i trong cc qu trnh bin ng ca h phn tn.

Lp in kp. Lp in kp gm hai lp tch in ngc du nhau, nhng cu to phc tp v lun

lun bin i di tc ng bn ngoi (mi trng, pH, lc ion, nhit .). Lp in kp c hnh thnh ch yu do s hp ph.

V d 1. Bng phn ng trong dung dch ca AgNO3 vi KI khi cho d lng ca mt trong hai cht, chng ta s nhn c cc h keo dng AgI v keo m AgI.

Phn ng trao i: AgNO3 + KI KNO3 + AgI To ra cht kh tan AgI, nhng phn t ny tp hp vi nhau thnh nhiu ht nh AgI.

Nu d AgNO3 th cc ht keo dng AgI c to thnh - xem hnh IV.1 cng thc cu to nh sau:

Ht nh AgI l nhn keo, hp ph chn lc i vi cc ion Ag+, mt s ion i (I-) ko

theo, to thnh ion keo. Lp ion i xung quanh ion keo c gi l tng khuch tn. Tng hp ph v tng khuch tn tch in ngc du nhau hp thnh lp in kp b mt ht keo. Ion Ag+ c gi l ion quyt nh th, ion keo mang in(+) nn gi l ht keo dng.

Nu d KI th cc ht nh AgI hp ph chn lc i vi ion I- to ra ht keo m AgI, cng thc cu to nh sau:

{[mAgI] nI- (n-x) K+ }x-xK+ Trng hp ny ion I- c gi l ion quyt nh th. Nu s mol AgNO3 v KI va phn ng vi nhau th trong h ch c rt nhiu ht

nh AgI ( c kch thc ca ht keo), chng s kt dnh vi nhau thnh cc ht ln hn v sa lng. V th ngi ta gi ion Ag+ hoc AgNO3 v ion I- hoc KI l nhng cht lm bn cho cc h keo AgI.

Tc dng lm bn cho h phn tn c th l cht v c, cht hu c, cng c th l dung mi hoc cc iu kin v pH, v nhit ca h. Ngi ta hay lm bn keo ght lu c phn tn cao bng cht prtit nh: gelatin, albumin

Tng hp ph c cu to dy c tng i n nh t khi b bin i bi ngoi cnh. Trong tng hp ph cc ion quyt nh th gn cht vi b mt nhn keo, c mt s ion i

nhn keo

tng hp ph

tng khuch tn

ion keo

ht keo hoc mi xen keo

[ ]{ } ++ 33 )( xNONOxnnAgmAgI x



{ } ++ xClClxnnFeOOHmFe x)(])([ 3

lin kt vi b mt rn bng lc tnh in v c th c lc hp ph (lc tng tc phn t), c th cn c cc phn t dung mi do tnh sonvt ca cc ion. Tng khuch tn c cu to gm nhng ion ngc du vi ion keo, phn b xung quanh ion keo v ch tng tc tnh in vi ion y. Tng khuch tn rt linh ng lun bin i theo tnh cht mi trng ca h.

V d 2: Nh tng git FeCl3 vo nc si, th FeCl3 thy phn:

Phn ng trn khng to ra Fe(OH)3 kt ta m iu ch c keo dng Fe(OH)3 mu , cng thc cu to ht keo nh sau:

Ion FeO+l ionquyt nh th, do FeOCl c to ra trong qu trnh phn ng. Qua hai v d trn chng ta nhn thy rng ion quyt nh th phi l ion ging vi

thnh phn ca nhn keo hoc ion tng t. Du in tch ca ion quyt nh th cng l du in tch ion keo. Nu l keo m th tng khuch tn gm nhng cation, keo dng th tng khuch tn gm cc anion, nu l keo lng tnh th tng khuch tn bao gm c cation v anion.

V d: keo Al(OH)3 l mt keo v c lng tnh. Trong dung dch nc phn t Al(OH)3 c trng thi cn bng axit baz:

cng thc ht keo nh sau:

[ ]{ } zAOHmAl OHxAl OOHyAl )( 22)( )(3 + Khi x > y th ion +2)(OHAl l ion quyt nh th, c ht keo dng Al(OH)3, trong

tng khuch tn gm z ion A trong a s l anion. Khi y > x th ion Al(OH)2O- l ion quyt nh th, c ht keo m Al(OH)3, trong tng khuch tn a s l cation.

Do trng thi cn bng (*) ph thuc pH mi trng nn du in tch hat keo Al(OH)3 s thay i khi pH mi trng bin i.

Ngoi nguyn nhn do hp ph ion nhng cng c trng hp ht phn tn mang in do b mt c nhm chc c kh nng in ly, do b mt hp ph nhng phn t cha cc nhm chc c kh nng ion ho. nhng trng hp y s phn chia nhn keo vi lp in kp khng r rng.

Keo t c thnh phn phc tp, bao gm keo dng, keo m, keo lng tnh. Nhng keo t phn ln l keo m, v b mt ht thng c hin tng in ly ra ion H+

III. Lp in kp v in th b mt ca ht keo. 1. Cu to lp in kp. Trong dung dch in ly b mt rn tch in xut hin mt lp in kp gm: lp

in tch b mt ca b mt rn v lp ion i c mt s l thuyt v lp in kp bt ngun t nhng quan nim khc nhau v cu to lp ion i. Sau y l nhng ni dung chnh ca thuyt Hemhn, thuyt Guy- Sepmen v thuyt Stc.

Thuyt Hemhn (Helm holtz) Lp in kp cu to theo kiu mt t in phng. Cc ion i nh nhng in tch

im, ch c tng tc tnh in vi b mt rn v cch b mt rn mt khong 0, gi l b

FeCl3 +3H2O Fe(OH)3 +3HCl

(*) Al(OH) Al(OH) )( 232++ ++ HOOHAlOH



dy lp in kp, bng hai ln bn knh ion i - xem hnh IV.2a. in th b mt hoc th ca lp kp chnh l in th ca t in phng (0):

: mt in tch b mt : hng s in mi ca mi trng 0: b dy lp in kp

in th bin i tuyn tnh theo b dy ca lp in kp - xem hnh IV.2b.

y l lp in kp n gin nht, tc gi b qua nhiu yu t tng tc v nhng

tnh cht khc nhau ca ion nh tnh sonvt, chuyn ng nhit Theo Hemhon th cc ion tng khuch tn ca ht keo khng linh ng, v vy thuyt Hemhn v cu to lp in kp khng gii thch c hin tng in ng h phn tn.

Thuyt Guy Sepmen (Gouy Chapman) Theo thuyt Guy Sepmen th: lp in kp cu to khuch tn thng gi tt l lp

in kp khuch tn, do cc ion i khng ch tng tc tnh in vi b mt rn m cn tham gia chuyn ng nhit.

T b mt rn vo dung dch mt ion gim dn, in th b mt bin i khng tuyn tnh theo b dy ca lp in kp m bin i theo hm m - xem hnh IV.3. Cc tc gi tnh ton v mt in tch ca dung dch, v s bin i ca in th b mt trong dung dch theo khong cch xv kho st s bin i in th b mt trong tng khuch tn ca lp in kp:

S phn b ion theo phng trnh Bonzman (Boltzmann):

Ci: nng ion (mol.l-1) v tr x tnh t b mt rn tch in Ci0: nng ion (mol.l-1) v tr x = ngha l trong th tch dung dch

+ + + + + + + + + + + +

+

0 0 x

0

a b

0 x

0

d

Hnh IV.2: a. lp in kp Hemhn b. in th b mt Hemhn

Hnh IV.3: S bin i th b mt theo khong cch (thuyt Guy Sepmen)

cht rn

(IV.1) 4 00 =

(IV.2) A-xp i0

=RT

eCC ii



Ai: cng vn chuyn 1mol ion i t v tr x = n v tr x, y in th b mt l .

Guy Sepmen coi Ai ch l cng ca lc tnh in Ai=ziF (IV.3)

zi: in tch ion i k c du F: s Faraay

Qua cc cng thc IV.2 v IV.3 suy ra:

v tnh c mt in tch () ca dung dch ti x (tnh bng tng i s in tch ca cc cation v anion:

Theo phng trnh Poaxng (Poisson):

y 2 l ton t Laplat (Laplace). Coi nh b mt rn phng v in th b mt thay i ch theo trc x th:

T Guy v Sepmen thit lp c quan h ca in th b mt vi khong cch x hoc cng thc v in th b mt ca lp in kp nh sau:

(IV.8) .0 dx

e=

: in th b mt ti x tnh t b mt rn 0: in th b mt Hemhon. d: b dy lp in kp (ln hn b dy lp in kp Hemhon).

khi dung dch ch c cht in ly tr(2) . Bin lun cng thc IV.1: - B dy lp in kp s l 0 khi x 0, th =0 - B dy lp in kp x=d th =0/e - Khi x- th = 0. Chng ta nhn thy: lp in kp kiu Hemhn ch l mt trng hp c bit ca

lp in kp Guy Sepmen.

(2) Trong thnh phn cht in ly tr khng cha ion c th xy dng li tinh th ca b mt rn.

(IV.4) exp0

=RTFzCC iii

(IV.5) xp0 == RTFzeCzFFCz iiiii

(IV.6) 42 =

== (IV.7) z-xp44 i02 RT

FeCzFx ii



Nhit ng hc chng t rng b dy lp in kp (d) t l nghch vi lc ion ca

dung dch (): y: l lc ion ca dung dch, k l hng s ph thuc nhit v hng s in mi

ca mi trng. i vi nc nhit thng (250C), cc cht in ly ho tr 1-1 th k = 3, cc ho tr cao hn th k nh hn.

Vy lc ion ca dung dch cng ln th b dy lp in kp cng nh, ngc li cng nh th d cng rng - xem hnh IV.4, tt nhin khi d > 0 th lp in kp Guy - Sepmen s tr thnh lp in kp Hemhn. Nhng mi im m xi = di , in th b mt khng bng 0, c th = 0/e, l im khc nhau c bn gia thuyt Guy Sepmen v thuyt Hemhn v cu to lp in kp.

Theo thuyt Guy Sepmen th trong dung dch c mt lp mng gn kt cht vi b rn ca mt ht phn tn mang in hoc ht c b mt trt (v d: b mt tip xc ca ion keo vi mi trng). v pha ngoi l lp dung mi (c cha cc ion ca tng khuch tn) rt linh ng. Chuyn ng ca ht trong cht lng l s dch chuyn (trt) tng i so vi lp dung mi linh ng.

Thuyt Guy - Sepmen l s m rng thuyt Hemhn, tng i ph hp vi cc dung dch keo long. Mt nhc im ca thuyt Guy - Sepmen l khng gii thch c cc trng hp i du in tch ca b mt ht trong dung dch in ly.

kd = (IV.9)

= 221 ii zC (IV.10)

(1)

0

d3

(2) (3)

d2 d1

x

e0

Hnh IV.4. S bin i b dy lp in kp (di) theo lu ion ca dung dch (i). Cc ng cong (1),(2) v (3) ng vi 1



Thuyt Stc. Tc gi b sung thm cho thuyt Guy v Sepmen rng: b mt trt ca ht phn

tn ngoi lc ht tnh in, cc ion i cn c th c lc hp ph (lc tng tc phn t) vi b mt rn. Trn c s nghin cu hin tng i du in tch b mt trong dung dch cht in ly.

Chng hn khi b mt rn tch in dng th tng i ca n l cc ion m, b dy lp in kp l d. Nu trong dung dch c nhng ion m ho tr cao nh SO42-, PO43- th lc ion ca dung dch tng c th rt ln v d 0 th mt lp in kp th 2 xut hin, c b dy lp l d. Du in tch v in th b mt i t dng thnh m - xem hnh IV.5.

Cc ion ho tr cng cao th ni chung cng km hydrrat ho, chng tng tc tnh in mnh hn vi b mt rn v c th b hp ph (bi lc tng tc phn t), dn n s qu in tch, do c hin tng i du in tch b mt rn. Lp in kp th 1 l lp in kp Guy - Sepmen, lp in kp th 2 c gi l lp in kp Stc.

Lp in kp Stc l s m rng lp in kp khuch tn Guy - Sepmen cho trng hp i du in tch b mt trong dung dch in ly. Trng hp ny t xy ra, thng ch xut hin cc dung dch c lc ion ln v nhiu ion ho tr cao. Ion hu c ho tr thp cng rt d b hp ph nn n c th gy ra hin tng i du in tch cc b mt.

2. in th b mt ca ht keo. Gm th nhit ng v th in ng Th nhit ng Th nhit ng xut hin b mt ca nhn keo v mi trng. in th ny tng

t in th in cc, l bc nhy th gia hai pha l kim loi (pha rn) v dung dch

x

0

d d'

(1)

(2)

0

e0

e0'

-0

Hnh IV.5: S bin i th b mt: (1) ca lp Guy Sepmen (2) ca lp Stc



in ly (pha lng). Nguyn nhn hnh thnh in th tng t nh in cc, do nhng yu t nhit ng gy nn.

Ni chung th nhit ng khng ph thuc b dy tng khuch tn, ch yu ph thuc hot ion quyt nh th c trong dung dch. S thay i nng cht in ly (khng cha ion quyt nh th hoc cht in ly tr) hu nh khng lm thay i th . Th nhit ng t nh hng n tnh cht h

Th in ng Th in ng (dzta) xut hin ranh gii gia tng hp ph v tng khuch tn

hoc lp in kp ca b mt ht keo. Lp in kp y gm tng hp ph tng i n nh gn cht vi nhn keo v tng khuch tn rt linh ng. Ion keo v tng khuch tn c chuyn ng trt tng i vi nhau. l s chuyn ng trt ca 2 lp in tch do tnh ng hc phn t , lm xut hin th .

Th in ng t l thun vi mt in tch b mt ht v b dy tng khuch tn ca ht keo:

= k. (IV.11) y: k l hng s ph thuc mt in tch b mt ht keo, :b dy tng khuch

tn ca ht keo. Lp in kp ca b mt ht keo c cu to tng t lp in kp khuch tn ca

thuyt Guy - Sepmen, nn th cng t l nghch vi lc ion ca dung dch: k: hng s ph thuc nhit , hng s in mi ca mi trng. Nng cc ion (Ci) cng ln, ho tr (zi) ca chng cng cao, th lc ion ca dung

dch cng ln, tng khuch tn cng b nn li lm cho b dy lp in kp cng nh. Ngc li nng cc ion cng nh, ha tr ca chng cng nh th b dy ca lp in kp cng rng, nn th in ng cng ln - xem hnh IV.6.

=

2

21

'

ii zC

k (IV.12)

Hnh IV.6. S ph thuc ca in th vo b dy lp in kp (hoc vo lc ion : cc ng cong (1), (2), (3) ng vi 3>2>1)

(1)

0

d3

(2) (3)

d2 d1

x

3 2

1



Do , th in ng c ngha ln n tnh bn ca mt h phn tn, v in th quyt nh tnh chuyn ng, tnh phn tn ca ht. Khi c th ln , c lc y ln gia cc ht, tnh chuyn ng, tnh phn tn ca chng tng ln.

3. Xc nh th in ng. Cc phng php xc nh in th da trn c s hin tng in ng ca cc h

keo nh in di, in thm Trong phng php in di, th xc nh theo cng thc: Trong v l tc in di hoc tc in thm: T cc cng thc IV.6 v IV.7 suy ra:

h(cm): cao ct nc dng ln trong mao qun t(s): thi gian in di (poa): nht ca dung mi : hng s in mi u(v.cm-1): cng in trng hay gradien in th trong in di.

Chuyn h n v CGS ra von(volt) chng ta c cng thc: V d: in di h keo Fe(OH)3 trong nc thc hin hiu s in th gia 2 in cc

l 30cm ht 20pht, ct nc dng ln trong mao qun l 24mm. Hy tnh th in ng ca ht keo?

Cho bit: 81 ,01,022 H

== OOH poa Gii: p dng cng thc IV.16: Vic s dng cng thc IV.8 tun theo mt s iu kin: - B dy ca lp in kp phi rt nh so vi kch thc ht. - Ht khng dn in.

uv

4= (IV.13)

thv = (IV.14)

th

u.4

= (n v CGS) (IV.15)

.12006020,4,2,.330

150 1 sxtcmhcmvu =====

mV8,55)300.(1200

4,2.581

01,01416,34 2 ==

2)300.(.4th

u = (IV.16)



IV. Hp ph trao i ion. 1. Khi nim. S hp ph trao i ion l s trao i ion tng khuch tn ca ht keo vi ion cng

du trong dung dch. K hiu: XA- v XM+ l cc ht keo , A- v M+l cc ion trao i trong dung dch,

chng ta m t s hp ph trao i ion gia chng bng cc s phn ng sau: Vy keo m c kh nng hp ph trao i cation, keo dng c kh nng hp ph trao

i amion, keo lng tnh l c kh nng trao i c cation v anion. c im: - Hp ph trao i ion c tnh cht mt phn ng thun nghch. S chuyn dch cc

cn bng (a) v (b) ph thuc nhiu yu t nh nng , ho tr, bn knh hyhat ca ion, nhit . C th vn dng nguyn l chuyn dch cn bng gii thch chiu hng ca s trao i.

V d: cc cation Ca2+ v Na+ cng c cu hnh electron ca v kh him, nhng ion Ca2+ c ho tr cao hn nn trng thi cn bng hp ph trao i:

X2Na+ + Ca2+ XCa2+ + 2Na+ (c) rt d chuyn dch theo chiu thun (y l mt c s ca bin php thm vi ci to t mn thnh t c nhiu Ca2+ thch hp cho cy trng). Hin tng trn chng t tnh trao i hp ph ca Ca2+mnh hn Na+. Trng hp cation trao i trong dung dch, nh M+ trong cn bng (a), l cc cation kim loi kim th th tnh trao i hp ph ca chng tng dn nh sau: Mg2+; Ca2+; Sr2+; Ba2+, do bn knh hydrt ca ion gim dn, nn cng in trng ca chng tng dn.

- Quan h cht ch vi pH mi trng: Do linh ln nn cc ion H+ v OH-c u tin trong qu trnh hp ph trao i:

S chuyn dch cc cn bng (d) v (e) u lm thay i pH ca h. V vy ni chung

s hp ph trao i ion ph thuc pH nhng cng lm bin i pH mi trng. Hin tng trn lun xy ra i vi keo lng tnh:

C th iu chnh mi trng n 1 gi tr pH sao cho s trao i cation v amion ca

keo lng tnh bng nhau. in tch b mt ht keo lc ny c trung ho, gi tr pH gi l im ng in ca keo. V d: im ng in ca keo Al(OH)3 l pH = 8,1; ca Fe(OH)3 l pH = 7,1.

nhng gi tr pH thp hn im ng in, b mt ht tch in dng, nn keo lng tnh u tin trao i anion, s trao i cation b hn ch. nhng gi tr pH cao hn

(a) ++++ ++ MXMMXM '' ++ AXAAXA '' (b)

++++ ++ MXHHXM (d) ++ AXOHOHXA (e)

X M+A- A- + H+ XH+A-A- + M+



im ng in, b mt ht mang in m, nn keo lng tnh u tin trao i cation, s trao i anion b hn ch.

2. Nha trao i. Cc cht ch c kh nng trao i cation c gi l cationit, cc cht ch c kh nng

trao i anion c gi anionit, cc cht c kh nng trao i c cation v anion c gi l cht trao i lng tnh hoc ionit lng tnh. Tn gi chung cc loi cht l ionit. Ionit cng c trong t nhin nh khong alumosilicat, cc polyme hu c ion ha trong t

Ngy nay ionit thng c tng hp di dng nha tng hp hoc nha trao i. l cc polyme hu c (tng hp t nhng monome) cu to mng khng gian; cha nhiu nhm chc sinh ion, nn thuc loi cht in ly cao phn t. u im ca nha trao i: thnh phn n gin nh trc c, khng tan trong nc, c tnh bn c hc, bn ho hc v c dung lng trao i cao.

Dung lng trao i tnh bng s mili ng lng gam ion b tch khi 1g ionit kh. l mt thng s c trng quan trng cho cht trao i ion, do n cho bit s nhm chc ti a m ionit c.

Nha cationt thng cha cc nhm hot ng l nhm sunfonic SO3H, nhm cacboxilic COOH, nhm OH (phenol)

Nha anionit thng cha cc nhm hot ng l nhm amin bc mt NH2, nhm amin bc hai = NH, nhm amin bc ba N

Nha ionit lng tnh bao gm c 2 loi nhm chc trn, n thuc loi polyme a in tch.

Gc hydrocacbon trong cc loi nha trao i c th l mnh h, mch vng thm. Cc loi nha trao i khng ch tch c cc cation v anion v c, m c cc hp cht hu c phn cc hoc ion ho.

Nha trao i c dng nhiu trong phn tch sc k tch cc nguyn t t him tch cc sn phm trong phn hu phng x, tch cc axit amin, cc hp cht hu c c hot tnh sinh hc , cc vitamin Trong i sng nha trao i s dng kh mui cng, tch cc cht c hi sinh hc ra khi nc thi.

3. Nhit ng hc v hp ph trao i ion Theo nhit ng hc th hp ph trao i ion l mt phn ng ha hc. S trao i ion

gia ionit XM1 vi M2 trong dung dch, c m t bng phn ng sau:

)(C )(G )(C )( 12211221

GMXMMXM ++

y: G1 v G2 l hp ph ca ionit i vi ion M1 v vi ion M2 , C1 v C2 l cc nng ca ion M1 v ca ion M2 trong dung dch.

Biu thc quan h gia hp th (G) ca ionit vi nng ion trao i (C) trong dung dch khi phn ng trng thi cn bng v ho tr (z) cc ion , c thit lp, c th gi l phng trnh hp ph trao i:

(IV.9) 2

1

2

1

1

2

1

11

2

1

1

z

z

z

z

C

CKG

G =

z1 v z2 l ho tr ln lt ca M1 v M2 K l hng s cn bng i vi 1 ionit xc nh v 1 dung dch xc nh (v ion trao i

v nng ca n) mt nhit khng i.



Thng gi K l h s chn lc, n cho thy s khc nhau tng i v i lc ho hc ca cc ion M2 v M1 i vi ionit. Tnh trao i hp ph ca ion trao i trong dung dch cng mnh th h s chn lc cng ln.

Phng trnh (IV.9) p dng vi dung dch long v ion trao i trong dung dch khng to thnh phc cht vi nhm chc ca ionit. Phng trnh IV.9 l mt dng ca phng trnh Nicnski (Nikolski)

Hin nay ngi ta hay dng cationit kh nc cng, v d nc cha nhiu ion Ca2+. C s ca phng php l phn ng (c), do ion Ca2+ c kh nng trao i hp ph mnh hn so vi ion Na+. Sau qu trnh c th phc hi cationit ban u theo phn ng:

XCa2+ + 2Na+ X2Na+ + Ca2+ bng dung dch NaCl m c.



CU HI V BI TP CHNG 4 1. Phn bit cc hin tng in di v in thm. 2. Cu to ca ht keo ght lu? c im cu to tng khuch tn ca ht keo? V s cu to ht keo AgI, iu ch t AgNO3 v KI (cc trng hp d AgNO3 v d KI), ht keo Fe(OH)3 t s phn tn h th Fe(OH)3 bng FeCl3 v s thu phn FeCl3 trong nc si. 3. c im cu to lp in kp theo thuyt Helmholtz v thuyt Gouy Chapman? Cng thc tnh in th b mt theo thuyt Gouy Chapman? nh hng ca lc ion n in th b mt. 4. Hin tng i du in tch v in th b mt theo thuyt Stec? 5. Cc in th ca b mt ht keo? nh hng ca lc ion n th in ng v nh hng ca th in ng n tnh bn ca h keo? 6. Cng thc tnh th in ng ca ht keo theo phng php in di? 7. S hp ph trao i v c im? Keo lng tnh v im ng in ca keo lng tnh? 8. Phn bit: cationit, anionit, iont? c im cu to v tnh cht cc loi nha trao i? Dung lng trao i? ngha? 9. Quan h gia hp ph vi nng v ho tr ca ion trong h khi c cn bng hp ph trao i ion (theo nhit ng hc)? ngha ca h s chn lc. 10. Vit cng thc cu to ht keo to thnh sau khi trn 100ml dung dch AgNO3 0,001M vi.

a/ 80ml dung dch KI 0,0015M b/ 80ml dung dch KI 0,001M

11. in thm thu ca h keo Fe(OH)3 tin hnh vi hiu s in th gia 2 in cc l 175V, khong cch gia 2 in cc l 34cm. Cc ht keo chuyn v cc m c mt on 28mm ht 26 pht 16 giy.

a/ Tnh th in ng ca ht keo? Bit: poaHH 01,0,81 00 22 == b/ Cho bit du in tch ca ht keo?

Tr li: 48mV. 12. Th in ng ca ht keo theo in di l 50mV, gradien in th s dng l 6Vcm-1. Tnh tc ca hin tng in di? Bit: poaHH 01,0,81 00 22 ==

Tr li: 0,00215cm.s-1



CHNG V

TNH BN CA CC H KEO V S KEO T Tnh bn v tnh keo t hoc tnh ng t l nhng tnh cht c bn ca cc h keo.

Cc tnh cht gn lin vi tnh siu vi d th v tnh cht ca b mt cc ht, t mi lin h gia b mt ht keo vi mi trng v vi cc yu t bn ngoi khc.

Cn thit phi hiu c yu t ch yu quyt nh tnh bn, nguyn tc keo t, c ch tc dng ca cc yu t lm bn hoc gy ng t Nghin cu cc tnh cht gip chng ta iu khin h s dng h c hiu qu.

I. Tnh bn ca cc h keo. 1. L thuyt v tnh bn. Tnh bn ca h phn tn gn lin vi tnh phn tn ca h. Thi gian m ht phn b

ng u trong h cng di th tnh bn ca h cng cao. Tri vi tnh bn l tnh keo t hoc ng t l tnh kt dnh, tnh tp hp ng vn

v sa lng. Khi xy ra keo t l lc tnh bn ca h bin mt, tnh phn tn khng cn. Kt qu l dung dch keo tch thnh 2 pha r rt: pha keo t ch gm cc ht ca cht phn tn v pha lng ch gm mi trng phn tn.

2. Tnh bn ng hc v tnh bn nhit ng hc Cn c vo tnh phn tn v tnh kt dnh, tnh bn ca h c 2 loi l tnh bn ng

hc v tnh bn nhit ng hc. Tnh bn ng hc hay tnh bn phn b: l kh nng chng li s sa lng ca ht. Cc ht keo c chuyn ng nhit, tnh cht ny chng li s sa lng ca ht. K hiu

l i lng c trng cho tnh bn phn b, th t l nghch vi tc sa lng ca ht (xem cng thc I.6):

(V.1) 1 2 th

rk

v===

v: tc sa lng r: bn knh ht phn tn h: cao m ht sa lng c trong thi gian t k: hng s i s mt h phn tn xc nh, nhit khng i

Vy: - Tnh bn ng hc t l nghch vi kch thc ht phn tn. Kch thc ht cng ln,

tnh bn ng hc ca ht cng nh. - Tnh bn ng hc cho bit thi gian sa lng () ca ht. l thi gian ht sa

lng c 1cm. Thi gian sa lng cng di th tnh bn ng hc cng cao xem bng V.1. Bng V.1: Thi gian sa lng ca cc ht SiO2 (khi lng ring 2,7gcm-3) trong nc

(=0,0115poa, khi lng ring 1g.cm-3).

Bn knh ht r (cm) 10-3 10-5 10-5 10-6 10-7 Tc

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