Tuyen tap BT Hoa Phan Tich

7/30/2019 Tuyen tap BT Hoa Phan Tich

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I. OLYMPIC HA HC VIT NAMOLYMPIC 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

-2

M v Fe

3+

10

-4

M. Thm dung dch SCN

-

vo to kt ta AgCN (coith tch khng i). Xc nh nng Ag+ cn li trong dung dch khi xut hin mu . BitTAgSCN = 10

-12

3. Thm 20cm3dung dch AgNO3 5.10-2

M vo 10cm3dung dch NaCl khng bit nng . Lng

d Ag+c chun bng dung dch KSCN vi s c mt ca Fe3+. im tng ng (khibt u xut hin mu ) c quan st thy khi thm 6cm3dung dch KSCN 10-1M. Tnh nng ca dung dch NaCl.

BI GII:

1. Fe3+

+ SCN- Fe(SCN)

2+

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

Ta c: 2

523

5

10)1010(

10

Fe

[Fe3+

] = 10-5

M Co = 2.10-5

M2. Khi xut hin mu th: [Fe(SCN)2+] = 10-5M. Vy nng Fe3+cn li l: 9.10-5M

Ta c:

MAgMSCN

SCN

103

2

5

5

10.1,910.1,1

1010.9

10

3 n(Ag+) = n(AgCl) + n(AgSCN)

20.10-3

.5.10-2

= 10.10-3

C + 6.10-3

.10-1

C = 4.10-2

M

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)2vo 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 ccphng trnh

phn ng (nu c).BI GII:

a) Tnh pH ca dung dch Na2S2 Na+ + S2-

0,01 0,01KI K+ + I-

0,06 0,06Na2SO4 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)


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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

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

0,09 0,010,08Pb2+ + SO4

2- PbSO4 (Ks-1) = 107,8.

0,08 0,050,03Pb2+ + 2 I- PbI2 (Ks

-1) = 107,6.0,03 0,06

Thnh phn hn hp: A : PbS , PbSO4 , PbI2

Dung dch B : K+

0,06M Na+

0,12MNgoi ra cn c cc ion Pb2+ ; SO42- ; S2- do kt ta tan ra.

tan ca

Bi v tan ca PbI2l ln nht nn cn bng ch yu trong dung dch l cn bng tan ca PbI2.PbI2 Pb

2+ + 2I- KsDo [Pb2+] = 10-47= 2 x 10-3M v [I-] = 4.10-3M.

107,8[SO4

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


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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 phnng (nu c).2.Dung dch bo ha H2S c nng 0,100 M. Hng s axit ca H2S: K1 = 1,0 x 10

-7v

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

Cho: TMnS= 2,5 x 10-10 ; TCoS= 4,0 x 10

21 ; TAg2S = 6,3 x 10-50

BI 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 AgNO3c kt ta mu nuAg+ + OHAgOH; (hoc 2Ag+ + 2OH Ag2O + H2O)

-Dung dch Mg(NO3)2c kt ta trng, keoMg2+ + 2OHMg(OH)2

- Cc dung dch AlCl3, Pb(NO3)2, Zn(NO3)2u c chung hin tng to ra kt ta trng,tan trong dung dch KOH (d).

Al3+ + 3OHAl(OH)3; Al(OH)3+ OHAlO2

+ 2H2OPb2+ + 2OH Pb(OH)2; Pb(OH)2+ OH

PbO2+ 2H2O

Zn2+ + 2OH Zn(OH)2; Zn(OH)2+ OHZnO2

+ 2H2O-Dung dch NaCl khng c hin tng g.

-Dng dung dch AgNO3nhn ra dung dch AlCl3do to ra kt ta trngAg+ + Cl AgCl -Dng dung dch NaCl nhn ra dung dch Pb(NO3)2do to ra kt ta trng

Pb2+ + 2 Cl PbCl2- cn li l dung dch Zn(NO3)2.2.a) Tnh nng ion S2trong 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 =

22

2

H S

H S

= Kl. K2

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

2

H S

H

= 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


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[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 K2CrO4vo dung dch A cho n d. C hin tng g xy ra?

b) Thm 50,0 ml K2CrO40,270 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 cationtrong 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

pKaca 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

4

24

Ba

)BaCrO(s

CrO C

KC (1)

bt u c Ag2CrO4:

Ag2

)CrOAg(s

CrO C

KC 42

24

(2)

tnh tch s tan Kscn t hp cn bng :BaCrO4 Ba

2+ + CrO42- Ks1

H2O H+ + OH- Kw

CrO42- + H+ HCrO4

- Ka-1

BaCrO4 + H2O Ba2+ + HCrO4

- + OH-

C K= Ks1 . Kw . Ka-1

Suy ra 93,914

50,643,17

w

a1s 10

10

10.10

K

K.KK

Ag2CrO4 2 Ag+ + CrO4

2- Ks2

H2O H+ + OH- Kw

CrO42- + H+ HCrO4

- Ka-1

Ag2CrO4 + H2O 2 Ag+ + HCrO4

- + OH

C K = 10-19,50


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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 Al

3+ 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 5.10

33; 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 bngK 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 ktta 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 CO3

2trong dung dch Na2CO3c 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 CaSO4vo 1 ml dung dch A.

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

HCO3 H+ + CO3

2 ; K = 1010,33 tan ca CO2trong nc bng 3,0.10

2 M.Tch s tan ca CaSO4bng 10

5,04; ca CaCO3bng 108,35

BI 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

K

KS;

NH3

Al(OH)3

NH3

Mg(OH)2

K

KS;

a1

a2


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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 cnbng sau c thit lp trong dung dch nc ca Cr(VI)

HCrO4- + H2O CrO4

2- + H3O+. pK1 = 6,50

2HCrO4- Cr2O72- + H2O pK2 = -1,361. 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- 2CrO4

2- + H2O

2. Tch s tan ca BaCrO4 l T = 1,2.10-10

. Ba2Cr2O7tan 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 cakali dicromat?

a) KOHb) HCl

c) BaCl2d) 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 0,010Mb) K2Cr2O7 0,010Mc) K2Cr2O7 0,010M + CH3COOH 0,100M

4. Hy tnh nng ti cn bng ca cc ion sau trong dung dch K2Cr2O7 0,010M + CH3COOH0,100M.

a) CrO42-.

b) Cr2O72-

.

BI GII:1) a) Hng s cn bng:K = [HCrO4

-][OH

-]/[CrO4

2-] = [H

+][OH

-]/([H

+][CrO4

2-]/[HCrO4

-]) = Kw/K1 = 3,2.10

-8

b) Hng s cn bng:K = ([CrO4

2-][H+]/[HCrO4-])2/([HCrO4

-]2/[Cr2O72-])/([H+][OH-])2 = 4,4.1013.

2) a) phib) Tric) BaCl2di 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 longdung 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 bngphi chuyn dch v bn phi.3) a) CrO4

2-+ H2O = HCrO4

-+ OH

-K = 3,16.10

-8.

CCr = [CrO42-

] + [HCrO4-] + 2[Cr2O7

2-] [CrO4

2-]

[HCrO4-] [OH

-]

Nh vy [OH-]2/CCr = K [OH-] = 1,78.10-5M nn [H+] = 5,65.10-10.

Vy pH = 9,25b) Cr2O7

2-+ H2O = 2HCrO4

-K = 1/K2 = 4,37.10

-2


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HCrO4-= H

++ CrO4

2-K = K1 = 3,16.10

-7.

CCr = 2,0.10-2

M = [CrO42-

] + [HCrO4-] + 2[Cr2O7

2-] [HCrO4

-] + 2[Cr2O7

2-]

[H+] [CrO4

2-] = x = (K1[HCrO4

-])

1/2

K2 = [Cr2O72-]/[HCrO4

-] = (CCrx)/2x2

iu ny dn n phng trnh: 2K2x2

+ xCCr = 0

Gii phng trnh trn ta thu c: x = 1,27.10-2

M [H+

] = 6,33.10-5

MVy 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) chothy nh hng ca K2Cr2O7trn pH c th an tm b qua c.

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

a) [HCrO4-] = 1,3.10

-2M (3b) [CrO4

2-] = K1[HCrO4

-]/[H

+] = 3,0.10

-6M

b) CCr = [CrO42-

] + [HCrO4-] + 2[Cr2O7

2-] [Cr2O7

2-] = 3,7.10

-3M

hoc [Cr2O72-

] = K2[HCrO4-] = 3,9.10

-3M

Cch 2:

a) [CrO42-

] = x; [HCrO4-

] = x[H+

]/K1[Cr2O7

2-] = K2[HCrO4

-] = x

2K2[H

+]

2/K1

2.

CCr = [CrO42-

] + [HCrO4-] + 2[Cr2O7

2-] = 2K2[H

+]2/K1

2x

2+ (1 + [H

+]/K1)x

K1 = 3,16.10-7; K2 = 22,9; [H

+] = 1,34.10-3.

8,24.108x

2+ 4,24.10

3x2,0.10-2 = 0

x = 3,0.10-6

Mb) [Cr2O7

2-] = K2 [HCrO4-] = K2[H

+]2/K12[CrO4

2-] = 3,7.10-3M

OLYMPIC HA HC QUC T LN TH 28:Cc phng php o hiuth 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 xcnh 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 chamt 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)6

4-]II v [Fe3+]I = [Fe(CN)6

3-]II. Th ca in cc platin nhng trong dung dch I l0,652V (so vi incc hydro tiu chun), trong khi th ca in cc platin nhng trong dung dch II l0,242V (so vi in cc hydro tiu chun). Phn trm truyn x ca dung dch II o c so vi dungdch I ti 420nm bng 10,7% (chiu di ng truyn quang l = 5,02mm). Gi thit rng phc[Fe(CN)6

4-] 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 thpphn ca phng trnh Nernst bng 0,0590 (v ghi trc logarit t nhin l 0,0256). Gi thit rng tt ccc 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)6

3-)/(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


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

BI GII:1) Phng trnh Nernst:

a) EI = Eo(Fe

3+/Fe

2+) + 0,0590lg([Fe

3+]/[Fe

2+]

b) EII = E

o

(Fe(CN)63-

/Fe(CN)64-

) + 0,0590lg([Fe(CN)63-

]/[Fe(CN)64-

])2) Volt (V)3) EII = E

o(Fe(CN)6

3-/Fe(CN)6

4-) + 0,0590lg([Fe(CN)6

3-]/[Fe(CN)6

4-]

= Eo(Fe

3+/Fe

2+) + 0,0590lg(1/2) + 0,0590lg([CN

-]

6/[CN

-]6) + 0,0590lg([Fe(CN)6

3-]/[Fe(CN)6

4-])

= 0,242

Trong 1 v 2ln lt l cc hng s bn vng ca [Fe(CN)64-] v [Fe(CN)6

3-].

[Fe(CN)63-

]/[Fe(CN)64-

] = Fe3+/Fe

2+nn E = EIIEI = 0,059lg(1/2) = 8,90.10

6.

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

5) a) Dng nh lut Bouger LambertBeerA = .l.C = .l.[Fe(CN)6

3-] = 0,971

[Fe(CN)63-

] = [Fe3+

] = 1,76.10-3

Mb) 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.Trs 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 0,648615 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.


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S mL HCl thm vo pH1,00 10,3310,00 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 ccho trong bng. Kaca CH3COOH l 1,75.10

-5.

Dung dch I Dung dch II Dung dch IIIN ng ton ph n caHIn

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

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 trnln 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 A

2-u b proton ha thnh HA-.

Do s mmol A2-c mt trong dung dch lc u = 3,00mmolTi 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 0,3000,300 = 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 cnbng ca cht ch th tu thuc vo [H+] ca dung dch

[H+

]III = (Ka.C)1/2

= 4,18.10-3

M

T HIn H+ + In- ta c:

HInInH

Ka

100,0

H

K

HIn

In In (1)

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


14/58

14

hp th ca dung dch III = 027,0300,0.10.00,1

10.091,05

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 gmCH3COO

-0,05M/CH3COOH 0,45M

5

3

3 10.75,15

COOCH

COOHCHKH a

V vy:

65,210.75,15

105

38,3

H

K

HIn

In 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 qatrnh vn chuyn st (III) trong cth ngi. Mi phn t transferin c th vn chuyn hai ion Fe3+ theophn ng:

FeIII

+ Tf (FeIII

)Tf K1 = 4,7.1020

M-1

.

FeIII + (FeIII)Tf (FeIII)2Tf K2 = 2,4.10

19M-1. phn t (FeIII)2Tf 2 ion Fe

3+lien kt tng t nhau nhng khng cng mt pha v 2 phn tst monotransferin (FeIII)Tf c th c biu th bng {FeIII.Tf} v {Tf.FeIII}. Bit K =[{Tf.Fe

III}].[{Fe

III.Tf}] = 5,9.

1. Tnh ga tr ca K1 = [{FeIII.Tf}].[FeIII]-1. [Tf]-1 v K1=[{Tf.Fe

III}].[FeIII]-1. [Tf]-1.

2. Tnh gi tr ca K2 = [(FeIII)2Tf].[Fe

III]-1.[{FeIII.Tf}]-1 v K2=[(FeIII)2Tf].[Fe

III]-1.[{Tf.FeIII}]-1.

Lin kt gia st (III) mi pha ca lin kt c bao quanh bi 6 nguyn t nhn t cc ligandkhc nhau. Theo cch ny, 2 nguyn t oxy ca anion CO 32-

phi tr vi kim loi v mi aminoaxit cutrc 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 mtion st (III) trong transferin.BI GII:

1. Nng ca phc monoferric transferin:[(Fe

III)Tf] = [{Fe

III.Tf}] + [{Tf.Fe

III}]

K1+ K1

= K1; K1

.K = K1

12020'

11

"

1

11920

1'

1

10.0,410).68,07,4(

10.8,69,51

10.7,4

1

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

MK

KKKKKKKKK

3. S nguyn t oxy = 2(CO32-

) + 1(Asp(O-)) + 2(2xTyr(O

-)) = 5


15/58

15

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 25

oC l: pKa1 = 2,12; pKa2 = 7,21; pKa3=12,32. Vit cng

thc baz lin hp ca H2PO4-v tnh ga tr Kbca n.

Mt lng nh H3PO4c s dng rng ri to v chua hay v cht cho nhiu thc ung nhcola v bia. Cola c t khi 1,00gmL-1 cha 0,05% H3PO4v 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. H3PO4c s dng lm phn bn trong nng nghip ; 1,00.10

-3M H3PO4c thm vo dungdch huyn ph ct v pH ca dung dch thu c l 7,00. Tnh nng phn mol ca cc loiphotphat 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 thc km dng dung dch nc. trong dung dch ncc ngm c pH = 7,0 ngi ta tm thyc Zn3(PO4)2. Tnh [Zn

2+] v [PO4

3-] trong dung dch bo ha. Bit T ca km photphat l9,1.10

-35.

BI GII:

1. Baz lin hp ca dihidro photphat (H2PO4-) l monohydrophotphat (HPO4

2-)

H2PO4- + H2O HPO4

2- + H3O+ K2a

HPO42-

+ H2O H2PO4-+ OH

-K2b

2H2O H3O+

+ OH-

Kw.pK2a + pK2b = pKw = 14

pK2b = 6,79

2. C(H3PO4) = 0,0051M

H3PO4 + H2O H2PO4- + H3O

+0,0051x x x

pKa1= 2,12. Vy Ka = 7,59.10-3

.Ta c:

46,2

10.49,3

10.59,70051,0

3

3

32

43

342

pH

OHx

x

x

POH

OHPOH

3. t:

C

Xf

C

HXf

CXHf

C

XHfo

3

3

2

2

21

3


16/58

16

k hu cc phn s nng ca cc loi photphat khc nhau; C l tng nng ban u caH3X (X = PO4):

OHf

f

HX

OHXK

OHf

f

XH

OHHXK

OHf

f

XH

OHXHK

ffff

a

a

o

a

o

3

2

3

2

3

3

3

3

1

2

2

3

2

2

3

1

3

32

1

321 1

Cc phng trnh trn y dn n:

D

KKKf

D

OHKKf

D

OHKf

D

OHf

aaa

aa

a

p

321

3

3122

2

31

1

3

3

.

Vi D = Ka1.Ka2.Ka3 + Ka1.Ka2.[H3O+] + Ka1[H3O

+] + [H3O+]3.

T cc ga tr Ka1; Ka2; Ka3v pH ta c c cc kt qa sau:Ka1 = 7,59.10

-3; Ka2 = 6,17.10

-8; Ka3 = 4,79.10

-13v [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 [PO4

3-] = f3.2S

f3c th c xc nh t cc quan h cu 3i vi pH = 7 th f3 = 1,83.10

-6.

T = [Zn2+]3[PO43-]2 = (3S)3.(f3.2S)

2 = 9,1.10-33


17/58

17

S = 3,0.10-5

M

[Zn2+

] = 9.10-5

M[PO4

3-] = 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 tnhkim ca nc l HCO3-, CO3

2- v OH-. ga tr pH di 7 th H+trong nc lm gim tnh kim canc. Chnh v vy, phng trnh nu kim ca nc khi c mt cc ion HCO3

-, CO3

2-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- CO3

2-+ 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 CO2ho 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 CO2trong kh quyn c th lin quan ti tnh kim ca ncdo n nm cn bng vi hm

lng CO2tan trong nc. Tnh nng ca CO2(mol/L) trn nc tinh khit nm cn bng vikhng kh khng b nhim p sut 1,01.105Pa v 298K cha 0,0360% (v s mol) CO2. Gis 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 CO2trong nc c th c nh ngha bng biu thc S=[CO2(aq)] + [H2CO3] +[HCO

3

-] + [CO

3

2-]. tan ca kh CO2trong 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 CaCO3ho 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 CaCO3ho tan trong nc nm cn bng vi CO2 trong khquyn.

Nu khng tnh c th ta c th gi s rng nng ca Ca2+

(aq)l 40,1mg/L tnh ton.7. Tnh kim cadung dch trn.8. mt h nc ngm cha CaCO3ho tan th nc c lng CO2rt cao. Nng ca Ca

2+-

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

2+ trn.

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


18/58

18

Ka1 = [HCO3-][H

+]/[H2CO3] = 2,23.10

-4 [HCO3

-]/[H2CO3] = 2,23.10

3

Ka2 = [CO32-

][H+]/[HCO3

-] = 4,69.10

-11 [CO3

2-]/[HCO3

-] = 4,69.10

-4

[H2CO3] : [HCO3-] : [CO3

2-] = 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-] + [CO3

2-]

= [CO2(aq)] + [HCO3-]

([H2CO3] = [CO2(aq)] . K(H2CO3) = 2,48.10-8

M 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

-7

T cu 6 2 [CO2(aq)]=1,24.10-5M ta tnh c [H+]=[HCO3

-]=2,35.10-6M

Vy tan ca CO2s bng 1,48.10-5

M.

b) S dng [CO2(aq)]=1,11.10-5

M tnh ton: tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO3

2-]

= [CO2(aq)] + [HCO3-]

([H2CO3] = [CO2(aq)] . K(H2CO3) = 2,48.10-8

M v[CO3

2-] = 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

-7

T cu 6 2 [CO2(aq)]=1,11.10-5M ta tnh c [H+]=[HCO3

-]=2,225.10

-6M

Vy tan ca CO2s 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 CO2phi 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) + CO3

2- + H2O 2HCO3- K = K(H2CO3).Ka1/Ka2 = 9,37.10

3

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-

] = 8,54.10-6

M

tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO3

2-]

[CO2(aq)] + [HCO3-] + [CO3

2-] = 1,02.10

-3M

b) S dng [CO2(aq)] = 1,11.10-5

M tnh ton:Trong dung dch NaOH 1,00.10-3M, tan ca CO2phi tng ln do phn ng sau:

(3)CO2(aq) + 2OH- CO3

2- + 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 = 4,43.10

7.

Do K rt ln nn ton b lng OH-u chuyn ht v HCO3-.

[HCO3-] = 1,00.10

-3M


19/58

19

[OH-] = 1,82.10

-6M

[H+] = 5,49.10

-9M

[CO32-] = 8,54.10-6M

tan = [CO2(aq)] + [H2CO3] + [HCO3-] + [CO3

2-]

[CO2(aq)] + [HCO3-] + [CO3

2-] = 1,02.10

-3M

5. Keq = Ksp.K(H2CO3).Ka1/Ka2 = 4,28.10-5

Nu 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[Ca

2+

]T cu 6 5: K = 5,00.10-5 = [Ca2+][HCO3-]2/[CO2(aq)]

= [Ca2+

](2[Ca2+

])2/[CO2(aq)]

T cu 6 2: [CO2(aq)] = 1,11.10-5

M

[Ca2+] = 0,5177.10-3M = 20,75mg/L

c) S dng Keq = 5,00.10-5 v [CO2(aq)] = 1,24.10

-5M tnh ton:Cn bng khi lng: [HCO3

-] = 2[Ca

2+]

T cu 6 5: K = 5,00.10-5 = [Ca2+][HCO3-]

2/[CO2(aq)]

= [Ca2+](2[Ca2+])2/[CO2(aq)]

T cu 6 2: [CO2(aq)] = 1,24.10-5M

[Ca2+

] = 0,5372.10-3

M = 21,53mg/L

d) S dng Keq = 4,28.10-5

v [CO2(aq)] = 1,11.10-5

M 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)] = 1,11.10-5M

[Ca2+] = 0,4916.10-3M = 19,70mg/L

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 Ka2l 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-

]


20/58

20

e) 2,00.10-3M (gi s [Ca2+(aq)] = 40,1mg/L)8.a) S dng Keq = 4,28.10

-5 tnh tonCn bng khi lng: [HCO3

-] = 2[Ca

2+]

[Ca2+

] = 100mg/L = 2,50.10-3

M

Thay vo biu thc Keq = 4,28.10-5

= [Ca

2+

][HCO3-

]

2

/[CO2(aq)]= 4[Ca2+]3/[CO2(aq)]

[CO2(aq)] = 1,46.10-3

M

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[Ca

2+]

[Ca2+

] = 100mg/L = 2,50.10-3

MThay vo biu thc Keq = 5,00.10

-5 = [Ca2+][HCO3-]2/[CO2(aq)]

= 4[Ca2+]3/[CO2(aq)]

[CO2(aq)] = 1,25.10-3

M

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 CO2trong khng kh ln tnh axit canc.

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 phikh 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% CO2trong Ar, chng t cn bng vi khng kh.i) Nc ct.ii) Dung dch HCl 1Miii) Dung dch CH3COONa 1M

d) Gi thit rng khng kh c cha 350ppm CO2(theo th tch), v t cn bng gia CO2 kh vtan (trong nc), hy tnh pH ca mtgit nc ma p sut khng kh. Cc hng s thch hpti 25oC l: kH(CO2) = 3,39.10

-2mol.L-1.atm-1; Kb(HCO3-) = 2,24.10-8; Kb(CO3

2-) = 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) CO3

2-(aq) + H

+(aq) (3)

l ta c thm cn bng:

CO2(aq) + H2O H2CO3(aq)


21/58

21

C th c gii thiu gii thch s tn ti ring bit ca CO2dng ho tan v ca axitcacbonic phn t trong dung dch nc nhng khng bt buc phi dng cn bng ny gii thch phnng 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 vdo ta bt u t CO2(k) v H2O nn dung dch thu c r rng phi c tnh axit.

b) NH3l 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 tngkh nng ho tan ca CO2c trong kh quyn.

Cl2l 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 CO2trong 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 CO2theo chiu thun.Dung dch HCls di cn bng ca CO2theo chiu nghch.Nh vy chiu hng CO2 hoa tan l: iii>i>ii

d) Nng ca CO2trong dung dch nc c tnh bi nh lut Henry:[CO2(aq)] = kH.P(CO2) = 1,187.10

-5M

Ka = Kw/Kb

Ka(CO

2(aq)) = 4,46.10

-7

Ka(HCO3-(aq)) = 4,67.10

-11

Do 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 = 5,64Nay, 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 CO3

2-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 = 3,91OLYMPIC HA HC QUC T LN TH 31:

a) Axit photphoric, H3PO4l mt axit ba chc. Nu chun mt dung dch H3PO40,1000M viNaOH 0,1000M. Hy c lng pH ti cc thi im sau:


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22

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 ngcong chun sau im tng ng th hai?K1 = 7,1.10

-3K2 = 6,2.10

-8K3 = 4,4.10

-13.

b) Mt dung dch cha 530mmol Na2S2O3v mt lng cha xc nh KI. Khi dung dch ny c

chun vi AgNO3th dng c 20,0mmol AgNO3trc khi bt u vn c v AgI ktta. 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

-

15,2

10.1,73

43

42

43

1

pH

MPOH

POH

POHKH

(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+ thmvo s to phc vi S2O3

2-v:

[Ag(S2O3)23-] = 0,100M

s mmol S2O32-

t do = 530 (2.20) = 490mmol.[S2O3

2-] = 2,450M

Nng ion Ag+t do c tnh t Kd

15

14

3

232

22

32

10.0,1

10.0,6)(

Ag

OSAg

OSAgKd

Ag+

+ I- AgI

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

[I-] = 8,5.10

-2M

mmol KI = 17,0mmol.


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23

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 cuvet1,00cm nh sau: (nm) Mt quang A

X (8,00.10-

M) Y (2,00.10-4

M)

400 0,077 0,555440 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 660nmb) 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. Hytnh 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 1,2.10

3,00.10

660nm 4,67.10

1,50.10

b) Ti 520nmA = AX + AY = 1,125

Ti 600nmA = 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 + 1,5.102CY

Gii 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 mnhhn. iu ny do mt s nguyn nhn trong c nhng nguyn nhn t nhin v nhng nguyn nhnxut pht t cc hot ng ca con ngi. Trong kh quyn SO2v NO b oxy ha theo th t thnh SO3v 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 SO2l mt axit hai chc trong dung dch nc. Ti 25

oC cc hng s axitbng:

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

+(aq) Ka1 = 10

-1,92M

HSO3-(aq) SO3

2-(aq) + H

+(aq) Ka2 = 10

-7,18M

Tt c cc cu hi sau u xt 25oC:a) Tnh tan ca SO2l 33,9L tng 1L H2O ti p sut ring phn ca lu hunh dioxit bng 1 bar.


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24

i) Hy tnh nng ton phn ca SO2trong nc bo ho kh SO2(b qua s thay ith 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) + SO3

2-(aq) + H2O(l).

i) Hy tnh hng s cn bng ca cn bng trn. ii) Hy tnh nng ca lu hunh dioxit trongdung 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 cabc sunfit bo ho (b qua tnh baz ca ion sunfit).

f) Tch s tan ca canxi sunfit bng 10-7,17

M2

. Hy tnh hng s cn bng ca phn ng:Ca

2+(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 dioxitb oxy ha thnh sunfat (VI). Brom d c tch ra bng cch sc vi kh nitVit mt phng trnh phn ng ca qa trnh v tnh nng ionhydro 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 tchdung dch. Tr s pKaca 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 tonphn ca axit sunfuric trong nc ma, gi thit rng s axit ho ch so axit sunfuric. Proton thnhn 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 Mx

x

x1224,010

368,1

99,12

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

10.57,210.89,310

10

10

01,0

c) Ta c:

i)

26,5

1

2

3

2

32

3

2

32 10.

a

a

K

K

H

H

HSO

SOSO

HSO

SOSOK

ii) [SO2] + [HSO3-] + [SO3

2-] = 0,01M v [SO2] = [SO3

2-]

Vy ta c:

MSO

SO

SO 52

26,5

2

2

2 10.33,210)201,0(


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25

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-] + [SO3

2-] = [Ba2+]

MSOMxx

x

42

3

5

82,6

4

2

10.26,710.0479,1

10)10.36,7(

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

f) K =

65,6

O2

3

2

3

2

2

2

2

10.

3

32

CaS

SOAg

T

T

SO

SO

Ca

Ag

Ca

Ag

g) Phn ng: 2H2O(l) + SO2(aq) + Br2(aq) SO42-

(aq) + Br-(aq) + 4H

+(aq)

Cn bng: HSO4-(aq) SO4

2-(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+] = 0,0324Mh) [H+] = 10-3,2M; Ka = 10

-1,99M; [HSO4

-] = 10

-1,28[SO4

2-]

[H+] = 10-3,2 = 10-1,28[SO42-] + 2[SO4

2-] + 10-10,8.

[SO42-] = 3,074.10-4M v [HSO4

-] = 1,613.10-5M

C(H2SO4) = [HSO4-] + [SO4

2-] = 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 9

3

4

pHMNH

NHKH a

b) [NH4+] = 0,124M; [Ba

2+] = 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 NH3trung ho. Gi thit rng th tch bng110mL, 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

MH

NHKNH a

94

3 10.9,2


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26

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)2v 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)2Mg2+ + 2OH-

pOH = 14,010,5 = 3,5 [OH-] = 10-3,5 = 3,2.10-4MTng ng vi [Mg2+] = [Mg(OH)2in ly] = tan ca Mg(OH)2 = 1,6.10

-4M hay 9,2.10-4g/100mL.b) Ksp = [Mg

2+][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 7

2

11

10.6,1)010,0(

10.6,1

tan bng 1,6.10-7M hay 9.10-7g/100mLd) Mg(OH)2c 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)2ho tan trong dung dch [Mg

2+] = 0,010 + x 0,050M

3,9)10.8,1lg(141410.8,1 552

pOHpHM

Mg

KOH

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 hh 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)2trong nc nguyn cht (b qua cn bng t proton phn) b) Hy tnh tan ca Cd(OH)2trong dung dch NaOH(aq) 0,010M

Ion Cd2+

c i lc mnh vi ion CN-:

Cd2+(aq) + CN-(aq) Cd(CN)

+(aq) K1 = 10

5,48M-1.

Cd(CN)

+

(aq) + CN

-

(aq)

Cd(CN)2(aq) K2 = 10

5,12

M

-1

.Cd(CN)2(aq) + CN

-(aq) Cd(CN)3

-(aq) K3 = 10

4,63M

-1.

Cd(CN)3-(aq) + CN

-(aq) Cd(CN)4

2-(aq) K4 = 10

3,65M-1.c) Hy tnh tan ca Cd(OH)2trong 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

-15M

3.


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27

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

C(Cd) = [Cd2+

] + [Cd(CN)+] + [Cd(CN)2] + [Cd(CN)3

-] + [Cd(CN)4

2-]

0,5[OH

-

] = [Cd

2+

](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

= 4,79.10-3

M

S = 2,4.10-3

Md) [OH-] = [2.T.(1 + K1K2K3K4 [CN

-]4)]3/2 = 4,47.10-3M

S = 2,24.10-3

M

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

to phcchng 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 oxalicln lt l: K1 = 5,6.10

-2v 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 = 6,76.10

3. Tnh tan ca bcoxalat trong dung dch cha 0,02M NH3 v c pH = 10,8.

BI GII:a) T = [Ag+]2[C2O4

2-]

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

2

42 1KK

H

K

HOC

SSKKHKH

KKOC ..

211

2

212

42

Ti pH = 7 th [H+] = 10-7 1T = 3,5.10

-11.

Ti pH = 5 th [H+] = 10-5 0,861S = 2,17.10-4.

b) [NH3] = 0,02M

Ti pH = 10,8 th [H+] = 1,585.10-11 1Tng nng [Ag+] trong dung dch c xc nh bi phng trnhCAg = 2S = [Ag

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

+]

Cc phn ng to phc:Ag

++ NH3 = Ag(NH3)

+ 1 = 1,59.10

3

Ag(NH3)+

+ NH3 = Ag(NH3)2+

2 = 6,76.103

T cc phng trnh trn ta d dng suy ra c biu thc sau:


28/58

28

CAg = 2S = [Ag+](1 + 1[NH3] + 12[NH3]

2)

SSNHNH

Ag

.1

12

32131

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 maxetat c tng nng axetat (HOAc+ OAc-) l 0,500M v c pH = 5. 300mL dung dch m cha0,01M RNO2em 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

OAc

HOAc

OAc

HOAc

pHpKa

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 RNO2b kh s l: 300.0,0100 = 3mmol

T phng trnh bn phn ng ta thy rng kh ha hon ton 3mmol hp cht nitro cn12mmol H

+. S mmol H+ny nhn c t s phn ly ca HOAc.Khi phn ng xy ra hon ton th:n(HOAc) = 54,5512,00 = 42,55mmoln(OAc

-) = 95,4512,00 = 83,45mmol

Vy

16,5

OAc

HOAcpKpH 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 tancng nh ca cc iu kin th nghim, v d nh nhit v p sut. pH v kh nng to phc cngnh hng n tan.

Mt dung dch cha BaCl2 v SrCl2u nng 0,01M. Khi ta thm mt dung dch bo honatri sunfat vo dung dch th 99,9% BaCl2s kt ta di dng BaSO4 v SrSO4ch c th kt ta nutrong dung dch cn di 0,1% BaSO4. Tch s tan ca cc cht c cho sau y: T(BaSO4) = 10

-10v

T(SrSO4) = 3.10-7

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


29/58

29

Tnh nng Ba2+cn li trong dung dch khi SrSO4bt u kt ta.Tnh %Ba

2+v Sr

2+sau khi tch ra.

S to phc gy nn mt nh hng ng k n tan. Phc l mt tiu phn tch in chamt 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 NH3l phi t.

tan ca AgCl trong nc ct l 1,3.10

-5

MTch s tan ca AgCl l 1,7.10-10MHng 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 NH31,0M th cao hn trongnc ct.

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

Ba2+ + SO42- = BaSO4

Sr2+ + SO42- = SrSO4

Kt ta BaSO4s xy ra khi [SO42-

] = T(BaSO4)/[Ba2+

] = 10-8

M

Kt ta SrSO4s xy ra khi [SO42-

] = 3.10-5

M

Nu khng xy ra cc iu kin v ng hc (chng hn nh s hnh thnh kt ta BaSO4 l vcng chm) th BaSO4s c to thnh trc, kt qa l s c s gim nng Ba2+. Khi nng

SO42-

tho mn yu cu kt ta SrSO4th lc ny nng cn li ca ion Ba2+

trong dung dch c thc tnh t cng thc:

T(BaSO4) = [Ba2+][SO4

2-] = [Ba2+].3.10-5 [Ba2+] = 0,333.10-5M

%Ba2+cn li tng dung dch = %033,010

10.333,02

5

2) Cn bng to phc gia AgCl v NH3c 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 NH3th cn bng: [Ag

+] = [Cl-]

S hnh thnh phc dn n: [Ag(NH3)2+] = [Cl

-]

Nh vy:

Mx

x

x

NH

ClNHAgK 046,010.6,2

20,1

)( 32

2

3

23

Kt qa ny c ngha l 4,6.10-2M AgCl tan trong dung dch NH31,0M, nhiu hn trong ncct 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 250cm3v dung dch baz mnh (chtchun) c cho vo buret. im tng ng ca php chun t c khi lng cht chun cnbng vi lng cht phn tch. Gin biu th s thay i ca pH nh l mt hm ca th tch chtchun 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 thc xc nh bng thc nghim. N ch c th c lng c bng cch xc nh s thay i ca mt


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30

vi tnh cht vt ltrong qa trnh chun . Trong phng php chun axit baz, im cui caphp 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 trongvic 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 bazMetyl da cam 3,24,4 - da camMatyl 4,26,2 - vngBromthymol xanh 6,07,6 Vng - xanhPhenol 6,88,2 Vng - Phenolphtalein 8,0 -9,8 Khng mu - Thymophtalein 9,310,5 Khng muxanh2) 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 C6H8O60,1000M 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 12,5cm3dung 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 26,00cm

3

cht chun l:a) 13,30 b) 11,30 c) 11,00 d) 11,42BI 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

CH3COOHCH3COO- + H+

T phng trnh phn ly:

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

pH = 2,87b) pH sau khi thm 10,00cm3cht 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:


31/58


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32

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 theophng trnh Henderson Hasselbalch:

OAc

HOAcpKpH 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 100,0cm3dung 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) 4,18c) 9,82d) 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 mnhc) Dung dch md) C ba u sai.ii) pH ca dung dch cui s l:a) 3,02

b) 4,44c) 3,17d) 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 mnhc) Dung dch md) C ba u sai.ii) pH ca dung dch cui s l:a) 12,00

b) 12,70c) 13,18d) 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 mnhc) Dung dch m


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33

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

d) 3,228) 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 mnhc) Dung dch md) C ba u sai.1. pH ca dung dch cui s l:a) 7,00b) 13,00c) 2,87

d) 3,02BI GII:1) pH = 3,552) Natri hydroxit s phn ng vi HCOOH:

HCOOH + OH- HCOO- + H2OPhn ng ny xy ra hon ton nn:[HCOOH] = 0,140M[HCOO-] = 0,160M

Vy pH = 3,60Lu 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 smmol CH

3COOH 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:

MV

VCOOCH

MV

COOHCH

0,100

200,0

0,100

200,000,15

3

3

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

3

5

5

3

3

21,4810.0,1

10.8,1

100

200,000,15

0,100

200,0

cmV

V

V

V

HK

COOHCHCOOCH a

OLYMPIC HA HC QUC T LN TH 37:


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34

Nha trao i ion c th c s dng hp th v phn lp cation v anion. Chng c thc iu ch t cc vt liu v c v hu c. Mt loi nha trao i cation hu c c th c tnghp bng s ng trng hp styren /divinyl benzen tip theo bi s sunfo ho bng H 2SO4nh s 1:

S 1Nha 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+bngphn 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 CaCl2trong 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 AlCl3v 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 ionR

+OH

-. Vit cc phng trnh phn ng xy ra.

H2SO4

C CCC

C CCC

SO3-H

[ cationic exchanger ]

R-H

+

C C

+

C C

C C

Na2S2O8

C CCC

C CCC

[ polymer ]


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35

b) Bng cch no ta c th loi c ion SO42-

trong nc my bng loi nha trao i ion R+OH-trn. Vit ccphng 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 ctnh 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 cth c xc nh t hng s cn bng KC, hng s phn b KDv 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 lpcc 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 tol 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 chiucao 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-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 nhbi cc h thc:

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

v = (t2

to) / (t1

to) (11)

a) Tnh s a l thuyt ca ctb) 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-

1.0 10.0 14.0

t0

t1

t2

tR

Retention Time / min

X-

Y-


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36

5) Mt s loi nha trao i ion c th nhn c t nhng vt liu v c. Zeolit[(M

2+)(Al2O3)m/(SiO2)n] (M

2+= Na

+; K

+hay Ca

2+; Mg

2+) l nhng v d in hnh v nhng nhatrao i ion v c. Mt s v d v zeolit c cho trong hnh 2.

Hnh 2: Mt s loi zeolitMt 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 H 2O v iso-butan. Nhvy, zeolit ng vai tr nh l mt ci ry phn t. Zeolit cng c th c s dng nh l mt cht xctc trong cng ngh ha du. V d: trong ha du iso-butan l kt qa ca s tng tc crackinh cctc 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+bngRNa 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

-+ SO4

2-= (R

+)2SO4

2-+ 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 = 1497b) H = L/N = 0,021cm

c) R = R = 2 (t2 - t1) / (1 + 2) = 3,2

d) = (t2 to) / (t1 to) = 1,445. 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:


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37

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

Bc 2: Thm ure (NH2)2CO v un si dung dch n khi ch th chuyn sang mu vng (vicny mt 15 pht). Kt ta CaC2O4xut hin.

Bc 3: Dung dch nng c lc v kt ta CaC2O4c ra bng nc lnh loi b lng

d ion C2O42-

.Bc 4: Cht rn khng tan CaC2O4c ho tan vo dung dch H2SO4 0,1M sinh ra ion Ca2+

v H2C2O4. Dung dch H2C2O4c chun vi dung dch chun KMnO4n khi dung dch c muhng 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

-14

1. 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 CaC2O4trong 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) + C2O42-

(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 caCaC2O4.

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+] = 6,85.10-3M3. [Ca2+] = [C2O4

2-] + [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+][C2O4

2-] ta tnh c [C2O42-

] = 1,92.10-4

M4. Ta c:


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38

MCaMOC

KTKOC

TOCd

dC

OCKKKOC

T

OCCaOCaCCaC

ff

Ca

fff

aqCa

6222

42

2122

42

2

42

2

422112

42

2

242)(42

2

10.3,110.0,1

0

1

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 H2C2O4c th b qua.Kt hp (1) v (2) ta c: [HC2O4

-] = Kw/[H+] - [H+] (3)

[C2O42-] = (K2Kw)/[H

+]2K2 (4)[Ca

2+] = T/[C2O4

2-] = T[H

+]2/(K2KwK2[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

OLYMPIC HA HC QUC T LN TH 37:

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 thu c L. Hc sinh ny xc nh c mt shp cht kt ta v mt sphc cht mu nhtrong Bng 1 di y:

Bng 1

X-

Y-

Z-

Cl-

OH- L

A+ *** *** *** *** kt ta

trng***

B + k t tavng

k t tatrng

*** *** *** Phc BLn +

C+ kt ta

trngkt ta

nukt ta

nukt tatrng

kt taen

Cc phc

CL2+

, CL22+

D2+

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

E2+

*** kt ta kt tatrng

*** *** ***

*** = khng phn ng,

1 Lp s tch cc cation A2+

, B2+

, C2+

, D2+

, E2+

trong dung dch nitrat bng cch sdng cc

dung dch thuc thkhc 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 sdng

cc dung dch thuc thkhc nhau cha cc cation A2+

, B2+

, C2+

, D2+

, E2+

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


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39

3 Kt ta trng BY2 v kt ta nu CY2tan t trong nc vi tch stan tng ng ti 25oC ln lt

l 3.20 10-8

v 2.5610-13

.

a) Tnh tan ca BY2.

b) Tnh tan ca CY2.

4 Chun btrong cc bnh nh mc 50 mL mt nhm cc dung dch cha B2+ v L bng cch thmvo mi bnh 2 mL dung dch B2+ 8,2 10

-3M. Thm vo mi bnh cc lng khc nhau ca dung

dch cha phi tL nng 1,0 10-2

M. Pha long dung dch trong mi bnh bng nc nvch mc (50 mL). o h shp th ca phc BLn ti 540 nm cho mi dung dch trong mtng

di 1,0 cm. Cc dliu thu c trong Bng 2. (C B2+

v phi tL khng hp th (A = 0) ti540 nm.) [Phng php tl mol]

a) Tnh gi tr n (sphi tr) trong phc BLn2+

.

b) Tnh hng sto thnh (Kf) ca phc BLn2+

.

Bng 2

L thm voVL (mL)

H shp th(A)

L thm voVL (mL)

H shp th(A)

1.00 0.14 2.00 0.263.00 0.40 4.00 0.485.00 0.55 6.00 0.607.00 0.64 8.00 0.669.00 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 tcc 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? (ChoKsp (BY2) = 3.20 10

-8v Ksp (CY2) = 2.5610

-13ti 25

oC.) [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)? Sdng tc

nhn Y-c thtch B

2+v C

2+bng phng php kt ta hay khng?

BI GII:

1.

cch tch:


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40

2. Cch tch:

3.a)

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

-3b)


41/58

41

4S23 = 2,56.10-13S2 = 4,0.10

-54) a) th ph thuc gia h shp th A v thtch VL ca cht L c dng nh sau:

Tthtch L im gy B (tt c cc ion B2+u to phc vi L) trn th th ta c thtnhn nh sau:

n = smol L / smol B2+ = 5,1.10-3.10-2 / 2.10-3 . 8,2.10-3 = 3iu c ngha l phc gia B2+ vi L c dng BL3

2+b) * Tnh h stt mol :

im gy A = 0,66 = .1. [BL32+] = 2,01.103

* Chn mt im bt ktrn 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:


42/58

42

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

Tnh tng tcho khi hnh thnh kt ta BY2ta c: [Y-] = 5,06.10-4M

Vy cht CY2 kt ta trcb. Cht C2+ skt ta hon ton di dng CY2 khi [C

2+] = 10-6MNh vy [C2+][Y-]2 = 2,56.10-3[Y-] lc ny sbng 5,06.10-4Miu ny c ngha l C2+ sb kt ta hon ton di dng CY2 khi [Y

-] = 5,06.10-4MLc ny th i vi kt ta BY2 th[B2+][Y-]2 = 2,56.10-8< Ksp(BY2) = 3,2.10

-8Nh vy lc [Y-] = 5,06.10-4M v [B2+] = 0,1M th cht BY2 vn cha c kt taiu ny c ngha l ta hon ton c thtch 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 ktqa sau:

Chun ho dung dch natri thiosunfat Na2S2O3: dng KIO3trong mi trng axit, khi ioniodat b kh thnh ion iodua. Vi 25,00mL dung dch KIO3 ((KIO3) = 174,8mg/L) phi dng ht12,45mL 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 Na2S2O3trn cho 103,50mL mu nc 20,0oC. Nng oxy bo hotrong nc 20,0oC l 9,08mg/L.

Mu th hai (V = 202,20mL, T = 20,0oC) c trong 5 ngy nhit 20,0oC, ng vi6,75mL dung dch Na2S2O3.iii)Vit phng trnh ion ca phn ng chun ho dung dch thiosunfat. iv)Tnh nng mol/L ca dung dch thiosunfatv) 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 + 4S2O3

2- = 2S4O62- + 4I-.

ii. a) IO3-+ 6S2O3

2-+ 6H

+= I

-+ 3H2O + 9S + 3SO4

2-

b) C(S2O32-

) = 9,841.10-3

M

c) n(O2) = 2,903.10-2mmol (O2) = 8,976mg/L.


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43

d) SSI = 98,9%

e) n(O2) = 0,0166mmol (O2) = 5,20mg/Lf) 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 tan2. Tnh hng s Henry ca CO2 25.0C.3. Tnh nng mol ca CO2ho tan trong nc ma, nu phn th tch ca CO2trong kh quyn

c hm lng 380 ppm mi ngy v p sut ca CO2c ga tr l 1.00 bar.Mt phn CO2 hotan s phn ng vi nc to thnh axit cacbonic. Hng s cn bng ca

phn ng ny l K=1.6710-3vi nng ca nc c a vo Ka4. Tnh nng ca axit cacbonic ho tan trong nc ma? Bit nng ca CO2l khng i.

i vi hng s phn li th nht th [H2CO3]* c s dng thay th cho nng ca axitcacbonic. [H2CO3]* l tng nng ca axit cacbonic v lng kh CO2ho 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 KA2caaxit cacbonic. Bit rng nng ca [H2CO3]* l khng i trong sut qa trnh.

Vo nm 1960 th phn th tch ca CO2trong kh quyn l 320ppm6. 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 muicacbonat 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 hiny 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 ngrng pH = 8.26 tin tnh ton11. S dng tt c nhng thng tin trn tnh tan ca vi.BI GII:

Nng ca CO2 ho tan:pV = nR.T

mol0335.0K15.29808314.0

8304.000.1

TR

Vpn

L/mol0335.01

0335.0c

2. Hng s Henry: ci = piKH

Lbar/mol0335.01

0335.0KH

3. C(CO2) trong nc ma:

2COc =3.810

-40.0335 = 1.2710

-5mol/L

Nng ca axit cacbonic ho tan trong nc ma:


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44

CO2 + H2O H2CO3

3

2

32 1067.1]CO[

]COH[

[H2CO3] = 1.6710-3[CO2] = 2.1310-8 mol/L

pH ca nc ma vo thi im ny:[H2CO3]* = [CO2] + [H2CO3] = 1.2810

-5mol/L

H2CO3 + H2O HCO3-+ H3O

+ 7

32

331S 1045.4

]COH[

]OH[]HCO[K

[H3O+] = [HCO3

-] = x 1S5

2

K10275.1

x

x = 2.38210-6

mol/L

pH = -log x = 5,62

Ga trpH ca nc ma vo nm 1960:[CO2] = 3.210

-4bar0.0335 mol/barL = 1.0710

-5mol/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

tan ca vi trong nc ct:[Ca

2+][CO3

2-] = KL = 4.710

-9

[Ca2+] = 9107.4 = 6.85610-5 mol/LS (CaCO3) = 6.8610

-5mol/L

Cc ion cha xc nh c nng :[Ca

2+], [H3O

+], [OH

-], [CO3

2-], [HCO3

-]

Cc phng trnh cn thit:(I) [Ca

2+][CO3

2-] = 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-

] Nhn c t phng trnh ca H3O

+:

]OH[

*]COH[K]HCO[

3

321S3

23

321S2S

3

32S23

]OH[

*]COH[KK

]OH[

]HCO[K]CO[


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45

*]COH[KK

]OH[K

]CO[

K]Ca[

321S2S

23L

23

L2

]OH[

K]OH[

3

W

Thay vo (V):

2*]COH[KK

]OH[K

321S2S

23L

+ [H3O

+] =]OH[

K

3

W

+

]OH[

*]COH[K

3

321S

+ 2

23

321S2S

]OH[

*]COH[KK

Chuyn v:

*]COH[KK

]OH[K2

321S2S

43L

+[H3O

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

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 amoniac1M.

BI GII:a) Ta c:

10

7

23

2

3

10.8,1

10.7,1)(

ClAgT

NHAg

NHAg

K

V [Ag+]


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46

BI GII:

CH3COOH CH3COO- + H+

5

3

3 10.8,1

COOHCH

HCOOCHKa

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

OAc

HOAcpKpH a

b) Khi thm baz mnh nng Cb th Cmui = Cmui + Cb; Caxit = Caxit - CbpH tng mt n v tng ng vi [H+] gim 10 ln:[H

+]2/[H

+]3 = [Caxit.(Cmui + Cb)]/[Cmui.(CaxitCb)]

[H+]3 = 1,8.10-5M; Cb = 0,045MmNaOH = 1,8g

c) [CH3COOH]1 = ([H+]1)

2/Ka 0,1M

[CH3COOH]2 = [H+].Cmui/Ka0,1M 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 dungdch 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; H2Od) HCN; H3O

+

BI GII: Cu dOLYMPIC HA HC ITALY 1999:

iu ch dung dch H2SO40,12M bng cch pha long H2SO4c (95%. d = 1,84g/mL), cth pha long vi nc.

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


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47

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 Na2CO3tan trong nc to thnh 870mL dung dchl:

a) 0,206M

b) 0,312Mc) 0,412Md) 0,103M

BI GII: Cu cOLYMPIC HA HC ITALY 1999:

Baz lin hp ca NH3khi phn ng vi axit l:a) NH3

-.

b) NH2-.

c) NH4+.

d) NH3+.

BI GII: Cu bOLYMPIC HA HC ITALY 1999:

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

a) H2SO4b) H3PO4.c) H2Od) CH3COOH

BI GII: Cu cOLYMPIC HA HC ITALY 1999:

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

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


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48

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 10-2md) HCl 3,7% m/V


Lng H2SO4trong mt dung dch nc (2000mL; 27,27%; d = 1,20g.cm-3) l:

a) 6,00 molb) 4,82 molc) 6,79 mol

d) 5,20 molBI GII: Cu cOLYMPIC HA HC ITALY 1999:

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


Trong s cc mui sau y th mui no l axit Bronsted?a) NaHSO

4.

b) Na3PO4.c) NaCN.d) Na2S.

BI GII: Cu aOLYMPIC 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 cOLYMPIC 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 cmt ca hipoclorit.


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49

OH

+ NH3 N OO

Xanh lam: max = 625nmTrong 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 mtbnh nh mc 500mL, cho thm vo 5mL dung dch phenol v 2mL dung dch hipoclorit , ri pha thnh50,0mL 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. Ngita cho 5,00mL dung dch vo mt bnh nh mc 50,0mL v sau vic phn tch c tin hnhnh m t trn.

Ngoi ra ngi ta cn o mt mu khng (mu m) vi nc nguyn cht trong ng cuvet:Mu tt ti 625nmKhng 0,132

i chng 0,278Cha 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 0,2780,132 = 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-]


50/58


51/58

51

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

cho (a) xy ra th nhm tc ng nh l axit trong NH3 phi c tc dng cho proton yu hnaxit 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 28,74cm3dung dch kali pemanganat.+ 25,00cm

3dung dch st (II) sunfat dng ht 24,03cm3dung dch kali pemanganat.

+ 25,00cm3dung 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:Fe

3++ e = Fe

2+E

o= 0,77V

Ce4+

+ e = Ce3+

Eo

= 1,61V3. Hy tnh KCca phn ng: Fe

2+ + 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

Fe

Feti im tngng.

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 xcca vic nhn bit im kt thc.

Nhng i vi cht ch th sau y th:InOx + 2e = In

2-kh E

o= 0,80V


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52

S chuyn mu s th hin r khi: 1

102

kh

Ox

In

In

6. Hy tnh

2

3

Fe

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

ln chun tin hnh.BI GII:1. 2MnO4

-+ 5C2O4

2-+ 16H3O

+= 2Mn

2++ 10CO2 + 24H2O

5Fe2+

+ MnO4-+ 8H3O

+= 5Fe

3++ Mn

2++ 12H2O

Fe2+ + Ce4+ = Fe3+ + Ce3+.

2. Chun 1: 0,2228g Na2C2O4tng 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:14

o

/Fe/ 10.61,1).(

lg2334

KRT

FEEK

Fe

o

CeCe

4. Ti im tng ng th lng cht cho vo n(Ce4+) = no(Fe2+

). Vi mi ion Ce3+mi hnhthnh th cng hnh thnh mt ion Fe3+, tc l [Ce3+] = [Fe3+] v c [Ce4+] = [Fe3+]Ta c:

7

2

3

22

23

24

33

10.27,1;

Fe

Fe

Fe

FeK

FeCe

FeCeK 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 nhvy ngi ta c th a ga tr [Ce4+]/[Ce3+] = (1,27.10-7)-1vo phng trnhNernst 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:

12,10

ln77,083,02

3

2

3

Fe

Fe

Fe

Fe

F

RT

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). NH3l 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 dOLYMPIC HA HC C 2000:


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53

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

a) 0,166Mb) 0,180Mc) 0,333M

d) 0,666Me) 1,136MBI GII: Cu aOLYMPIC 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 baonhiu?

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

BI GII: Cu aOLYMPIC HA HC C 2000:Tch s tan ca mt s mui sunfat c cho bng sau:

Mui T1 CaSO4 9.10

-

2 SrSO4 3.10-

3 PbSO4 2.10-

4 BaSO4 1.10-1

Khi cho dung dch Na2SO40,0001M vo dung dch cc mui tan ca cc cation trn th mui nos kt ta.

a) 1, 2 v 3.

b) 1 v 2c) 1 v 3d) 2 v 4e) ch 4

BI GII: Cu eOLYMPIC HA HC C 2000:

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

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

4) 0,6M KOHa) 1, 2 v 3.b) 1 v 2c) 1 v 3d) 2 v 4e) ch 4



54/58

54

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

e) 12,40BI GII: Cu eOLYMPIC HA HC C 2001:

Cht no l axit lin hp ca HPO42-

?a) H3PO4(aq).b) H2PO4

-(aq).

c) H3O+

(aq)d) H+(aq).e) PO4

3-(aq)

BI GII: Cu bOLYMPIC HA HC C 2001:

Dung dch axit yu HA 0,075M c [H

+

] bng bao nhiu nu Ka(HA) = 4,8.10-8

a) 6,1.10-4Mb) 2,2.10-4Mc) 6,0.10-5Md) 4,8.10-8Me) 3,1.10-9M

BI GII: Cu cOLYMPIC HA HC C 2001:

Nu trn cng mt lng th tch ca BaCl2v NaF th nng no ca mi cht th kt tac hnh thnh?. Bit T(BaF2) = 1,7.10

-7.

a) 0,020M BaCl2 v 0,0020M NaF.b) 0,015M BaCl

2v 0,010M NaF.

c) 0,010M BaCl2 v 0,015M NaF.d) 0,0040M BaCl2 v 0,020M NaF.e) Tt c u khng th to kt ta.

BI GII: Cu aOLYMPIC HA HC C 2002:

Nu trn cng mt lng th tch ca cc cht sau th hn hp no s hnh thnh dung dch m?1) 0,1M HCl v 0,1M NH3.2) 0,1M HNO2 v 0,05M NaOH.3) 0,05M HNO2 v 0,05M NH3.

a) 1b) 2c) 3d) 1 v 3e) 2 v 3



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55

S mol KOH trong 500mL dung dch c xc nh bng cch chun 10,00mL dung dchKOH ny vi dung dch HCl 0,115M . Nu s chun trn cn 18,72mL HCl th s mol KOH trong500mL dung dch u s l bao nhiu?

a) 0,00215 mol.b) 0,00430 mol.

c) 0,108 mold) 0,215 mole) 0,115 mol

BI GII: Cu cOLYMPIC HA HC C 2002:

tan ca bc sunfat l 1,5.10-5mol-3.L-3. Trong mt dung dch m nng ion SO42-

l 2,4.10-

2M th nng Ag+cc i s l:a) 0,025Mb) 6,25.10-4Mc) 3,125.10-4Md) 2,5.10-4M

e) 6,25.10

-2

MBI GII: Cu aOLYMPIC HA HC C 2003:

Tch s tan ca Ag2CrO4(r) 25oC l 2,6.10-12. Vy tan ca Ag2CrO4trong 1L nc l bao

nhiu?

a) 1,6.10-6Mb) 2,6.10-12Mc) 2,1.10-8Md) 1,4.10-4Me) 8,7.10-5M


Mt axit mnh hai nc H2A ho tan trong nc 25oC cho mt dung dch c pH = 1,85. Nng

ca axit trong dung dch (gi s n phn li hon ton thnh A2-) lc ban u s nm trong khong:a) 0,05 v 0,1b) 0,01 v 0,05c) 0,1 v 0,5d) 0,001 v 0,005e) 0,005 v 0,01


Mt dung dch axit axetic 0,100M c chun bng dung dch NaOH 0,05M. Khi 60% axitc trung ho th pH ca dung dch lc ny l bao nhiu?

a) 2,38b) 4,56c) 4,74d) 4,92e) 7,00

BI GII: Cu dOLYMPIC HA HC C 2003:


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56

a) Tnh pH cui c cc h sau y khi ho tan:i) 2,00M HClii) 0,500M NaOHvo 500,0mL nc 25oC (pH ca nc iu kin ny l 7,00)b) Mt lnh vc quan trng trong ho hc

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