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PHN NG TO KT TA

TS Vi Anh TunKhoa ha hc Trngi hc KHTN -HQG H Ni

Phn ng to kt ta l phn ng to thnh cht rn t cc cht tan trong dung dch.

Th d:

Ag+

+ Cl- AgCl (r)

Ca2+

+ C2O42-

CaC2O4 (r)

Trong ho phn tch, phn ng to kt ta c s dng :

Tch cht cn xc nh khi cc cht cn tr. Phn tch khi lng. Phn tch gin tip. Chun kt ta.

1. Tch s tan v tan

1.1 Tch stan

Qu trnh ho tan l qu trnh thun nghch, do cng tun theo nh lut tc dng

khi lng. Xt cn bng ha tan (Mn+ l ion kim loi, Xm- l gc axit hoc OH-):

MmXn mMn+

+ nXm-

T = [M]m[X]

n

(*)

T c gi l tch s tan (solubility product).

Tch s tan c s dng :

So snh tan ca cc cht t tan "ng dng". Xem mt dung dch bo ho hay cha: n

X

m

MCCQ = > T: dung dch qu bo ho => xut hin kt ta.

nX

m

MCCQ = = T: dung dch bo ho.

nX

m

MCCQ = < T: dung dch cha bo ho => khng xut hin kt ta.

Tnh tan ca cc cht t tan (mui, hidroxit).

Cu 1.1. So snh tan ca AgCl v AgBr trong nc ct. Bit TAgCl = 10-10

, TAgBr = 10-

13.

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3

Hng dn gii (AgCl > AgBr)

*Ch : Mc d TAgCl= 10-10

> TMg(OH)2= 1,2.10-11

, nhng trong nc ct, tan ca

Mg(OH)2 li ln hn tan ca AgCl.

Cu 1.2. (a) Trn 1 ml dung dch K2CrO4 0,12M vi 2 ml dung dch Ba(OH)2 0,009M.

C kt ta BaCrO4 to thnh khng? Bit TBaCrO4= 1,2. 10-10

.

(b) Tnh nng cn bng ca cc cu t sau khi trn.

Hng dn gii

(a. Q= 0,04 0,006 = 2,4.10-4 > T => c kt ta to thnh;

(b) TPGH: CrO42-

: 0,034 M

BaCrO4 Ba2+

+ CrO42-

Cb x 0,034 + x

T = x (0,034 + x) = 1,2.10-10

x = 3,53. 10-9

M.

[CrO42-

] = 0,034 M;

[Ba2+

] = 3,53.10-9

M)

Cu 1.3. Metylamin, CH3NH2, l mt bazyu phn li trong dung dch nh sau:

CH3NH2 + H2O CH3NH3+

+ OH-

(a) 25C, phn trm ion ho ca dung dch CH3NH2 0,160M l 4,7%. Hy tnh [OH-],

[CH3NH3+], [CH3NH2], [H3O

+] v pH ca dung dch.

(b) Hy tnh Kb ca metylamin.

(c) Nu thm 0,05 mol La(NO3)3 vo 1,00 L dung dch cha 0,20 mol CH3NH2 v 0,20

mol CH3NH3Cl. C kt ta La(OH)3 xut hin khng? Cho tch s tan ca La(OH)3 l

1.10-19

.

Hng dn gii

(a) [CH3NH2]= 0,152 M; [CH3NH3+

]=[OH

-

]= 7,5.10

-3

; pH= 11,9(b) 3,7.10

-4

(c) Q = 2,56.10-12

> T, c kt ta)

Cu 1.4. MgF2(r) Mg2+

(aq) + 2 F-(aq)

Trong dung dch bo ho MgF218 C, nng ca Mg2+

l 1,21.10-3

M.

(a) Hy vit biu thc tch s tan, T, v tnh gi tr ny 18 C.

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(b) Hy tnh nng cn bng ca Mg2+ trong 1,000 L dung dch MgF2 bo ho 18C

cha 0,100 mol KF.

(c) Hy don kt ta MgF2 c to thnh khng khi trn 100,0 mL dung dch Mg(NO3)2

3.10-3

M vi 200,0 mL dung dch NaF 2,00.10-3 M 18C.

(d) 27C nng ca Mg2+ trong dung dch bo ho MgF2 l 1,17.10-3

M. Hy cho bit

qu trnh ho tan MgF2 l to nhit hay thu nhit? Gii thch.

Hng dn gii

(a) 7,09.10-9

(b) 7,09.10-7M

(c) Q < T, khng c kt ta

(d) To nhit)

Cu 1.5. Dung dch bo ha H2S c nng 0,100 M.

Hng s axit ca H2S: K1 = 1,0 10-7

v K2 = 1,3 10-13

.

(a) Tnh nng ion sunfua trong dung dch H2S 0,100 M khi iu chnh pH = 2,0.

(b) Mt dung dch A cha cc cation Mn2+, Co2+, v Ag+ vi nng ban u ca mi ion

u bng 0,010 M. Ho tan H2S vo A n bo ho v iu chnh pH = 2,0 th ion no to

kt ta?

Cho: TMnS = 2,5 10-10

; TCoS = 4,0 10-21

; TAg2S = 6,3 10-50

.

(c) Hy cho bit c bao nhiu gam kt ta ch(II) sunfua c tch ra t 1,00 lit dung dch

bo ha ch(II) sunfat? bit nng sunfua c iu chnh n 1,00 .10-17 M? Cho cc

gi tr tch s tan: TPbSO4 = 1,6 10-8

v TPbS = 2,5 10-27

.

Hng dn gii

a) 17

211

2

212 10.3,1][][

][2

++

=++

= SHaaa

aa CKKKHH

KKS

b) C: [Mn2+

] [S2-

] = 10-2

1,3 .10-17 = 1,3 .10-19 < TMnS= 2,5 .10-10

;

khng c kt ta

[Co2+

] [ S2-

] = 10-2

1,3 .10-17= 1,3 .10-19 > TCoS= 4,0 .10-21

; c kt ta

CoS

[Ag+]

2[S

2-] = (10

-2)21,3 .10-17 = 1,3 .1021 > TAg2S= 6,3 .10

-50; c kt ta

Ag2S

c) C: [Pb2+

][SO42-

] = 1,6.10-8

.

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[Pb2+

] = [SO42-

] = 1,265.10-4

.

Khi nng sunfua t 1,00.10-17M th nng Pb2+ cn li trong dung dch l:

[Pb2+

] = 2,5.10-27

/ 1,00.10-17

= 2,5.10-10

.

mggammPbS 3,3010.03,312,239)10.5,210.265,1(

2104 === )

1.2 Quan h gia tan v tch stan

tan (S, solubility) ca mt cht l nng ca cht trong dung dch bo ho.

tan thng c biu din theo nng mol/l.

tan v tch s tan l nhng i lng c trng cho dung dch bo ho ca cht t

tan. Do , tch s tan v tan c mi quan h vi nhau, iu c ngha l ta c th tnh

c tan ca mt cht t tan t tch s tan ca n v ngc li.

MmXn m Mn+

+ n Xm-

mS nS

C: T = [M]m[X]

n= [mS]

m[nS]

n nmnmnm

TS

+

=

1

*Nhn xt: Cng thc trn chng nu Mn+ v Xm- khng tham phn ng no khc.

Cu 1.6. Cho tch s tan ca Ag2CrO425oC l 2,6.10

-12.

(a) Hy vit biu thc tch s tan ca Ag2CrO4.

(b) Hy tnh [Ag+] trong dung dch bo ha Ag2CrO4.

(c) Hy tnh khi lng Ag2CrO4 c th tan ti a trong 100 ml nc 25oC.

(d) Thm 0,1 mol AgNO3 vo 1,0 lit dung dch bo ha Ag2CrO4. Gi thit th tch dung

dch khng thay i. Hy cho bit [CrO42-

] tng, gim hay khng i? Gii thch.

Trong dung dch bo ha Ag3PO425oC, nng Ag+ l 5,3.10-5 M.

(e) Hy tnh tch s tan ca Ag3PO425oC.

(g) Lm bay hi 1,00 lit dung dch bo ha Ag3PO425oC n cn 500 ml. Hy tnh

[Ag+] trong dung dch thu c.

p s

b. 8,66.10-5

M.

c. 2,88.10-3

gam;

d. gim;

e. 2,63.10-18

.

g. khng i, 5,3.10-5 M)

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2. Kt ta phn on

Nu trong dung dch c cha hai hay nhiu ion c kh nng to kt ta vi cng mt

ion khc, nhng cc kt ta hnh thnh c tan khc nhau nhiu th khi thm cht to kt

ta vo dung dch, cc kt ta s ln lt c to thnh. Hin tng to thnh ln lt

cc kt ta trong dung dch c gi l kt ta phn on.

*iu kin kt ta hon ton:

[X] < 10-6M, hoc %X cn li trong dung dch < 0,1%

Cu 2.1. Thm AgNO3 rn vo dung dch NaCl 0,10 M v Na2CrO4 0,0010 M. Cho tch

s tan ca AgCl l 1,8.10-10 v ca Ag2CrO4 l 2,4.10-12

.

(a) Hy tnh nng Ag+

cn thit bt u xut hin kt ta AgCl.(b) Hy tnh nng Ag+ cn thit bt u xut hin kt ta Ag2CrO4.

(c) Kt ta no c to thnh trc khi cho AgNO3 vo dung dch trn?

(d) Hy tnh phn trm ion Cl- cn li trong dung dch khi Ag2CrO4 bt u kt ta?

p s

(a) 1,8.10-9M

(b) 4,9.10-5M

(c) AgCl

(d) 3,7.10-3

%)

Cu 2.2. ttaann ll mmtt yyuu tt qquuaann ttrrnnggdng nh gi mc gy nhim mi

trng ca mui. tan ca mui ph thuc nhiu vo bn cht ca mui, dung mi v

cc iu kin th nghim nh nhit , pH v s to phc.

Mt dung dch cha BaCl2 v SrCl2 c cng nng l 0,01 M. Cu hi t ra l liu

c th tch hon ton hai mui ny ra khi nhau bng cch thm dung dch bo ha natri

sunfat hay khng. Bit iu kin tch hon ton l t nht 99,9% Ba2+ b kt ta

dng BaSO4 v SrSO4 chim khng qu 0,1 % khi lng kt ta. Bit cc gi tr tch s

tan nh sau: TBaSO4 = 1 10-10

v TSrSO4 = 3 10-7

.

(a) Hy tnh nng ca Ba2+ cn li trong dung dch khi 99,9% Ba2+ b kt ta v cho

bit phng php ny c dng c tch hon ton hai mui ra khi nhau hay khng?

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S to phc c th lm tng ng k tan. Bit tch s tan ca AgCl l 1,7 10-10,

hng s bn tng cng ca phc Ag(NH3)2+

l 1,5 107.

(b) Hy chng minh (bng php tnh c th) tan ca AgCl trong dung dch amoniac 1,0

M cao hn so vi tan trong nc ct.

Hng dn gii

a. MBa52 10.0,101,0

100

9,99100][ + =

=

Sau khi 99,9% Ba2+

b kt ta th nng SO42-

trong dung dch l:

MBa

TSO

BaSO 5

5

10

2

2

410

10.0,1

10.1

][][ 4

+

===

MMSO

TSr

SrSO01,010.3

10.0,1

10.3

][][ 2

5

7

2

4

2 4 >===

+

Sr2+

cha kt ta. Vy c thsdng phng php ny tch hon ton hai mui ra

khi nhau.

b. tan ca AgCl trong nc ct:

MTAgS AgCl5

1 10.30,1][+ ===

Tnh tan ca AgCl trong dung dch amoniac 1,0 M.

AgCl + 2 NH3Ag(NH3)2+

+ Cl3107 10.55,210.7,110.5,1 ==K

b 1,0

cb 1,0 - 2x x x

3

2

2

10.55,2)20,1(

=

=x

xK x = 4,59.10

-2M

S2 = x = 4,59.10-2

M; lanS

S 3

1

2 10.6,4= )

3. Cc yu tnh hng n tanTrong thc t, ion kim loi ca kt ta c th to phc vi OH- v anion ca kt ta

c th phn ng vi H+ trong dung dch. Ngoi ra, nhng cu t khc c trong dung dch

cng c th tham gia phn ng vi cc ion ca kt ta hoc t nht cng lm bin i h s

hot ca chng. Nhng yu t u nh hng n tan ca kt ta.

3.1 nh hng ca pH

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Cu 3.1. (a) Hy cho bit dung dch ca cc mui sau c tnh axit, bazhay trung tnh?

Gii thch. Natri photphat, ng (II) nitrat v xesi clorua.

(b) Hy tnh khi lng bc photphat cn dng pha 10 lit dung dch bo ha. Khi tnh

b qua s thy phn ca ion photphat.

Bit bc photphat c T = 1,3 .1020.

(c) Hy cho bit trong thc t nu ha tan lng bc photphat tnh c phn (b) vo 10

lit nc th dung dch thu c bo ha hay cha? Gii thch.

Hng dn gii

a. Na3PO4: baz; Cu(NO3)2: axit; CsCl: trung tnh;

b. Ag3PO4 3 Ag+

+ PO43-

3S S

SST

3

)3(=

MT

S 6420

4 10.68,427

10.3,1

27

===

mAg3PO4 = 4,68.10-610419 = 1,96.10-2 gam

c. Cha, v PO43-

b thy phn lm tng tan ca mui)

Cu 3.2. Tnh tan ca AgOCN trong dung dch HNO3 0,001M.

Cho TAgOCN= 2,3.10-7

; HOCN c Ka=3,3.10-4

.

Hng dn gii

AgOCN Ag+

+ OCN-

T = [Ag+][OCN

-] (1)

OCN-

+ H+ HOCN

][

]][[

HOCN

OCNHKa

+

= (2)

Lp phng trnh

[Ag+] = [OCN

-] + [HOCN] (3)

[H

+

] + [HOCN] = 10

-3

(4)Gii h:

(2, 4) ][

]])[[10(10.3,3

34

HOCN

OCNHOCN

=

][10.3,3

].[10][

4

3

+=

OCN

OCNHOCN (5)

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(3, 5) ][10.3,3

][10][][

4

3

+

++=

OCN

OCNOCNAg (6)

t [OCN-]= x

(1,6) 7

4

3

10.3,2)10.3,3

10(

=+

+ xx

xx

x3

+ 1,33.10-3x

2- 2,3.10

-7x - 7,59.10

-11= 0

x= 2,98.10-4

= [OCN-]

(5) [HOCN]= 4,75.10-4

(4) [H+]= 5,25.10

-4

(1) => [Ag+]= 7,72.10

-4= S.

*Nhn xt: v nng ca ion cc ion v phn tgn bng nhau nn khng thgii gn

ng c)

Cu 3.3. ((aa)) 110000 mmll nncc2255ooCC hhaa ttaanncc ttiiaa 444400 mmll kkhhHH22SS ((kkttcc)).. HHyy ttnnhh

nnnngg mmooll ccaa HH22SS ttrroonngg dduunngg ddcchh bboo hhaa.. GGii tthhiitt rrnngg qquu ttrrnnhh hhaa ttaann HH22SS kkhhnngg

llmm tthhaayyii tthh ttcchh ccaa dduunngg ddcchh..

((bb)) DDuunngg ddcchh FFeeCCll22 00,,001100 MMcc bboo hhaa HH22SS bbnngg cccchh xxcc lliinn ttcc ddnngg kkhhHH22SS vvoo

dduunngg ddcchh.. CChhoo TTFFeeSS == 88,,00 ..1100--1199

.. HH22SS cc KKaa11 == 99,,55 ..1100--88

vv KKaa22 == 11,,33 ..1100--1144

.. HHnngg ss iioonn ccaa

nncc KKww == 11 ..1100--1144

.. HHyy cchhoo bbiitt tthhuucc nnhhiiuu kktt ttaa FFeeSS hhnn tthh ccnn pphhii ttnngg hhaayy

ggiimm ppHH ccaa dduunngg ddcchh??

((cc)) HHyy ttnnhh ppHH ccnn tthhiitt llpp nnnngg FFee22++ ggiimm tt 00,,001100 MM xxuunngg ccnn 11,,00 ..1100--88 MM..

((dd)) NNggii ttaa tthhmm aaxxiitt aaxxeettiicc vvoo dduunngg ddcchhpphhnn ((bb)) nnnngguu ccaa aaxxiitt aaxxeettiicctt

00,,1100 MM.. HHyy ttnnhh nnnngguu ccaa nnaattrrii aaxxeettaatt ccnn tthhiitt llpp nnnngg FFee22++ ttrroonngg dduunngg

ddcchh tthhuucc ll 11,,00..1100--88 MM.. KKhhii ttnnhh cchh ss ttoo tthhnnhh HH++ ddoo pphhnnnngg:: FFee22++ ++ HH22SS

FFeeSS ((rr)) ++ 22HH++.. BBiitt aaxxiitt aaxxeettiicc cc KKaa == 11,,88 ..1100

--55.. GGii ss vviicc tthhmm aaxxiitt aaxxeettiicc vv nnaattrrii

aaxxeettaatt kkhhnngg llmm tthhaayyii tthh ttcchh ccaa dduunngg ddcchh..((ee)) HHyy ttnnhh ppHH ccaa dduunngg ddcchhmm ttrrcc kkhhii xxcc kkhhHH22SS..

Hng dn gii

((aa.. MCSH SH 196,01,0

4,22

44,0

][22

=== ((HH22SSpphhnn llii kkhhnnggnngg kk))

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bb.. TTnngg ppHH..

cc.. CC::11

8

19

2

2 10.0,810.0,1

10.0,8

][][

+

===Fe

TS FeS

MMttkkhhcc:: 22122

][

][][

+

=H

KKSHS aa

MS

KKSHH aa

6

11

148

2

212 10.77,110.8

10.3,110.5,9196,0

][

][][

+ =

==

ppHH== 55,,7755;;

dd.. FFee22++

++ HH22SS FFeeSS((rr)) ++ 22 HH++

00,,0011 00,,0022

CCHH33CCOOOO--

++ HH++ CCHH33CCOOOOHH

bb aa 00,,0022 00,,11

ccbb aa--00,,0022 -- 00,,11 ++ 00,,0022

CC::][

][log

3

3

COOHCH

COOCHpKpH a

+=

12,0

02,0log74,475,5

+=

a

aa == 11,,2255 MM

ee.. 84,51,0

25,1log74,4

][

][log

3

3 =+=+=

COOHCH

COOCHpKpH a ))

Cu 3.4. ((QQGG 22000077)) MMtt dduunngg ddcchh cc bbaa cchhtt HHCCll,, BBaaCCll22,, FFeeCCll33 ccnngg nnnngg

00,,00115500MM.. SScc kkhhCCOO22 vvoo dduunngg ddcchh nnyy cchhoonn bboo hhoo.. SSaauu tthhmm tt tt NNaaOOHH vvoo

dduunngg ddcchhnn nnnngg 00,,112200MM.. CChhoo bbiitt:: nnnngg CCOO22 ttrroonngg dduunngg ddcchh bboo hhoo ll 33..1100--

22MM;; tthh ttcchh ccaa dduunngg ddcchh kkhhnngg tthhaayyii kkhhii cchhoo CCOO22 vv NNaaOOHH vvoo;; cccc hhnngg ss:: ppKKaa

ccaa HH22CCOO33 ll 66,,3355 vv 1100,,3333;; ppKKss ccaa FFee((OOHH))33 ll 3377,,55 vv ccaa BBaaCCOO33 ll 88,,3300;; ppKKaa ccaa FFee33++

ll 22,,1177.. HHyy ttnnhh ppHH ccaa dduunngg ddcchh tthhuucc..

Hng dn gii

HH++

++ OOHH-- HH22OO

00,,001155 00,,001155

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GGii tthhiitt rrnngg cchh cc HHUUrr vv UUrr-- llnnhh hhnnggnn ggii ttrr ppHH ccaa dduunngg ddcchh.. SSii tthhnn

tthhnngg cc aaxxiitt uurriicc.. NNgguuyynn nnhhnn ll nnnngg qquu ccaaoo ccaa aaxxiitt uurriicc vv uurraatt cc ttrroonngg nncc

ttiiuu vv ppHH tthhpp ccaa nncc ttiiuu ((ppHH == 55 -- 66))..

((ee)) HHyy ttnnhh ggii ttrr ppHH ttii ssii ((cchhaa aaxxiitt uurriicc kkhhnngg ttaann))cc hhnnhh tthhnnhh tt nncc ttiiuu

ccaa bbnnhh nnhhnn.. GGii tthhiitt rrnngg nnnngg ttnngg ccnngg ccaa aaxxiitt uurriicc vv uurraatt ll 22,,00 mmmmooll//LL..

Hng dn gii

aa.. 66,,441100--55;;

bb.. 44,,991100--44 MM;;

cc.. NNhhiitt ggiimm tthh ttcchh ss ttaann ggiimm..

dd.. CC][

][log

HUr

UrpKpH a

+=

24,54,7][][log ===

apKpHHUrUr

10010][

][ 2 ==

HUr

Ur

VV ttrroonngg mmuu kkhhnngg cc kktt ttaa NNaaUUrr nnnn [[UUrr--]]

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Cu+

+ NH3 [Cu(NH3)]+

lg1 = 6,18

[Cu(NH3)]+

+ NH3 [Cu(NH3)2]+

lg2 = 4,69

(b) Hy tnh th tch dung dch amoniac 0,1 M ti thiu cn dng ha tan hon ton 1

gam CuBr.

(c) Biu thc tnh tch s tan iu kin ca CuBr nh sau:][Br)])[Cu(NH)][Cu(NH]([CuT'

-

233

+++ ++=

Hy tnh gi tr T' ca dung dch thu c phn (b).

Hng dn gii

a. CuBr Cu+

+ Br-

C:4-7,4 10.00,210][CuS + ===

Mt khc:410.00,2

143,35V1S ==

=> V = 34,9 lit;

b. CuBr Cu+

+ Br-

pT = 7,4

Cu+

+ NH3 [Cu(NH3)]+

lg1 = 6,18

[Cu(NH3)]+

+ NH3 [Cu(NH3)2]+

lg2 = 4,69

C: [Br-] = [Cu

+] + [Cu(NH3)

+]

+ [Cu(NH3)2

+] (1)

[NH3] + [Cu(NH3)+]

+ 2[Cu(NH3)2

+] = 0,1 (2)

Gi s: [Cu(NH3)2+] >> [Cu

+], [Cu(NH3)

+]

(1) [Br-] = [Cu(NH3)2+]

(2) [NH3] + 2[Cu(NH3)2+] = 0,1

C:])2[Br-(0,1

]r[

10

]r[

])2[Br-](0,1[Cu

]r[

]][NH[Cu

])[Cu(NH

-

-

7,4-

-

-

-

3

232,1

B

BB===

++

+

[Br ] = 0,05 ; [Cu+] = 1,99.10-6 ; [Cu(NH3)2+] = [Br

-] = 0,05

7

87,106

2,1

233 10.39,3

1010.99,1

05,0

][Cu

])[Cu(NH][NH

+

+

=

==

[Cu(NH3)+] = 1[Cu

+][NH3] = 10

6,181,99.10-63,39.10-7 = 1,02.10-6

KTGT: tha mn;

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14

05,0][35,143

1

2

=== BrV

S V2 = 0,140 lit

c. T= ([Cu+]+[Cu(NH3)

+] + [Cu(NH3)2

+]) [Br]

= (1,99106 +3,39107 +0,05) 0,05 = 2,5103

Cu 3.7.Bi

t tch s

tan c

a Zn(OH)2 l 1,80 10

-17

.(a) Hy tnh tan ca Zn(OH)2 trong nc.

(b) Hy tnh pH ca dung dch Zn(OH)2 bo ha.

Cho cc gi tr th kh chun:

[Zn(OH)4]2-

+ 2 e Zn (r) + 4 OH-

E = -1,285 V

Zn2+

+ 2e Zn (r) E = - 0,762 V

(c) Hy tnh hng s bn tng cng ca phc tetrahidroxozincat(II).

(d) Hy tnh tan ca Zn(OH)2 trong dung dch m c pH = 9,58. B qua s to phc

[Zn(OH)4]2-

.

(e) Hy tnh tan ca Zn(OH)2 trong dung dch m c pH = 9,58 v c tnh n s to

thnh phc [Zn(OH)4]2-

.

(g) Hy so snh kt qu tm c (d) v (e) v rt ra nhn xt.

Hng dn gii

a. b qua c sphn li ca nc; S = 1,65.10-6;

b. 8,52;

c. Cch 1: Thit lp cng thc tnho

ZnOHZnE

/)( 24 theo

o

ZnZnE

/2+ .

C 4

2

4

/

2

// ][

])([lg

2

0592,0]lg[

2

0592,0222

+ +=+= +++

OH

OHZnEZnEE o

ZnZn

o

ZnZnZnZn

4

2

4

/ ][

])([lg

2

0592,0lg

2

0592,02

+= +OH

OHZnE

o

ZnZn

Khi [Zn(OH)42-

] = [OH-] = 1 M th:

lg2

0592,0//)(/ 224

2 == ++o

ZnZn

o

ZnOHZnZnZnEEE

= 4,67.1017;

Cch 2:

7/23/2019 file546_7022

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15

Zn(r) + 4 OH- [Zn(OH)4]

2-+ 2 e E1 = +1.285 V

G1 = -zFE1 = -247.97 kJ/mol

Zn2+

+ 2e- Zn(r) E2 = -0.762 V

G2 = -zFE2 = 147.04 kJ/mol

Zn

2+

+ 4 OH

-

[Zn(OH)4]

2-

G =G1 +G2 = -100.92 kJ/mol

17298314,8

100920

10.90,4===

eeK RTG

d. MOH

TZn

8

2

2 10.25,1][

][

+ == .

e. S = [Zn2+

] + [Zn(OH)42-

] = [Zn2+

] + [Zn2+][OH-]4

= MOH

OH

T 842

10.56,2)][1(

][

=+ .

g. Kt qu khc nhau: (2,56- 1,25)/2,56 = 51%; rt ln; nhvy sto phc nh hng

ng kn tan)

Cu 3.8. (IChO 43) PbO l mt oxit lng tnh. Khi ha tan vo nc xy ra cc cn

bng:

PbO (r) + H2O Pb2+

(aq) + 2 OH-(aq) T = 8,010

-16

PbO (r) + 2 H2O Pb(OH)3-(aq) + H3O

+(aq) Ka = 1,010

-15

(a) Hy tnh gi tr pH ca dung dch ti dung dch Pb2+ 1,0010-2 M bt u c kt ta

PbO xut hin?

(b) T gi tr pH tnh c phn (a), ngi ta tng pH ca dung dch n mt gi tr

nht nh th kt ta bt u tan hon ton. Hy tnh gi tr pH ny?

(c) Hy vit biu thc tnh tan ca PbO.

(d) tan ca PbO t gi tr cc tiu ti pH =9,40. Hy tnh nng ca cc cu t v

tan ca PbO ti gi tr pH ny.(e) Hy tnh khong pH ti tan ca PbO nh hn 1,010-3 M.

Hng dn gii

a. [Pb2+

][OH-]

2= 8.10

-16;

[OH-] = 2,83.10

-7 pH = 7,45;

b. [Pb(OH)3-][H3O

+] = 1.10

-15

7/23/2019 file546_7022

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16

[H3O+]= 1.10

-13 pH = 13;

c. S = [Pb2+

] + [Pb(OH)3-];

d. [Pb2+

]= 8.10-16

/ [OH-]

2= 1,27.10

-6M;

[Pb(OH)3-] = 10

-15/ [H3O

+]= 2,51.10

-6M;

S = 3,78.10-6M;

Mrng: chng minh rng Smin ti gi tr pH = 9,40;

][

10][10.8

][

10

][OH

8.10][Pb(OH)][PbS

15212

15

2-

-16-

3

2

+

+

+

+ +=+=+=

HH

H

0][

10][10.16'S

2

1512 ==

+

+

HH

[H+]= 3,97.10

-10(pH = 9,40);

e.315212 10

][10][10.8S

+

+ =+=H

H

010][10][10.8153312 = ++ HH

[H+]1 = 1,12.10

-8; pH1 = 7,95;

[H+]2 = 1,0.10

-12; pH2 = 12,00;

7,95 pH12,00)

3.3. nh hng ng thi ca pH v phn ng to phc

Cu 3.9. Tnh tan ca AgI trong dung dch NH3 0,1M. Bit TAgI = 8,3.10-17

; NH3 c Kb

= 1,75.10-5

v:

Ag+

+ 2NH3 Ag(NH3)2+

; 1,2 = 1,7.107

Hng dn gii

Cc cn bng xy ra:

AgI Ag+

+ I-

Ag+ + 2 NH3 Ag(NH3)2+

NH3 + H2O NH4+

+ OH-

Thit lp cc phng trnh:

T = [Ag+][I

-] = 8,3.10

-17(1)

7

2

3

232,1 10.7,1

]][[

])([==

+

+

NHAg

NHAg (2)

7/23/2019 file546_7022

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17

5

3

4 10.75,1][

]][[ +

==NH

OHNHKb (3)

S = [I-] = [Ag

+] + [Ag(NH3)2

+] (4)

[NH3] + 2 [Ag(NH3)2+] + [NH4

+] = 0,1 M (5)

[NH4+] = [OH

-] (6)

Gi s [NH4+]

7/23/2019 file546_7022

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18

C: [AgL2] + [Ag+] = [Cl] + [Br] (1)

[NH3] + [NH4+] + 2[AgL2] = 0,02 (2)

Gi s: [Ag+]

7/23/2019 file546_7022

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19

CCuu((OOHH))22 ttrroonngg 2255,,0000 mmLL dduunngg ddcchh NNHH33.. BBiitt nnnngg ccnn bbnngg ccaa NNHH33 ttrroonngg dduunngg ddcchh

tthhuucc ll 11,,0000 ..110033 MM,, hhnngg ss bbnn ttnngg ccnngg ccaa pphhcc CCuu((NNHH33))4422++

ll11,,44 == 11001111,,7755

..

ii.. HHyy ttnnhh nnnngg mmooll ttnngg ccnngg ccaanngg ttrroonngg dduunngg ddcchh tthhuucc..

iiii.. HHyy ttnnhh nnnngg ccnn bbnngg ccaa cccc ccuu tt cchhaanngg ttrroonngg dduunngg ddcchh..

iiiiii.. HHyy ttnnhh nnnngg ccnn bbnngg ccaa NNHH44++..

iivv.. HHyy ttnnhh ppHH ccaa dduunngg ddcchh..

vv.. HHyy ttnnhh nnnngg ccaa dduunngg ddcchh NNHH33 bbaannuu..

Hng dn gii

aa.. ii.. CCuu((OOHH))22 CCuu22++

++ 22 OOHH--

SS 22SS

CC::21222

10.50,4)2(]][[+ === SSOHCuT

MS73

21

10.04,14

10.50,4

==

mlgS 100/10.01,159,971,010.04,1'67 ==

iiii.. CC:: [[OOHH--]]== 22SS == 2211,,0044 ..1100

--77== 22,,0088 ..1100

--77

ppHH == 1144 ++ lloogg[[OOHH--

]] == 1144 ++ lloogg ((22,,0088 ..1100--77

)) == 77,,3322;;

bb.. ii.. MCCu3

3

10.05,2025,0

59,97

10.00,5

2

==+

iiii.. CCuu((OOHH))22 CCuu22++

++ 22 OOHH--

TT == 44,,5500..1100--2211

CCuu22++

++ 44 NNHH33 CCuu((NNHH33))4422++

11,,44 == 11001111,,7755

CC:: CCCCuu22++ == [[CCuu22++

]] ++ [[CCuu((NNHH33))4422++

]] == 22,,0055..1100--33

((11))

4

3

2

243

4,1]][[

])([

NHCu

NHCu+

+

= ((22))

((22)) ])([778,1)10(10

])([

][

])([][ 2434375,11

2

43

4

34,1

2

432 +

+++ =

== NHCu

NHCu

NH

NHCuCu

((33))

((11,, 33)) [[CCuu22++

]] == 11,,3311..1100--33

[[CCuu((NNHH33))4422++

]] == 77,,3388..1100--44

7/23/2019 file546_7022

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20

iiiiii.. NNHH33 ++ HH22OO NNHH44++

++ OOHH--

KKbb ==1100--44,,7766

CC::][

][

][

]][[

3

2

4

3

4

NH

NH

NH

OHNHKb

++

==

MNHKNH b4376,4

34 10.32,110.00,110][][+ ===

iivv.. CC:: [[OOHH

--

]] == [[NNHH44++

]]== 11,,3322..1100

--44

ppHH == 1144 ++ lloogg[[OOHH-

-]] == 1144 ++ lloogg((11,,3322..1100

--44)) == 1100,,1122

vv.. CCNNHH33 == [[NNHH33]] ++ [[NNHH44++]] ++ 44[[CCuu((NNHH33))44

22++]] == 11..1100

--33++ 11,,3322..1100

--44++ 4477,,3388..1100

--44==

44,,0088..1100--33

MM))

4. Xc nh tch s tan

4.1. Da vo tan

Cu 4.1. Thm t t dung dch bari nitrat 0,0010 M vo 200 ml dung dch NaF 0,040 M.

Khi 35 ml dung dch bari nitrat c thm vo th thy kt ta BaF2 bt u xut hin.

Hy tnh tch s tan ca BaF2.

p s: (1,72.10-7)

Cu 4.2. Dung dch bo ha Cd(OH)2 c pH = 9,56. Hy tnh tch s tan ca Cd(OH)2.

p s: (2,39.10-14)

Cu 4.3. Bit 1 lit dung dch NH3 1M ha tan c ti a 0,33 gam AgBr. Hy tnh

TAgBr. Bit phc Ag(NH3)2+

c 1,2 = 5,88.106.

Hng dn gii

[Ag(NH3)2+] = [Br

-] = 0,33/188 = 1,76.10

-3M.

[NH3] = 1 2[Ag(NH3)2+] = 0,996 M

10

2

32,1

23 10.02,3][

])([][

++ ==

NH

NHAgAg

T = [Ag+][Br

-]= 5,32.10

-13)

Cu 4.4. Tnh tch s tan ca Ca(IO3)2Th nghim 1. Chun ha dung dch natri thiosunfat.

Ly 10,0 ml dung dch KIO3 0,0120 M cho vo bnh nn. Thm 2 gam KI v 10 ml

dung dch HCl 1M. Dung dch c mu nu thm. Chun bng dung dch Na2S2O3n

mu vng rm. Thm 5 ml h tinh bt v tip tc chun n mt mu xanh ca phc

tinh bt vi I3-thy ht 20,55 ml.

7/23/2019 file546_7022

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21

Th nghim 2. Tnh tan ca Ca(IO3)2 trong nc ct.

Ly 10,0 ml dung dch bo ha Ca(IO3)2 cho vo bnh nn. Thm 2 gam KI v 10 ml

HCl 1M. Tin hnh chun dung dch thu c bng dung dch Na2S2O3trn thy ht

19,20 ml. Hy:

(a) vit cc phn ng c m t trong th nghim.

(b) tnh nng dung dch Na2S2O3.

(c) tnh nng ca IO3-.

(d) tnh tan ca Ca(IO3)2 trong nc.

(e) tnh tch s tan ca Ca(IO3)2.

p s: a. IO3-

+ 5 I + 6 H+ 3 I2 + 3 H2O

I2 + 2 S2O32-

2 I + S4O62-

b. 0,0350M;c. 0,0122 M.

d. 5,6.10-3

M.

e. 7,1.10-7

)

4.2. Da vo gi trthkhchun

Cu 4.5. Cho VEo

HgHg789,0

/22=+ ; VE

o

HgClHg 268,0/22 = . Hy tnh tch s tan v tan ca

Hg2Cl2.

Hng dn gii

Hg22+

+ 2 e 2 Hg

Hg22+

+ 2 Cl- Hg2Cl2 ; T = [Hg2

2+][Cl

-]

2

Thit lp cng thc tnho

HgClHgE /22 theoo

HgHgE

/22+ .

C 2/2

2// ][lg

2

0592,0]lg[

2

0592,022

22

22

+ +=+= +++Cl

TEHgEE o

HgHg

o

HgHgHgHg

]lg[0592,0lg2

0592,0/22

+= + ClTEo

HgHg

Khi [Cl] = 1 M th:

TEEE oHgHg

o

HgHgClHgHgCllg

2

0592,0///

2222

+== +

T = 2,51.10-18

7/23/2019 file546_7022

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22

C S(2S)2

= T MT

S73 10.56,8

4

== )

Cu 4.6. Cho 3 pin in ha vi cc sc in ng tng ng 298K:

(1) Hg/HgCl2, KCl (bo ha ) // Ag+

(0,0100 M)/Ag E1= 0,439 V

(2) Hg/HgCl2, KCl (bo ha ) // AgI (bo ha)/Ag E2= 0,089 V

(3) Ag/AgI (bo ha), PbI2 (bo ha ) // KCl (bo ha), HgCl2/Hg E3= 0,230 V

a) Hy tnh tch s tan ca bc idodua.

b) Hy tnh tch s tan ca ch (II) iodua.

Cho VEo

AgAg799,0

/=+ , R = 8,314 J/mol/K, F = 96487 C/mol.

p s: a) 1,37.10-16;

b) [Ag+] = 4,58.10

-14; [I] = 2,99.10

-3;

[Pb2+]= 0,5 ([I] [Ag+]) = 1,5.10-3; T = 1,34.10-8

Cu 4.7. ((IICChhOO 4422)) CChhoo cccc ggii ttrr tthh kkhh cchhuunn ssaauu::

BBnn pphhnnnngg EE00,, VV ((229988KK))

SSnn22++

++ 22ee SSnn --00,,1144

SSnn44++

++ 22ee SSnn22++

++00,,1155

HHgg2222++

++ 22ee 22 HHgg ++00,,7799

HHgg22CCll22 ++ 22ee 22 HHgg

++ 22 CCll

++00,,2277

((aa)) HHyy ttnnhh hhnngg ss ccnn bbnngg ccaa pphhnnnngg ssaauu229988 KK::

SSnn ((rr)) ++ SSnn44++

((aaqq)) 22 SSnn22++

((aaqq))

((bb)) HHyy ttnnhh ttaann ccaa HHgg22CCll22 ttrroonngg nncc229988 KK ((tthheeoonn vv mmooll//ll))..

((cc)) HHyy ttnnhh ssuuttiinnnngg cchhuunn,, EE,, ccaa ppiinn nnhhiinn lliiuu ss ddnngg pphhnnnngg ssaauu::

HH22((kk)) ++ 11//22 OO22((kk)) HH22OO ((ll)) GG == 223377,,11 kkJJ..mmooll11

Hng dn gii

aa..90592,0

)14,015,0(2

10.27,610 ==

+

K

bb.. XXccnnhh ttcchh ss ttaann::

CCcchh 11::

--11 HHgg2222++

++ 22ee 22 HHgg

7/23/2019 file546_7022

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23

GG00

11 == --nnFFEE00

11 == --22 996644885500,,7799 == --115522,,44..110033

JJ

11 HHgg22CCll22 ++ 22ee 22 HHgg ++ 22 CCll--

GG0022 == --nnFFEE

0022 == --22 996644885500,,2277 == --5522,,11..1100

33JJ

HHgg22CCll22

HHgg2222++

++ 22 CCll

--

GG00

33 == --GG00

11++GG

0022 == 110000,,33..1100

33JJ == --RRTTllnnTT

48,40298314,8

10.3,100ln

30

3 =

=

=RT

GT

TT == 22,,6622..1100--1188

CCcchh 22::

Hg22+

+ 2 e 2 Hg

Hg22+

+ 2 Cl- Hg2Cl2 ; T = [Hg2

2+][Cl

-]

2

Thit lp cng thc tnh o HgClHgE /22 theoo

HgHgE

/22+ .

C2/

2

2// ][lg

2

0592,0]lg[

2

0592,022

22

22

+ +=+= +++Cl

TEHgEE

o

HgHg

o

HgHgHgHg

]lg[0592,0lg2

0592,0/22

+= + ClTEo

HgHg

Khi [Cl-] = 1 M th:

TEEEo

HgHg

o

HgHgClHgHgCllg

2

0592,0///

2222

+== +

T = 2,71.10-18

Tnh tan:

C HHgg22CCll22 HHgg2222++

++ 22 CCll--

SS 22SS S(2S)2 = T

MT

S73

18

3 10.68,84

10.62,2

4

===

cc.. CC:: GG00

== --nnFFEE00

ppiinn

7/23/2019 file546_7022

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24

VnF

GEpin 23,1

964852

10.1,237 30

0 =

=

=

Cu 4.8. K thut in ho thng c dng xc nh tan ca cc mui kh tan.

Do sc in ng l hm bc nht theo logarit ca nng nn c th xc nh c cc

nng d rt nh.

Bi tp ny s dng mt pin in ho gm hai phn, c ni vi nhau bng cu

mui. Phn bn tri ca s pin l mt thanh Zn nhng trong dung dch Zn(NO3)2

0,200M; cn phn bn phi l mt thanh Ag nhng trong dung dch AgNO3 0,100M. Mi

dung dch c th tch 1,00L 250C.

(a) V s pin v vit cc bn phn ng xy ra mi cc.

(b) Hy tnh sc in ng ca pin v vit phng trnh phn ng xy ra khi pin phng

in.

Gi s pin phng in hon ton v lng Zn c d.

(c) Hy tnh in lng c phng thch trong qu trnh phng in.

Trong mt th nghim khc, KCl c thm vo dung dch AgNO3pha bn phi

ca pin ban u. Xy ra phn ng to kt ta AgCl v lm thay i sc in ng. Sau khi

thm xong. Sc in ng bng ca pin bng 1,04V v [K+] = 0,300M.

(d) Hy tnh [Ag+] ti trng thi cn bng.

(e) Hy tnh [Cl

-

] ti trng thi cn bng v TAgCl.Cho: E

oZn2+/Zn= -0,76V; E

oAg+/Ag= 0,80V.

p s: a. Zn| Zn2+||Ag+ |Ag b. 1,52V

c. 9649C d. 7,3.10-10

M

e. [Cl-] = 0,2M; T = 1,5.10

-10)

Cu 4.9. XXeemm xxtt ppiinniinn hhaa ssaauu::

PPtt ||HH22((pp == 11 aattmm))||HH22SSOO44 00,,0011 MM||PPbbSSOO44((rr))||PPbb((rr))..

((aa)) HHyy ttnnhh nnnngg ccnn bbnngg ccaa SSOO4422-- vv ppHH ccaa dduunngg ddcchh ttrroonngg ppiinn ttrrnn..

((bb)) HHyy vviitt pphhnnnngg xxyy rraa kkhhii ppiinn pphhnnggiinn..

SSuuttiinnnngg ccaa ppiinn ttrrnn229988,,1155 KK ll 00,,118888 VV.. GGii tthhiitt rrnngg ttrroonngg pphhnn ((cc)) vv

((dd)) nnnngg ccnn bbnngg ccaa SSOO4422--

ll 55 1100--33

MM vv ccaa HH33OO++

ll 1155 1100--33

MM ((cccc ggii ttrr nnyy cc

tthh kkhhcc ggii ttrr ttnnhhccpphhnn ((aa))))..

((cc)) HHyy ttnnhh ttcchh ss ttaann ccaa PPbbSSOO44..

7/23/2019 file546_7022

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25

((dd)) HHyy cchhoo bbiitt ssuuttiinnnngg ccaa ppiinn ttrrnn ttnngg hhaayy ggiimm bbaaoo nnhhiiuu VV kkhhii pp ssuutt ccaa

hhiiddrroo ggiimm mmtt nnaa??

VVnngg kkiimm llooii kkhhnngg ttaann ttrroonngg dduunngg ddcchh aaxxiitt nniittrriicc nnhhnngg ttaann cc ttrroonngg nncc

ccnngg ttooaann ((ll hhnn hhpp ggmm aaxxiitt cclloohhiiddrriicccc vv aaxxiitt nniittrriicccc cc tt ll tthh ttcchh ttnnggnngg

ll 33 :: 11)).. VVnngg pphhnnnngg vvii nncc ccnngg ttooaann ttoo tthhnnhh iioonn pphhcc [[AAuuCCll44]]--..

((ee)) SS ddnngg cccc ggii ttrr tthh kkhh cchhuunn cchhoo ddiiyy,, hhyy ttnnhh hhnngg ss bbnn ttnngg ccnngg ccaa

pphhcc [[AAuuCCll44]]--..

CChhoo:: ppKKaa22 ((HH22SSOO44)) == 11,,9922;; EE((PPbb22++//PPbb)) == -- 00,,112266 VV

EE((AAuu33++//AAuu)) == ++ 11,,5500 VV EE(([[AAuuCCll44]]

--//AAuu ++ 44 CCll

--)) == ++ 11,,0000 VV

Hng dn gii

aa.. HH22SSOO44 HH++

++ HHSSOO44--

00,,0011 00,,0011HHSSOO44-- HH

++++ SSOO44

22--

ccbb 00,,0011 -- xx 00,,0011 ++ xx xx

92,1

2 1001,0

)01,0( =

+=

x

xxKa

xx == 44,,5533..1100--33

MM

[[SSOO4422--

]]== xx == 44,,5533..1100--33

MM

[[HH++]] == 00,,0011 ++ xx == 00,,00114455 MM

ppHH == 11,,8844;;

bb.. CCaattoott ((++)) PPbbSSOO44 ++ 22ee PPbb ++ SSOO4422--

AAnnoott ((--)) HH22 22 HH++

++ 22ee

PPbbSSOO44 ++ HH22 PPbb ++ 22 HH++

++ SSOO4422--

cc.. CC:: ]lg[2

0592,0 20/2

++= + PbEEPbPbcatot

Vp

HEE

H

HHanot108,0

1

)10.15(lg

2

0592,00

][lg

2

0592,0 2320/

2

2

=+=+=+

+

188,0108,0]lg[2

0592,0 20/2

=++== ++ PbEEEEPbPbanotcatotpin

[[PPbb22++]]== 11,,8811..1100--66 MM

7/23/2019 file546_7022

26/26

TT == [[PPbb22++

]][[SSOO4422--

]] == 11,,8811..1100--66

55..1100== 99,,0055..1100

--99

dd.. KKhhii pp ssuutt ccaa hhiiddrroo ggiimm mmtt nnaa::

Vp

HEE

H

HHanot099,0

5,0

)10.15(lg

2

0592,00

][lg

2

0592,0 2320/

2

2

=+=+=+

+

EEccaattoott kkhhnnggii,, EEaannoott ttnngg ((--00,,009999 ++ 00,,110088)) == 00,,000099VV,, vvyy EEppiinn ggiimm 00,,000099 VV;;

ee.. ++11 AAuu33++

++ 33ee AAuu

GG00

11 == --nnFFEE0011 == --33 996644885511,,5500 == --443344,,22..1100

33JJ

--11 AAuuCCll44--

++ 33ee AAuu ++ 44 CCll--

GG00

22 == --nnFFEE0022 == --33 996644885511,,0000 == --228899,,55..1100

33JJ

AAuu33++

++ 44CCll-- AAuuCCll44

--

GG0033 ==GG0011--GG0022 == --114444,,77..110033 JJ == --RRTTllnn11,,44

4,54298314,8

10.7,144ln

30

34,1 =

=

=

RT

G

11,,44 == 22,,3311..11002255

))