bai tap sac ky co giai

7/27/2019 Bai Tap Sac Ky Co Giai

1/18

1

Bi 1 [2_422] :Thi gian cn i qua bi pha di ng chuyn qua ct bng 25

pht. Gi tr R i vi cht tan c thi gian gi l 261 pht phi bng bao nhiu ?

Cht tan c bao nhiu thi gian s trong cc pha di chuyn v pha tnh trc khi

ra khi ct ?

Bi gii :

Ta c : R= M

M S

t

t t

25

L ;

261

L

Cht tan c 9,6% thi gian pha ng

9,6 261

100Mt

= 25 pht ; tS= 26125 = 236 pht

Bi 2 [3] :Tnh s a l thuyt trung bnh v chiu cao a l thuyt trung bnh

ca mt ct sc k c chiu cao 3,2m. Bit t lc bt u bm mu vo u ct th

pic ca cu t A t gi tr cc i : 350s ; cu tB l 375s. Cho bit WA= 14s ;

WB=15s Bi gii:

a> S a l thuyt:2

16.W

RtN

- i vi cu t A:

2 2

A

35016. 16. 10000

W 14

R

A

tN

- i vi cu t B:

2 2

B

37516. 16. 10000

W 15

BR

B

tN

S a l thuyt trung bnh : 100002

A BN NN

b> Chiu cao a trung bnh:

43, 2 3, 2.10 0, 03210000

LH m cm

N

Bi 3 [2_422] :R i vi 1 cht tan cho khi dng ct sc k xc nh bng 0,1.

Th tch ca pha di chuyn trong ct VM l 2,0 ml. Gi tr tSi vi cht tan bng

bao nhiu khi tc dng ca pha di ng bng 10 ml/pht ? Tnh KDnu VS =

vpha ngvchttan

vpha ng vchttan


2/18

2

0,5ml

Bi gii :

Ta c : R M

M S

t

t t

Vi tM = 2

10= 0,2

0.2 0,02 0,1 St

tS = 1,8 pht = 108 giy

Mt khc :1

1 SD

M

RV

K

V

= 0,1

1

0,10,5

12

DK

KD= 36

Bi 4 [2_422] :H s phn b K i vi cht A trong ct sc k cho bi trn

ln hn so vi cht B. Hp cht no trong cc hp cht c gi mnh hn trong

ct sc k?

Bi gii :

Ta c :

1A

SA

M

UU

VK

V

v

1B

SB

M

UU

VK

V

M : VS= const ; VM= const

Nu KA> KBth UA < UB

Cht A s c gi mnh hn cht B trong ct sc k.

Bi 5 [2_422] :Tc di chuyn ca pha di ng trong ct c chiu di 10 cm

bng 0,01 cm/giy. ra cu t A cn 40 pht. Phn no ca thi gian chung cas ra cn c mt ca cu t A trong pha di ng ? Gi tr R i vi hp cht

0,1


3/18

3

ny bng bao nhiu?

Bi gii :

Ta c : R =

Vi : vpha ng= 0,01 cm/giy

Vcht tan =10

40 60= 4,16.10-3 cm/giy

Suy ra : R M

M S

t

t t

= 0,4167

V : tM + tS= 40 pht

Bi 6 [2_422] :Trong sc k khi tc di chuyn ca pha ng c th o trc tip,

nu a vo mt lng bo ca cht tan tng t mtan, n khng c gi bi

pha tnh. Trong ct mao qun c chiu di 50m thi gian gi ca metan bng 71,5

giy, cn thi gian gi ca n-heptaecan l 12,6 pht.

a>Tc di chuyn ca pha ng bng bao nhiu ?

b>Tc di chuyn ca di n-heptanecan bng bao nhiu ?

c>Gi tr R i vi di ca n-heptanecan bng bao nhiu ?

Bi gii :a>Tc di chuyn ca pha ng :

vM=50

71,5= 0,6993 m/giy = 69,93 cm/giy

b>Tc di chuyn ca di n-heptaecan :

vS=50

0,066112,6 60

m/giy = 6,61 cm/giy.

c>Gi tr R ca di n-heptaecan :

R =6.61

0.094569.93

S

M

v

v

Bi 7 [2_423] :Trong nhng iu kin xc nh, trong ct sc k phn b kh-lng

cho ngi ta ra gii cht A, c R=0,5 , cn VR= 100ml. Tc dng ca pha

di ng cn phi hng nh nhng VS(lng ca pha tnh lng) c th thay i so

vi i lng ban u ca nbng 1,5 ml. Cn phi thay i i lng VSbao

vpha ngvchttan

R= 0,4167

tM = =16,668 pht

= 1000 giy


4/18

4

nhiu ln VR tng gp hai? C th lun lun dng tha s ny khng khi cn

tng gp 2 ln i lng VR hay n ch c dng trong trng hp cho?

Bi gii :

Ta c : VR= VM+ KD.VS

100 = VM+ 1,5.KD (1)

Mt khc : 0,5M

M D S

VR

V K V

VM = 0,5.VM+ 0,5.1,5.KD

VM = 0,5.VM+ 0,75.KD (2)

T (1) v (2) : VM+ 1,5.KD = 100 VM= 50

0,5.VM0,75.KD= 0 KD= 33,33

Nu VRtng gp 2 : VR = VM+ KD.V

S

200 = 50 + 33,33.VS

VS= 4,5 ml

VS= 3.VS

Vy tng i lng VRln gp 2 ln th phi tng VSln 3 ln.

Ta khng th dng tha s ny cho tt c cc trng hp mun tng V Rln 2 ln

v vi mi php sc k s c cc i lng VMv KDkhc nhau.

Vy thi gian lu hiu chnh ca cu t th 2 trn ct sc k th 2 l : 29,33 pht

Bi 9 [2_424]:Trong mt ct sc k phn b lng cho hp cht A c K=10 cn

hp cht B c K=15. i vi ct cho VS= 0,5ml , VM= 1,5ml v tc di

chuyn ca pha di ng bng 0,5ml/pht. Hy tnh VR, tRv R ca mi cu t.

Bi gii :Ta c : V = VM + K .VS Tng t : V = 1,5 + 15.0,5

= 1,5 + 10.0,5 = 9 ml

= 6,5 ml

1 10,231

0,51 101

1,5A

AS

D

M

RV

KV

RA DA RB


5/18

5

1 10,167

0,51 151

1,5B

BS

D

M

RV

KV

Ta lic : tM

=1,5

30,5

M

M

V

v pht

313

0,231ARt pht

M M MR

M S R

t t tR t

t t t R

318

0,167BRt pht

Bi 10 [2_425]:Khi kim tra ct sc k thy rng : pic c dng ng phn b

Gauss v b rng 40 giy thi gian gi 25 pht. Ct c s a l thuyt l bao

nhiu? Bi gii :

Ta c :2

16 Rt

NW

Thay tR= 25 pht = 25.60 = 1500 giy

W = 40 giy

21500

16 2250040

N

a

Bi 12 [2_426]:Ct sc k lng c chiu di 2m c hiu qu 2450 a l thuyt tc ca dng 15ml/pht v hiu qu 2200 a l thuyt tc dng

40ml/pht. Vy nng ti u ca dng phi bng bao nhiu v hiu qu tc

ca dng s gn bng bao nhiu ? Bi gii :

Ta c :B

H C uu

Vi u1= 15 ml/pht 11

200 0,08162450

LHN

cm

Vi u2= 40 ml/pht 22

2000,0909

2200

LH

N cm

Ta c h : 0,0816 1515

BC B = 0,829

C = 1,754.10-3

Ut = 30,829

21,741,754 10

B

C

(ml/pht)

0,0909 40

40

BC


6/18

6

Hmin=32 2 0,829 1,754 10 0,076B C (cm)

Bi 13 [2_427] :Ngi ta th nghim ct sc k kh - lng c chiu di 2m bao

tc khc nhau ca dng, mt khc, lm pha di ng ngi ta dng hli.

Kt qu th nghim tm thy rng ct c cc c trng sau :Mtan(pha di ng) n-octaecan

tR tR W

18.2 giy 2020 giy 223 giy

8.0 giy 888 giy 99 giy

5.0 giy 558 giy 68 giy

a>Hy xc nh tc di chuyn ca pha ng i vi mi dng.b>Hy xc nh s a l thuyt v gi tr H i vi mi dng.c>Bng cch gii ng thi cc phng trnh cn thit hy tm cc gi tr ca

cc hng s trong phng trnh sau :B

H A C uu

d>Tc ti u ca s di chuyn ca pha di ng bng bao nhiu ?Bi gii :

a> Tc di chuyn ca pha di ng

u1=200

1118,2

cm/s ; u2=200

258

cm/s ; u3=200

405

cm/s

b> S a l thuyt v gi tr H :

N1=1

2 2

1

202016 16 1310

223

Rt

W

H1 =

2 1000,1526

1310

(cm)

N2=2

2 2

2

88816 16 1287

99

Rt

W

H2=

2 1000,1554

1287

(cm)

N3=3

2 2

3

55816 16 1076

68

Rt

W

H3=

2 1000,1859

1067

(cm)

c> Xc nh A, B, C :

Ta c h :


7/18

7

0,1526 1111

0,1554 2525

0,1859 4040

BA C

BA C

B

A C

3

0,059

0,695

2,729 10

A

B

C

d> Tc ti u :

Ut= 30,695

15,962,729 10

B

C

cm/s

Bi 15 [3] :Tin hnh sc k hn hp 2 cht A v B trn ct sc k c chiu di

L=4m c s a l thuyt n = 800 a. Tc tuyn tnh ca 2 cu t A v B trong

pha ng ln lt l 2 cm/s ; 1,6 cm/s, tm= 10s.

a> Tnh tRA v tRB

b> C th tch A v B ra khi nhau c khng ?

c> Tnh phn gii ca php sc k.

Bi gii:

a> Ta c :R

Lt

U Vi : L - chiu cao ct sc k

U - Tc tuyn tnh ca cu t

- Cu t A :A

400200

2AR

Lt s

U

- Cu t B:B

400250

1, 6BR

Lt s

U

b> Tnh h s tch :'

'200 10 0,8 1250 10

A A

B B

R R m

R R m

t t tt t t

Vy c th tch 2 cht A v B ra khi nhau.

Bi 16 [2_429] :Cc thi gian gi ca - cholestan v - cholestan trong h cht

lng pha rn trn ct sc k c chiu di 1m v vi hiu qu 104a l thuyt

tng ng bng 4025 v 4100 giy. Nu 2 hp cht ny cn c phn chia vi

phn gii bng mt th cn bao nhiu a l thuyt t c mc ch ny ?Chiu di no ca dng ny cn nhn c phn gii ch ra, nu H=


8/18

8

0,1mm?

Bi gii :

Ta c :B

2W

A BR R

A

t tR

W

=1

Vi t = 4025 s

t = 4100 s

Mt khc : A BR R

A B

t t

W W 0

B AR A R Bt W t W (2)

T (1) v (2) ta c : WA = 74,31 giy

WB= 75,69 giy

Vi H = 0,1 mm L = 4,6942 m

Bi 17 [5_525] :Pic sc k ca hp cht c pht hin sau 15 pht khi a mu

vo (lc pic ca hp cht Y khng c gi bi vt liu ca ct xut hin qua

1,32 pht). Pc ca cht X c dng ng phn b Gauss vi b rng ca y l

24,2s. di ca ct l 40,2 cm.

a>Tnh s a l thuyt trong ct.b>Tnh H ca ctc>Tnh T v ca ct.d>Tnh ch s lu gi ca X .e>T phng php chun b bit rng th tch ca cht lng gi trn b mt

ca cht mang ca ct bng 9,9. Th tch ca pha ng bng 12,3 ml. Tnh hng s

phn b KD.

Bi gii :

a> Tnh N :

2

16 Rt

NW

=

2

15.6016

24,2

=22130 a

b> Tnh H :

40,20,0181

22130

LH

N mm

RA

RB

WA+ WB= 150 s (1)

N=46942 a


9/18

9

c> Tnh T v :

24,26,05

4 4

WT giy

2

2N H

3

1,81.10 . 22130 0,27H N

cm

d> Tnh R :1,32

0,08815

R

e> Tnh KD:1

1 SD

M

RV

KV

=1

9,91

12,3DK

= 0,088 KD= 12,9

Bi 18 [5_525] : Trn sc k ngi ta tm thy 3 pic 0,84 pht, 10,6 pht v

11,08 pht tng ng vi cc hp cht A, B v C. Hp cht A khng c lu gi

bi pha tnh lng. Cc pc ca cc hp cht B v C c dng ng Gauss c chiu

rng 0,56 pht v 0,59 pht tng ng. di ct bng 28,3 cm.

a>Tnh gi tr trung bnh N v H theo cc pic B v C.b>Tnh gi tr trung bnh T v .c>Tnh ch s lu gi i vi B v C.d>Th tch ca cht lng c gi trn b mt ca cht mang ca ct bng

12,3 ml cn th tch ca pha ng bng 17,6 ml. Hy tnh hng s phn b ca tp

cht B v C.

Bi gii :

a> Tnh N v H trung bnh :

Ta c :

2

16 R

t

N W

Vi B : N

1= 5733 a H1= 0,494 mm

Vi C : N2= 5643 a H2= 0,5 mm

N = 5688 a

H = 0,497 mm

b> Tnh T v trung bnh :

T1=

0,56

0,144 4

W

pht

T = 0,14375 pht


10/18

10

T2=0,59

0,14754 4

W pht

Ta c :2

2N

H

0,497. 5688 37,48H N

c> Tnh ch s lu gi ca B v C :

0,840,079

10,6BR ;

0,840,076

11,08CR

d> Tnh :

1

1B

BS

D

M

RV

KV

=1

12,31

17,6BD

K

= 0,079 16,682BD

K

Tng t : 17,4CD

K

Do :16,682

0,9617,4

B

C

D

D

K

K

Bi 20 [3] :Trong qu trnh tin hnh sck cacc cht trong ctsck c chiu

di 45 cm. Ngita thu ctrn sc3 pic tng ng: 60s, 360s, 375s ngvi

3 cutX, Y, Z. Cc pic c dngngGauss caY v Z c WY= 24s, WZ = 25scn chtX khng clu gibi vtliuct.

a> Tnh sal thuyttrung bnh v chiucao atheo pic Y v Z.

b> Tnh sal thuythiulccactsck.

c> Tnh hslu gitng igiaY v Z.

e>Thtch cacht lngcgitrn bmtchtmang cactl 8,7ml. Thtch pha ngbng11,5ml. Tnh hsphn bcaY v Z.

Bi gii:

a> N v H theo Y v Z:

* Sal thuyt:

2

16.W

RtN

- ivicu tY:Y

2 2

Y

36016. 16. 3600

W 24

R

Y

tN


11/18

11

' . SDm

VK K V

- ivicu tZ:

2 2

Z

37516. 16. 3600

W 25

ZR

Z

tN

Sal thuyttrung bnh : 36002

Y ZN NN

* Chiucao atrung bnh:

450,0125

3600

LH cm

N

b> Sal thuythiulc:

2

16.W

R mf

t tn

- Cu tY: Y2 2

Y

360 6016. 16. 2500W 24Y

R mf

t tn

- Cu tZ: Z2 2

Z

375 6016. 16. 2540

W 25ZR m

f

t tn

Sal thuythiulctrung bnh:2500 2540

25202

fn

c> Tnh hslu gitng igiaA v B:'

'

360 600,952

375 60

Y Y

Z Z

R R m

R R m

t t t

t t t

e>Ta c :2

16.W

R mf

t tn

(1)

Ta lic : tR= tm .(1 + K) '1R

m

t

t K (2) (Vi : )

Thay (2) vo (1) ta c:

2

22 ''

'

116. 16.W W 1

RR

Rf

tt

t KKnK

2'

'.

1f

Kn N

K

- Xt cutY:

2'

'. 1YY

f

Y

Kn NK


12/18

12

'

'

25000,8333

1 3600

YfY

Y

nK

K N

' 5YK

M :' ' 11,5. 5. 6, 61

8,7Y YS m

Y D D Y

m S

V VK K K K

V V

Tnh ton tng tivicu tZ ta c: 6,94ZD

K

Mt s bi tp lin quan AASBi tp v phng php ng chunBi 1: xc nh hm lng Cu trong mt mu phn tch, ngi ta cn 10 gng mu

v x l mu bng cc dung dch thch hp, axit ho a dung dch v pH


13/18

13

0,05 0,1 0,15 0,2 0,25 0,3 0,35

Ao A1 A2 A3 A4 A5 A6

0,2450 0,4825 0,7200 0,9575 1,1950 1,4325 1,6700

1. Xy dng phng trnh ng chun2. Ly 3 lit nc c cn c 4 mg cht rn. Ho tan trong dung dch HCl

1% ri kho st cc iu kin ti u nh dy dung dch chun v o A bcsng 422,7 nm thu c A=2,1450. Xc nh hm lng Ca2+trong 1 lt nc.

Gii1.Xydngphngtrnhngchun

0.05 0.10 0.15 0.20 0.25 0.30 0.35

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

A

C

B

Vy phng trnh ng chun l: A = 0,0075 + 4,75.Cx2. Khi A = 2,145 t phng trnh bng chun ta tnh c nng canxi

trong nc cng l:

2

2,145 0,00750,45

4,75CaC

g/ml

Vy hm lng ca Can xi trong 1 lt nc cng l:0,45 g/mlBi 3: Ngy nay xc nh s nhim Hg ca cc dung dch nc bng phng

php hp th nguyn t ngi ta dng phng php khng ngn la mi ca sphun m.Thit b gm mt bnh kh Hg ni vi mt cu vt hp th. 10 mlmu nc vo bnh kh Hg v pha long n 100 ml, sau thm vo 25 mlH2SO4m c v 10 ml SnSO410 %, H2SO40,25 M (dung dch cui ny dnglm cht kh). Thu ngn b kh n trng thi nguyn t (nguyn t)v c

chuyn vo cu vt hp th bi dng khng kh, ngi ta cho dng khng kh nyi qua dng dung dch trong bnh kh Hg. Cui cng, dng n catt rnglm ngun, ngi ta o s hp th ca cc nguyn t Hg bc sng 2537 Ao ,s hp th t c mc cc i gn 3 pht.

Ngi ta nhn c cc gi tr sau ca hp th i vi dy cc dung dchchun ca Hg(II):

Hm lng Hg trongdung dch chun, g

hp th

0,00 0,002

0,30 0,090

0,60 0,1751,00 0,268

2,00 0,440


14/18

14

Cc gi tr ca hp th ca hai mu nc bng 0,040 v 0,305 tngng.Vy hm lng ca Hg trong tng mu bng bao nhiu? Nng ( g/ml)ca Hg trong tng mu bng bao nhiu?

GiiS dng phng php hi quy tuyn tnh tm A theo CX. Ta xy dng

phng trnh tuyn tnh c dng A= a + b.CX (*)p dng phng php bnh phng ti thiu ta c:2

m in).( ii CbaAQ

0).(.2

ii CbaA

a

Q

22 )(

.

ii

iiii

CCn

ACACnb

(1)

0)(2

iii bCaAC

b

Q

n

Cb

n

Aa ii

(2)

Trong : n l s ln th nghim, n=5Cil hm lng Hg trong mu dung dch chun Ail hp th quang tng ng vi CiThay s vo (1) v (2) ta c:

b = 0,21574

a = 0,02672

Thay gi tri a,b vo (*) ta c: A= 0,02672 + 0,21574CXKhi A=0,04 th CX = 0,0616 g tc l hm lng Hg trong mu chun ny l0,0616 g. V nng Hg l 0,0616/10 = 0,00616 g/ mlKhi A=0,305 th C= 1,2902 g.V nng ca Hg l 1,2902/10=0,129 g /ml

Bi 4: C th dng php o ph hp th nguyn t xc nh cc vt cc kim loi nngtrong du mazut. phn tch 5,000 g mu ca loi du mazut dng, ngi ta t vo

mt bnh nh mc c dung tch 25,00 ml, ho tan vo 2- metyl-4-pentanol v bng dungmi ny a th tch trong bnh n vch. Sau phun m dung dch nhn c trongngn la khng kh axetilen. xc nh Cu v Pb cn dng cc n catot rng vi ccvch pht x 324,7 v 283,3 nm tng ng. nhn c cc thchun cn mt dycc dung dch chun cha nhng lng bit ca Cu v Pb trong hn hp tng ngvi du mazut cha dng v 2-metyl-4-pentanol.Tnh hm lng % ca Cu v Pb trong5,000g mu du mazut dng theo cc s liu sau:

Dungdch chun( g/ml )

hp th

Pb Cu 283,3nm (Pb) 324,7nm (Cu)19,5 5,25 0,356 0,5144,00 4,00 0,073 0,392

12,1 6,27 0,220 0,612

8,50 0 0,155 0,101

15,2 2,4 0,277 0,232

Cha bit Cha bit 0,247 0,371

GiiS dng phng php hi quy tuyn tnh tm A theo CX. Ta xy dng

phng trnh tuyn tnh c dng A= a + b.CX (*)p dng phng php bnh phng ti thiu ta c:

2

m in).( ii CbaAQ


15/18

15

0).(.2

ii CbaA

a

Q

22 )(

.

ii

iiii

CCn

ACACnb

(1)

0)(2

iii bCaAC

b

Q

n

Cb

n

Aa ii

(2)

Trong : n l s ln th nghim, n=5(vi Pb), n=4 (vi Cu)

Cil nng ca Pb hoc Cu trong mu dung dch chun Ail hp th quang tng ng vi CiThay cc gi tr Ci, Aitng ng ca Cu v Pb vo (1) v (2) ta c:

Phng trnh tuyn tnh ca Pb dng A= a + bCXta c:b=0,01825, a= -0,00024Vy phng trnh tuyn tnh ca Pb l: A= 0,01825CX - 0,00024Vi A= 0,247 th Cx=13,55 g /mlHm lng ca Pb trong 5g mu du mazut l:

% Pb = %1005

10.25.55,13 6= 0,00678%

Phng trnh tuyn tnh ca Cu dng A= a + bCXta c:b=0,09822, a= -0,00253

Vy phng trnh tuyn tnh ca Cu l: A= 0,09822.CX - 0,00253Vi A= 0,371 th Cx=3,8g /mlHm lng ca Cu trong 5g mu du mazut l:

%0019,0%1005

10.25.8,3%

6

Cu

Bi 5: xc nh hm lng ca mt kim loi trong mt mu phn tch bng ph hpth nguyn t, ngi ta s dng phng php ng chun. Dy mu chun c chun

b trong nhng iu kin nh nhau, em o ph AAS v xy dng ng chun ngi tac phng trnh tuyn tnh A= 0,4342.Cx+0,0009(CXtnh bng ppm)

a, Xc nh nng ca mu chun khi A= 0,682; 0,245b, Tnh hm lng kim loi trong mu khi A = 0,565

Gii

T phng trnh: A= 0,4342.Cx+0,0009 4342,0

0009,0

ACx

a. Khi A= 0,682 th Cx= 1,5686

Khi A= 0,245 th Cx= 0,5622 ppm

b. Khi A= 0,565 th Cx= 1,2992 ppmPhn trm kim loi trong mu l:

%kim loi = %10.2992,1%1001

10.2992,1 46

Bi tp v phng php thm:Bi 1: xc nh Mg trong mt mu nc cng, ngi ta ly 5 lt nc em c cn thuc 5g cht rn. Ho tan cht rn vo 100 ml dung dch HCl 1% v kho st cc iukin ti u ri tin hnh hai th nghim sau:

Th nghim 1: ly 25 ml dung dch trn em o ph hp th nguyn t bc sng285,2 nm thu c cng ph hp th A1 = 0,3420

Th nghim 2: Thm 3 ml dung dch MgCl21g/ml vo 25 ml dung dch trn v ophng php ph hp th nguyn t c A2 =0,3817


16/18

16

Xc nh hm lng % ca Mg trong mu nc cng. Cho phng trnh chun cdng A=K.C (vi C mol/l)Gii

Phng trnh ng chun ca ph hp th : A = K.CGi nng ca Mg trong 25 ml dd l CX

Ta c: A1= K. Cx

A2= K_ 6.25 3.10

[ ]25 3 24.(25 3)

xC

Lp t l ta thu c: 1_ 6

2 .25 3.1.10

25 3 24.(25 3)

x

x

CA

CA

Cx= 2.10-8 MVy hm lng Mg trong mu nc cng l:

=> %Mg =_8

_ 72.10 .24.100 .100% 9,6.10 %1000.5

Bi 2:Xc nh hm lng ca Mn bng phng php AAS, ngi ta em o dung dchA bc sng 4033 nm th cng vch ph hp th l 0,45. Dung dch B c hmlng Mn nh dung dch A cng thm mt lng 100 g/ml Mn, c cng vch hpth l 0,835. Xc nh hm lng Mn trong dung dch A.

Giip dng cng thc : A = K.CTa c: Vi dung dch X : Ax= K. CX (1)

Vi dung dch Y : AY= K. CY (2)Ly (2) chia cho (1) ta c:

X

X

X

Y

CC

AA 100

XY

X

X

AA

AC

.100

Thay s vo ta c: CX= 116,88 mLg/ Bi 3: xc nh hm lng Pb trong nc tiu nh php o ph hp th nguyn t

c th dng phng php thm chun. Tng 50,00 ml nc tiu c chuyn vo miphu di c dung tch 100 ml, thm vo mt phu 300 l dung dch chun cha 50,0

mg/l Pb. Sau pH ca dung dch c a n 2,8 bng cch thm tng git dung dchHCl. Trong mi phiu ngi ta a vo 500 l dung dch amonipyroliinithiocacbaminat mi chun b 4% trong metyl-n-amylxeton, trn cn thn ccpha nc v pha hu c chit Pb. Hm lng ca Pb trong pha hu c c xc nhbng phng php hp th nguyn t, mt khc ngi ta dng n catot rng vi nghp th 283,3 n.m. Nng Pb bng bao nhiu mg/l trong mu nc tiu ban u, nu shp th ca dch chit ca mu khng thm Pb l 0,325, cn dch chit ca mu pha thmlng Pb bit l 0,670.

GiiGi nng Pb trong 50 ml nc tiu l C xta c:

A1= K. Cx

A2= K. C2 = K._ 350. 0,3.50.10

( )50 0, 3 207.(50 0, 3)

xC


17/18

17

Ta c: 1_ 3

2

0,325

50 0,3.50.100,670

50 0, 3 207(50 0, 3)

x

x

CA

CA

Gii phng trnh trn ta c Cx= 4,15.10-6

MNng Pb tnh theo n v mg/l trong mu nc tiu ban u l:

4,15.10_6

.207.1000 = 0,85905 mg/lBi 4: xc nh Cu2+ trong mu phn tch bng phng php AAS, ngi ta ch

ho 0,628g mu vo bnh nh mc 50 ml v nh mc n vch. Ly 25ml dung dchny em c cn ri bm ton b mu vo my o AAS khe o 424,7 nm th gi tr A xo c l 0,246. Ly 25 ml dung dchcn li thm vo 2 ml dung dch chun Cu2+10

-4M, ri cng tin hnh c cn v chuyn ton b mu vo my o AAS v cng o

khe o trn th gi tr A o c l 0,312. Tnh hm lng phn trm Cu 2+ trong muphn tch?

GiiGi Cxl nng ca Cu

2+c trong mu phn tch

Theo nh lut Beer ta c: Ax = K. Cx (1)aTng t A1 = K. C1 (2) vi C1=

_ 425 2.10

27

xC

=>_ 4

1 1

27. 0,246

25 2.10 0,312

x x x

x

A C C

A C C

=> Cx = 2,1636.10

-5M

S mol Cu2+c trong mu phn tch l:

nCu =_ 5

_ 650.2,1636.10 1,0818.101000

mol

S gam Cu2+c trong mu phn tch l:

mCu= 1,0818.10

_6

.63,5= 68,6943.10

_6

gamVy % ca Cu trong mu phn tch l:

% Cu =_ 668,6943.10

.100 0,011%0,6218

Phng php vi saiBi 1: Xc nh hm lng Fe trong mt mu qung ngi ta tin hnh th nghim nh

sau:

Ly 5gam mu qung ho tan trong dung dch HNO31M c 100ml dung dch(dd x) .

Chun b hai dung dch Fe(NO3)3kho st iu kin chun c nng ln lt l

C1= 0,01M(dd1) v C2=0,013M(dd2)Tin hnh hai th nghim o ph hp th nguyn t nh sau:o cng ph hp th ca dung dch 2 so vi dung dch 1 thu c At =0,30

Ly 25 ml mu trn ri kho st iu kin chun v tin hnh o cng hp thnguyn t ca ddx so vi dd1 ta c Atx=0,50

Tnh % Fe trong mu qung?.Gii

Theo phng php vi sai nng ln ta c:At = K.(C2- C1)

Atx = K.(Cx- C1)

Lp t l ta thu c:


18/18

1 2 1

2 1x

A C C

A C C

Thay s vo v gii ta c: Cx=0,015 M

% Fe =0,015.100.56

.100%1000.5

= 0,168 %

Bi 2: xc nh hm lng As trong nc ngm mt vng cao nguyn, ngi ta ly4 lt nc em c cn thu c 6g cht rn. Ho tan lng cht rn trn bng dung dchHNO32M thu c 100ml dung dch.

Chun b hai dung dch Hg(NO3)2 kho st iu kin ti u c nng lnlt l C1= 0,001 M v C2=0,0015 M.

Tin hnh o ph hp th nguyn t nh sau: Ly 25 ml mu trn kho st iukin ti u v tin hnh o ph hp th nguyn t nh sau:

-o cng ph hp th nguyn t ca dd C1so vi dd Cx c A1=0,175-o cng ph hp th nguyn t ca dd C 2so vi dd Cx c A2=0,286

Xc nh hm lng As trong mt lt nc?

Gii.p dng phng php vi sai nng b ta c:A1= k.(C1- CX) (1)

A2= k.(C2- CX) (2)

Ly(1) chia cho (2) ta c:

X

X

CC

CC

A

A

2

1

2

1

Thay s vo ta c:

X

X

C

C

0015,0

001,0

268,0

175,0CX= 5,914.10

-5M

Khi lng Asen c trong mu phn tch l:5,914.10

-5.0,1.75 = 44,355.10

-5g

Vy hm lng A sen trong mt lit nc l:

%As = 1004/6

10.355,44 5% =0,02957%

bai tap sac ky co giai

Documents