Chuyên đề Hóa Phân Tích

I. OLYMPIC HA HC VIT NAM OLYMPIC HA HC SINH VIN VIT NAM 2005 (Bng A):

Ion Fe(SCN)2+ c mu nng bng hoc ln hn 10-5M. Hng s in li ca n l 10-2. 1. Mt dung dch cha vt Fe3+. Thm vo dung dch ny mt dung dch KSCN 10-2M (coi th tch

khng i). Xc nh nng ti thiu ca Fe3+ dung dch xut hin mu . 2. Mt dung dch cha Ag+ 10-2M v Fe3+ 10-4M. Thm dung dch SCN- vo to kt ta AgCN (coi

th tch khng i). Xc nh nng Ag+ cn li trong dung dch khi xut hin mu . Bit TAgSCN = 10-12

3. Thm 20cm3 dung dch AgNO3 5.10-2M vo 10cm3 dung dch NaCl khng bit nng . Lng d Ag+ c chun bng dung dch KSCN vi s c mt ca Fe3+. im tng ng (khi bt u xut hin mu ) c quan st thy khi thm 6cm3 dung dch KSCN 10-1M. Tnh nng ca dung dch NaCl.

BI GII: 1. Fe3+ + SCN- Fe(SCN)2+

Nng cn bng: Co x 10-2 x x = 10-5

Ta c: [ ] 25235

10)1010(

10 +

=Fe

[Fe3+] = 10-5M Co = 2.10-5M 2. Khi xut hin mu th: [Fe(SCN)2+] = 10-5M. Vy nng Fe3+ cn li l: 9.10-5M Ta c:

[ ][ ] [ ] MAgMSCNSCN 1032

5

5

10.1,910.1,1

1010.9

10

+

===

3 n(Ag+) = n(AgCl) + n(AgSCN) 20.10-3.5.10-2 = 10.10-3C + 6.10-3.10-1 C = 4.10-2M

K THI CHN HC SINH GII QUC GIA NM 2002 (BNG A) Dung dch X gm Na2S 0,010M, KI 0,060M, Na2SO4 0,050M.

(a) Tnh pH ca dung dch X. (b) Thm dn Pb(NO3)2 vo dung dch X cho n nng 0,090M th thu c kt ta A v dung

dch B. i Cho bit thnh phn ho hc ca kt ta A v dung dch B. ii Tnh nng cc ion trong dung dch B (khng k s thu phn ca cc ion, coi th tch

dung dch khng thay i khi thm Pb(NO3)2). iii Nhn bit cc cht c trong kt ta A bng phng php ho hc, vit cc phng trnh

phn ng (nu c). BI GII: a) Tnh pH ca dung dch Na2S 2 Na+ + S2- 0,01 0,01 KI K+ + I-

0,06 0,06 Na2SO4 2Na+ + SO42- 0,05 0,05

S2- + H2O HS- + OH- Kb(1) = 10-1,1 (1) SO42- + H2O H SO4- + OH- Kb(2) = 10-12 (2)

Kb(1) >> Kb(2) nn cn bng (1) quyt nh pH ca dung dch: S2- + H2O HS- + OH- K = 10-1,1

[ ] (0,01 -x) x x

010x0794,0x10x01,0

x 1,321,12 =+=

x = 8,94. 10-3 [OH-] = 8,94.10-3 pH = 11,95 b) Pb2+ + S2- PbS (Ks-1) = 1026. 0,09 0,01 0,08

Pb2+ + SO42- PbSO4 (Ks-1) = 107,8. 0,08 0,05 0,03

Pb2+ + 2 I- PbI (Ks2 -1) = 107,6. 0,03 0,06

Thnh phn hn hp: A : PbS , PbSO4 , PbI2Dung dch B : K+ 0,06M Na+ 0,12M Ngoi ra cn c cc ion Pb2+ ; SO42- ; S2- do kt ta tan ra. tan ca

13-26 1010 S :PbS ==9,3-7,84 1010 S :PbSO

==7,23 6,72 104/10:PbI =Bi v tan ca PbI2 l ln nht nn cn bng ch yu trong dung dch l cn bng tan ca PbI2.

PbI2 = Pb2+ + 2I- Ks Do [Pb2+] = 10-47 = 2 x 10-3M v [I-] = 4.10-3M. 107,8 [SO4

2 102-] = = 5. 105,8 = 7,9.106M

K THI CHN HC SINH GII QUC GIA NM 2003 (BNG A) 1. Trong phng th nghim c cc dung dch b mt nhn: AlCl3, NaCl, KOH, Mg(NO3)2, Pb(NO3)2, Zn(NO3)2, AgNO3. Dng thm mt thuc th, hy nhn bit mi dung dch. Vit cc phng trnh phn ng (nu c). 2. Dung dch bo ha H2S c nng 0,100 M. Hng s axit ca H2S: K1 = 1,0 x 10-7 v K2 = 1,3 x 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 x 10-10 ; TCoS = 4,0 x 10 21 ; TAg2S = 6,3 x 10-50BI GII: 1. C th dng thm phenolphtalein nhn bit cc dung dch AlCl3, NaCl, KOH, Mg(NO3)2, Pb(NO3)2, Zn(NO3)2, AgNO3. * Ln lt nh vi git phenolphtalein vo trong dung dch. - Nhn ra dung dch KOH do xut hin mu ta. * Ln lt cho dung dch KOH vo mi dung dch cn li: - Dung dch AgNO3 c kt ta mu nu Ag+ + OH AgOH ; (hoc 2Ag+ + 2OH Ag2O + H2O) - Dung dch Mg(NO3)2 c kt ta trng, keo Mg2+ + 2OH Mg(OH)2 - Cc dung dch AlCl3, Pb(NO3)2, Zn(NO3)2 u c chung hin tng to ra kt ta trng, tan trong dung dch KOH (d). Al3+ + 3OH Al(OH)3 ; Al(OH)3 + OH AlO2 + 2H2O Pb2+ + 2OH Pb(OH)2 ; Pb(OH)2 + OH PbO2 + 2H2O Zn2+ + 2OH Zn(OH)2 ; Zn(OH)2 + OH ZnO2 + 2H2O - Dung dch NaCl khng c hin tng g. - Dng dung dch AgNO3 nhn ra dung dch AlCl3 do to ra kt ta trng Ag+ + Cl AgCl - Dng dung dch NaCl nhn ra dung dch Pb(NO3)2 do to ra kt ta trng Pb2+ + 2 Cl PbCl2 - cn li l dung dch Zn(NO3)2. 2. a) Tnh nng ion S2 trong dung dch H2S 0,100 M; pH = 2,0. CH2S = [H2S] = 0,1 M H2S (k) H2S (aq) [H2S] = 10-1 H2S (aq) H+ + HS K1 = 1,0 x 10-7

[H+] = 10-2 HS H+ + S2- K2 = 1,3 x 10-13

H2S (aq) 2H+ + S2- K = [ ]2 2

2

H S

H S

+ = Kl. K2

[S2- ] = 1,3 x 10-20 x [ ]2 2H SH +

= 1,3 x 10-20 x ( )1

22

10

10

= 1,3 x 10-17 (M)

b) [Mn2+] [S2- ] = 10-2 x 1,3 x 10-17 = 1,3 x 10-19 < TMnS = 2,5 x 10-10 khng c kt ta

[Co2+] [ S2- ] = 10-2 x 1,3 x 10-17 = 1,3 x 10-19 > TCoS = 4,0 x 10-21 to kt ta CoS [Ag+]2[S2- ] = (10-2)2x 1,3 x 10-17 = 1,3 x 1021 > TAg2S = 6,3 x 10-50 to kt ta Ag2S K THI CHN HC SINH GII QUC GIA NM 2004 (BNG B) 1. Dung dch A gm Ba(NO3)2 0,060 M v AgNO3 0,012 M.

a) Thm tng git K2CrO4 vo dung dch A cho n d. C hin tng g xy ra? b) Thm 50,0 ml K2CrO4 0,270 M vo 100,0 ml dung dch A. Tnh nng cc ion trong hn hp thu c.

2. Trnh by s nhn bit v phng trnh ion ca cc phn ng xy ra khi nhn bit cc cation trong dung dch X gm Ba2+, Fe2+, Pb2+, Cr3+, NO3-.

Cho: BaCrO4 + H2O Ba2+ + HCrO4- + OH - ; K = 10-17,43

Ag2CrO4 + H2O 2Ag+ + HCrO4- + OH - ; K = 10-19,50 pKa ca HCrO4- bng 6,50. BI GII: 1. a) Hin tng: C kt ta BaCrO4 v Ag2CrO4.

Xt th t xut hin cc kt ta:

bt u c BaCrO4 : +

>2

424

Ba

)BaCrO(s

CrO CK

C (1)

bt u c Ag2CrO4 : + > Ag2)CrOAg(s

CrO CK

C 4224

(2)

tnh tch s tan Ks cn t hp cn bng :

BaCrO4 Ba2+ + CrO42- Ks1 H2O H + + OH - Kw CrO42- + H+ HCrO4- Ka-1

BaCrO4 + H2O Ba2+ + HCrO4- + OH - C K= Ks1 . Kw . Ka-1

Suy ra 93,91450,643,17

w

a1s 1010

10.10K

K.KK

===

Ag2CrO4 2 Ag + + CrO42- Ks2 H2O H + + OH - Kw

CrO42- + H+ HCrO4- Ka-1

Ag2CrO4 + H2O 2 Ag + + HCrO4- + OH C K = 10-19,50

1214

50,650,19

2s 101010.10K

==

T (1) M10.96,1060,010C 9

93,9

CrO24

=>

T (2) M10.94,6)012,0(10C 92

12

CrO24

=> < nhng khng nhiu, v vy s c hin tng kt ta vng ca BaCrO4

xut hin trc mt t, sau n kt ta vng nu ca Ag2CrO4 ( gch ) v BaCrO4 vng cng xut

hin.

CCrO42-(BaCrO4) CCrO42-(Ag2CrO4)

b) Sau khi thm K2CrO4:

M090,0000,150

00,50x270,0C24CrO

== ; M040,0000,15000,100x060,0C

2Ba==+

M0080,0000,150

00,100x0120,0C2Ag

== Cc phn ng: Ba2+ + CrO42- BaCrO4 0,046 0,090

- 0,050

2 Ag + + CrO42- Ag2CrO4 0,0080 0,050

- 0,046 Thnh phn sau phn ng :

BaCrO4 ; Ag2CrO4 ; CrO42- (0,046 M ). Ag2CrO4 2 Ag + + CrO42- 10-12

BaCrO4 Ba2+ + CrO42- 10-9,93 Nng CrO42- d kh ln, c th coi nng CrO42- do 2 kt ta tan ra l khng ng k.

CrO42- + H2O HCrO4- + OH - Kb = 10-7,5

C 0,046

[ ] (0,046 x ) x x

5,72

10x046,0

x = x = 3,8.10-5

Cr(OH)3 + OH - CrO2- + 2 H2O 2 CrO2- + 3 H2O2 + 2 OH- 2 CrO42- + 4 H2O

Fe2+ + 2 OH - Fe(OH)2 2 Fe(OH)2 + H2O2 Fe(OH)3

K THI CHN HC SINH GII QUC GIA NM 2005 (BNG A) Bng dung dch NH3, ngi ta c th lm kt ta hon ton ion Al3+ trong dung dch nc

dng hydroxit, nhng ch lm kt ta c mt phn ion Mg2+ trong dung dch nc dng hydroxit. Hy lm sng t iu ni trn bng cc php tnh c th. Cho bit: Tch s tan ca Al(OH)3 l 5.1033; tch s tan ca Mg(OH)2 l 4.1012; hng s phn

ly baz ca NH3 l 1,8.105. BI GII: Tnh hng s cn bng K ca phn ng kt ta hidroxit:

3 NH3 + H2O NH4+ + OH ; K = 1,8.105 Al(OH)3 Al3+ + 3 OH ; KS, = 5. 1033

Al3+ + 3 NH3 + 3 H2O Al(OH)3 + 3 NH4+ ; K = = 1,17.1018

Tng t nh vy, i vi phn ng:

Mg2+ + 2 NH3 + 2 H2O Mg(OH)2 + 2 NH4+ ; K = = 81

Phn ng thun nghch, Mg2+ khng kt ta hon ton di dng magie hidroxit nh Al3+. K THI CHN HC SINH GII QUC GIA NM 2005 (BNG B) 1. Tnh in li ca ion CO32 trong dung dch Na2CO3 c pH =11,60 (dung dch A). 2. Thm 10,00 ml HCl 0,160 M vo 10,00 ml dung dch A. Tnh pH ca hn hp thu c. 3. C hin tng g xy ra khi thm 1 ml dung dch bo ho CaSO4 vo 1 ml dung dch A.

Cho: CO2 + H2O HCO3 + H+ ; K = 106,35 HCO3 H+ + CO32 ; K = 1010,33

tan ca CO2 trong nc bng 3,0.102 M. Tch s tan ca CaSO4 bng 105,04; ca CaCO3 bng 108,35BI GII: 1. CO32 + H2O HCO3 + OH ; Kb1 = 10-14/10-10,33

= 103,67 (1) HCO3 + H2O ( H2O.CO2) + OH ; Kb2 = 10-14/10-6.35

= 107,65 (2) Kb1 >> Kb2 , cn bng (1) l ch yu.

CO32 + H2O HCO3 + OH ; 103,67

C C

NH3

Al(OH)3

K3

KS;

NH3

Al(OH)3

NH3

Mg(OH)2

2

KS;

2

K

a1 a

[ ] C 102,4 102,4 102,4

( )4,2

24,2

1010

C = 103,67 C = 102,4 + (10-4,8/10-3,67) = 0,0781 M

= = 5,1 % 102,4 10 0,0781

2

CO32 2.

CHCl = 0,16/2 = 0,08 M ; C = = 0,03905 M 0,0781

CO32 + 2 H+ CO2 + H2O Na2CO3 2

[ ] 0,03905 0,08 1,9. 103 0,03905 C > L

CO2 + H2O H+ + HCO3 ; 106,35 (do Ka1 >> Ka2) CO2 CO2

C 3,0 102 1,9. 103 3,0 102 x 1,9. 103 + x x = 106,35 x = 7,05.106 > Kb2) C 0,0391 [ ] 0,0391 x x x

367,32

10.89,2100391,0

== xxx

C = 0,0391 2,89.103 = 0,0362 M C . CCa2+ = 0,0362 102,82 = 5,47.105 > 108,35

Kt lun: c kt ta CaCO3

x(1,9. 103+x) + 3,0 102 x

CO32

CO32

CO3 2

II. OLYMPIC HA HC QUC T: OLYMPIC HA HC QUC T LN TH 28:

Kali dicromat l mt trong nhng tc nhn to kt ta c s dng rng ri nht. Nhng cn bng sau c thit lp trong dung dch nc ca Cr(VI)

HCrO4- + H2O CrO42- + H3O+. pK1 = 6,50 2HCrO4- Cr2O72- + H2O pK2 = -1,36

1. Tch s ion ca nc KW = 1,0.10-14 Tnh hng s cn bng ca cc phn ng sau:

a) CrO42- + H2O HCrO4- + OH- b) Cr2O72- + 2OH- 2CrO42- + H2O

2. Tch s tan ca BaCrO4 l T = 1,2.10-10. Ba2Cr2O7 tan dn dng trong nc. Cn bng ca phn ng (1b) s di chuyn theo chiu no khi thm cc tc nhn sau vo dung dch tng i m c ca kali dicromat?

a) KOH b) HCl c) BaCl2 d) H2O (xt tt c cc cn bng trn).

3. Hng s phn ly ca axit axetic l Ka = 1,8.10-5. Hy tnh tr s pH ca cc dung dch sau:

a) K2CrO4 0,010M b) K2Cr2O7 0,010M c) K2Cr2O7 0,010M + CH3COOH 0,100M

4. Hy tnh nng ti cn bng ca cc ion sau trong dung dch K2Cr2O7 0,010M + CH3COOH 0,100M.

a) CrO42-. b) Cr2O72-.

BI GII: 1) a) Hng s cn bng:

K = [HCrO4-][OH-]/[CrO42-] = [H+][OH-]/([H+][CrO42-]/[HCrO4-]) = Kw/K1 = 3,2.10-8b) Hng s cn bng: K = ([CrO42-][H+]/[HCrO4-])2/([HCrO4-]2/[Cr2O72-])/([H+][OH-])2 = 4,4.1013.

2) a) phi b) Tri c) BaCl2 di cn bng qua phi do ion cromat lin kt to thnh hp cht kh tan: Ba2+ + CrO42- = BaCrO4 d) H2O di cn bng qua phi do khi thm nc vo dung dch dicromat dn n vic lm long

dung dch v lm cho cn bng phn ly ca ion dicromat qua bn phi. Theo bi th pH ca dung dch phi b hn 7. Vi s pha long ny th pH ca dung dch s tng ln nn cn bng phi chuyn dch v bn phi.

3) a) CrO42- + H2O = HCrO4- + OH- K = 3,16.10-8. CCr = [CrO42-] + [HCrO4-] + 2[Cr2O72-] [CrO42-] [HCrO4-] [OH-] Nh vy [OH-]2/CCr = K [OH-] = 1,78.10-5M nn [H+] = 5,65.10-10. Vy pH = 9,25

b) Cr2O72- + H2O = 2HCrO4- K = 1/K2 = 4,37.10-2

HCrO4- = H+ + CrO42- K = K1 = 3,16.10-7. CCr = 2,0.10-2M = [CrO42-] + [HCrO4-] + 2[Cr2O72-] [HCrO4-] + 2[Cr2O72-] [H+] [CrO42-] = x = (K1[HCrO4-])1/2 K2 = [Cr2O72-]/[HCrO4-] = (CCr x)/2x2iu ny dn n phng trnh: 2K2x2 + x CCr = 0 Gii phng trnh trn ta thu c: x = 1,27.10-2M [H+] = 6,33.10-5M Vy pH = 4,20

c) Trong CH3COOH 0,10M th [H+] = (KaC)1/2 = 1,34.10-3 pH = 2,87 y l tr s cn thit. So snh tr s ny vi pH ca dung dch dicromat 0,1M cho trn (b) cho thy nh hng ca K2Cr2O7 trn pH c th an tm b qua c.

4) C th tnh bng hai cch: Cch 1:

a) [HCrO4-] = 1,3.10-2M (3b) [CrO42-] = K1[HCrO4-]/[H+] = 3,0.10-6M b) CCr = [CrO42-] + [HCrO4-] + 2[Cr2O72-] [Cr2O72-] = 3,7.10-3M

hoc [Cr2O72-] = K2[HCrO4-] = 3,9.10-3M Cch 2:

a) [CrO42-] = x; [HCrO4-] = x[H+]/K1 [Cr2O72-] = K2[HCrO4-] = x2K2[H+]2/K12. CCr = [CrO42-] + [HCrO4-] + 2[Cr2O72-] = 2K2[H+]2/K12x2 + (1 + [H+]/K1)x K1 = 3,16.10-7; K2 = 22,9; [H+] = 1,34.10-3. 8,24.108x2 + 4,24.103x 2,0.10-2 = 0 x = 3,0.10-6M

b) [Cr2O72-] = K2 [HCrO4-] = K2[H+]2/K12[CrO42-] = 3,7.10-3M OLYMPIC HA HC QUC T LN TH 28:

Cc phng php o hiu th v quang ph c s dng rng ri xc nh cc nng cn bng v hng s cn bng trong dung dch. C hai phng php thng xuyn c dng kt hp xc nh ng thi nhiu tiu phn.

Dung dch nc axit ha I cha mt hn hp FeSO4 v Fe2(SO4)3, v dung dch nc II cha mt hn hp K4[Fe(CN)6] v K3[Fe(CN)6]. Nng ca cc tiu phn c cha st tho mn cc quan h [Fe2+]I = [Fe(CN)64-]II v [Fe3+]I = [Fe(CN)63-]II. Th ca in cc platin nhng trong dung dch I l 0,652V (so vi in cc hydro tiu chun), trong khi th ca in cc platin nhng trong dung dch II l 0,242V (so vi in cc hydro tiu chun). Phn trm truyn x ca dung dch II o c so vi dung dch I ti 420nm bng 10,7% (chiu di ng truyn quang l = 5,02mm). Gi thit rng phc [Fe(CN)64-] Fe3+(aq); Fe2+(aq) khng hp th nh sng ti 420nm. hp th mol (Fe(CN)63-) = 1100L/mol.cm ti bc sng ny. Th kh chun ca Fe3+/Fe2+ l 0,771V. Yu t ghi trc logarit thp phn ca phng trnh Nernst bng 0,0590 (v ghi trc logarit t nhin l 0,0256). Gi thit rng tt c cc h s hot u bng 1. 1) Vit phng trnh Nernst ca h thng oxy ha - kh ca:

a) Dung dch 1. b) Dung dch 2 (ngoi tr phc xiano, b qua mi dng khc c trong dung dch)

2) n v ca yu t ghi trc logarit trong phng trnh Nernst c n v l g? 3) Tnh t s cc hng s bn vng (Fe(CN)63-)/(Fe(CN)64-) 4) Khong bin thin tuyt i trong ln (bin ) ca cc tham s vt l sau l bao nhiu.

a) truyn x (T)% b) Mt quang (A) %.

5) Tnh nng ca

a) Fe3+ trong dung dch I b) Fe2+ trong dung dch I

BI GII: 1) Phng trnh Nernst:

a) EI = Eo(Fe3+/Fe2+) + 0,0590lg([Fe3+]/[Fe2+] b) EII = Eo(Fe(CN)63-/Fe(CN)64-) + 0,0590lg([Fe(CN)63-]/[Fe(CN)64-])

2) Volt (V) 3) EII = Eo(Fe(CN)63-/Fe(CN)64-) + 0,0590lg([Fe(CN)63-]/[Fe(CN)64-]

= Eo(Fe3+/Fe2+) + 0,0590lg(1/2) + 0,0590lg([CN-]6/[CN-]6) + 0,0590lg([Fe(CN)63-]/[Fe(CN)64-]) = 0,242

Trong 1 v 2 ln lt l cc hng s bn vng ca [Fe(CN)64-] v [Fe(CN)63-]. [Fe(CN)63-]/[Fe(CN)64-] = Fe3+/Fe2+ nn E = EII EI = 0,059lg(1/2) = 8,90.106.

4) a) T 0 n 100% b) T 0 n

5) a) Dng nh lut Bouger Lambert Beer A = .l.C = .l.[Fe(CN)63-] = 0,971 [Fe(CN)63-] = [Fe3+] = 1,76.10-3M b) Dng phng trnh Nernst:

EI = Eo(Fe3+/Fe2+) + 0,0590lg([Fe3+]/[Fe2+]) = 0,652V T : [Fe3+]/[Fe2+] = 9,62.10-3M [Fe2+] = 0,183M

OLYMPIC HA HC QUC T LN TH 29:

HIn l mt cht ch th c tnh axit yu: Hin + Na+OH- Na+In- + H2O nhit thng, hng s phn li axit ca cht ch th ny l 2,93.10-5. Tr s bc sng di hp th (cuvet 1,00cm) cho cc dung dch 5,00.10-4M (mol.dm-3) ca cht

ch th ny tng cc dung dch axit mnh v kim mnh c cho trong bng sau: Tr s bc sng di hp th

(nm) pH = 1,00 pH = 13,00 400 0,401 0,067 470 0,447 0,050 485 0,453 0,052 490 0,452 0,054 505 0,443 0,073 535 0,390 0,170 555 0,342 0,342 570 0,303 0,515 585 0,263 0,648 615 0,195 0,816 625 0,176 0,823 635 0,170 0,816 650 0,137 0,763 680 0,097 0,588

1) D on mu ca: a) Dng axit.

b) Dng baz ca cht ch th. 2) Knh lc c mu no l thch hp nht phn tch bng quang k cht ch th ny trong mi trng

axit mnh?. Bit knh lc c t gia ngun sng v mu cht ch th. 3) Khong no ca bc sng l thch hp nht phn tch bng quang k cht ch th ny trong mi

trng baz mnh? 4) Cc dung dch ca cht ch th ny c pha ch trong dung dch HCl 0,1M v trong dung dch

NaOH 0,1M. xc nh c cc biu thc hon ton tuyn tnh gia bc sng di hp th v nng ti 490nm v 625nm cho tng mi trng tng ng trn. ln ca hng s phn li axit cho thy rng cht ch th ny hon ton khng phn li trong HCl 0,1M m li phn li hon ton trong NaOH 0,1M

490nm 625nmHIn (M-1.cm-1) 9,04.102 3,52.102

In-(NaOH) 1,08.102 1,65.103

Hy tnh bc sng di hp th (cuvet 1,00cm) hai bc sng trn ca mt dung dch khng cha cht m ca cht ch th ny c nng 1,80.10-3M BI GII:

1) Mu quan st c l mu kt hp vi mu ca s hp th ti a a) iu kin axit (pH = 1): Mu hp th 490 25 (xanh lam - lc) v nh vy s truyn mu kt

hp v c mu vng cam (625 25 nm) b) iu kin baz (pH = 13): Mu hp th 625 25 (vng cam) v nh vy s truyn mu kt

hp v c mu xanh lam - lc (490 25nm). 2) Knh lc cn truyn mu m mu s hp th hiu qa nht. Mu axit hp th mnh nht trong

khong xanh (490 25nm) v nh vy mt knh lc mu tng t s thch hp nht cho s phn tch bng quang k ca mu th.

3) Khong bc sng c dng vi nhy cao nht s tng ng vi bc sng m mu th hp th mnh nht. hp th ti a trong dng baz ca cht ch th trong dung dch xy ra 625 25 nm v y l bc sng thch hp nht cho s phn tch.

4) Hng s in li ca cht ch th l: [ ][ ][ ]HIn

InHK a+

= (1) Ta li c: [H+] = [In-] (2) [HIn] + [In-] = 1,80.10-3M (3) Thay (3) v (2) vo (1) ta c [In-] = 2,15.10-4M [HIn] = 1,58.10-3M Khi y ta c th tnh c mt quang hai bc sng l: A490 = 1,45 A625 = 0,911

OLYMPIC HA HC QUC T LN TH 31: PHN A:

Mt axit hai cha H2A tham gia vo cc phn ng phn li sau: H2A HA- + H+ K1 = 4,50.10-7

HA- A2- + H+ K2 = 4,70.10-11. Mt mu 20,00mL dung dch cha hn hp Na2A v NaHA c chun vi axit clohydric

0,300M. Qa trnh chun c thc hin vi mt pH - k in cc thy tinh. Hai im tng ng trn ng cong chun nh sau:

S mL HCl thm vo pH1,00 10,33

10,00 8,34 1) Khi thm 1,00mL HCl, tiu phn no phn ng trc ht v to sn phm g? 2) Lng sn phm to thnh (mmol) cu 1 l bao nhiu? 3) Vit cn bng chnh ca sn phm cu 1 tc dng vi dung mi 4) Lng (mmol) Na2A v NaHA c mt lc u? 5) Tnh tng th tch ca HCl cn thit t n im tng ung th hai. PHN B:

Cc dung dch I, II v III c cha cht ch th pH HIn (KIn = 4,19.10-4) v cc tc nhn khc ghi trong bng. Cc ga tr hp th ti 400nm ca cc dung dch c o trong cng mt cuvet cng c cho trong bng. Ka ca CH3COOH l 1,75.10-5. Dung dch I Dung dch II Dung dch III Nng ton phn ca HIn

1,00.10-5M 1,00.10-5M 1,00.10-5M

Cc tc nhn khc 1,00M HCl 0,100M NaOH 1,00M CH3COOH hp th ti 400nm 0,000 0,300 ? 6) Hy tnh hp th ti 400nm ca dung dch III. 7) Ngoi H2O, H+ v OH- cn c tt c nhng tiu phn no c mt trong dung dch thu c t s trn

ln dung dch II v dung dch III theo t l th tch 1:1 8) hp th ti 400nm ca dung dch cu 7 l bao nhiu? 9) truyn x ti 400nm ca dung dch cu 7 l bao nhiu? BI GII: PHN A:

1) Tiu phn phn ng trc ht l: A2- Sn phm l HA-

2) S mmol sn phm = 1,00.0,300 = 0,300mmol. 3) HA- + H2O H2A + OH- 4) Ti pH = 8,34 = (pKa1 + pKa2)/2 tt c A2- u b proton ha thnh HA-.

Do s mmol A2- c mt trong dung dch lc u = 3,00mmol Ti pH = 10,33 h l mt dung dch m vi t l [A2-]/[HA-] = 1. Nh vy: [HA-]lc u + [HA-]to thnh = [A2-]lc u - [HA-]to thnh Nh vy s mmol HA lc u = 3,00 0,300 0,300 = 2,40mmol.

5) VHCl = [(2.3,00) + 2,40]/0,300 = 28,00mL PHN B:

6) Dung dch III l dung dch ch th ti 10-5M trong dung dch c cha CH3COOH 1,0M. c c hp th hay mt quang ca dung dch, cn thit phi tnh nng dng cn bng ca cht ch th tu thuc vo [H+] ca dung dch [H+]III = (Ka.C)1/2 = 4,18.10-3M

T HIn H+ + In- ta c: [ ][ ][ ]HInInHK a+

= [ ][ ] [ ] 100,0== +

HK

HInIn In (1)

Ta li c: [HIn] + [In-] = 10-5 (2) T (1) v (2) ta tnh c [In-] = 0,091.10-5M

hp th ca dung dch III = 027,0300,0.10.00,110.091,0

5

5

=

7) CH3COOH, CH3COO-; Na+; HIn; In-. 8) Khi cc dung dch II v III c trn ln theo t l th tch 1:1 thu c mt dung dch m gm

CH3COO- 0,05M/CH3COOH 0,45M

[ ] [ ][ ] 533 10.75,15 + == COOCHCOOHCH

KH a

V vy: [ ][ ] [ ] 65,210.75,1510 538,3

===

+

HK

HInIn In (3)

Ta li c: [HIn] + [In-] = 10-5 (2) T (2) v (3) ta tnh c [In-] = 0,726.10-5M

hp th ca dung dch = 218,0300,0.10.0,1

10.726,05

5

=

9) truyn x ca dung dch = 10-( hp th) = 0,605 OLYMPIC HA HC QUC T LN TH 32:

Transferin (Tf) - mt loi huyt thanh - l mt n protein c chc nng chnh l tham gia qa trnh vn chuyn st (III) trong c th ngi. Mi phn t transferin c th vn chuyn hai ion Fe3+ theo phn ng:

FeIII + Tf (FeIII)Tf K1 = 4,7.1020M-1. FeIII + (FeIII)Tf (FeIII)2Tf K2 = 2,4.1019M-1. phn t (FeIII)2Tf 2 ion Fe3+ lien kt tng t nhau nhng khng cng mt pha v 2 phn t

st monotransferin (FeIII)Tf c th c biu th bng {FeIII.Tf} v {Tf.FeIII}. Bit K = [{Tf.FeIII}].[{FeIII.Tf}] = 5,9. 1. Tnh ga tr ca K1 = [{FeIII.Tf}].[FeIII]-1. [Tf]-1 v K1=[{Tf.FeIII}].[FeIII]-1. [Tf]-1. 2. Tnh gi tr ca K2 = [(FeIII)2Tf].[FeIII]-1.[{FeIII.Tf}]-1 v K2=[(FeIII)2Tf].[FeIII]-1.[{Tf.FeIII}]-1.

Lin kt gia st (III) mi pha ca lin kt c bao quanh bi 6 nguyn t nhn t cc ligand khc nhau. Theo cch ny, 2 nguyn t oxy ca anion CO32- phi tr vi kim loi v mi aminoaxit cu trc bc 1 ca protein l: 1 Aspartic, 1 Histidin, 2 Tyrosin cng phi tr vi st (III). 3. C bao nhiu nguyn t oxy xung quanh mt ion st (III) trong transferin. BI GII:

1. Nng ca phc monoferric transferin: [(FeIII)Tf] = [{FeIII.Tf}] + [{Tf.FeIII}] K1 + K1 = K1; K1.K = K1

12020'11

"1

11920

1'1

10.0,410).68,07,4(

10.8,69,51

10.7,41

====+=+=

MKKK

MK

KK

2. Ta c:

119"1

21"221

"2

'1

120'1

21'221

"2

"1

'2

'1

10.8,2

10.7,1

===

====

MK

KKKKKKK

MKKKKKKKKKK

3. S nguyn t oxy = 2(CO32-) + 1(Asp(O-)) + 2(2xTyr(O-)) = 5

OLYMPIC HA HC QUC T LN TH 33: Axit Photphoric l mt loi phn bn quan trng. Bn cnh axit photphoric v mui ca n c

nhiu ng dng trong x l kim loi, thc phm, cht ty ra v cng nghip ch to thuc nh rng. 1. Ga tr pK ca ba nc phn ly ca H3PO4 25oC l: pKa1 = 2,12; pKa2 = 7,21; pKa3=12,32. Vit cng

thc baz lin hp ca H2PO4- v tnh ga tr Kb ca n. Mt lng nh H3PO4 c s dng rng ri to v chua hay v cht cho nhiu thc ung nh

cola v bia. Cola c t khi 1,00gmL-1 cha 0,05% H3PO4 v khi lng. 2. Tnh pH ca cola (b qua nc phn li th 2 v 3). Gi s rng nguyn nhn gy ra tnh axit ca cola

l do H3PO4. 3. H3PO4 c s dng lm phn bn trong nng nghip ; 1,00.10-3M H3PO4 c thm vo dung

dch huyn ph ct v pH ca dung dch thu c l 7,00. Tnh nng phn mol ca cc loi photphat khc nhau trong t bit rng trong t khng c cht no phn ng vi photphat.

4. Km l nguyn t vi lng quan trng cn cho s pht trin cy trng. Cy trng c th hp th c km dng dung dch nc. trong dung dch ncc ngm c pH = 7,0 ngi ta tm thy c Zn3(PO4)2. Tnh [Zn2+ ] v [PO43-] trong dung dch bo ha. Bit T ca km photphat l 9,1.10-35.

BI GII: 1. Baz lin hp ca dihidro photphat (H2PO4-) l monohydrophotphat (HPO42-)

H2PO4- + H2O HPO42- + H3O+ K2aHPO42- + H2O H2PO4- + OH- K2b2H2O H3O+ + OH- Kw. pK2a + pK2b = pKw = 14 pK2b = 6,79

2. C(H3PO4) = 0,0051M H3PO4 + H2O H2PO4- + H3O+0,0051 x x x pKa1 = 2,12. Vy Ka = 7,59.10-3. Ta c: [ ][ ]

[ ][ ]46,2

10.49,3

10.59,70051,03

3

32

43

342

===

==+

+

pHOHx

xx

POHOHPOH

3. t: [ ][ ][ ][ ]

CXf

CHXf

CXH

f

CXH

f o

=

=

=

=

3

3

2

2

21

3

k hu cc phn s nng ca cc loi photphat khc nhau; C l tng nng ban u ca H3X (X = PO4):

[ ][ ][ ] [ ][ ][ ][ ] [ ][ ][ ][ ] [ ]+

+

+

+

++

==

==

===+++

OHff

HXOHX

K

OHff

XHOHHX

K

OHff

XHOHXH

K

ffff

a

a

oa

o

32

323

3

3

31

2

2

32

2

31

3

321

321 1

Cc phng trnh trn y dn n: [ ]

[ ][ ]

DKKK

f

DOHKK

f

DOHK

f

DOH

f

aaa

aa

a

p

3213

3122

231

1

33

.

=

=

=

=

+

+

+

Vi D = Ka1.Ka2.Ka3 + Ka1.Ka2.[H3O+] + Ka1[H3O+] + [H3O+]3. T cc ga tr Ka1; Ka2; Ka3 v pH ta c c cc kt qa sau: Ka1 = 7,59.10-3; Ka2 = 6,17.10-8; Ka3 = 4,79.10-13 v [H3O+] = 10-7. V cc phn s nng ca cc loi photphat khc nhau s l: H3PO4 (fo) = 8,10.10-6. H2PO4- (f1) = 0,618 HPO42- (f2) = 0,382 PO43- (f3) = 1,83.10-6.

4. t S(M) l tan ca km photphat trong nc ngm: [Zn2+] = 3S Tng nng ca cc dng khc nhau ca photphat = 2S [PO43-] = f3.2S f3 c th c xc nh t cc quan h cu 3 i vi pH = 7 th f3 = 1,83.10-6. T = [Zn2+]3[PO43-]2 = (3S)3.(f3.2S)2 = 9,1.10-33

S = 3,0.10-5M [Zn2+] = 9.10-5M [PO43-] = 1,1.10-10M

OLYMPIC HA HC QUC T LN TH 37: Kh nng nhn ion H+ ca nc c gi l tnh kim. Tnh kim rt quan trng i vi vic x

l nc, tnh cht ho hc v sinh hc ca nc. Ni chung, cc thnh phn ch yu nh hng n tnh kim ca nc l HCO3-, CO32- v OH-. ga tr pH di 7 th H+ trong nc lm gim tnh kim ca nc. Chnh v vy, phng trnh nu kim ca nc khi c mt cc ion HCO3-, CO32- v OH- c th c biu din bi:

kim = [HCO3-] + 2[CO32- ] + [OH-] - [H+]. Cc cn bng v hng s cn bng ( 298K) c cho sau y: CO2(k) CO2(aq) K(CO2) = 3,44.10-2. CO2 + H2O H2CO3 K(H2CO3) = 2,00.10-3. H2CO3 HCO3- + H+ Ka1 = 2,23.10-4. HCO3- CO32- + H+ Ka2 = 4,69.10-11

CaCO3 Ca2+ + CO32- Ksp = 4,50.10-9. H2O H+ + OH- Kw = 1,00.10-14

1. Nc t nhin (nc sng hay h) lun cha CO2 ho tan. T l [H2CO3] : [HCO3-] : [CO32-] = a : 1,00 : b. Xc nh a, b trong nc c nng [H+] = 1,00.10-7M.

2. Kh CO2 trong kh quyn c th lin quan ti tnh kim ca nc do n nm cn bng vi hm lng CO2 tan trong nc. Tnh nng ca CO2 (mol/L) trn nc tinh khit nm cn bng vi khng kh khng b nhim p sut 1,01.105Pa v 298K cha 0,0360% (v s mol) CO2. Gi s p sut tiu chun l 1,01.105Pa. Nu bn khng lm c cu ny th c th gi s rng nng CO2(aq) = 1,11.10-5M. tan ca CO2 trong nc c th c nh ngha bng biu thc S=[CO2(aq)] + [H2CO3] +

[HCO3-] + [CO32-]. tan ca kh CO2 trong nc nm cn bng vi khng kh khng b nhim 298K v 1,01.105Pa lun khc vi kim

3. Tnh tan ca CO2(k) tring nc tinh khit (mol/L). B qua s phn li ca nc. 4. Khi trong nc c 1,00.10-3M NaOH th tan ca CO2(k) lc ny s l bao nhiu?

298K, 1,01.105Pa th kh khng nhim s nm cn bng vi nc thin nhin cha CaCO3 ho tan. Cn bng sau y c th tn ti:

CaCO3(r) + CO2(aq) + H2O Ca2+ + 2HCO3-. 5. Tnh hng s cn bng ca phn ng trn.

Nu khng tnh c th ta c th gi s K = 5,00.10-5 tnh ton cho cu tip theo. 6. Tnh nng Ca2+ (mg/L) trong CaCO3 ho tan trong nc nm cn bng vi CO2 trong kh

quyn. Nu khng tnh c th ta c th gi s rng nng ca Ca2+(aq) l 40,1mg/L tnh ton.

7. Tnh kim ca dung dch trn. 8. mt h nc ngm cha CaCO3 ho tan th nc c lng CO2 rt cao. Nng ca Ca2+ -

trong h cao n 100mg/L. Gi thit rng h nc v khng kh bn trn l mt h kn, tnh hot p ca CO2 (Pa) trong khng kh nm cn bng vi Ca2+ trn.

BI GII: 1 [H+] = 1,00.10-7M

Ka1 = [HCO3-][H+]/[H2CO3] = 2,23.10-4 [HCO3-]/[H2CO3] = 2,23.103Ka2 = [CO32-][H+]/[HCO3-] = 4,69.10-11 [CO32-]/[HCO3-] = 4,69.10-4[H2CO3] : [HCO3-] : [CO32-] = 4,48.10-4 : 1,00 : 4,69.10-4

(a) (b) 2. P(CO2) = 1,01.105.3,60.10-4 = 36,36Pa [CO2(aq)] = K(CO2).P(CO2) = 1,24.10-5mol/L

Nu khng lm c cu 6 2 th c th gi s [CO2(aq)]=1,11.10-5M tnh cc cu tip theo. 3. a) tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO32-] = [CO2(aq)] + [HCO3-] ([H2CO3] = [CO2(aq)] . K(H2CO3) = 2,48.10-8M v [CO32-] = Ka2/([H+].[HCO3-] = Ka2 = 4,69.10-11M u qa nh nn ta b qua).

[H+].[HCO3-]/[CO2(aq)] = Ka1.K(H2CO3) = 4,46.10-7T cu 6 2 [CO2(aq)]=1,24.10-5M ta tnh c [H+]=[HCO3-]=2,35.10-6M Vy tan ca CO2 s bng 1,48.10-5M.

b) S dng [CO2(aq)]=1,11.10-5M tnh ton: tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO32-]

= [CO2(aq)] + [HCO3-] ([H2CO3] = [CO2(aq)] . K(H2CO3) = 2,48.10-8M v [CO32-] = Ka2/([H+].[HCO3-] = Ka2 = 4,69.10-11M u qa nh nn ta b qua).

[H+].[HCO3-]/[CO2(aq)] = Ka1.K(H2CO3) = 4,46.10-7T cu 6 2 [CO2(aq)]=1,11.10-5M ta tnh c [H+]=[HCO3-]=2,225.10-6M Vy tan ca CO2 s bng 1,34.10-5M.

4. a) S dng [CO2(aq)] = 1,24.10-5M tnh ton:

Trong dung dch NaOH 1,00.10-3M, tan ca CO2 phi tng ln do phn ng sau: (1) CO2(aq) + 2OH- CO32- + H2O K = K(H2CO3).Ka1.Ka2/(1,00.10-14)2 = 2,09.1011 (2) CO2(aq) + CO32- + H2O 2HCO3- K = K(H2CO3).Ka1/Ka2 = 9,37.103 Kt hp (1) v (2): CO2(aq) + OH- HCO3- K = 4,43.107. Do K rt ln nn ton b lng OH- u chuyn ht v HCO3-. [HCO3-] = 1,00.10-3M [OH-] = 1,82.10-6M [H+] = 5,49.10-9M [CO32-] = 8,54.10-6M tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO32-]

[CO2(aq)] + [HCO3-] + [CO32-] = 1,02.10-3M b) S dng [CO2(aq)] = 1,11.10-5M tnh ton:

Trong dung dch NaOH 1,00.10-3M, tan ca CO2 phi tng ln do phn ng sau: (3) CO2(aq) + 2OH- CO32- + H2O K = K(H2CO3).Ka1.Ka2/(1,00.10-14)2 = 2,09.1011 (4) CO2(aq) + CO32- + H2O 2HCO3- K = K(H2CO3).Ka1/Ka2 = 9,37.103 Kt hp (1) v (2): CO2(aq) + OH- HCO3- K = 4,43.107. Do K rt ln nn ton b lng OH- u chuyn ht v HCO3-. [HCO3-] = 1,00.10-3M

[OH-] = 1,82.10-6M [H+] = 5,49.10-9M [CO32-] = 8,54.10-6M tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO32-]

[CO2(aq)] + [HCO3-] + [CO32-] = 1,02.10-3M 5.

Keq = Ksp.K(H2CO3).Ka1/Ka2 = 4,28.10-5Nu khng tnh c cu 6 5 th ta c th gi s rng Keq = 5,00.10-5 tnh ton.

6. a) S dng Keq = 4,28.10-5 v [CO2(aq)] = 1,24.10-5M tnh ton:

Cn bng khi lng: [HCO3-] = 2[Ca2+] T cu 6 5: K = 4,28.10-5 = [Ca2+][HCO3-]2/[CO2(aq)]

= [Ca2+](2[Ca2+])2/[CO2(aq)] T cu 6 2: [CO2(aq)] = 1,24.10-5M [Ca2+] = 0,510.10-3M = 20,5mg/L

b) S dng Keq = 5,00.10-5 v [CO2(aq)] = 1,11.10-5M tnh ton: Cn bng khi lng: [HCO3-] = 2[Ca2+] T cu 6 5: K = 5,00.10-5 = [Ca2+][HCO3-]2/[CO2(aq)]


c) S dng Keq = 5,00.10-5 v [CO2(aq)] = 1,24.10-5M tnh ton: Cn bng khi lng: [HCO3-] = 2[Ca2+] T cu 6 5: K = 5,00.10-5 = [Ca2+][HCO3-]2/[CO2(aq)]


d) S dng Keq = 4,28.10-5 v [CO2(aq)] = 1,11.10-5M tnh ton: Cn bng khi lng: [HCO3-] = 2[Ca2+] T cu 6 5: K = 4,28.10-5 = [Ca2+][HCO3-]2/[CO2(aq)]


7. HCO3- l thnh phn ch yu trong dung dch: pH ca dung dch ny c th c tnh bng cng thc: pH = (pKa1 + pKa2)/2 = 6,99 7,00 Vi Ka1 v Ka2 l hng s axit ca H2CO3. Ti pH = 7,00 th [OH-] v [H+] ta c th b qua. Bn cnh theo cu 6 1 th:[CO32-]

e) 2,00.10-3M (gi s [Ca2+(aq)] = 40,1mg/L) 8. a) S dng Keq = 4,28.10-5 tnh ton Cn bng khi lng: [HCO3-] = 2[Ca2+] [Ca2+] = 100mg/L = 2,50.10-3M

Thay vo biu thc Keq = 4,28.10-5 = [Ca2+][HCO3-]2/[CO2(aq)] = 4[Ca2+]3/[CO2(aq)]

[CO2(aq)] = 1,46.10-3M P(CO2) = {[CO2(aq)]/K(CO2).1,01.105 = 4,28.103Pa

b) S dng Keq = 5,00.10-5 tnh ton: Cn bng khi lng: [HCO3-] = 2[Ca2+] [Ca2+] = 100mg/L = 2,50.10-3M

Thay vo biu thc Keq = 5,00.10-5 = [Ca2+][HCO3-]2/[CO2(aq)] = 4[Ca2+]3/[CO2(aq)]

[CO2(aq)] = 1,25.10-3M P(CO2) = {[CO2(aq)]/K(CO2).1,01.105 = 3,67.103Pa III. BI TP CHUN B CHO OLYMPIC HA HC QUC T:

OLYMPIC HA HC QUC T LN TH 30: Tnh axit ca mt mu nc ty thuc s hp th kh. Ni chung, kh quan trng nht gy nn

tnh axit l cacbon dioxit. a) Vit ba phng trnh phn ng minh ha nh hng ca CO2 trong khng kh ln tnh axit ca

nc. b) Xp cc hn hp kh sau theo th t tng dn kh nng ha tan ca CO2(k) trong dung dch nc

(tnh theo % s mol) i) 90% Ar; 10% CO2. ii) 80% Ar; 10% CO2; 10% NH3. iii) 80% Ar; 10% CO2; 10%Cl2.

Vit cc phng trnh ca bt k phn ng ho hc no xy ra trong dung dch nc khi phi kh cc hn hp kh trn. c) Xp cc h sau (trong nc) theo th t kh nng ho tan ca CO2. Gi thit rng trc khi phi

di hn hp 10% CO2 trong Ar, chng t cn bng vi khng kh. i) Nc ct. ii) Dung dch HCl 1M iii) Dung dch CH3COONa 1M

d) Gi thit rng khng kh c cha 350ppm CO2 (theo th tch), v t cn bng gia CO2 kh v tan (trong nc), hy tnh pH ca mt git nc ma p sut khng kh. Cc hng s thch hp ti 25oC l: kH(CO2) = 3,39.10-2mol.L-1.atm-1; Kb(HCO3-) = 2,24.10-8; Kb(CO32-) = 2,14.10-4.

e) Tnh pH ca mt chai nc c ga (P(CO2(k)) = 1atm) BI GII:

a) Cc phn ng: CO2(k) CO2(aq) (1) CO2(aq) + H2O HCO3-(aq) + H+(aq) (2) HCO3-(aq) CO32-(aq) + H+(aq) (3) l ta c thm cn bng: CO2(aq) + H2O H2CO3(aq)

C th c gii thiu gii thch s tn ti ring bit ca CO2 dng ho tan v ca axit cacbonic phn t trong dung dch nc nhng khng bt buc phi dng cn bng ny gii thch phn ng ho hc phn ng ca cacbonat trong nc.

Do cn bng c thit lp vi s c mt ng thi ca cc cht hai v ca mi phn ng v do ta bt u t CO2(k) v H2O nn dung dch thu c r rng phi c tnh axit.

b) NH3 l mt kh c tnh baz: NH3(k) NH3(aq)NH3(aq) + H2O NH4+(aq) + OH-(aq)Nn s xy ra phn ng axit baz, ko cn bng (2) v (3) theo chiu thun. iu ny lm tng

kh nng ho tan ca CO2 c trong kh quyn. Cl2 l mt kh c tnh axit: Cl2(k) Cl2(aq)Cl2(aq) + H2O Cl-(aq) + H+(aq) + HOCl(aq)HOCl(aq) H+(aq) + OCl-(aq)S gia tng [H+] sinh ra t cc phn ng ny s di cc cn bng (2) v (3) theo chiu nghch.

iu ny lm gim kh nng ho tan ca CO2 trong kh quyn. Nh vy chiu hng CO2 ho tan l: ii>i>iii.

c) Axetat CH3COO- l baz lin hp ca mt axit yu: CH3COO-(aq) + H2O CH3COOH(aq) + OH-(aq)Dung dch natri axetat c tnh kim v s di mi cn bng ca CO2 theo chiu thun. Dung dch HCl s di cn bng ca CO2 theo chiu nghch. Nh vy chiu hng CO2 hoa tan l: iii>i>ii

d) Nng ca CO2 trong dung dch nc c tnh bi nh lut Henry: [CO2(aq)] = kH.P(CO2) = 1,187.10-5M Ka = Kw/Kb Ka(CO2(aq)) = 4,46.10-7Ka(HCO3-(aq)) = 4,67.10-11Do Ka(CO2(aq)) >> Ka(HCO3-(aq)) ta gi s rng trong dung dch axit ch c cn bng ca qa trnh

tch loi proton H+ th nht l ng k (c th kim tra li iu ny mt khi tm c [H+]). Do : [H+] = [HCO3-] = 2,30.10-6M Vy pH = 5,64 Nay, vi [H+] = [HCO3-] = 2,30.10-6M ta c th thy [CO32-] = 4,67.10-11M. Do mc phn

ly ca HCO3- thnh H+ v CO32- rt nh v gi thit nu trn l ng. e) Thy ngay l 1atm CO2(k) s to dung dch axit hn l 350ppm CO2(k): Vy vi cc l do nh

trnh by cu d ta ch cn xt cn bng: CO2(k) CO2(aq) CO2(aq) + H2O HCO3-(aq) + H+(aq) gii quyt cu hi [CO2(aq)] = kH.P(CO2) = 3,39.10-2M v [H+] = [HCO3-] = (Ka[CO2(aq)])0,5 = 1,23.10-4M Vy pH = 3,91

OLYMPIC HA HC QUC T LN TH 31: a) Axit photphoric, H3PO4 l mt axit ba chc. Nu chun mt dung dch H3PO4 0,1000M vi

NaOH 0,1000M. Hy c lng pH ti cc thi im sau:

i) Gia im bt u v im tng ng th nht. ii) Ti im tng ng th hai. iii) Ti sao rt kh xc nh ng cong chun sau im tng ng th hai? K1 = 7,1.10-3 K2 = 6,2.10-8 K3 = 4,4.10-13. b) Mt dung dch cha 530mmol Na2S2O3 v mt lng cha xc nh KI. Khi dung dch ny c

chun vi AgNO3 th dng c 20,0mmol AgNO3 trc khi bt u vn c v AgI kt ta. C bao nhiu mmol KI?. Bit th tch sau cng l 200mL. Ag(S2O3)23- Ag+ + 2S2O32-(aq) Kd = 6,0.10-14. AgI(r) Ag+(aq) + I-(aq) T = 8,5.10-17.

BI GII: a) (i) C dung dch m H3PO4 v H2PO4-

[ ] [ ][ ][ ]

15,210.1,7 343

42

431

==

=

+

pHMPOH

POHPOH

KH

(ii) Ti im tng ng th hai, c HPO42- nn: [H+] = (K2K3)0,5 = 1,7.10-10M pH = 9,77 (iii) HPO42- (K3 = 4,4.10-13) c tnh axit khng mnh hn H2O bao nhiu (Kw = 1,00.10-14). Thm

baz mnh vo dung dch HPO42- tng t nh thm baz mnh vo nc. b) Do hng s to phc ca Ag(S2O3)23-, Kf = (Kd)-1 = 1,667.1013 l rt ln nn hu ht Ag+ thm

vo s to phc vi S2O32- v: [Ag(S2O3)23-] = 0,100M s mmol S2O32- t do = 530 (2.20) = 490mmol. [S2O32-] = 2,450M Nng ion Ag+ t do c tnh t Kd[ ][ ][ ][ ] 15

143232

2232

10.0,1

10.0,6)(

+

+

===

AgOSAgOSAg

K d

Ag+ + I- AgI T = [Ag+][I-] = 8,5.10-17

[I-] = 8,5.10-2M mmol KI = 17,0mmol.

OLYMPIC HA HC QUC T LN TH 31: Cc dung dch X, Y tun theo nh lut Beer trn mt khong nng kh rng. S liu ph ca

cc tiu phn ny trong cuvet 1,00cm nh sau: Mt quang A (nm) X (8,00.10-5M) Y (2,00.10-4M)

400 0,077 0,555 440 0,096 0,600 480 0,106 0,564 520 0,113 0,433 560 0,126 0,254 600 0,264 0,100 660 0,373 0,030 700 0,346 0,063

a) Hy tnh hp th mol ca X v Y ti 440 v 660nm b) Hy tnh mt quang ca mt dung dch 3,00.10-5M theo X v 5,00.10-4M theo Y ti 520 v

600nm. c) Mt dung dch cha X v Y c mt quang 0,400 v 0,500 theo th t ti 440 v 660nm. Hy

tnh nng ca X v Y trong dung dch. Gi s khng xy ra phn ng gia X v Y. BI GII:

a) T nh lut Beer A = .l.C Thay s t bng s liu ta c bng sau: X (cm-1.mol-1.L) Y (cm-1.mol-1.L) 440nm 1,2.103 3,00.103

660nm 4,67.103 1,50.102

b) Ti 520nm A = AX + AY = 1,125 Ti 600nm A = AX + AY = 0,349

c) Ti 440nm ta c: 0,400 = 1,2.103CX + 3,0.103CY Ti 660nm ta c: 0,500 = 4,67.103CX + 1,5.102CYGii h phng trnh trn ta c: CX = 1,04.10-4M v CY = 9,17.10-5M

OLYMPIC HA HC QUC T LN TH 32: Tr s pH ca nc nguyn cht l 7,0; trong khi nc ma t nhin c tnh axit yu do s

ho tan ca cacbon dioxit trong kh quyn. Tuy nhin trong nhiu khu vc nc ma c tnh axit mnh hn. iu ny do mt s nguyn nhn trong c nhng nguyn nhn t nhin v nhng nguyn nhn xut pht t cc hot ng ca con ngi. Trong kh quyn SO2 v NO b oxy ha theo th t thnh SO3 v NO2, chng phn ng vi nc chuyn thnh axit sunfuric v axit nitric. Hu qa l to thnh ma axit vi pH trung bnh khong 4,5. Tuy nhin cng o c cc tr s thp n mc 1,7.

Lu hunh dioxit SO2 l mt axit hai chc trong dung dch nc. Ti 25oC cc hng s axit bng:

SO2(aq) + H2O(l) HSO3-(aq) + H+(aq) Ka1 = 10-1,92M HSO3-(aq) SO32-(aq) + H+(aq) Ka2 = 10-7,18M Tt c cc cu hi sau u xt 25oC:

a) Tnh tan ca SO2 l 33,9L tng 1L H2O ti p sut ring phn ca lu hunh dioxit bng 1 bar.

i) Hy tnh nng ton phn ca SO2 trong nc bo ho kh SO2 (b qua s thay i th tnh xy ra do s ho tan SO2)

ii) Hy tnh thnh phn phn trm ca ion hydrosunfit. iii) Tnh pH ca dung dch.

b) Hy tnh [H+] trong dung dch nc ca Na2SO3 0,0100M c) Cn bng chnh trong dung dch nc ca NaHSO3.

2HSO3-(aq) SO2(aq) + SO32-(aq) + H2O(l). i) Hy tnh hng s cn bng ca cn bng trn. ii) Hy tnh nng ca lu hunh dioxit trong dung dch nc ca natri hydrosunfit

0,0100M nu ch xt cn bng ghi trn. d) Tnh tan ca bari sunfit trong nc bng 0,016g/100mL

i) Hy tnh nng ion Ba2+ trong nc bo ho. ii) Hy tnh nng ca ion sunfit trong nc bo ho. iii) Tnh T ca bari hydrosunfit.

e) Tch s tan ca bc sunfit bng 10-13,82M3. Hy tnh nng ion bc trong dung dch nc ca bc sunfit bo ho (b qua tnh baz ca ion sunfit).

f) Tch s tan ca canxi sunfit bng 10-7,17M2. Hy tnh hng s cn bng ca phn ng: Ca2+(aq) + Ag2SO3(r) CaSO3(r) + 2Ag+(aq)

g) Nh tng git brom n d vo dung dch lu hunh dioxit 0,0100M. Ton b lu hunh dioxit b oxy ha thnh sunfat (VI). Brom d c tch ra bng cch sc vi kh nit Vit mt phng trnh phn ng ca qa trnh v tnh nng ion hydro tng dung dch thu

c. Gi s cc qa trnh ho hc cng nh cc thao tc th nghim u khng lm thay i th tch dung dch. Tr s pKa ca ion hydrosunfat bng 1,99.

h) Sau mi t phun tro ni la, tr s pH ca nc ma o c bng 3,2. Hy tnh nng ton phn ca axit sunfuric trong nc ma, gi thit rng s axit ho ch so axit sunfuric. Proton th nhn trong axit sunfuric c th c xem nh phn li hon ton.

BI GII: a) i) pV = nRT n = 1,368 mol C(SO2) = 1,368M

ii) SO2(aq) + H2O HSO3-(aq) + H+(aq)

vi [H+] = [HSO3-] = x th Mxxx 1224,010

368,199,1

2

==

Vy %HSO3- = 8,95% iii) pH = 0,91

b) SO32-(aq) + H2O(l) OH-(aq) + HSO3-(aq) Vi [OH-] = [HSO3-] = x th:

[ ] MHMxx

x 10582,618,7

142

10.57,210.89,3101010

01,0+

==== c) Ta c:

i) [ ][ ][ ] [ ][ ][ ] [ ][ ] 26,5123

232

3

232 10. +

+

====

a

a

KK

HH

HSOSOSO

HSOSOSO

K

ii) [SO2] + [HSO3-] + [SO32-] = 0,01M v [SO2] = [SO32-]

Vy ta c: [ ]

[ ] [ ] MSOSOSO 5

226,5

2

22 10.33,210

)201,0( ==

d) M(BaSO3) = 217,39g.mol-1. i) [Ba2+] = 7,36.10-4M ii) SO32-(aq) + H2O(l) OH-(aq) + HSO3-(aq)

[OH-] = [HSO3-] = x [HSO3-] + [SO32-] = [Ba2+]

[ ] MSOMx xx

423

5

82,64

2

10.26,710.0479,1

10)10.36,7(

===

e) [Ag+] = 3,927.10-5M

f) K = [ ][ ] [ ][ ] [ ][ ] 65,6O23

23

2

2

2

2

10.3

32

+

+

+

+===

CaS

SOAg

TT

SOSO

CaAg

CaAg

g) Phn ng: 2H2O(l) + SO2(aq) + Br2(aq) SO42-(aq) + Br-(aq) + 4H+(aq) Cn bng: HSO4-(aq) SO42-(aq) + H+(aq) Ka = 10-1,99M [SO42-] = [HSO4-] = 0,01M v [H+] + [HSO4-] = 0,04M [HSO4-] = 0,04 - [H+] v [SO42-] = [H+] 0,03M [H+] = 0,0324M

h) [H+] = 10-3,2M; Ka = 10-1,99M; [HSO4-] = 10-1,28[SO42-] [H+] = 10-3,2 = 10-1,28[SO42-] + 2[SO42-] + 10-10,8. [SO42-] = 3,074.10-4M v [HSO4-] = 1,613.10-5M C(H2SO4) = [HSO4-] + [SO42-] = 3,24.10-4M.

OLYMPIC HA HC QUC T LN TH 32: Ho tan 1,00NH4Cl v 1,00g Ba(OH)2.8H2O vo 80mL nc. Pha long dung dch thu c

bng nc n 100mL ti 25oC. a) Tnh pH ca dung dch (pKa(NH4+) = 9,24 b) Hy tnh nng ca tt c cc ion trong dung dch. c) Hy tnh pH sau khi thm 10,0mL dung dch HCl 1,00M vo dung dch trn. d) Hy tnh [NH3] ca dung dch mi.

BI GII: a) NH4+(aq) + OH-(aq) NH3(aq) + H2O(aq)

18,7mmol NH4Cl v 3,17 mmol Ba(OH)2.8H2O (6,34mmol OH-) to ra 6,34mmol NH3 v 12,4mmol NH4+ cn li khng i.

[ ] [ ][ ] 95,810.13,1 934 === +

+ pHMNHNH

KH a

b) [NH4+] = 0,124M; [Ba2+] = 0,0317M; [H+] = 1,13.10-9M; [Cl-] = 0,187M; [OH-] = 8,85.10-6M c) Thm 10,0mmol HCl, trong c 6,34mmol c NH3 trung ho. Gi thit rng th tch bng

110mL, v b qua axit yu NH4+ ta c: [H+] = 0,0333M pH = 1,48 d) Trong dung dch axit mnh [NH3] s rt nh: [NH4+] = 0,170M

[ ] [ ][ ] MHNHKNH a 943 10.9,2 ++==

OLYMPIC HA HC QUC T LN TH 32: Mt hc sinh iu ch dung dch bo ho magie hydroxit trong nc tinh khit ti 25oC. Tr s

pH ca dung dch bo ho c tnh bng 10,5. a) Dng kt qa ny tnh tan ca magie hydroxit trong nc. Phi tnh tan theo mol.L-1

cng nh g/100mL. b) Hy tnh tch s tan ca magie hydroxit. c) Hy tnh tan ca magie hydroxit trong dung dch NaOH 0,010M ti 25oC.

Khuy trn mt hn hp gm 10g Mg(OH)2 v 100mL dung dch HCl 0,100M bng my khuy t tnh trong mt thi gian ti 25oC.

d) Hy tnh pH ca pha lng khi h thng t cn bng. BI GII: a) Mg(OH)2 Mg2+ + 2OH-pOH = 14,0 10,5 = 3,5 [OH-] = 10-3,5 = 3,2.10-4M Tng ng vi [Mg2+] = [Mg(OH)2 in ly] = tan ca Mg(OH)2 = 1,6.10-4M hay 9,2.10-4g/100mL. b) Ksp = [Mg2+][OH-]2 = 1,6.10-11M3

c) Mg(OH)2(r) Mg2+ (aq) + 2OH- (aq) [Mg2+] = x; [OH-] = 0,010 + 2x 0,010M Ksp = [Mg2+][OH-]2 = x[OH-]2 = 1,6.10-11 Mx 72

11

10.6,1)010,0(

10.6,1 == tan bng 1,6.10-7M hay 9.10-7g/100mL d) Mg(OH)2 c rt d v axit clohydric b trung ho hon ton theo phn ng: Mg(OH)2 (r) + 2H+ (aq) Mg2+ (aq) + 2H2O (l) Gi s th tch khng i v bng 100mL, phn ng ny to ra Mg2+ c nng 0,050M. Ri Mg(OH)2 ho tan trong dung dch [Mg2+] = 0,010 + x 0,050M [ ] [ ] 3,9)10.8,1lg(141410.8,1 552 =+==== + pOHpHMMgKOH sp OLYMPIC HA HC QUC T LN TH 32:

Cadimi l mt trong nhng kim loi rt c c tm thy vi nng cao trong cht thi t s luyn km, m in v x l nc thi. Ht phi cadimi dng ht nh s nhanh chng nh hng n h h hp ri sau l thn. Cadimi cho thy s cnh tranh vi km ti cc vng hot ng ca enzym.

Cadimi to thnh hydroxit hi kh tan l Cd(OH)2. a) Hy tnh tan ca Cd(OH)2 trong nc nguyn cht (b qua cn bng t proton phn) b) Hy tnh tan ca Cd(OH)2 trong dung dch NaOH(aq) 0,010M

Ion Cd2+ c i lc mnh vi ion CN-: Cd2+(aq) + CN-(aq) Cd(CN)+(aq) K1 = 105,48M-1. Cd(CN)+(aq) + CN-(aq) Cd(CN)2(aq) K2 = 105,12M-1. Cd(CN)2(aq) + CN-(aq) Cd(CN)3-(aq) K3 = 104,63M-1. Cd(CN)3-(aq) + CN-(aq) Cd(CN)42-(aq) K4 = 103,65M-1.

c) Hy tnh tan ca Cd(OH)2 trong nc c cha ion CN-. Nng cn bng l [CN-]=1,00.10-3M

d) Gi thit rng ch to thnh phc Cd(CN)42-, hy tnh phn trm sai lch tan so vi tan tm c cu c. Bit T(Cd(OH)2) = 5,9.10-15M3.

BI GII: a) S = 1,14.10-5M b) S = 5,9.10-11M c) S = 0,5[OH-] = C(Cd)

C(Cd) = [Cd2+] + [Cd(CN)+] + [Cd(CN)2] + [Cd(CN)3-] + [Cd(CN)42-] 0,5[OH-] = [Cd2+](1 + K1[CN-] + K1K2[CN-]2 + K1K2K3[CN-]3 + K1K2K3K4 [CN-]4) [OH-] = [2.T(1 + K1[CN-] + K1K2[CN-]2 + K1K2K3[CN-]3 + K1K2K3K4 [CN-]4)]3/2 = 4,79.10-3M S = 2,4.10-3M

d) [OH-] = [2.T.(1 + K1K2K3K4 [CN-]4)]3/2 = 4,47.10-3M S = 2,24.10-3M Phn trm sai lch = 6,7%

OLYMPIC HA HC QUC T LN TH 33: Hai yu t quan trng nht nh hng ln tan ca cc mui kh tan l pH v s c mt ca tc

nhn to phc. Bc oxalat l mt v d in hnh: Tch s tan ca n trong nc l T = 2,06.10-4 ti pH=7. tan ca n b nh hng bi pH khi anion oxalat phn ng vi ion hydroni v bng tc nhn to phc chng hn nh amoniac to phc vi cation bc.

a) Tnh tan ca bc oxalat trong dung dch axit c pH = 5,0. Hai hng s phn li ca axit oxalic ln lt l: K1 = 5,6.10-2 v K2 = 6,2.10-6.

b) Vi s c mt ca amoniac th ion bc to thnh hai dng phc Ag(NH3)+ v Ag(NH3)2+. Cc hng s to phc tng nc tng ng s l 1 = 1,59.103 v 2 = 6,76.103. Tnh tan ca bc oxalat trong dung dch cha 0,02M NH3 v c pH = 10,8.

BI GII: a) T = [Ag+]2[C2O42-]

Ta c: [Ag+] = 2S C(C2O42-) = S = [C2O42-] + [HC2O4-] + [H2C2O4] H2C2O4 = H+ + HC2O4- K1 = 5,6.10-2. HC2O4- = H+ + C2O42- K2 = 6,2.10-6.

Ta c kt qa sau: S = [ ] [ ] [ ]

++

++

21

2

2

242 1 KK

HKHOC

[ ] [ ] [ ] SSKKHKHKK

OC ..211

2212

42 =++= ++

Ti pH = 7 th [H+] = 10-7 1 T = 3,5.10-11. Ti pH = 5 th [H+] = 10-5 0,861 S = 2,17.10-4.

b) [NH3] = 0,02M Ti pH = 10,8 th [H+] = 1,585.10-11 1 Tng nng [Ag+] trong dung dch c xc nh bi phng trnh CAg = 2S = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] Cc phn ng to phc: Ag+ + NH3 = Ag(NH3)+ 1 = 1,59.103Ag(NH3)+ + NH3 = Ag(NH3)2+ 2 = 6,76.103T cc phng trnh trn ta d dng suy ra c biu thc sau:

CAg = 2S = [Ag+](1 + 1[NH3] + 12[NH3]2) [ ] [ ] [ ] SSNHNHAg =++= + .1 1 232131 Thay vo biu thc ca T ta tnh c S = 5,47.10-2.

OLYMPIC HA HC QUC T LN TH 33: Mt hp cht nitro hu c (RNO2) c kh bng phng php in ha trong dung dch m

axetat c tng nng axetat (HOAc+ OAc-) l 0,500M v c pH = 5. 300mL dung dch m cha 0,01M RNO2 em kh in ha hon ton. Axit axetic c Ka = 1,75.10-5 25oC. Phn ng kh in ha hp cht nitro xy ra nh sau:

RNO2 + 4H+ + 4e RNHOH + H2O Tnh pH ca dung dch sau khi kt thc phn ng.

BI GII: RNO2 + 4H+ + 4e RNHOH + H2O Ta c: [ ][ ][ ][ ] 5715,0=

=

OAcHOAc

OAcHOAcpHpK a

Mt khc ta c: [HOAc] + [OAc-] = 0,500 [HOAc] = 0,1818 [OAc-] = 0,3182 Nh vy s mmol cc cht lc ban u l: n(OAc-) = 95,45 n(HOAc) = 54,55 S mmol RNO2 b kh s l: 300.0,0100 = 3mmol T phng trnh bn phn ng ta thy rng kh ha hon ton 3mmol hp cht nitro cn

12mmol H+. S mmol H+ ny nhn c t s phn ly ca HOAc. Khi phn ng xy ra hon ton th: n(HOAc) = 54,55 12,00 = 42,55mmol n(OAc-) = 95,45 12,00 = 83,45mmol

Vy [ ][ ] 16,5=+= OAcHOAcpKpH a OLYMPIC HA HC QUC T LN TH 34:

tan l mt thng s quan trng xc nh c s nhim mi trng do cc mui gy ra. tan ca mt cht c nh ngha l lng cht cn thit c th tan vo mt lng dung mi to ra c dung dch bo ho. tan ca cc cht khc nhau tu thuc vo bn cht ca dung mi v cht tan cng nh ca cc iu kin th nghim, v d nh nhit v p sut. pH v kh nng to phc cng nh hng n tan.

Mt dung dch cha BaCl2 v SrCl2 u nng 0,01M. Khi ta thm mt dung dch bo ho natri sunfat vo dung dch th 99,9% BaCl2 s kt ta di dng BaSO4 v SrSO4 ch c th kt ta nu trong dung dch cn di 0,1% BaSO4. Tch s tan ca cc cht c cho sau y: T(BaSO4) = 10-10 v T(SrSO4) = 3.10-7.

1) Vit cc phng trnh phn ng to kt ta.

Tnh nng Ba2+ cn li trong dung dch khi SrSO4 bt u kt ta. Tnh %Ba2+ v Sr2+ sau khi tch ra.

S to phc gy nn mt nh hng ng k n tan. Phc l mt tiu phn tch in cha mt ion kim loi trung tm lin kt vi mt hay nhiu phi t. V d Ag(NH3)2+ l mt phc cha ion Ag+ l ion trung tm v hai phn t NH3 l phi t.

tan ca AgCl trong nc ct l 1,3.10-5M Tch s tan ca AgCl l 1,7.10-10M Hng s cn bng ca phn ng to phc c ga tr bng 1,5.107.

2) S dng tnh ton cho thy rng tan ca AgCl trong dung dch NH3 1,0M th cao hn trong nc ct.

BI GII: 1) Cc phn ng to kt ta:

Ba2+ + SO42- = BaSO4Sr2+ + SO42- = SrSO4Kt ta BaSO4 s xy ra khi [SO42-] = T(BaSO4)/[Ba2+] = 10-8M Kt ta SrSO4 s xy ra khi [SO42-] = 3.10-5M Nu khng xy ra cc iu kin v ng hc (chng hn nh s hnh thnh kt ta BaSO4 l v

cng chm) th BaSO4 s c to thnh trc, kt qa l s c s gim nng Ba2+. Khi nng SO42- tho mn yu cu kt ta SrSO4 th lc ny nng cn li ca ion Ba2+ trong dung dch c th c tnh t cng thc:

T(BaSO4) = [Ba2+][SO42-] = [Ba2+].3.10-5 [Ba2+] = 0,333.10-5M %Ba2+ cn li tng dung dch = %033,0

1010.333,02

5

=

2) Cn bng to phc gia AgCl v NH3 c th c xem nh l t hp ca hai cn bng: AgCl(r) Ag+(aq) + Cl-(aq) T = 1,7.10-10. Ag+(aq) + 2NH3(aq) Ag(NH3)2+ Kf = 1,5.107

AgCl(r) + 2NH3(aq) Ag(NH3)2+ + Cl-(aq) K = T.Kf = 2,6.10-3Cn bng: (1,0 2x) x x

Do K rt b nn hu ht Ag+ u tn ti dng phc: Nu vng mt NH3 th cn bng: [Ag+] = [Cl-] S hnh thnh phc dn n: [Ag(NH3)2+] = [Cl-] Nh vy: [ ][ ]

[ ] Mxxx

NHClNHAg

K 046,010.6,220,1

)( 322

3

23 ====

+

Kt qa ny c ngha l 4,6.10-2M AgCl tan trong dung dch NH3 1,0M, nhiu hn trong nc ct l 1,3.10-5M. Nh vy s to thnh phc Ag(NH3)2+ dn n vic lm tng tan ca AgCl. OLYMPIC HA HC QUC T LN TH 35:

Cc axit yu c chun vi dung dch baz mnh bit trc nng (dung dch chun). Dung dch axit yu (cht phn tch) c chuyn vo bnh nn 250cm3 v dung dch baz mnh (cht chun) c cho vo buret. im tng ng ca php chun t c khi lng cht chun cn bng vi lng cht phn tch. Gin biu th s thay i ca pH nh l mt hm ca th tch cht chun c thm vo c gi l ng cong chun .

im tng ng ca php chun ch c th c xc nh bng l thuyt, n khng th c xc nh bng thc nghim. N ch c th c lng c bng cch xc nh s thay i ca mt

vi tnh cht vt l trong qa trnh chun . Trong phng php chun axit baz, im cui ca php chun c xc nh bng cch s dng cht ch th axit baz. 1) Xy dng ng cong chun bng cch tnh mt vi im c trng v chn cht ch th thch

hp trong vic chun 50,00cm3 CH3COOH 0,1000M (Ka = 1,8.10-5) bng dung dch NaOH 0,1000M. V cht ch th c th tham kho bng 1:

Tn ch th Khong chuyn mu Mu dng axit baz Metyl da cam 3,2 4,4 - da cam Matyl 4,2 6,2 - vng Bromthymol xanh 6,0 7,6 Vng - xanh Phenol 6,8 8,2 Vng - Phenolphtalein 8,0 -9,8 Khng mu - Thymophtalein 9,3 10,5 Khng mu xanh 2) Axit ascorbic (Vitamin C) l mt axit yu v chu s phn ly theo phng trnh:

Chnh v vy axit ascorbic c th chun c nc 1 bng NaOH 50,00cm3 dung dch C6H8O6 0,1000M c chun bng 0,2000M:

(i) pH ca dung dch lc u l: a) 7,00; b) 2,58 c) 4,17 d) 1,00 (ii) Th tch ca cht chun cn t n im tng ng l: a) 50,00cm3 b) 35,00cm3 c) 25,00cm3 d) 20,00cm3(iii) Sau khi thm 12,5cm3 dung dch chun th pH ca dung dch s l: a) 4,17 b) 2,58 c) 7,00 d) 4,58 (iv) pH im tng ng s l: a) 7,00 b) 8,50 c) 8,43 d) 8,58 (v) Cht ch th c s dng trong phn ng ny s l (xem bng 1) a) bromthymol xanh b) phenol c) phenolphthalein d) thymolphtalein (vi) pH ca dung dch sau khi thm 26,00cm3 cht chun l: a) 13,30 b) 11,30 c) 11,00 d) 11,42 BI GII: Phn ng chun : CH3COOH + OH- = CH3COO- + H2O

a) pH trc khi tin hnh chun Do trc khi chun th trong bnh nn ch c CH3COOH nn pH ca dung dch s c tnh

t phng trnh phn ly CH3COOH CH3COOH CH3COO- + H+T phng trnh phn ly:

th nng H+ c th c tnh theo biu thc:

pH = 2,87 b) pH sau khi thm 10,00cm3 cht chun:

Trong dung dch lc ny cha mui natri axetat v axit axetic cn d nn n l dung dch m: Nng ca mi cht trong dung dch c tnh nh sau:

Nng H+ lc ny c th tnh c bng cch p dng phng trnh Henderson

Hasselbatch:

pH = 4,14 c) pH im tng ng

Lc ny th ton b lng axit axetic phn ng ht vi lng NaOH thm vo nn trong dung dch lc ny ch cn li anion axetat. Lc ny pH c quyt nh bi s phn ly ca anion ny:

CH3COOH + H2O CH3COOH + OH-Th tch cht chun cn t n im tng ng (Vep) c tnh nh sau:

Vo lc ny th tng th tch dung dch l 100cm3. Vo thi im ny ca vic chun th

[CH3COOH] = [OH-] v:

pOH = 5,28 pH = 8,72 d) pH sau khi thm 50,10cm3 cht chun

Vo thi im ny th ton b axit axetic chuyn ht thnh natri axetat nn pH ca dung dch lc ny s c quyt nh bi lng d dung dch natri hydroxit thm vo. Nh vy:

pOH = 4,00 pH = 10,00 Lc ny ng nh phn s c dng

Do pH im tng ng l 8,72 nn ch th ph hp lc ny l phenolphthalein

2) (i) b, (ii) c, (iii) a, (iv) b, (v) c, (vi) d

OLYMPIC HA HC QUC T LN TH 35: Dung dch m l dung dch c kh nng chng li s thay i pH. Thng thng dung dch m

gm mt axit yu v baz lin hp ca n (v d: CH3COOH/CH3COO-) hay mt baz yu v axit lin hp ca n (V d NH3/NH4+). Dung dch m c to thnh khi trung ho mt phn axit yu bi baz mnh hay baz yu v axit mnh. Chnh v vy ta c th chun b dung dch m bng cch trn mt lng axit (hay baz) yu tnh trc vi phn lin hp ca n.

pH ca dung dch m c to thnh bi axit yu HA v baz lin hp A- c tnh theo phng trnh Henderson Hasselbalch:

[ ][ ]+= OAcHOAcpKpH a 3) Tnh pH ca dung dch m cha 0,200M axit fomic (Ka = 2,1.10-4) v 0,150M natri fomiat. 4) Tnh pH ca dung dch khi thm 0,01000M dung dch NaOH vo dung dch m cu 1 5) Tnh th tch ca dung dch NaOH 0,200M cn thm vo 100,0cm3 dung dch CH3COOH 0,150M

(Ka = 1,8.10-5) thu c dung dch m c pH = 5,00 6) pH ca dung dch m cha 0,0100M axit benzoic (Ka = 6,6.10-5) v C6H5COONa 0,0100M s l:

a) 5,00 b) 4,18 c) 9,82 d) 9,0

7) Khi trn cng mt th tch 0,100 CH3COOH (Ka = 1,8.10-5) v 0,0500M NaOH th: i) Dung dch sau cng s l: a) D axit yu. b) D baz mnh c) Dung dch m d) C ba u sai. ii) pH ca dung dch cui s l: a) 3,02 b) 4,44 c) 3,17 d) 7,00

6) Khi trn cng mt th tch dung dch CH3COOH 0,100M v NaOH 0,150M th: i) Dung dch cui cng s l: a) D axit yu. b) D baz mnh c) Dung dch m d) C ba u sai. ii) pH ca dung dch cui s l: a) 12,00 b) 12,70 c) 13,18 d) 12,40

7) Khi trn cng mt th tch dung dch CH3COOH 0,150M v NaOH 0,100M th: i) Dung dch cui cng s l: a) D axit yu. b) D baz mnh c) Dung dch m

d) C ba u sai. ii) pH ca dung dch cui s l: a) 3,17 b) 3,02 c) 2,78 d) 3,22

8) Khi trn cng mt th tch dung dch CH3COOH 0,100M v NaOH 0,100M th: i) Dung dch cui cng s l: a) D axit yu. b) D baz mnh c) Dung dch m d) C ba u sai. 1. pH ca dung dch cui s l: a) 7,00 b) 13,00 c) 2,87 d) 3,02

BI GII: 1) pH = 3,55 2) Natri hydroxit s phn ng vi HCOOH:

HCOOH + OH- HCOO- + H2O Phn ng ny xy ra hon ton nn: [HCOOH] = 0,140M [HCOO-] = 0,160M Vy pH = 3,60 Lu rng ta thm mt baz mnh nh NaOH m pH ch thay i 0,05 n v

3) Gi V l th tch ca dung dch NaOH. Nh vy th tch cui ca dung dch s l (100,0 + V) v s mmol CH3COOH v OH- l 100,0.0,150 = 15,00mmol v V.0,200 = 0,200V mmol tng ng. T phn ng: CH3COOH + OH- CH3COO- + H2O

Nh vy lng CH3COO- sinh ra s l 0,200V mmol v lng CH3COOH cha phn ng s l (15,00-0,200V)mmol. Nh vy nng ca cc tiu phn trong dung dch m s l:

[ ][ ] M

VVCOOCH

MV

COOHCH

+=+

=

0,100200,0

0,100200,000,15

3

3

T biu thc hng s phn li ca axit axetic ta c th nhn c: [ ][ ] [ ]

35

5

3

3

21,4810.0,110.8,1

100200,000,15

0,100200,0

cmV

V

VV

HK

COOHCHCOOCH a

==++

=

+


Nha trao i ion c th c s dng hp th v phn lp cation v anion. Chng c th c iu ch t cc vt liu v c v hu c. Mt loi nha trao i cation hu c c th c tng hp bng s ng trng hp styren /divinyl benzen tip theo bi s sunfo ho bng H2SO4 nh s 1:

H 2SO 4

C CCC

C CCC

SO 3- H +

[ c a tio n ic ex ch an g e r ]

R -H +

C

+

C C

C C

N a2S 2O 8

C CCC

C CCC

[ po lym er ]

C

S 1

Nha trao i cation (k hiu l R-H+) c th c s dng hp th cc cation. Phn ng c th c biu din nh sau:

R-H+ + M+ = RM + H+ KC = [RM][H+]/[R-H+][M+] (1) KD = [RM]/[M+] (2) Nha trao i cation R-H+ c th c chuyn ha thnh cht trao i ion R-M+ hay R-M2+ bng

phn ng gia R-H+ vi mt hydroxit kim lai M(OH)z. Phng trnh phn ng s l: R-H+ + MOH = R-M+ + H2O (3)

V zR-H+ + M(OH)z = (R-)zM+ + zH2O (4) 1) Mt loi nha trao i cation R-Na+ c s dng loi CaCl2 trong nc my. a) Vit phng trnh phn ng. b) Nu mt loi nha trao i khc R-H+ c s dng thay th cho R-Na+.

i) Vit phn ng xy ra. ii) Cho bit loi nha trao i ion no R-H+ hay R-Na+ th ph hp hn trong vic loi Ca2+ ra khi

nc thi v cho bit l do. 2) Mt loi nha trao i anion hu c (R+Cl-) c th c tng hp bng s ng trng hp styren

/divinyl benzen vi xc tc l mt axit Lewis nh AlCl3 v mt amin bc 3 NR3, nh s sau:

Loi nha trao i ion R+OH- c th nhn c t phn ng: R+Cl- + NaOH = R+OH- + NaCl (5)

a) Bng cch no m H+ sinh ra t dung dch HCl c th b loi b vi mt loi nha trao i ion R+OH-. Vit cc phng trnh phn ng xy ra.

b) Bng cch no ta c th loi c ion SO42- trong nc my bng loi nha trao i ion R+OH- trn. Vit cc phng trnh phn ng xy ra. Dung lng (S) ca nha trao i cation R-H+ i vi mt ion b hp th c th c xc nh

bng s mol ca ion hp th/gam ca mt nha trao i ion tng 1,0mL dung dch nc v c th c tnh bng cch s dng phng trnh sau:

S = ([RM] + [RH]).10-3 (6) Dung lng (S) ca mt nha trao i cation R-H+ i vi ion M+ trong mt dung dch nc c

th c xc nh t hng s cn bng KC, hng s phn b KD v nng ca cc ion M+ v H+ trong dung dch. 3) Hy chng minh phng trnh sau: 1/Kd = [M+]/1000S + [H+]/KC.S.1000 (7) 4) Nha trao i ion c th c s dng lm pha tnh trong php sc k lng hp th v phn

lp cc loi ion khc nhau. V d: nha trao i anion R+OH- c th c s dng phn lp cc ion X- v Y- c th c ra gii bng NaOH. Php sc k lng trong vic phn tch cc anion X- v Y- s dng 30cm ct nha trao i ion nh hnh 1. Vi t1, t2 v to l thi gian duy tr (Retention time)(tR) i vi X- v Y- v dung mi ra gii tinh khit NaOH c th i qua ct, 1 v 2 l chiu rng pic ca X- v Y-. S a l thuyt N v chiu cao a H (chiu cao ca tng s a l thuyt) ca ct c th c tnh bi cc biu thc sau:

N = 16(tR/)2. (8) v H = L/N (9)

Hnh 1: Sc k ph ca ion X- v Y-

1.0 10. 14.

t0

t1 t2

1.0 1.5 tR

Retention Time / min

X-Y-

vi L l chiu di ct. phn gii (R) ca ct v h s phn ly i vi X- v Y- c th c xc nh bi cc h thc:

R = 2 (t2 - t1) / (1 + 2) (10) v = (t2 to) / (t1 to) (11)

a) Tnh s a l thuyt ca ct b) Tnh chiu cao a. c) Tnh phn gii (R) ca ct i vi hai anion X- v Y-. d) Tnh h s phn ly i vi X- v Y-

5) Mt s loi nha trao i ion c th nhn c t nhng vt liu v c. Zeolit [(M2+)(Al2O3)m/(SiO2)n] (M2+ = Na+; K+ hay Ca2+; Mg2+) l nhng v d in hnh v nhng nha trao i ion v c. Mt s v d v zeolit c cho trong hnh 2.

Hnh 2: Mt s loi zeolit

Mt loi Na* - zeolit (k hiu l Z-Na*) vi kch thc l hng l 13 l mt loi nha trao i ion quan trng loi cc ion Ca2+ v Mg2+ ra khi nc my. Cc loi zeolit vi kch thc l hng xc nh th c chn lc hp th rt cao i vi cc phn t khc nhau. V d H2O v iso-butan. Nh vy, zeolit ng vai tr nh l mt ci ry phn t. Zeolit cng c th c s dng nh l mt cht xc tc trong cng ngh ha du. V d: trong ha du iso-butan l kt qa ca s tng tc crackinh cc tc nhn hp ph chn lc.

a) Vit phng trnh phn ng loi Ca2+ ra khi nc my vi zeolit Z-Na*. b) Vit phng trnh phn ng ca vic hp th K+ vi zeolit Z-Na*.

BI GII: 1) a) 2RNa + Ca2+ = (R)2Ca + 2Na+

hay 2RNa + CaCl2 = (R)2Ca + 2NaCl b) i) 2RH + Ca2+ = (R)2Ca + 2H+ hay 2RH + CaCl2 = (R)2Ca + 2HCl

ii) S dng RNa th kh thi hn so vi vic s dng RH bi v sn phm ca s hp th Ca2+ bng RNa l NaCl l sn phm t c hi hn l HCl (lm gim pH)

2) a) R+OH- + HCl = R+Cl- + H2O b) Bc u tin s xy ra phn ng: 2R+OH- + SO42- = (R+)2SO42- + 2OH- Sau th thm HCl vo trung ho lng OH- sinh ra bc 1: H+ + OH- = H2O

3 Thay phng trnh (1), (2) vo (6) v s dng mt s bin i ton hc n gin ta nhn c: 1/Kd = [M+]/1000S + [H+]/KC.S.1000

4 a) N1 = 16(t1/1)2 = 1600 N2 = 16(t2/2)2 = 1394 N = (N1 + N2)/2 = 1497 b) H = L/N = 0,021cm c) R = R = 2 (t2 - t1) / (1 + 2) = 3,2 d) = (t2 to) / (t1 to) = 1,44

5. a) Z-Na+ + Ca2+ = Z-Ca + Na+ b) Z-Na+ + K+ = Z-K+ + Na+ OLYMPIC HA HC QUC T LN TH 37:

Lng canxi trong mu c th c xc nh bi cch sau:

Bc 1: Thm mt vi git ch th metyl vo dung dch mu c axit ha v sau l trn vi dung dch Na2C2O4.

Bc 2: Thm ure (NH2)2CO v un si dung dch n khi ch th chuyn sang mu vng (vic ny mt 15 pht). Kt ta CaC2O4 xut hin.

Bc 3: Dung dch nng c lc v kt ta CaC2O4 c ra bng nc lnh loi b lng d ion C2O42-.

Bc 4: Cht rn khng tan CaC2O4 c ho tan vo dung dch H2SO4 0,1M sinh ra ion Ca2+ v H2C2O4. Dung dch H2C2O4 c chun vi dung dch chun KMnO4 n khi dung dch c mu hng th ngng.

Cc phn ng xy ra v cc hng s cn bng: CaC2O4(s) Ca2+(aq) + C2O42-(aq) T = 1.30x10-8

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) T = 6.50x10-6 H2C2O4(aq) HC2O4-(aq) + H+(aq) Ka1 = 5.60x10-2

HC2O4-(aq) C2O42-(aq) + H+(aq) Ka2 = 5.42x10-5

H2O H+(aq) + OH-(aq) Kw = 1.00x10-141. Vit v cn bng cc phng trnh phn ng xy ra bc 2. 2. 25,00mL dung dch mu canxi c xc nh bng phng php trn v s dng ht 27,41mL

dung dch KMnO4 2,50.10-3M bc cui cng. Xc nh nng Ca2+ trong mu. 3. Tnh T ca CaC2O4 trong mt dung dch m c pH = 4. (B qua h s hot )

Trong php phn tch trn th ta b qua mt nguyn nhn quan trng gy nn sai s. S kt ta CaC2O4 bc 1 s khng hon ton nu ta thm mt lng d C2O42- do cc phn ng sau:

Ca2+

(aq) + C2O42-

(aq) CaC2O4(aq) Kf1 = 1.0 x 103 CaC2O4(aq) + C2O4

2-(aq) Ca(C2O4)22-(aq) Kf2 = 10

4. Tnh nng cn bng ca Ca2+ v C2O42- trong dung dch sau khi to thnh lng kt ta ti a ca CaC2O4.

5. Tnh nng ion H+ v Ca2+ trong dung dch bo ho CaC2O4 (B qua h s hot ). BI GII:1. (NH2)2CO + H2O 232 CONH +2. [Ca2+] = 6,85.10-3M 3. [Ca2+] = [C2O42-] + [HC2O4-] + [H2C2O4]

= [C2O42-](1 + [H+]/K1 + [H+]2/K1K2) Vy [C2O42-] = [Ca2+]/(1 + [H+]/K1 + [H+]2/K1K2) (1) Thay (1) vo biu thc tch s tan: T = [Ca2+][C2O42-] ta tnh c [C2O42-] = 1,92.10-4M

4. Ta c:

[ ] [ ] ( )[ ][ ] [ ]

[ ] [ ][ ] MCaMOCKTK

OCT

OCddC

OCKKKOC

T

OCCaOCaCCaC

ffCa

fff

aqCa

622242

212242

242

2422112

42

2242)(42

2

10.3,110.0,1

01

1

+

+

==

=+=

++=

++=

5. Cn bng in tch: 2[Ca2+] + [H+] = 2[C2O42-] + [HC2O4-] + [OH-] (1) Cn bng khi lng: [Ca2+] = [C2O42-] + [HC2O4-] + [H2C2O4] (2) V Kb2 rt nh nn nng ca H2C2O4 c th b qua. Kt hp (1) v (2) ta c: [HC2O4-] = Kw/[H+] - [H+] (3) [C2O42-] = (K2Kw)/[H+]2 K2 (4) [Ca2+] = T/[C2O42-] = T[H+]2/(K2Kw K2[H+]2) (5) Thay (3), (4), (5) vo (2) v gii phng trnh sinh ra ta c: [H+] = 5,5.10-8M [Ca2+] = 1,04.10-4M


Mt hc sinh nghin cu phn ng ha hc gia cc cation A2+

, B2+

, C2+

, D2+

, E2+

trong dung dch nitrat v cc anion X

-, Y

-, Z

-, Cl

-, OH

- trong dung dch cha cation natri ng thi c mt

phi t hu c L. Hc sinh ny xc nh c mt s hp cht kt ta v mt s phc cht mu nh trong Bng 1 di y:

Bng 1

X-

Y-

Z-

Cl-

OH- L

A2+ ***

*** *** *** kt ta

trng ***

B2+ kt ta

vng kt ta trng

*** *** *** Phc BLn2+

C

2+ kt ta trng

kt ta nu

kt ta nu

kt ta trng

kt ta en

Cc phc CL

2+, CL2

2+

D

2+ ***

kt ta *** *** *** ***

E2+ ***

kt ta kt ta

trng *** *** ***

*** = khng phn ng,

1 Lp s tch cc cation A2+

, B2+

, C2+

, D2+

, E2+

trong dung dch nitrat bng cch s dng cc dung dch thuc th khc nhau cha cc anion X

-, Y

-, Z

-, Cl

-, OH

-. Ghi r sn phm cc sn phm

hnh thnh trong mi bc.

2 Lp s tch cc anion X-, Y

-, Z

-, Cl

-, OH

- trong dung dch cha cation natri bng cch s dng

cc dung dch thuc th khc nhau cha cc cation A2+

, B2+

, C2+

, D2+

, E2+

. Ghi r sn phm cc sn phm hnh thnh trong mi bc.

3 Kt ta trng BY2 v kt ta nu CY2 tan t trong nc vi tch s tan tng ng ti 25oC ln lt l 3.20 10-8 v 2.56 10-13.

a) Tnh tan ca BY2. b) Tnh tan ca CY2. 4 Chun b trong cc bnh nh mc 50 mL mt nhm cc dung dch cha B2+ v L bng cch thm

vo mi bnh 2 mL dung dch B2+ 8,2 10-3 M. Thm vo mi bnh cc lng khc nhau ca dung dch cha phi t L nng 1,0 10-2 M. Pha long dung dch trong mi bnh bng nc n vch mc (50 mL). o h s hp th ca phc BLn ti 540 nm cho mi dung dch trong mt ng di 1,0 cm. Cc d liu thu c trong Bng 2. (C B

2+ v phi t L khng hp th (A = 0) ti

540 nm.) [Phng php t l mol]

a) Tnh gi tr n (s phi tr) trong phc BLn2+

.

b) Tnh hng s to thnh (Kf) ca phc BLn2+

. Bng 2

L thm vo VL (mL)

H s hp th (A)

L thm vo VL (mL)

H s hp th (A)

1.00 0.14 2.00 0.26 3.00 0.40 4.00 0.48 5.00 0.55 6.00 0.60 7.00 0.64 8.00 0.66 9.00 0.66 10.00 0.66

5 Thm rt chm cht rn NaY (tan) vo mt dung dch cha B2+

0,10M v C2+ 0,05 M c pha t cc dung dch mui nitrat tng ng ca chng.

a) Cation no kt ta trc (B2+

hay C2+

)? Nng [Y-] bng bao nhiu khi ion ny kt ta? (Cho

Ksp (BY2) = 3.20 10-8 v Ksp (CY2) = 2.56 10-13 ti 25oC.) [Tch bng kt ta] b) Nng ca ion Y

- v cation cn li bng bao nhiu khi cation u tin kt ta hon ton (gi

thit rng sau khi kt ta hon ton nng cation u tin trong dung dch 10-6 M)? S dng tc nhn Y

- c th tch B

2+ v C

2+ bng phng php kt ta hay khng?

BI GII: 1. cch tch:

2. Cch tch:

3. a)

4S13 = 3,2.10-8 S1 = 2,0.10-3b)

4S23 = 2,56.10-13 S2 = 4,0.10-54) a) th ph thuc gia h s hp th A v th tch VL ca cht L c dng nh sau:

T th tch L im gy B (tt c cc ion B2+ u to phc vi L) trn th th ta c th tnh

n nh sau: n = s mol L / s mol B2+ = 5,1.10-3.10-2 / 2.10-3 . 8,2.10-3 = 3 iu c ngha l phc gia B2+ vi L c dng BL32+

b) * Tnh h s tt mol : im gy A = 0,66 = .1. [BL32+] = 2,01.103

* Chn mt im bt k trn th, v d: Ti im P: l im m 2,0mL L c thm vo; A = 0,26

Nh vy:

5 a) Khi bt u hnh thnh kt ta CY2 th:

Ksp = [C2+][Y-]2 = 2,56.10-13

Tnh tng t cho khi hnh thnh kt ta BY2 ta c: [Y-] = 5,06.10-4M

Vy cht CY2 kt ta trc b. Cht C2+ s kt ta hon ton di dng CY2 khi [C2+] = 10-6M

Nh vy [C2+][Y-]2 = 2,56.10-3 [Y-] lc ny s bng 5,06.10-4M iu ny c ngha l C2+ s b kt ta hon ton di dng CY2 khi [Y-] = 5,06.10-4M Lc ny th i vi kt ta BY2 th [B2+][Y-]2 = 2,56.10-8 < Ksp(BY2) = 3,2.10-8

Nh vy lc [Y-] = 5,06.10-4M v [B2+] = 0,1M th cht BY2 vn cha c kt ta iu ny c ngha l ta hon ton c th tch cc ion B2+ v C2+ ra khi dung dch bng phng

php kt ta phn on vi Y- l tc nhn. IV. OLYMPIC HA HC CC NC TRN TH GII:

OLYMPIC HA HC O 1999: Lng oxi trong mu c xc nh bng php phn tch iot nh sau (phng php Winkler): Bc 1: Oxi trong dung dch oxi ho Mn2+ thnh Mn(IV) trong mi trng kim to thnh

MnO(OH)2. Bc 2: Thm axit vo hp cht ca mangan ni trn phn ng vi lng d Mn2+ to thnh ion

Mn3+. Bc 3: Ion Mn3+ ny oxi ha thuc th iodua to thnh iot v Mn3+ b kh thnh Mn2+. Bc 4: Lng iot sinh ra trong bc 3 c chun bng dung dch thiosunfat.

2) Vit phng trnh ion ca 4 phn ng trn. 2) Phn tch nhng mu nc sng Schwechat cho kt qa sau:

Chun ho dung dch natri thiosunfat Na2S2O3: dng KIO3 trong mi trng axit, khi ion iodat b kh thnh ion iodua. Vi 25,00mL dung dch KIO3 ((KIO3) = 174,8mg/L) phi dng ht 12,45mL dung dch Na2S2O3.

Ngay sau khi ly mu nc, lng oxy ca n c xc nh theo phng php Winkler. phi dng 11,80mL dung dch Na2S2O3 trn cho 103,50mL mu nc 20,0oC. Nng oxy bo ho trong nc 20,0oC l 9,08mg/L.

Mu th hai (V = 202,20mL, T = 20,0oC) c trong 5 ngy nhit 20,0oC, ng vi 6,75mL dung dch Na2S2O3. iii) Vit phng trnh ion ca phn ng chun ho dung dch thiosunfat. iv) Tnh nng mol/L ca dung dch thiosunfat v) Tnh hm lng oxy (mg/L) ca mu nc ngay sau khi ly mu. vi) Tnh ch s bo ho oxy ca mu nc ny. vii) Tnh hm lng oxy ca mu nc ny sau khi 5 ngy. viii) T cc kt qa trn c th xc nh c cc thng s c trng no? Gi tr ca n l

bao nhiu? BI GII: i. Bc 1: 2Mn2+ + O2 + 4OH- = 2MnO(OH)2. Bc 2: 2MnO(OH)2 + 2Mn2+ + 8H+ = 4Mn3+ + 6H2O Bc 3: 4Mn3+ + 4I- = 2I2 + 4Mn2+. Bc 4: 2I2 + 4S2O32- = 2S4O62- + 4I-. ii. a) IO3- + 6S2O32- + 6H+ = I- + 3H2O + 9S + 3SO42-

b) C(S2O32-) = 9,841.10-3M c) n(O2) = 2,903.10-2mmol (O2) = 8,976mg/L.

d) SSI = 98,9% e) n(O2) = 0,0166mmol (O2) = 5,20mg/L f) BSB5 = 3.78mg/L

OLYMPIC HA HC O 2005: 25.0C v p sut ring phn ca CO2 l: p(CO2) = 1.00 bar. 0.8304 lt kh CO2 ho tan trong

1.00 lt nc. 1. Tnh nng mol ca CO2 ho tan 2. Tnh hng s Henry ca CO2 25.0C. 3. Tnh nng mol ca CO2 ho tan trong nc ma, nu phn th tch ca CO2 trong kh quyn

c hm lng 380 ppm mi ngy v p sut ca CO2 c ga tr l 1.00 bar. Mt phn CO2 ho tan s phn ng vi nc to thnh axit cacbonic. Hng s cn bng ca

phn ng ny l K=1.6710-3 vi nng ca nc c a vo Ka 4. Tnh nng ca axit cacbonic ho tan trong nc ma? Bit nng ca CO2 l khng i.

i vi hng s phn li th nht th [H2CO3]* c s dng thay th cho nng ca axit cacbonic. [H2CO3]* l tng nng ca axit cacbonic v lng kh CO2 ho tan trong nc.

Cc ga tr l: KA1 = 4.4510-7 and KA2 = 4.8410-11. 5. Tnh pH ca nc ma. B qua cn bng t proton phn ca nc v hng s phn li KA2 ca

axit cacbonic. Bit rng nng ca [H2CO3]* l khng i trong sut qa trnh. Vo nm 1960 th phn th tch ca CO2 trong kh quyn l 320ppm

6. Tnh pH ca nc ma vo thi im ny (tt c cc iu kin khc u nh cu 3.5.). vi (CaCO3) c tch s tan T = 4.7010-9.

7. Tnh tan ca vi trong nc tinh khit. Gi thit rng c mui hydrocacbonat v mui cacbonat u khng phn ng sinh ra axit cacbonic. Tnh tan ca CaCO3 trong nc ma vo thi im ny. Nh ni trn [H2CO3]* lun l

hng s. tr li cu hi ny phi lm nhng vic sau: 8. Hy xc nh nhng ion cha bit nng . 9. Vit cc phng trnh cn thit tnh nng cc ion ny. 10. Xc nh phng trnh cui cng vi [H3O+] l n s.

Vi nhng phng trnh bc cao th ta kh lng gii c chnh xc. Ta c th gi thit gn ng rng pH = 8.26 tin tnh ton 11. S dng tt c nhng thng tin trn tnh tan ca vi. BI GII: 1. Nng ca CO2 ho tan:

pV = nR.T

mol0335.0K15.29808314.08304.000.1

TRVpn =

==

L/mol0335.010335.0c ==

2. Hng s Henry: ci = piKH

Lbar/mol0335.010335.0KH ==

3. C(CO2) trong nc ma: 2COc =3.810

-4 0.0335 = 1.2710-5 mol/L 4. Nng ca axit cacbonic ho tan trong nc ma:

CO2 + H2O H2CO3 3

2

32 1067.1]CO[]COH[ =

[H2CO3] = 1.6710-3[CO2] = 2.1310-8 mol/L 5. pH ca nc ma vo thi im ny:

[H2CO3]* = [CO2] + [H2CO3] = 1.2810-5 mol/L

H2CO3 + H2O HCO3- + H3O+ 732

331S 1045.4]COH[

]OH[]HCO[K +

==

[H3O+] = [HCO3-] = x 1S52

K10275.1

x = x = 2.38210-6 mol/L pH = -log x = 5,62

6. Ga tr pH ca nc ma vo nm 1960: [CO2] = 3.210-4 bar0.0335 mol/barL = 1.0710-5 mol/L

[H2CO3] = 1.6710-3[CO2] = 1.7910-8 mol/L [H2CO3]* = [CO2] + [H2CO3] = 1.0710-5 mol/L L/mol1019.2K*]COH[]OH[ 61S323 + == pH = -log [H3O+] = 5,66 7. tan ca vi trong nc ct:

[Ca2+][CO32-] = KL = 4.710-9

[Ca2+] = 9107.4 = 6.85610-5 mol/L S (CaCO3) = 6.8610-5 mol/L 8. Cc ion cha xc nh c nng :

[Ca2+], [H3O+], [OH-], [CO32-], [HCO3-] 9. Cc phng trnh cn thit:

(I) [Ca2+][CO32-] = KL(II) [H3O+][OH-] = KW

(III) 1S32

33 K*]COH[]OH[]HCO[ =

+

(IV) 2S3

323 K

]HCO[]OH[]CO[ =

+

a. 2[Ca2+] + [H3O+] = [OH-] + [HCO3-] + 2[CO32-] 10. Nhn c t phng trnh ca H3O+:

]OH[*]COH[K]HCO[

3

321S3 + =

23

321S2S

3

32S23 ]OH[

*]COH[KK]OH[

]HCO[K]CO[ ++

==

*]COH[KK]OH[K

]CO[K]Ca[

321S2S

23L

23

L2

==

+

+

]OH[K]OH[3

W+

= Thay vo (V):

2*]COH[KK

]OH[K321S2S

23L

+ + [H3O+] = ]OH[

K3

W+ + ]OH[

*]COH[K3

321S+

+ 2 23

321S2S

]OH[*]COH[KK

+

Chuyn v:

*]COH[KK]OH[K2321S2S

43L

+ +[H3O+]3 (KW+KS1[H2CO3]*)[H3O+]2KS1KS2[ H2CO3]* = 0

11. tan ca vi: S (CaCO3) = 4

321S2S

23L2 1017.5

*]COH[KK]OH[K]Ca[

++ =

= mol/L OLYMPIC HA HC BUNGARI 1998:

Bc clorua d dng ho tan trong dung dch amoniac trong nc v to ion phc: AgCl(r) + 2NH3 [Ag(NH3)2]+ + Cl-.

a) Mt lt dung dch amoniac 1M ho tan c bao nhiu gam AgCl? Bit: AgCl(r) Ag+ + Cl- T = 1,8.10-10. [Ag(NH3)2]+ Ag+ + 2NH3 K = 1,7.10-7.

b) Xc nh tch s tan T ca AgBr. Bit rng 0,54g AgBr c th tan c trong dung dch amoniac 1M.

BI GII:a) Ta c: [ ][ ][ ][ ][ ] 10

7

23

23

10.8,1

10.7,1)(

+

+

+

====

ClAgTNHAg

NHAgK

V [Ag+]

BI GII: CH3COOH CH3COO + H- +

[ ][ ][ ] 53

3 10.8,1==+

COOHCHHCOOCH

K a

CH3COONa = CH3COO- + Na+CH3COOH + NaOH = CH3COONa + H2O i vi dung dch axit axetic (tinh khit) ban u: [CH3COO-] = [H+]; [CH3COOH]1 = Caxit 0,1M [H+] = (0,1Ka)1/2 = 1,34.10-3M

a) Hn hp axit yu v mui ca n l dung dch m nn: [ ][ ] 74,3=+= OAcHOAcpKpH a b) Khi thm baz mnh nng Cb th Cmui = Cmui + Cb; Caxit = Caxit - Cb pH tng mt n v tng ng vi [H+] gim 10 ln: [H+]2/[H+]3 = [Caxit.(Cmui + Cb)]/[Cmui.(Caxit Cb)] [H+]3 = 1,8.10-5M; Cb = 0,045M mNaOH = 1,8g c) [CH3COOH]1 = ([H+]1)2/Ka 0,1M [CH3COOH]2 = [H+].Cmui/Ka 0,1M hoc chnh xc hn

[CH3COOH]2 = Caxit - [H+]2 = 0,0986M [CH3COOH]3 = [H+].(Cmui + Cb)/Ka = 0,055M [CH3COOH]2/[CH3COOH]1 1 [CH3COOH]3/[CH3COOH]1 0,55

OLYMPIC HA HC ITALY 1999: Phi iu ch mt dung dch m (pH = pKa) t mt dung dch axit n chc. Phi thm vo dung

dch ny mt lng cht theo s mol l: a) Bng (s mol ca baz lin hp) b) Gp i (s mol ca baz lin hp) c) Khng (s mol ca baz lin hp). d) Bng (s mol ca baz mnh)

BI GII: Cu a. OLYMPIC HA HC ITALY 1999:

Trong phn ng cn bng: HCN + H2O = H3O+ + CN-Nhng phn t no l axit theo nh ngha ca Bronsted v Lowry:

a) HCN; CN- b) H2O; H3O+ c) HCN; H2O d) HCN; H3O+

BI GII: Cu d OLYMPIC HA HC ITALY 1999:

iu ch dung dch H2SO4 0,12M bng cch pha long H2SO4 c (95%. d = 1,84g/mL), c th pha long vi nc.

a) 5,00mL axit thnh 500mL. b) 11,00mL axit thnh 1000mL.

c) 15,00mL axit thnh 2000mL. d) 7,00mL axit thnh 1000mL.


Trong phn ng: NH3 + HCl = NH4+ + Cl- th NH3 l:

a) Axit Arrhenius. b) Baz Bronsted. c) Baz Arrhenius. d) Cht trung tnh.


Nng ion Na+ trong dung dch do 19,0g Na2CO3 tan trong nc to thnh 870mL dung dch l:

a) 0,206M b) 0,312M c) 0,412M d) 0,103M

BI GII: Cu c OLYMPIC HA HC ITALY 1999:

Baz lin hp ca NH3 khi phn ng vi axit l: a) NH3-. b) NH2-. c) NH4+. d) NH3+.

BI GII: Cu b OLYMPIC HA HC ITALY 1999:

Trong s cc axit sau y, cht no to c baz lin hp mjanh nht khi n phn ng nh mt axit?

a) H2SO4 b) H3PO4. c) H2O d) CH3COOH


Nu trn hai dung dch (trong nc) m mt cha NH3 (20mL; 0,5M) cn dung dch kia cha HCl (20mL; 0,5M) th pH ca dung dch to thnh s l:

a) 7 b) 1 c) 10 d) 5


Cht in ly lng tnh l nhng cht m trong dung dch: a) C th phn ng nh cht oxy ho hoc cht kh.

b) C th phn ng nh axit hoc baz. c) C th phn ng theo kiu ng ly v d ly. d) Th hin l mt phn t c mt phn a nc v mt phn k nc.


Trong chc dung dch HCl sau y dung dch no c hn? a) HCl 10-2M. b) HCl 3,6% c) HCl 10-2m d) HCl 3,7% m/V


Lng H2SO4 trong mt dung dch nc (2000mL; 27,27%; d = 1,20g.cm-3) l: a) 6,00 mol b) 4,82 mol c) 6,79 mol d) 5,20 mol


chun CH3COOH bng NaOH th trong cc cht ch th sau y th cht no tt nht? a) Metyl da cam pKa = 3,7 b) Metyl pKa = 5,1 c) Bromthymol xanh pKa = 7,0 d) Phenolphtalein pKa = 9,4


Trong s cc mui sau y th mui no l axit Bronsted? a) NaHSO4. b) Na3PO4. c) NaCN. d) Na2S.

BI GII: Cu a OLYMPIC HA HC ITALY 1999:

Cht phi thm vo dung dch nc lm thay i pH t 12 thnh 10 l: a) Nc ct. b) Natri hydroxit. c) Hidro clorua. d) Natri axetat.

BI GII: Cu c OLYMPIC HA HC C 1999 (Vng 3):

Ngi ta c th xc nh amoniac bng phng php quang k t phn ng vi phenol vi s c mt ca hipoclorit.

OH

+ NH3 N OO

Xanh lam: max = 625nm Trong 7,56mg mt mu th mioglobin ca b c, ngi ta chuyn ha nit c trong thnh

amoniac, sau mu th c pha long thnh 10,0mL. Sau ngi ta cho 10,0mL dung dch vo mt bnh nh mc 500mL, cho thm vo 5mL dung dch phenol v 2mL dung dch hipoclorit , ri pha thnh 50,0mL dung dch c ng yn 30 pht. Sau ngi ta o tt ti 625nm trong cuvet 1,00cm.

Bn cnh , ngi ta pha ch mt dung dch chun gm 0,0154g NH4Cl trong 1L nc. Ngi ta cho 5,00mL dung dch vo mt bnh nh mc 50,0mL v sau vic phn tch c tin hnh nh m t trn.

Ngoi ra ngi ta cn o mt mu khng (mu m) vi nc nguyn cht trong ng cuvet: Mu tt ti 625nm Khng 0,132 i chng 0,278 Cha bit 0,711

a) Hy tnh h s tt mol (mt quang mol ca sn phm mu xanh). b) Hy tnh phn khi lng (bng phn trm) ca nit trong mioglobin.

BI GII: a) i vi mu th i chng th trong tng s tt c 0,132 c quy nh cho nc nguyn

cht, phn cn li 0,278 0,132 = 0,146 l do hp cht ca nit gy ra. Ta bit A = .l.C vi l = 1,00cm Tnh C trong mu i chng: n(NH4Cl) trong 5,0mL = (0,0154/53,50).0,005 = 1,439.10-6M. Khi pha ln 50mL ta c dung dch c nng C = 2,88.10-5M. 0,146 = .1,00.2,88.10-5 = 5072L/cm.

b) Vi ngi ta c th tnh c nng trong dung dch cha bit. C y cng phi ch n mu khng.

OLYMPIC HA HC C 1999 (Vng 3): C mt s thuyt v nh ngha khc nhau v axit v baz. Mt trong s cc nh ngha c

lin quan n s t phn li ca dung mi: 2HB H2B+ + B-. Theo l thuyt ny th cht no lm tng phn cation ca dung mi (H2B) l mt axit v cht no

lm gim phn (hoc tng phn anion) l mt baz. Chng hn nc t phn ly: 2H2O H3O+ + OH-Axit l nhng cht no lm tng [H3O+] v baz l nhng cht no lm tng [OH-] Trong etanol th: 2C2H5OH C2H5OH2+ + C2H5O-Axit l nhng cht no lm tng nng [C2H5OH2+] v baz l nhng cht no lm

tng[C2H5O-]

Khi phn ng trung ho l phn ng trong mt axit phn ng vi mt baz to thnh mt mui v mt dung mi.

Theo l thuyt ny th pH = -lg[H2B+] (L thuyt ny cng c th p dng c cho cc dung mi phi proton).

a) Hy n c mt v d v mt axit v mt baz trong dung mi amoniac lng. b) Tch s ion ca amoniac l 1,0.10-29 (mol/L)2. Hi amoniac lng nguyn cht c pH no? c) Nc l mt axit hay l baz trong amoniac lng?. Gii thch. d) Hy l gii ti sao CH3COOH l mt axit tng amoniac lng. N mnh hn hay yu hn tng

dung dch nc. e) Mt hp cht l mt axit mnh trong nc c th l mt baz yu trong amoniac lng hay

khng? Nu c th hy cho v d cn nu khng th hy gii thch. f) Hy ch ra rng NaOH l mt mui trong NH3 lng. Hy ch v d v mt phn ng m n

c to ra trong mi trng amoniac lng. g) C hp cht no l mt baz trong nc m li l mt axit trong NH3 lng khng? Nu c th

hy cho v d cn nu khng th hy gii thch. h) Hy t b NH3. Liu c mt dung mi no m n l mt baz khng?. Nu c th hy cho v

d cn nu khng th hy gii thch. i) Trong CCl4 c axit hay baz khng? Nu c th hy cho v d cn nu khng th hy gii thch.

Lu : Tt c cc khi nim dng trong bi tp u lin h vi l thuyt v cc h dung mi c gii thch trn. BI GII:

a) Trong amoniac lng din ra qa trnh t phn ly nh sau: 2NH3 NH4+ + NH2-. Nh vy axit l cc cht lm tng nng NH4+ cn baz l cc cht lm tng nng NH2-. V d v mt axit: NH4Cl V d v mt baz: KNH2.

b) Theo nh ngha u bi th pH = -lg[NH4+] Ta bit: Kamoniac = [NH4+][NH2-] = 1,0.10-29

[NH4+] = 1,0.10-14,5 pH = 14,5 c) Nc phn ng nh l mt axit v n lm tng nng NH4+:

H2O + NH3 NH4+ + OH-. d) CH3COOH + NH3 NH4+ + CH3COO-

Axit axetic lm tng nng NH4+ nn n l mt axit V rng NH3 l cht cho cp in t mnh hn nc cho nn s ho tan axit axetic tng amoniac

th ln hn trong nc v nh vy th tnh axit mnh hn. e) NH3 l phn t cho cp in t mjanh hn H2O (NH4+ hnh thnh d hn H3O+). Nh vy th s

ho tan mi axit trong amoniac u mnh hn nc. V vy mt axit trong h nc khng th l mt baz trong h amoniac.

f) Ch cn chng minh rng NaOH c hnh thnh trong mt phn ng trung ho l : NaNH2 + H2O NH3 + NaOH

(axit + baz = dung mi + mui) g) Mt hp cht nh vy cn phi to thnh OH- trong nc v NH4+ trong amoniac. C th

l mt hp cht c hai chc nng, vi mt chc nng baz yu hn amoniac trong nc v mt nhm axit lin hp vi chc nng baz trong dung dch nc. Mt v d y l hydroxilamin NH2OH trong nc s hnh thnh cn bng:

H2O + H2NOH H3NOH+ + OH-. trong amoniac th cn bng chim u th l: H2NOH + NH3 NH4+ + H2NO-. (Gii thch b sung nhng trong phn bi tp khng yu cu: Cht cha bit cn cha t nht l

mt H c kh nng tch ra thnh proton, tc l nn vit tt HnX) trong nc th HnX tc dng nh l mt baz: HnX + H2O Hn+1X+ + OH- (a) V trong amoniac nh l mt axit: HnX + NH3 NH4+ + Hn-1X-. (b) cho (b) xy ra th nhm tc ng nh l baz trong nc phi l tc nhn nhn proton km

hn NH3. iu c ngha l nhm phi l mt baz yu hn NH3. Qa thc pKb(NH3) = 4,75 v pKb(H2NOH) = 8,2.

cho (a) xy ra th nhm tc ng nh l axit trong NH3 phi c tc dng cho proton yu hn axit Hn+1X+, Hn+1X+ l axit lin hp ca nhm HnX, nhm ny tc dng nh l mt baz. Qa thc pKa(NH3+OH) = 5,4 v pKa(NH2OH) = 13,2.

h) C. V d nh axit sunfuric: 2H2SO4 H3SO4+ + HSO4-

H2O + H2SO4 H3O+ + HSO4-. i) Khng, do CCl4 khng phn li.

OLYMPIC HA HC C 1999 (Vng 4): Mt dung dch ceri (IV) sunfat cn c chun ha, Cho cc dung dch v cc cht sau y: Natri oxalat rn, dung dch kali pemanganat v dung dch st (II) sunfat, c hai u khng bit

nng . Ngi ta tin hnh ba ln chun trong dung dch axit (mi ln u i vi mt lng d axit

sunfuric) v thu c nhng kt qa sau y: + 0,2228g natri oxalat dng ht 28,74cm3 dung dch kali pemanganat. + 25,00cm3 dung dch st (II) sunfat dng ht 24,03cm3 dung dch kali pemanganat. + 25,00cm3 dung dch st (II) sunfat dng ht 22,17cm3 dung dch ceri (IV) sunfat.

1. Vit cc phng trnh phn ng ca ba ln chun . 2. Hy tnh nng ca dung dch ceri (IV) sunfat.

Ngi ta p dng cc th in cc tiu chun sau y: Fe3+ + e = Fe2+ Eo = 0,77V Ce4+ + e = Ce3+ Eo = 1,61V

3. Hy tnh KC ca phn ng: Fe2+ + Ce4+ = Fe3+ + Ce3+. (i vi phn cn li ca bi tp cn gi thit cc iu kin l tiu chun)

4. Hy tnh t s: [ ][ ]++

2

3

FeFe ti im tng ng.

5. Hy tnh th ca dung dch ti im tng ng. Nu nh ngi ta s dng mt cht ch th oxi ha - kh (In) vi Eo = th ca dung dch ti im

tng ng nhn bit im kt thc ca vic chun th s khng c vn g v chnh xc ca vic nhn bit im kt thc.

Nhng i vi cht ch th sau y th: InOx + 2e = In2-kh Eo = 0,80V

S chuyn mu s th hin r khi: [ ][ ] 1102 =khOxInIn

6. Hy tnh [ ][ ]++

2

3

FeFe ti im chuyn mu ca cht ch th ny v cho bit sai s phn trm trong

ln chun tin hnh. BI GII: 1. 2MnO4- + 5C2O42- + 16H3O+ = 2Mn2+ + 10CO2 + 24H2O

5Fe2+ + MnO4- + 8H3O+ = 5Fe3+ + Mn2+ + 12H2O Fe2+ + Ce4+ = Fe3+ + Ce3+.

2. Chun 1: 0,2228g Na2C2O4 tng ng 1,66.10-3 mol C2O42-. (2/5).1,66.10-3 = [MnO4-].V(MnO4-) [MnO4-] = 0,0023M Chun 2: [MnO4-].V(MnO4-) = (1/5)[Fe2+]V(Fe2+) [Fe2+] = 0,111M Chun 3: [Ce4+] = [Fe2+].V(Fe2+)/V(Ce4+) = 0,125M

3. Ta c:

14o

/Fe/ 10.61,1).(

lg2334 == ++++ K

RTFEE

K Feo

CeCe

4. Ti im tng ng th lng cht cho vo n(Ce4+) = no(Fe2+). Vi mi ion Ce3+ mi hnh thnh th cng hnh thnh mt ion Fe3+, tc l [Ce3+] = [Fe3+] v c [Ce4+] = [Fe3+] Ta c: [ ][ ][ ][ ] [ ][ ] [ ][ ] 72

3

22

23

24

33

10.27,1; === ++

+

+++++

FeFe

FeFeK

FeCeFeCeK CC

5. a ga tr mi tm c vo phng trnh Nernst i vi th ca st ngi ta thu c: E = 1,19V (Cng tng t nh vy ngi ta c th a ga tr [Ce4+]/[Ce3+] = (1,27.10-7)-1 vo phng trnh Nernst i vi th ca ceri).

6. Th ca dung dch ti im chuyn mu l: E = 0,80 + RT/2F(ln10) = 0,83V a ga tr ny vo phng trnh Nernst i vi st: [ ][ ] [ ][ ] 12,10ln77,083,0 2

3

2

3

=+= ++

+

+

FeFe

FeFe

FRT

Nh vy sai s s l: (11,2)-1.100% = 8,95% OLYMPIC HA HC C 2000:

HCN l mt axit yu (Ka = 6,2.10-10). NH3 l mt baz yu (Kb = 1,8.10-5). Mt dung dch NH4CN 1,0M s c tnh cht:

a) Axit mnh. b) Axit yu. c) Trung tnh. d) Baz yu. e) Baz mnh.

BI GII: Cu d OLYMPIC HA HC C 2000:

20,00mL mu dung dch Ba(OH)2 c chun bng 0,245M. Nu s dng 27,15mL HCl th nng mol ca Ba(OH)2 lc ny s l bao nhiu?

a) 0,166M b) 0,180M c) 0,333M d) 0,666M e) 1,136M

BI GII: Cu a OLYMPIC HA HC C 2000:

Tch s ion ca nc 45oC l 4,0.10-14. Vy pH ca nc tinh khit thi im ny l bao nhiu?

a) 6,7 b) 7,0 c) 7,3 d) 8,5 e) 13,4

BI GII: Cu a OLYMPIC HA HC C 2000:

Tch s tan ca mt s mui sunfat c cho bng sau: Mui T 1 CaSO4 9.10-6

2 SrSO4 3.10-7

3 PbSO4 2.10-8

4 BaSO4 1.10-10

Khi cho dung dch Na2SO4 0,0001M vo dung dch cc mui tan ca cc cation trn th mui no s kt ta.

a) 1, 2 v 3. b) 1 v 2 c) 1 v 3 d) 2 v 4 e) ch 4

BI GII: Cu e OLYMPIC HA HC C 2000:

Khi trn dung dch NaOH 0,5M vi mt lng bng nhau ca dung dch no sau y th s xy ra s gim pH.

1) H2O. 2) 0,25M Na2CO3. 3) 0,5M HCl. 4) 0,6M KOH

a) 1, 2 v 3. b) 1 v 2 c) 1 v 3 d) 2 v 4 e) ch 4


Dung dch KOH 0,025M c pH bng bao nhiu? a) 1,60 b) 3,69 c) 7,00 d) 10,31 e) 12,40


Cht no l axit lin hp ca HPO42-? a) H3PO4(aq). b) H2PO4-(aq). c) H3O+(aq) d) H+(aq). e) PO43-(aq)

BI GII: Cu b OLYMPIC HA HC C 2001:

Dung dch axit yu HA 0,075M c [H+] bng bao nhiu nu Ka(HA) = 4,8.10-8a) 6,

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