hóa phân tích (tập 1) - pgs.ts. võ thị bạch huệ (chủ biên)

B Y T

HA PHN TCH TP 1

(DNG CHO O TO DC S I HC) M S : .20.Y.14

NH XUT BN GIO DC H NI 2007

Ch o bin son : V KHOA HC V O TO B Y T

Ch bin :

Page 1 of 239Bo Y te - Hoa phan tich

30/09/2009file://C:\Windows\Temp\nrlmohbynp\hoa_phan_tich.htm

PGS. TS. V TH BCH HU

Nhng ngi bin son : PGS. TS. V TH BCH HU ThS. PHNG TH NG ThS. TRN TH TRC THANH ThS. PHAN THANH DNG ThS. NGUYN HU LC THU

Tham gia t chc bn tho : ThS. PH VN THM TS. NGUYN MNH PHA

Li gii thiu Thc hin mt s iu ca Lut Gio dc, B Gio dc & o to v B Y t ban hnh chng

trnh khung o to Dc s i hc. B Y t t chc bin son ti liu dy - hc cc mn c s v chuyn mn theo chng trnh trn nhm tng bc xy dng b sch t chun chuyn mn trong cng tc o to nhn lc y t.

Sch HO PHN TCH c bin son da trn chng trnh gio dc ca Trng i hc Y Dc Thnh ph H Ch Minh trn c s chng trnh khung c ph duyt. Sch c cc tc gi PGS.TS. V Th Bch Hu, ThS. Phng Th ng, ThS. Trn Th Trc Thanh, ThS. Phan Thanh Dng, ThS. Nguyn Hu Lc Thy bin son theo phng chm : Kin thc c bn, h thng ; ni dung chnh xc, khoa hc ; cp nht cc tin b khoa hc, k thut hin i v thc tin Vit Nam.

Sch HO PHN TCH c Hi ng chuyn mn thm nh sch v ti liu dy - hc chuyn ngnh Dc s i hc ca B Y t thm nh nm 2007. B Y t quyt nh ban hnh l ti liu dy -

Bn quyn thuc B Y t (V Khoa hc v o to) 874 - 2007/CXB/ 5 - 1918/GD M s : 7K725 M7 - DAI



hc t chun chuyn mn ca ngnh trong giai on hin nay. Trong thi gian t 3 n 5 nm, sch phi c chnh l, b sung v cp nht.

B Y t xin chn thnh cm n cc tc gi v Hi ng chuyn mn thm nh gip hon thnh cun sch ; Cm n GS.TS. T Vng Nghi, PGS.TS. Trn T An c v phn bin cun sch sm hon thnh kp thi phc v cho cng tc o to nhn lc y t.

Ln u xut bn, chng ti mong nhn c kin ng gp ca ng nghip, cc bn sinh vin v cc c gi ln xut bn sau sch c hon thin hn.

V KHOA HC V O TO - B Y T

LI NI U HO PHN TCH l mt mn hc dnh cho cc hc vin thuc ngnh Ho. khoa Dc ca

Trng i hc Y Dc Thnh ph H Ch Minh, mn ny c ging dy cho:

- sinh vin Dc nm th 3 (h chnh quy 5 nm) - sinh vin Dc nm th 2 (h tp trung 4 nm)

nhm trang b cc kin thc c bn v ho hc phn tch t sinh vin Dc hiu r hn cc mn Bo ch, Dc liu, Ho dc, Dc lc, Kim nghim thuc l cc mn nghip v s c hc tp cc nm k tip.

Ho phn tch cng l mt trong nhng mn hc nhm cung cp cc kin thc nn tng rt quan trng sau khi tt nghip, d lm vic bt k cc v tr no th Dc s cng xc nh c nhim v ca mnh l trc tip hay gin tip gp phn lm cho thuc phi ng v cht lng v v s lng khi ti tay ngi s dng.

Vi mc ch trn, b mn Ho Phn tch Kim nghim bin son sch Ho phn tch, tp 1, cho sinh vin ngnh Dc theo ch o ca B Y t v B gio dc o to.

Ni dung sch Ho phn tch, tp 1, gm cc phn sau: Phn I: l phn i cng gm cc chng tng qut cp n:

- i tng, chc nng v phn loi ca ho hc phn tch.

- Cc loi phn ng ho hc, khi nim v cc yu t nh hng n cn bng ho hc, hng s cn bng v nh ngha v hot , nng .

- Cc khi nim v sai s v ng dng ton thng k x l cc kt qu thu c t thc nghim nhm m bo chnh xc theo yu cu ca ngnh Dc.

Phn II: l phn nh lng gm cc chng vit v:



- Cch biu th v tnh ton nng dung dch ca mt cht c s dng thng xuyn khi phn tch.

- Nguyn tc v cch th hin cc bc tin hnh phn tch khi lng. - Nguyn tc, phn loi, gii thch cc khi nim, c ch cng nh cc ng dng ca cc phng

php phn tch th tch: acid base; oxy ho kh; kt ta v to phc ch yu s dng trong ngnh Dc.

Cc cn b ging ca b mn c gng cp nht ho kin thc bin son v vn lun mong gio trnh s mang li cho hc vin kin thc b ch, chnh xc.

Chng ti knh mong qu thy c, qu ng nghip, qu c gi v cc sinh vin c k v nu nhng kin ng gp, ch gip cc thiu st chng ti hon thin hn ti liu ging dy ny.

Chn thnh cm n v mong nhn c nhng kin ng gp qu bu ca qu v.

Chng 1 I CNG HO PHN TCH

Ch bin

PGS.TS. V TH BCH HU

Phn I

I CNG

MC TIU

Trnh by c i tng, chc nng v phn loi ca ho hc phn tch.

Nu c ni dung ca ho hc phn tch.

Gii thch c cc bc thc hin ca quy trnh phn tch.



NI DUNG

1. M U

Ho phn tch l khoa hc v cc phng php xc nh thnh phn ho hc ca cht v cu trc ca cc hp phn c trong cht phn tch.

Cc phng php ca Ho phn tch cho php xc nh nh tnh mt cht, ngha l c th xc nh xem cht phn tch c cu to bi nhng nguyn t no, nhm chc no, v phn tch xem cc nguyn t v cc nhm chc c sp xp v lin kt vi nhau nh th no (phn tch cu trc).

Ho phn tch nghin cu cc phng php phn tch nh lng, dng cc phng php xc nh thnh phn nh lng ca cc nguyn t, cc hp cht ho hc trong cht phn tch.

Ni v Ho phn tch, ngi ta cho rng y l khoa hc v cc phng php v phng tin ca phn tch ho hc v trong mc nht nh xc nh cu trc ho hc. V phng tin c th hiu l: dng c, thuc th, cht chun, ... Phng php v phng tin ca phn tch thay i thng a ra nhng hng mi, s dng nhng nguyn tc mi, c nhng hin tng t nhng lnh vc c t xa xa. Th d, trong thc nghim phn tch ho hc hin nay phng php vt l gi vai tr quan trng l phng php quang ph v vt l ht nhn.

Cn phn bit khi nim "ho hc phn tch" v "phn tch ho hc". Ho phn tch l khoa hc v cc phng php phn tch, cn phn tch ho hc l nhng phng php c dng trong thc t xc nh thnh phn ho hc ca cht phn tch.

Ngi ta cn phn bit khi nim k thut phn tch v phng php phn tch. K thut phn tch l da trn cc hin tng khoa hc thu thp thng tin v thnh phn ho hc

ca cht phn tch. Th d nh k thut sc k lng hiu nng cao, k thut cc ph. Phng php phn tch l ng dng c th ca mt k thut phn tch gii quyt mt vn phn

tch. Phng php ca phn tch l phng cch (cch) xc nh, lun chng r rng, nh gi ton din c cn c xc nh thnh phn ca i tng phn tch. Th d phng php phn tch vitamin C bng k thut cc ph, phng php xc nh phenol trong nc sng dng k thut sc k.

2. CHC NNG CA HO PHN TCH HIN I C th chia ra lm 3 chc nng (lnh vc khoa hc):

2.1. Gii quyt cc vn chung ca phn tch

Th d: pht trin v hon thin nhng lun thuyt v cc phng php phn tch.

2.2. Nghin cu cc phng php phn tch

2.3. Gii quyt cc nhim v c th ca phn tch

Th d: s thnh lp chuyn ngnh Ho phn tch v thuc phng chng dch bnh. Cu trc su ca mn hc ny c th chia ra phn tch nh tnh v phn tch nh lng. Phn tch

nh tnh gii quyt vn l cc hp phn no c trong i tng phn tch, phn tch nh lng cho bit v hm lng ca tt c hay ca tng hp phn.

3. PHN LOI



3.1. Phn loi theo ng li phn tch

3.2. Phn loi da theo th tch v khi lng cht phn tch

- Phn tch th: lng mu th cht rn 0,1 - 1 g, lng mu th dung dch t 1 - 100 ml.

- Phn tch bn vi lng: lng mu th t 0.01 - 0,1 g, dung dch t 0,1- 0,3 ml.

- Phn tch vi lng: lng mu th t 10-3 -10-2 g, dung dch t 10-2 - 10-1 ml.

- Phn tch siu vi lng: lng mu th t 10-6 - 10-12 g, dung dch t 10-3 - 10-6 ml.

3.3. Phn loi da trn bn cht ca cc hp phn ca cht cn xc nh

- Phn tch ng v: chuyn ngnh trong phn tch t s dng, nhng phn tch ny thng dng trong vt l, m, sinh hc. Th d nh: xc nh nc deuterium trong nc thng, cng nh ca oxy "nng" (ng v oxy 18) trong hn hp vi s ng v ph bin oxy 16. Phn tch ng v cn thit khi nghin cu cc nguyn t nhn to.

- Phn tch nguyn t (nguyn t - ion): phn tch nhng nguyn t no c trong i tng nghin cu, hm lng l bao nhiu.

- Phn tch phn t: y l s pht hin v xc nh hp cht ho hc c c trng bng khi lng phn t xc nh. Th d: phn tch hn hp kh, xc nh trong khng kh nhng thnh phn chnh (N2, O2, CO2, O3, kh tr). Mt trong nhng phng php phn tch phn t ngi ta s dng phng php sc k.

- Phn tch nhm chc: i vi phn tch ho hu c cn c mt dng phn tch gia phn tch nguyn t v phn t - l phn tch nhm chc. Phn tch ny trc ht xc nh nhm chc, ngha l xc nh tng nhm hu c ring bit nh nhm carboxyl, hydroxyl, amin,...

- Phn tch cht: Trong phn tch cht ngi ta xc nh trong dng no c mt hp phn ta quan tm trong i tng phn tch ny v hm lng ca nhng dng ny. Th d: trong mc no ca s oxy ho c mt nguyn t As(III) hay As (V), trong trng thi ho hc no c mt nguyn t (th d ng trong khong cht c th dng oxyd hay sulfit hay hn hp nhng hp cht ny). Phn tch cht c nhiu ci chung vi phn tch phn t hay tng.

- Phn tch tng (pha): l phn tch i tng trong h d th. Th d: sunfit v oxyd km phn b trong khong cht khng ng th m to nhng tng khc nhau. Ngi ta dng cc phng php khc nhau tch v xc nh cc h d th tham gia trong thnh phn cu trc (cc tng) ring bit, khc nhau v tnh cht, v cu trc vt l v phn tch nhau bi gii hn b mt.

3.4. PHN LOI THEO BN CHT CA PHNG PHP Xc nh theo c tnh ca tnh cht o hay theo kh nng ghi nhn tn hiu tng ng.

C th chia ra:

- Phng php ho hc: l phng php da trn nhng phn ng ho hc. Th d: phn ng acid-

- Tng th - Cc b - Trc tip - Gin tip - Phn hy - Khng phn hy - Gin on - Lin tc



base, oxy ho - kh, kt ta - ha tan, to phc. - Phng php ho l, vt l: phng php vt l da trn nhng hin tng v qu trnh vt l

nh phng php quang ph, phng php so mu, vt l ht nhn,... - Phng php sinh hc: da trn nhng hin tng ca cuc sng (trao i cht, tng trng, c

ch ca vi sinh vt...), th d nh phng php phn tch vi sinh vt l da vo qu trnh trao i cht ca vi sinh vt. Bng phng php vi sinh vt ngi ta nh lng cc thuc khng sinh, khng nm, vitamin,...

4. PHN TCH NH TNH V PHN TCH NH LNG 4.1. Phn tch nh tnh

Phn tch nh tnh l xc nh cc nguyn t, cc ion, cc phn t c trong thnh phn cht phn tch.

tin hnh phng php nh tnh ngi ta dng nhiu phng php c bn cht khc nhau, cc k thut, cc ng li khc nhau: phng php ho hc, vt l v ho l.

4.1.1. Phng php ho hc Phng php ho hc l nhng phng php nh tnh da trn cc phn ng ho hc. Nhng

phng php trong phn tch c s dng rng ri trong thc t, phng php ny khng i hi trang b phc tp nhng cn c mt s nhc im l: trong trng hp cn phi tch cht ra khi cc tp cht v phi tch dng tinh khit thng rt kh khn, i khi khng thc hin c, khng pht hin c nhng lng cht rt nh.

4.1.2. Phng php vt l v ho l Phng php vt l v ho l l nhng phng php da trn cc tnh cht vt l ca cc cht cn

phn tch.

Th d: S nhum mu ngn la n kh khng mu thnh mu c trng do cc hp cht bay hi ca mt s nguyn t ho hc Na+, K+,...

Phng php so mu ngn la: Ngn la Na+ c mu vng, K+ c mu tm, Ba2+ c mu xanh lc nht.

Phng php phn tch quang ph hng ngoi, phng php o quang (quang ph hp th), phng php cc ph, phng php sc k v.v...

Dng phng php vt l v ho l c th tch c cc nguyn t m chng kh b tch bi cc phng php ho hc, phng php vt l v ho l c nhy v chnh xc cao nhng phi c trang b phc tp, my mc t tin khng phi tt c cc c s u trang b c nn phng php ho hc vn ng vai tr quan trng trong cng tc phn tch.

4.2. Phn tch nh lng

Phn tch nh lng cho php xc nh thnh phn v lng cc hp phn ca hp cht cho hoc ca hn hp cc cht. Khc vi phn tch nh tnh, phn tch nh lng cho kh nng xc nh hm lng ca cc hp phn ring r ca cht phn tch hoc hm lng chung ca cht cn xc nh trong cht nghin cu.

Cc phng php dng trong phn tch nh lng:



4.2.1. Phng php phn tch ho hc - Phn tch trng lng (phn tch khi lng) Phng php ny da vo phn ng kt ta cc cht cn nh lng vi thuc th. Kt ta c tch

ra khi dung dch, ra tht sch ri nung hoc lm kh sau em cn. T khi lng ca kt ta ta xc nh c khi lng ca cht cn nh lng.

Phng php ny mt nhiu thi gian nhng kh chnh xc.

- Phn tch th tch (chun ) Phng php ny da trn s o th tch dung dch thuc th bit chnh xc nng cn dng

cho phn ng vi cht cn nh lng. Phng php ny rt ph bin, da trn cc phn ng trung ha, to phc, to kt ta, oxy ho - kh,...

4.2.2. Phng php phn tch vt l v ho l (phng php phn tch dng c) - Phng php phn tch vt l Cc phng php phn tch nh lng cho php xc nh thnh phn ca cht cn phn tch khng

dng n cc phn ng ho hc. xc nh thnh phn cht cn phn tch c th o cc ch s v cc tnh cht vt l nh: h s khc x, dn in, nhit,...

- Phng php ho l Cc phng php phn tch da trn s thay i tnh cht vt l ca mt h cn phn tch, s thay

i xy ra do cc phn ng ho hc xc nh. Cc phng php vt l, ho l thng dng: cc phng php sc k, cc phng php o quang, phng php in ho, cc ph, in di,...

5. CC BC THC HIN CA MT QUY TRNH PHN TCH 5.1. Mu th - xc nh i tng

u tin phi xc nh r mc tiu (cn nhng thng tin g) v yu cu phn tch (nh tnh hay nh lng). Thu thp thng tin v mu th: bn cht, ngun gc, cch ly mu, tnh trng mu v bo qun mu.

5.2. La chn phng php

La chn phng php phn tch da trn nhng thng tin c trc nh: c mu phn tch, phng tin phn tch, yu cu phn tch,.. t kt qu phn tch tt ph thuc nhiu vo s la chn phng php.

5.3. Ly mu th

y l bc quan trng nht trong c quy trnh phn tch. Chn mu i din c thnh phn phn nh ng thnh phn mu cn phn tch. T mu i din chn v chun b mu lm th nghim v phn tch theo yu cu.

5.4. X l mu th phn tch, mu th phi c x l l tch cc cht cn tr ra khi hn hp trc khi o. y l

giai on quan trng trong phn tch.



5.5. Tin hnh o cc cht phn tch

S dng nhng dng c, my mc thch hp o cht cn phn tch.

5.6. Tnh ton - x l kt qu phn tch

Cc d liu thu c x l theo ton thng k nh gi tin cy ca kt qu o c. Cc bc trn lin quan mt thit vi nhau v nh hng ln nhau. Trong thc t, tu theo tng trng hp c th, cc bc tin hnh trn c n gin ho hoc b qua mt s bc, hoc thc hin ng cc bc trn.

6. HO PHN TCH LIN QUAN TI CC KHOA HC KHC Ho phn tch khng th v khng ch l mt phn ca ngnh ho hc m n lin quan mt thit vi

cc ngnh khc nh: vt l v k thut. Phn tch ho hc phn ln da trn cc thnh tu ca quang ph (quang hc, phng x...), vt l ht nhn v nhiu phn khc ca vt l.

Cc phng php phn tch ho hc c s dng trn nn tng cc thnh tu ca cc ngnh ho khc nh: l thuyt v cn bng ho hc, in ho, ng ho hc, ho v c, ho hu c, ho keo. Ngoi ra Ho phn tch cn lin quan ti ton hc v sinh hc.

Nh vy c th ni rng Ho phn tch l c trng ca khoa hc gm nhiu ngnh, khoa hc lin quan.

7. HO PHN TCH VI NGNH DC Trong ngnh Dc, Ho hc phn tch gi vai tr quan trng. N c mt trong sut qu trnh sn

xut (kim nghim nguyn liu, bn thnh phm, v sn phm cui cng), bo qun, lu thng v s dng thuc. Cc phng php phn tch s dng rt phong ph v a dng thuc cc lnh vc: vt l, ho hc v sinh hc. Hin nay ngi ta cn quan tm nhiu ti cc phng php kim nghim dng trong i tng sinh hc, (th d nh phn tch mu, nc bt, dch no ty, gp phn vo cc nghin cu sinh dc hc v dc ng hc).

8. NG DNG V HNG PHT TRIN CA HO PHN TCH 8.1. ng dng

Ho phn tch ng vai tr quan trng trong s pht trin khoa hc k thut v nhiu mn khoa hc t nhin: ho hc, a cht, a l, khong vt hc, vt l, sinh hc, sinh ho, nng ho, y dc hc. Trong nhng lnh vc ny thc hin cc cng trnh nghin cu khoa hc u i hi phi p dng cc phng php phn tch. Da vo cc thng tin ca phn tch ho hc m cc nh a cht tm kim cc khong cht. Da vo kt qu phn tch mu cc thy thuc chn on c tnh trng sc khe ca con ngi. S nh lng cc ion K+, Ca2+, Na+ trong dch t bo ng vt cho php cc nh sinh l hc nghin cu vai tr ca cc ion ny trong s dn truyn lung thn kinh cng nh trong c ch co v dui ca cc c. Cc nh ho hc gii thch cc c ch phn ng ho hc nh vo vic nghin cu vn tc phn ng, nh c cc phng php phn tch hin i m ngi ta tng hp c cc cht ho hc mi, ...

8.2. Hng pht trin

Ngy nay vi s pht trin ca khoa hc k thut tin tin i hi chuyn ngnh Ho phn tch ngy mt hon thin, p ng nhu cu trn Ho phn tch tin ti phi gii quyt c 5 vn sau:



8.2.1. Xc nh v gii quyt nhng kh khn trong Ho phn tch Cn phi xc nh cc thng tin c th (nh tnh, nh lng, tnh cht, chc nng). Xc nh tnh hung c th.

8.2.2. Thit k mt quy trnh phn tch Thit lp tiu chun cho mt quy trnh phn tch ( chnh xc, ng, nhy, chi ph, quy m

thc hin, tin thc hin...) Xc nh cc yu t cn tr.

La chn phng php.

Thit lp cc tiu chun thm nh.

Cch ly mu.

8.2.3. Thc hin v thu thp d liu Kim nh dng c v trang thit b.

Chun ho thuc th.

Thu thp d liu. 8.2.4. X l d liu

8.2.5. Thc hin nh gi ca c s bn ngoi Kim tra nh gi li kt qu phng th nghim ca c s bn ngoi. Tu iu kin thc t, tnh hnh c th ca c s c th khng gii quyt c ngay ht cc vn

m tng bc gii quyt tin ti hon chnh nh cc yu cu t ra. Hin nay nc ta ngnh Dc ang pht trin, cc x nghip Dc c phn trong nc v lin

doanh ra i, bn cnh cc phng bo ch cc khoa Dc cc bnh vin cng pht trin khng ngng, ngoi ra cc ngun thuc nhp khu cng rt phong ph t cc nc pht trin v ang pht trin. m bo cht lng thuc sn xut v thuc xut nhp khu n tay ngi tiu dng, cng tc kim nghim thuc ngy cng c quan tm, v vy ngi Dc s cn trang b nhng kin thc vng vng v Ho hc phn tch. Ngoi cng tc kim nghim thuc, cc cng tc khc trong cc phng th nghim ho dc, dc l, dc liu, dc lm sng, sinh ho, c cht,... cng i hi nhng kin thc v Ho hc phn tch.

T KHO

- Gim hoc bin i d liu. - nh gi kt qu. - Phn tch thng k. - Biu th kt qu.

- Ho hc phn tch - Phn tch cht

- Phn tch nh lng - Phn tch vi lng, bn vi lng, siu vi lng - Phn tch phn t - Phn tch nguyn t

- Phn tch nh tnh - Quy trnh phn tch



CU HI LNG GI 1. Cho bit i tng ca ho hc phn tch. S lin quan ca ho hc phn tch vi ngnh Dc v

cc mn khoa hc khc.

2. Th no l phn tch nh tnh? Nu cc phng php phn tch trong phn tch nh tnh.

3. Nu cc phng php dng phn tch nh lng.

4. Trnh by cc bc thc hin ca mt quy trnh phn tch.

CU HI TRC NGHIM Hy chn t ng thch hp in vo ch trng trong cc cu 1 v 2 sao cho ph hp:

1. Ho phn tch l khoa hc v s xc nh......... ca cht phn tch

2. Phn tch nh lng cho php xc nh..... ca cc hp phn trong cht nghin cu:

3. Phng php ho hc l phng php da trn

4. Ho phn tch ng vai tr quan trng trong s pht trin ca mn khoa hc

5. C my bc ch yu ca mt quy trnh phn tch

Chng 2

a. phn ng ho hc c. thnh phn e. tnh cht ho hc

b. thnh phn ho hc d. nhm chc

a. cu trc c. th tch e. hm lng

b. thnh phn d. trng lng

a. tnh cht ho hc c. phn ng ho hc e. cu trc ho hc

b. thnh phn ho hc d. hin tng ho hc

a. y dc hc c. khong vt hc e. ch cu a,b

b. a cht d. tt c cc cu trn

a. 3 c. 5 e. 7 b. 4 d. 6



CN BNG HO HC

NI DUNG

1. PHN NG HO HC 1.1. nh ngha

Phn ng ho hc l phn ng khi lin kt mt hay nhiu hn 2 nguyn t th cc lin kt ho hc trong cht tham gia phn ng thay i v to ra cht mi (sn phm). Qu trnh ny lun km theo 1 s thay i nng lng v tun theo nh lut bo ton nng lng. Phn ng ho hc kt thc khi c s cn bng phn ng ho hc hay cc cht tham gia phn ng c chuyn i hon ton.

1.2. Cc loi phn ng ho hc

Cc ti liu v phn ng ho hc hin nay phn loi nhiu kiu phn ng theo cc c ch nhng ch yu tp trung cc loi sau:

1.2.1. Phn ng kt hp (synthesis reaction)

Trong phn ng ny, 2 hay nhiu cht n gin s kt hp to nn cht phc tp hn.

Dng c bn: A + X AX

Th d: kh hydro kt hp vi kh oxy c th to nn cht phc tp hn l nc

2H2 + O2 H2O

Th d 1: kim loi + oxy oxid kim loi*:

2Mg(r) + O2(k) 2MgO(r)

Th d 2: phi kim + oxy oxid phi kim: C(r) + O2(k) CO2(k)

MC TIU

Trnh by c nh ngha v phn loi phn ng ho hc.

Trnh by c khi nim v s cn bng ho hc.

Trnh by c cch tnh cc hng s cn bng ho hc.

Din t c s khc bit gia hot v nng .

Gii c mt s bi tp cn bng ho hc.



Th d 3: oxid kim loi + nc hydroxid kim loi

MgO(r) + H2O(l) Mg(OH)2(r)

Th d 4: oxid phi kim + nc acid: CO2(k) + H2O(l) H2CO3(nc)

Th d 5: kim loi + phi kim mui: 2 Na(r) + Cl2(k) 2NaCl(r)

Th d 6: Vi phi kim kt hp vi nhng cht khc: 2P(r) + 3Cl2(k) 2PCl3(k)

C 2 phn ng thuc loi ny cn ghi nh N2(k) + 3H2(k) 2NH3(k)

NH3(k) + H2O(l) NH4OH(nc)

1.2.2. Phn ng phn hy (decomposition reaction)

Phn ng phn hy l mt phn ng ho hc tch ri do mt cht t hy hay chia ra thnh cc n cht*

Dng c bn: AX A + X

2H2O 2H2 + O2

Th d 1: carbonat kim loi, khi un nng to thnh cc oxid kim loi v kh CO2(k).

CaCO3(r) CaO(r) + CO2(k)

Th d 2: Hu ht cc hydroxid kim loi, khi un nng phn hy thnh oxid kim loi v nc. Ca(OH)2(r) CaO(r) + H2O(k)

Th d 3: Cc clorat kim loi khi un nng phn hy thnh clorid kim loi v oxy. 2KClO3(r) 2KCl(r) + 3O2(k)

Th d 4: Mt s acid khi un nng phn hy thnh oxid phi kim v nc

H2SO4 H2O(l) + SO3(k)

Th d 5: Mt s oxid khi un nng phn hy:

2HgO(r) 2Hg(l) + O2(k)

* K hiu (k = th kh = gaz); (l = th lng = liquid); (r = th rn = solid); (nc = dung mi, nc = aqueous)



Th d 6: Mt s phn ng phn hy c sn xut do in phn 2H2O(l) 2H2(k) + O2(k)

2NaCl(l) 2Na(r) + Cl2(k)

1.2.3. Phn ng trao i (replacement reaction)

Dng c bn: A + BX AX + B hay AX + Y AY + X

Th d 1: Trao i 1 kim loi ca 1 cht bng kim loi c hot tnh hn Fe(r) + CuSO4(nc) FeSO4(nc) + Cu(r)

Th d 2: Trao i nguyn t hydro c trong nc bng kim loi c hot tnh 2Na(r) + 2H2O(l) 2NaOH(nc) + H2(k)

Mg(r) + H2O(k) MgO(r) + H2(k)

Th d 3: Trao i nguyn t hydro c trong acid bng kim loi c hot tnh

Zn(r) + 2HCl(nc) ZnCl2(nc) + H2(k)

Th d 4: Trao i phi kim bng kim loi c hot tnh hn. Cl2(k) + 2NaBr(nc) 2NaCl(nc) + Br2(l)

a) Phn ng trao i n (single replacement reaction) Phn ng trao i n l mt phn ng ho hc ca mt n cht vi acid. Trong phn ng trao i

n th mt nguyn t khng kt hp s thay th nguyn t khc trong mt hp cht. Hai cht tham gia phn ng nhng ch cho 2 sn phm.

Th d: Natri kt hp vi acid hydrochloric th natri s thay th hydro

2Na + 2HCl 2NaCl + H2

b) Phn ng trao i kp (double replacement reaction) Trong phn ng trao i kp th 2 hp cht chuyn v tr to thnh 2 hp cht mi. Hai cht tham

gia phn ng nhng ch cho 2 sn phm mi.

Th d: Bc nitrat kt hp vi natri clorid s to 2 hp cht mi l bc clorid v natri nitrat

AgNO3 + NaCl AgCl + NaNO3

Fe2O3 + 6HCl 2FeCl3 + 3H2O

c) Phn ng trao i ion Phn ng trao i ion l phn ng ho hc xy ra gia cc ion trong dung dch nc. Mt phn ng

s xy ra khi cp ion cng nhau trao i to thnh t nht mt trong nhng cht sau: kt ta, kh, nc hay cht khng ion ho khc.



Dng c bn: AX + BY AY + BX

Th d 1: Phn ng to ta

NaCl (nc) + AgNO3(nc) NaNO3(nc) + AgCl(r)

BaCl2(nc) + Na2 SO4(nc) 2NaCl(nc) + BaSO4(r)

Th d 2: Phn ng to kh

HCl(nc) + FeS(r) FeCl2(nc) + H2S(k)

Th d 3: Phn ng to nc. (Nu phn ng gia acid v base th gi l phn ng trung ha) HCl(nc) + NaOH(nc) NaCl(nc) + H2O(l)

Th d 4: Phn ng to ra mt sn phm phn hy CaCO3(r) + HCl(nc) CaCl2(nc) + CO2(k) + H2O(l)

1.2.4. Phn ng oxy ho kh Cht tham gia phn ng b oxy ho (mt mt hay nhiu electron) v cc cht tham gia khc b kh

(nhn mt hay nhiu electron). Cc th d chung ca cc phn ng oxy ho - kh lin quan n oxy l s st r ca kim loi nh st

(cc kim loi b oxy ho bi oxy ca kh quyn), s t chy v chuyn ho nng lng trong c th. C10H8+ 12O2 10CO2 + 4H2O + nhit

CH2S + 6F2 CF4 + 2HF + SF6 + nhit

Mt th d ca phn ng oxy ho - kh khng lin quan n oxy ca kh quyn l phn ng to in trong bnh acquy:

Pb + PbO2 + 4H

+ + 2SO4

2 2PbSO4 + 2H2O

2. KHI NIM V S CN BNG HO HC Khi kho st cc phn ng ho hc ngi ta nhn thy c rt t phn ng ho hc di chuyn ch theo

mt chiu. a s cc phn ng ho hc u di chuyn thun nghch. Khi bt u c s to thnh sn phm l c phn ng theo chiu ngc li hay c th ni ngay t lc c nhng phn t sn phm xut hin l bt u c phn ng nghch.

Khi tc ca phn ng di chuyn theo chiu thun v theo chiu nghch bng nhau v nng ca cht tham gia phn ng v ca nng sn phm khng thay i theo thi gian na l cn bng ho hc t n.

Cn bng ho hc l mt qu trnh cn bng ng. Hy tng tng nh nhng ngi ang trt tuyt trn 1 dc thoi. Lng ngi ang i ln v trt xung lin tc ni tip nhau thay i nhng tng s ngi tham d trt tuyt t nh ti chn dc khng thay i.

Khi ni n cn bng ho hc l ni n c t nht 2 cht khc nhau: cht tham gia phn ng v sn phm.



Khi ni n cn bng vt l l ch cp n mt cht c trong 2 pha (th) khc nhau v s thay i sn phm ch l thay i v l tnh. Th d trong s bay hi nc s mol H2O thot ra khi th lng bng vi s mol H2O chuyn thnh th kh:

H2O (l) H2O (k)

Mt th d in hnh ca cn bng ho hc l phn ng:

y l phn ng to thnh kh nit dioxyd (NO2) t kh dinit tetraoxyd (N2O4). C th theo di din bin ca phn ng ny d dng do N2O4 l kh khng c mu trong khi NO2 c mu nu m. Nu bm mt lng chnh xc N2O4 vo mt bnh cu trong l chn khng th mu nu ca s xut hin ngay chng t l NO2 c mt. Mu nu s m dn do N2O4 t phn ly. n khi t cn bng th mu khng i na.

i vi phn ng ngc vi phn ng (2.1)

Trong cc hnh minh ha 2.1 v 2.2 vng mu nu en i din cho kh NO2 v vng trng i din cho kh N2O4.

Nu bt u vi NO2, c th minh ha cn bng t n nh sau:

Hnh 2.1. Chuyn 2NO2 (k) N2O4 (k), h thng cn bng cc 4, 5 v 6

Nu bt u vi N2O4, cng nhit c th minh ha cn bng t n l

Hnh 2.2. Chuyn N2O4 (k) 2NO2 (k), h thng cn bng cc 10, 11 v 12

Tu theo nhit v khi lng ban u ca N2O4 (c th c ln NO2 hay khng) s dng m nng ca NO2 v N2O4 trng thi cn bng ca h ny s khc h khc.

Ch : - Mt h thng cn bng khng cn bt c nng lng no duy tr s cn bng.

N2O4 (k) 2NO2 (k) (2.1)

2NO2(k) N2O4 (k) (2.2)



- Khng nn ln ln trng thi cn bng v trng thi tnh. Trng thi tnh l trng thi m cc nng u khng i theo thi gian.

- Cn bng ho hc l trng thi ng trong cht tham gia phn ng v sn phm chuyn i lin tc cho nhau. Tc mt i v xut hin ca chng bng nhau.

Bng 2.1 nu vi d liu thc nghim lin quan n phn ng ny 25 C. Nng ca kh c din t bng nng mol v c tnh t s phn t ca kh c mt t lc phn ng bt u cho n lc phn ng cn bng xy ra trong 1 th tch ca bnh cu (tnh bng lt).

Phn tch cc d liu t s cn bng s cho thy trong khi t s cho nhng gi tr thay

i th t s li gn nh hng nh v trung bnh l 4,63 x 10-3.

Gi tr trung bnh ny gi l hng s cn bng K i vi phn ng ny 25 C. Biu thc ton hc ca hng s cn bng i vi h NO2 - N2O4 l

Cng cn ch l s m 2 ca [NO2] trong biu thc trn bng h s cn bng t lng ca NO2trong phng trnh (2.1).

Khi nim ny thu c khi xt phng trnh ca phn ng thun nghch:

A, B: Cc cht tham gia phn ng

C, D: Sn phm to thnh.

Phng trnh (2.3) l biu thc ton hc tun theo nh lut tc dng khi lng.

A + B C + D (2.3)

Bng 2.1. Gi tr cc t s nng ca NO2 v N2O4

Nng ban u (M) Nng cn bng (M) Gi tr cc t s nng

[NO2] [N2O4] [NO2] [N2O4]

0,000 0,670 0,0547 0,643 0,0851 4,65 10-3

0,0500 0,446 0,0457 0,448 0,102 4,66 10-3

0,0300 0,500 0,0475 0,491 0,0967 4,60 10-3

0,0400 0,600 0,0523 0,594 0,0880 4,60 10-3

0,200 0,000 0,0204 0,0898 0,227 4,63 10-3



Hnh 2.3. Bin i nng ca NO2 v N2O4 theo thi gian

a) Lc u ch c NO2; b) Lc u ch c N2O4; c) Lc u c hn hp NO2 v N2O4

3. HOT V NNG Khi ni n hot v nng cn xc nh r 2 loi nng :

- Nng mol/l (phn t gam / l) ch s phn t gam cht tan (cht tan c th l phn t, ion) c trong mt 1000 ml dung dch.

- Nng molan ch s phn t gam cht tan c trong mt 1000 gam dung dch. Khi nim hot v h s hot do nh ho l ngi Hoa K Lewis a ra (1907). hiu r

hn s xt th d bng 2.2

Biu thc tnh pH l pH= - lg[H+] (l mt cng thc rt quen thuc din t s lin quan gia pH v nng mol ca ion H+).

Theo bng 2.2, nu so snh pH tnh ton theo cng thc v pH o c qua thc nghim th s nhn thy c sai s gia gi tr pH l thuyt v thc nghim.

L do c sai s nh thy l do biu thc pH= - lg[H+] ch chnh xc trong mt gii hn pha long v pH tht s tnh theo hot ion hydro ch khng tnh theo nng .

Ngha l pH= -lg aH+

Hot aH+ ca ion hydro c trong mt dung dch cho c th c xc nh bng cch n gin l o pH.

Bng 2.2. S khc bit v pH tnh theo l thuyt v thc nghim

Mt cch tng qut, gia hot v nng c mi lin quan theo cng thc:

a = C hay a = f C

vi a: hot ; (hoc f): h s hot ; C: nng mol;

Nng mol ca HCl pH theo l thuyt pH theo thc nghim Sai s pH 0,00050 3,30103 3,31 -0,00897

0,0100 2 2,04 -0.04

0,100 1 1,10 -0,1

0,4 0,39794 0,52 -0,12206



Hot da trn cc nng cho c p dng tt i vi cc dung dch khng in ly (hay cc hp cht phn t). Trong dung dch pha long, h s hot thng = 1 cho nn hot = nng .

Trong dung dch in ly, do cc tc ng tng h ca in tch nn cn quan tm n vi yu t c bit tnh c h s hot nh:

- Lc ion: Khi bn n dung dch ion th t nht phi bit v lc ion ca dung dch ny v ngi ta tin rng lc ion s nh hng n hot ion. i lng ny c k hiu l I

Mt cht in ly MxXm c phn ly thnh:

MxXm = x Mm+ + m Xx-

Cc ion dng Mm+ v cc ion m Xx- phi cng c mt trong mt dung dch v khng c cch no tch ring hot ca cc ion dng v m. Do vy, thng dng i lng hot ion trung bnh (a)

a = C (MmXx)-(m+x)

i vi c 2 ion dng v m, C l nng t lng ca cht in ly (stoichiometric concentration). Trng hp ny h s hot trung bnh c tnh theo biu thc:

Trong + v - ln lt l s cation v anion trong 1 n v cng thc ca cht in ly.

Ngoi ra, phng trnh Debye Huckel m rng (ci tin t nh lut gii hn ca Debye Huckel) cng l biu thc c c s chc chn lm nh i cch tnh nng khi tnh theo h s hot trung bnh:

Trong A l hng s tu thuc vo tnh cht ca dung mi (A= 0,5085 i vi nc 25oC), z+ v z- l cc in tch ion, I l lc ion c xc nh theo phng trnh:

I = zi2mi

trong mi l nng ca ion i. Tm li, I c tnh n tt c cc ion c th c trong dung dch ny.

i vi cc dung dch m c, vic s dng nng tnh theo trng lng ca dung mi (nng molan) thng c th cung cp mt c lng tt hn l dng nng tnh theo th tch (nng mol).

4. HNG S CN BNG (HSCB)

4.1. Cch tnh hng s cn bng

nh lut tc dng khi lng l nh lut biu din mi lin quan gia nng (hot ) ca cc cht phn ng v ca sn phm phn ng trng thi cn bng.

nh lut tc dng khi lng ch p dng c vi cc dung dch l tng l dung dch ca cc



cht khng in ly v cc cht in ly yu vi nng rt long (hot ). Trong thc t, thng thng nng ca cc dung dch khng tun theo pha long mt cch chnh xc.

Trong trng hp cc cht in ly mnh hoc cc cht in ly yu nhng nng cao th s sai lch rt ln, y l gii hn ca nh lut tc dng khi lng. V th khi p dng nh lut tc dng khi lng ch nhn c cc kt qu gn ng. Tuy vy, trong nhiu trng hp chnh xc ny vn chp nhn c vi cc mc ch thc tin.

Th d: i vi phn ng trng thi cn bng

Nu tnh theo hot Khi mt phn ng t trng thi cn bng th t s gia tch cc hot ca cc sn phm phn ng

(c nng lu tha vi s m bng h s t lng trong phng trnh phn ng) v tch tng ng ca cc cht phn ng l mt hng s ( nhit cho sn).

Khi t cn bng th t s

Ka c gi l HSCB, trong aA, aB, aC, aD l hot cn bng ca cc cht phn ng (A, B) v cc sn phm phn ng (C, D); m, n, p, q l cc h s t lng tng ng.

Nu tnh theo nng Thay cho hot , HSCB cng c th biu din qua nng Ci, p sut ring Pi hoc phn mol Xi

ca cc cu t i tng ng. Trong cc trng hp ny HSCB c k hiu Kc, Kp, Kx.

Kc - hng s cn bng:

Trong phng trnh 2.6 [A], [B], [C], [D]: nng cn bng tng ng ca cht phn ng v sn phm, m, n, p, q: cc h s cn bng (cc h s t lng tng ng). Nu coi (2.4) l phn ng c bn (xy ra trong mt giai on, mt va chm) theo chiu t tri sang

phi th tc phn ng theo chiu thun c biu din bng phng trnh:

Trong , V1 l tc phn ng; m, n l cc h s t lng;

K1: hng s tc ; [A] v [B] l nng tng ng ca cht phn ng A v B thi im kho st.

Nu coi (2.4) l phn ng c bn hai chiu, thun v nghch, V2 l tc phn ng xy ra theo chiu ngc li. Phn ng c c trng bng hai hng s tc K1 v K2 vi

mA + nB pC +qD (2.4)

(2.5)

(2.6)

(2.7)

(2.8)



Khi phn ng t ti trng thi cn bng th V1 = V2

Ngha l K1[A]m[B]n = K2[C]p[D]q

Trn thc t, (2.4) thng biu din mt phn ng phc tp (bao gm nhiu giai on, qua nhiu va chm) th biu thc (2.6) ni chung khng ng. Trong trng hp ny hm tc phn ng ph thuc nng c dng phc tp. Ch trong mt s trng hp tc phn ng c th biu din bng hm m:

Trong n1 v n2 tng ng l bc phn ng theo cc cu t 1 v 2. C: nng ca cc cht tham gia hoc sn phm to thnh sau phn ng.

Nh vy, n1 m v n2 n. Bc phn ng l i lng c xc nh bng thc nghim, khng lin quan n cc h s t lng. Bc c th l s nguyn, phn s hay s khng.

4.2. Hng s cn bng ca cc phn ng hay s dng trong ho hc phn tch

4.2.1. Hng s cn bng ca cc phn ng phn ly (HSPL) L hng s cn bng ca phn ng phn ly mt cht. Th d: i vi phn ng phn ly acid yu HA:

HA = H+ + A-

Ka c gi l HSPL nhit ng; a: hot ca ion hoc phn t; g: h s hot . Trong nhng trng hp, khi c th chp nhn h s hot khng i:

K'a c gi l HSPL iu kin. Ka ph thuc nhit , khng ph thuc nng ; cn K'a ph thuc c nng v nhit . Bn cnh Ka ngi ta cn dng khi nim pKa = - lgKa.

Th d. i vi acid acetic CH3COOH 25

oC, Ka = 1,754. 10-5

v pKa = 4,76.

Trong hng s phn ly ngi ta c th lm r hn vi hng s khc l:

a) Hng s phn ly acid (phn ng acid base) Phn ng ca mt acid vi dung mi (nh nc) c gi l phn ng phn ly acid. Acid chia lm 2

loi:

- Acid mnh nh l HCl chuyn ht proton ca n cho dung mi.

(2.9)

(2.10)



- Acid yu nh acid acetic khng th cho ht proton acid ca n cho dung mi. Thay vo hu ht acid cn li khng phn ly, ch mt phn nh hin din di dng base lin hp

Hng s cn bng ca phn ng ny c gi hng s phn ly acid Ka. Ka ca acid acetic l 1,75 x 10-5

Ch : nng ca nc c th b qua trong biu thc tnh Ka v gi tr qu ln khng b nh hng bi phn ng phn ly. ln ca Ka cung cp thng tin v lc tng i ca acid vi Ka cng nh th acid cng yu. (trnh by trong chng 7)

b) Hng s phn ly base (phn ng acid base): Hng s phn ly base l hng s cn bng gia base yu v proton m base nhn to nn acid v

acid .

Th d gia NH3 v proton to nn acid NH4+ v acid , c cn bng:

NH3 + H+ NH4

+

Vi hng s cn bng

Kb c gi l hng s phn ly base. V cng ca base c biu th bng Kb nn Kb cng ln, base kt hp vi proton cng nhiu, ngha l base cng mnh. Th d NH3 c Kb = 1,76.10

-5 mnh hn

base CH3COO-

c Kb = 1,74.10-9

.

c) Hng s khng bn ca phc cht (HSKB) Hng s cn bng gia phc cht v cc tiu phn (ion, phn t) to nn phc cht . Th d: gia ion phc Ag(NH3)2+ v Ag+, NH3 (do ion phc phn ly) c cn bng:

Hng s cn bng Kb ny c gi l HSKB ca ion phc Ag(NH3)2+; v Kb cng ln, ion phc phn li cng nhiu nn Kb cn c gi l hng s phn ly ca phc cht. (trnh by trong chng 10).

4.2.2. Hng s cn bng ca cc phn ng kt hp (HSKH) L hng s cn bng gia ion t do v cp ion (do hai ion khc du kt hp vi nhau to thnh).

CH3COOH (nc) + H2O (l) H3O+ (nc) + CH3COO- (nc) (2.11)



Th d: i vi KCl (kali clorid) gia cc ion t do K+, Cl- v cp ion K+Cl- c cn bng:

HSKH ph thuc vo hng s in mi D ca dung mi v nng cht tan: Kkh ca KCl 0,1 M trong acid acetic vi D = 6,2 l 106,9 v trong metanol vi D = 32,6 l 101,15. Trong dung mi c D cng nh, Kkh cng cao; trong benzen, clorofom hu nh khng c ion t do, v D rt nh.

Trong hng s kt hp ngi ta c th lm r hn hng s khc nh:

a) Hng s bn (HSB = K) (Phn ng to phc) Hng s bn l hng s cn bng gia cc tiu phn (ion, phn t) to nn phc cht v phc cht

.

Th d: gia Ag+ v NH3 to nn ion phc Ag(NH3)2+ v ion phc c cn bng

Ag+ + 2HN3 Ag(NH3)2+

vi hng s cn bng

Hng s cn bng K cng c gi l HSB ca ion phc Ag(NH3)2+; v K cng ln, Ag(NH3)2+c to nn cng nhiu nn K cn c gi l hng s to thnh phc cht. (trnh by trong chng 10)

b) Hng s bn iu kin (Phn ng to phc) (trnh by trong chng 10) 4.2.3. Hng s cn bng ca cc phn ng trao i Tch s tan (Phn ng to ta) Phn ng to ta xy ra khi c 2 hay nhiu cht ha tan kt hp vi nhau to thnh cht khng

tan gi l ta. Phn ng kt ta chung nht l phn ng hon v m trong 2 hp cht ion ho tan trao i cc phn.

Th d: Pb2+ (nc) + 2Cl (nc) PbCl2 (r)

Tuy nhin, trong vic x l s cn bng ca s kt ta, cng thng xuyn gp phn ng ngc chiu m t s phn ly ca ta ny.

PbCl2 (r) Pb2+ (nc) + 2Cl (nc)

Trong phn ng ny c mt hng s c gi l tch s tan (solubility product, Ksp) l tch nng cc ion ca hp cht t tan trong dung dch bo ho cht

TST ca PbCl2 = Ksp = [Pb2+] [Cl-]2 = 1.7 10-5

i vi cc cht tan kh, c nng bo ho ln, khng p dng tch s nng , m phi tnh theo tch s hot a (trnh by trong chng 9)



4.2.4. Hng s cn bng ca cc phn ng oxy ho - kh Khng ging nh cc phn ng c xt, v tr cn bng ca phn ng oxy ho - kh him c

din t bng hng s cn bng. V phn ng oxy ho - kh lin quan n s di chuyn electron t tc nhn kh sang tc nhn oxy ho nn xt nhit ng hc ca phn ng v mt electron li thch hp hn (trnh by trong chng 8).

Nhit ng hc l khoa hc nghin cu s tng tc ca nhit vi s bin thin nng lng v quan trng hn c i vi lnh vc ho hc l nghin cu s thay i nng lng trong qu trnh phn ng. Nhit ng hc ho hc phn tch cc yu t xc nh im cn bng ca phn ng. Hng s cn bng K c th suy ra t l thuyt nhit ng hc. K ph thuc vo bn cht ca cht phn ng, nhit v p sut.

nhit v p sut hng nh, nng lng t do Gibb biu din theo cng thc G = H T. S

Trong :

- T l nhit Kelvin; G: l bin thin v nng lng t do Gibb - H: l bin thin enthalpy; S: l bin thin entropy

Bin thin nng lng G lin quan n hng s cn bng K ca phn ng ho hc

G = Go + RTlnK

4.3. ng dng ca hng s cn bng K

Hng s K cho bit v chiu mc hon ton ca phn ng. Nu K ln rt nhiu hn 1 (K>>1) th phn ng s chim u th to ra C v D; K cng ln, phn ng theo chiu thun cng hon ton.

Nu K nh rt nhiu so vi 1 (K< 0,1) th phn ng khng hon ton, K qu nh th phn ng to C v D hu nh khng xy ra.

V K l mt hng s do nu thay i mt tha s [A] hoc [B] hoc [C] hoc [D] th s ko theo s thay i mt hay nhiu tha s khc. Nh vy, mun quyt nh chiu ca phn ng ta c th thay i nng ca cc cht.

Khi bit hng s cn bng ho hc K, ta c th tnh ton: - Nng cn bng ca cc ion cht in ly yu, nu bit nng ban u ca chng. - Nng cn bng ca ca cc cht tham gia phn ng hoc sn phm to thnh ca cc phn ng

ho hc.

- Nng cn bng cc ion hydro, hydroxyd, in ly ca cht in ly trong cc dung dch nc ca cc acid hoc baz yu, cc mui thu phn, dung dch m...

- Nng cn bng ca cc cation, anion v tan ca cht in ly t tan trong nc.

5. CC YU T NH HNG N CN BNG HO HC



Theo nguyn l Le Chatelier Khi tc ng ln mt h th bn thn h s t iu chnh hn ch nh hng ca tc ng . C rt nhiu yu t nh hng n hng s cn bng ca phn ng.

5.1. nh hng ca nhit

Xt phn ng thun nghch sau y:

Phn ng thun to nhit, phn ng nghch thu nhit. Do , nu ta tng nhit ca h thng th cn bng s dch chuyn theo hng lm gim nhit tc l chuyn v pha phn ng nghch, v nu gim nhit ca h thng th phn ng s xy ra theo chiu thun.

5.2. nh hng ca p sut

S tng hay gim p sut ring phn ca cht kh cng tng t nh s thay i nng . Xt phn ng th kh:

H2 + Cl2 2HCl

S phn t kh hai v bng nhau do khi thay i v p sut s khng lm thay i trng thi cn bng. Nhng i vi phn ng tng hp NH3 (phng trnh 2.12) trn th hn hp s gim th tch, do khi tng p sut th h thng cn bng s dch chuyn theo chiu thun.

V nng ca cc cc cht kh tu thuc p sut ring phn ca chng v khng ph thuc vo tng p sut ca h thng, do khi thm hay b mt lng kh tr s khng lm thay i cn bng phn ng ca cht kh .

5.3. nh hng ca dung mi

Hu ht phn ng c s tham gia ca cht phn ng v sn phm dng phn tn trong dung mi. Nu lng dung mi thay i th nng ca tt c cc cht phn ng v sn phm s thay i.

Th d: phn ng to phc Bc-amoniac trong nc:

Ag+ + 2NH3 Ag(NH3)2+

Hng s cn bng ca phn ng ny l

Nu mt phn ca phn ng ny c pha long vi ng th tch nc, nng ca cc cht trong phng trnh (2.13) u gim mt na. Do h s phn ng s l

Hoc c th vit l

N2 + 3H2 2NH3 (2.12)

(2.13)



V Q ln hn 2, cn bng phi c ti lp bng cch dch chuyn sang tri v lm gim nng ca Ag(NH3)2+. Hn na, v tr cn bng mi s hng v pha c s cht tan nhiu nht. Nu dung dch Ag(NH3)2+ c c m c, bng cch lm bay hi th cn bng c ti lp theo hng ngc li. Kt lun tng qut ny c th p dng vi bt k phn ng no, bt k pha rn, lng hay kh. Vic gia tng th tch lun lm phn ng din ra u tin theo hng c s phn t nhiu nht v s gim th tch th thc y phn ng theo hng c t phn t nht. Nu s phn t c hai v phng trnh l nh nhau cn bng ca phn ng khng b nh hng bi s thay i th tch.

5.4. nh hng ca nng cc cht tham gia phn ng v sn phm to thnh

- H thng cn bng ca phn ng b ph v trc tip Vi phn ng phn ly ca acid acetic, (phng trnh 2.11) mt nhit hng nh, cc dung dch khc nhau ca acid acetic s c cc gi tr khc nhau cho

nng H3O+, acetat, v acid acetic, nhng lun c cng gi tr Ka. Ti thi im cn bng nu thm

vo dung dch natri acetat, nng acetat tng ln, c kh nng lm tng gi tr ca Ka. Tuy nhin, v gi tr Ka gi nguyn khng i, do nng ca c ba phn t trong phng trnh (2.11) phi thay i phc hi Ka v gi tr nguyn thu ca n. Trong trng hp ny, cn bng c ti lp bng phn ng mt phn ca acetat v H3O

+ to thm acid acetic.

Nh vy, h thng cn bng ca phn ng s b ph v khi thay i nng ca cht tham gia phn ng hay sn phm to thnh.

- H thng cng c th ph v mt cch gin tip Th d s phn ly ca AgCl

nh hng trn tan ca AgCl bng cch thm AgNO3 th r rng, nu thm vo mt cht c th to phc bn v tan c vi Ag+ nh amoniac th s xy ra phn ng sau:

Thm amoniac lm gim nng Ag+ di dng ion phc . V s gim nng Ag+lm tng tan ca AgCl v phn ng (2.15) ti lp im cn bng. T cc phn ng (2.14) v (2.15) s lm r tc ng ca amoniac trn tan ca AgCl v amoniac cng l mt cht tham gia phn ng.

Th d: Nu thm HNO3 vo th cn bng ca phng trnh (2.16) s thay i nh th no?

Gii:

HNO3 l acid mnh, acid ny s phn ng vi NH3

AgCl Ag+ + Cl- (2.14)

Ag+ + 2NH3 (2.15)

AgCl + 2NH3 + Cl-

(2.16)



Khi thm t HNO3 vo s lm gim NH3. Vic gim nng NH3 s lm phn ng (2.16) xy ra theo chiu to ra cht phn ng, gim tan AgCl.

6. BI TP V CN BNG HO HC 6.2. X l cc hng s cn bng

thun tin cho vic tnh ton cc hng s cn bng ngi ta c th x l cc hng s cn bng theo cc cch sau y:

- T phn ng cho nu ta o ngc phn ng th hng s cn bng ca phn ng mi s l nghch o ca phn ng u. Th d, hng s cn bng cho phn ng

L nghch o ca phn ng

- Nu cng hai phn ng vi nhau to ra phn ng mi th hng s cn bng ca phn ng mi l tch s ca hng s cn bng ca cc phn ng ban u.

6.2. Bi tp v tnh p sut cn bng ring phn

Hng s cn bng KP i vi phn ng sau PCl5(k) PCl3(k) + Cl2(k)

l 1,05 250 C. Nu p sut ring phn trng thi cn bng ca PCl5 v PCl3 ln lt l 0, 875 atm v 0,463 atm th p sut ring phn trng thi cn bng ca Cl2 250 C l bao nhiu?

Gii: trc tin tnh KP

6.3. Bi tp v chuyn i KC thnh KP



Kc : hng s cn bng theo nng .

Kp : hng s cn bng theo p sut tng phn ca kh .

Ch : theo phng trnh: aA (k) bB (k) th KC = v KP =

Theo phn ng N2 (k) + 3H2(k) 2NH3(k)

Gi tr ca Kp = 4,3.10

-4 375 C. Hy tnh Kc.

v T0 = 293 Kelvin v n = s mol sn phm kh s mol cht kh tham gia.

Nn ta c T0 = 648 Kelvin v n = 2- 4 = -2

6.4. Bi tp v d on chiu ca phn ng

Mt phn ng cho lc bt u c 0,249 mol kh N2; 3,21.10-2

mol kh H2 v 6,42.10-4

mol NH3trong mt bnh cha 3,5 lt 375 C. Nu hng s cn bng KC i vi phn ng N2 (k) +3 H2 (k) 2NH3 (k) c gi tr l 1,2 nhit ny th:

a) Hy gii thch l h thng ny c cn bng hay khng? b) Nu khng th phn ng s c khuynh hng di chuyn theo chiu no? Cc bc tin hnh Gi QC (Quotient ractionnel) l i lng thu c khi s dng cc nng ban u trong biu

thc hng s cn bng. xc nh chiu phn ng th so snh QC v KC. C 3 trng hp xy ra: - QC > KC: T s gia cc nng ban u ca sn phm v cht tham gia qu ln. t c cn

bng, mt lng no ca sn phm phi chuyn thnh cht tham gia. Phn ng di chuyn theo chiu tri.

- QC > KC: Phn ng ang thi im cn bng .

- QC < KC: T s gia cc nng ban u ca sn phm v cht tham gia qu nh. t c cn bng, mt lng no ca cht tham gia phi chuyn thnh sn phm. Phn ng di chuyn theo chiu phi.

Nu trng hp hng s s dng l KP th tnh QP.

Gii



- QC < KC H khng cn bng v phn ng di chuyn theo chiu phi.

6.5. Bi tp v tnh hng s cn bng

Mt h gm mt cp ng phn hnh hc trong mt dung mi hu c m hng s cn bng KC = 24,0 200 C

Cis stilben Trans - stilben Gi d rng lc u ch c Cis stilben c mt v nng l 0,850 mol/l. Hy tnh nng ca 2

ng phn ny thi im cn bng. Cc bc tin hnh: Bc 1: Din t nng cn bng ca tt c cc cht theo nng ban u v ch bng mt n s.

Bc 2: Din t hng s cn bng theo nng cn bng. Gi tr ca hng s cn bng bit c s dng gii phng trnh theo x.

Bc 3: Sau khi xc nh gi tr x, tnh nng cn bng ca tt c cc cht

Thay i (+) chng t s tng ln v thay i (-) chng t s gim nng . Tip n l vit biu thc hng s cn bng theo nng cn bng:

x = 0,816 M

Cis stilben Trans stilben

Nng ban u (M) 0,850 0 Thay i v nng (M) -x +x Nng cn bng 0,850 -x x



p s: a) [Cis stilben] = 0,034M; b) [Trans stilben] = 0,816M

6.6. Bi tp v tnh tan ca mt cht trong nc

Khi mt hp cht khng tan nh Pb(IO3)2 c cho vo dung dch, mt phn cht rn s tan ra. Cn bng t c khi nng Pb2+ v tha mn tan sn phm i vi Pb(IO3)2. im cn bng, dung dch b bo ha bi Pb(IO3)2. Lm th no xc nh c nng ca Pb2+ v v tan ca Pb(IO3)2 trong dung dch bo ha khi cho Pb(IO3)2 vo nc ct.

tin hnh gii bi ton ny ta vit phn ng cn bng:

V hng s cn bng ca n

Khi cn bng c thit lp, hai ion c to ra tng ng vi mi ion Pb2+. Nu chng ta gi

nh nng mol ca Pb2+ im cn bng l x th nng ca l 2x.

Th nng cn bng vo phng trnh (2.18) (x)(2x)2 = 2,5 x 10-13

4x3 = 2,5 x 10-13

x = 3,97 x 10-5

Do nng cn bng ca Pb2+ v l

[Pb2+] = x = 4,0 x 10-5 M

[I-] = 2x = 7,9 x 10-5 M

V mt mol Pb(IO3)2 cha mt mol Pb2+, tan ca Pb(IO3)2 bng nng ca Pb2+; theo

tan ca Pb(IO3)2 l 4,0 105 M.

6.7. Bi tp tnh tan ca mt cht trong dung dch c ion cng tn

Trng hp trn l tnh ton tan ca Pb(IO3)2 trc tip trong nc ct v cht rn l ngun duy nht cung cp Pb2+ v cho dung dch. tan ca Pb(IO3)2 s b nh hng nh th no nu chng

Pb(IO3)2 Pb2+ + 2 (2.17)

(2.18)

PbI2

Pb2+ + 2

Nng ban u Cht rn 0 0 Thay i v nng Cht rn +x +2x

Nng cn bng Cht rn 0 + x = x 0 + 2x = 2x



ta thm n vo dung dch Pb(NO3)2 0,1M? Thit lp bng theo di nng ca Pb2+

v trong h thng.

Th nng cn bng vo biu thc tan sn phm ca phng trnh (2.18)

hay 4x3 + 0,40x2 = 2,5 x 10-13

Chng ta s vn dng hiu bit v ho hc n gin ho phng trnh. T nguyn l Le Chatelier, on rng nng Pb2+ ln ban u s lm gim tan ca Pb(IO3)2. Trong trng hp ny c th suy on nng cn bng ca Pb2+ s rt gn vi nng ban u ca n; theo s c tnh sau y v nng cn bng ca Pb2+ l hp l

[Pb2+] = 0,10 + x 0,10 M

th vo phng trnh (2.19) (0,10)(2x)2 = 2,5 x 10-13 0,4x2 = 2,5 x 10-13

v gii thu c x l

x = 7,91 x 10-7

Trc khi chp nhn nghim ny, hy kim tra xem s c lng c hp l khng. Trong trng hp ny c lng 0,1 + x = 0,1 dng nh hp l v s khc nhau gia hai gi tr c th b qua. Do nng cn bng ca Pb2+ v l

[Pb2+] = 0,10 + x 0,10 M

[I] = 2x = 1,6 x 10-6

tan ca Pb(IO3)2 bng s b sung ca Pb2+ trong dung dch hay 7,9 10-7 mol/l. ng nh suy

on, tan ca Pb(IO3)2 b gim trong dung dch c cha mt trong nhng ion ca n. iu ny c gi l hiu ng ion chung.

7. KT LUN

Chng ny trnh by cc loi phn ng, khi nim v s cn bng ho hc, hng s cn bng, hot , nng m rng hiu bit v cn bng ho hc. T chng ny hc vin s c th hiu mt cch tng qut cn bng ho hc ca cc phng php s dng trong ho hc phn tch vn dng trong tnh ton.

T KHO

PbI2

Pb2+ + 2

Nng ban u Cht rn 0,10 0

Thay i v nng Cht rn +x +2x

Nng cn bng Cht rn 0,10 + x 0 + 2x = 2x

(0,10 + x)(2x)2 = 2,5 x 10-13 (2.19)



CU HI LNG GI 1. Nu nh ngha phn ng ho hc v trnh by cc loi phn ng ho hc?

2. Th no l cn bng ho hc? Cn bng vt l? Nu th d c th v gii thch. 3. Ti sao ngi ta cho rng cn bng ho hc l cn bng ng? 4. Nng l g? Hot l g? Trong trng hp no 2 i lng ny ging nhau?

5. Nu ngha ca h s hot . 6. Trnh by cc cng thc tnh hng s cn bng theo hot , theo nng . 7. Nu cc yu t nh hng n h thng cn bng ho hc.

CU HI TRC NGHIM 1. Phn ng thun nghch:

2. Cch thng thng biu din nng lng t do ca phn ng bng hm s

3. Cc yu t nh hng n hng s cn bng ca phn ng l: a. nng cc cht tham gia phn ng.

b. nhit , p sut. c. dung mi. d. nng ca sn phm to thnh.

e. tt c u ng

4. x l cc hng s cn bng ta c th lm theo cc cch: a. o ngc phn ng th hng s cn bng ca phn ng mi s l nghch o ca phn ng u.

b. cng hai phn ng vi nhau to ra phn ng mi th hng s cn bng ca phn ng mi l tch s ca hng s cn bng ca cc phn ng ban u.

Phn ng ho hc Nng

Cn bng ho hc H s hot Hng s cn bng Hng s phn ly Hot Hng s bn

a. c th xy ra th kh hoc lng d. ch a v c ng b. c th xy ra th rn hoc lng e. c ba cu u ngc. nhit , p sut nht nh th thnh phn hn hp trng thi cn bng l khng i

a. nng lng t do Gibb c. bin thin entropy

b. bin thin enthalpy d. T student



c. Nhn hai phn ng vi nhau th hng s cn bng mi s l tng ca cc hng s cn bng ban u.

d. a v b 5. Cn bng ho hc l trng thi.......... m trong cht tham gia phn ng v sn phm chuyn i

lin tc cho nhau. Tc mt i v xut hin ca chng bng nhau.

BI TP

Nu hng s cn bng KC i vi phn ng 2NO(k) + Cl2(k) 2NOCl(k) c gi tr l 6,5 104 35oC. Gi s ngi ta trn 2,0 x 10-2 mol ca NO; 8,3 10-3 mol ca Cl2 v 6,8 mol NOCl trong mt bnh cu 2 lt th phn ng t n cn bng di chuyn theo chiu no?

Chng 3 S DNG THNG K

TRONG PHN TCH S LIU

NI DUNG

1. M U

Khi tin hnh php phn tch vi mt mu bt k, qua bao gi chng ta cng c c cc kt qu (s liu) t thc nghim. Cu hi c t ra l: s liu thu c c gn vi gi tr thc c ca mu khng? Qu trnh phn tch c mc phi sai s khng? Gi tr thc nm khong no ca cc s liu thc nghim? v.v

Sai s trong phn tch l khng th trnh khi, tuy nhin trong qu trnh phn tch cn khng ch sai s mc thp nht c th, c ngha l kt qu thu c khi tin hnh mt php phn tch lun lun mc phi sai s. Gi tr thc ca mu o l i lng khng th xc nh, nhng nu xc nh c cc sai s

a. ng c. tnh e. lng

b. ng yn d. kh

MC TIU

Trnh by c khi nim v sai s, cc i lng c trng ca ton thng k.

ng dng ton thng k x l cc kt qu thu c t thc nghim. Trnh by kt qu phn tch m bo c mc chnh xc theo yu cu.



trong qu trnh phn tch th c th c lng gi tr thc nm trong khong no ca gi tr thc nghim. V vy vic x l kt qu sau qu trnh phn tch l lun lun cn thc hin, qua nh gi kt qu thu c ng v chnh xc n mc no. Thng thng cn tin hnh phn tch nhiu ln v p dng ton thng k nh gi tin cy ca s liu vi mc xc sut c n nh trc.

2. MT S I LNG C TRNG TRONG THNG K

2.1. Gi tr trung bnh (Xtb)

Trong : n: l tng s ln o; i

: gi tr o c th i (i = 1, 2, 3, , n)

Th d 1: hm lng ca dung dch KMnO4 c xc nh sau 7 ln chun :

3,080; 3,094; 3,107; 3,056; 3,112; 3,174; 3,198 (g/l)

2.2. Phng sai S2 (VARIANCE)

Mi ln o chng ta s c 1 gi tr xi v gi tr ny c th nh hn hay ln hn so vi gi tr trung bnh Xtb. Hiu ca hai gi tr ny c th l mt s m hay dng v khi tng cc hiu li c th bng s khng. Nu gi tr tuyt i ca hiu cng ln th s bnh phng s cng ln v cng cho thy lch cng nhiu. Do vy i lng phng sai c cp

2.3. lch chun (sd: standard deviation)

lch chun l i lng cho bit mc dao ng gia cc gi tr xi so vi gi tr trung bnh Xtb. y l i lng c trng cho phn tn ca s liu (xi) v th hin mc ca sai s ngu nhin.

Cng thc tnh:

SD: lch chun; n: s ln thc nghim; (n 1) c gi l bc t do.

2.4. lch chun tng i (RSD: relative standard deviation)

lch chun tng i l t s gia lch chun so vi gi tr trung bnh. RSD th hin gi tr

(3.1)

(3.2)

(3.3)



lch r hn SD v c so snh vi gi tr trung bnh Xtb.

Cng thc tnh:

RSD c th c tnh bng t l phn trm RSD% hay cn gi l h s bin thin CV (Coefficient of Variation):

Nh vy, ta c th tnh CV da trn lch chun v ngc li. Th d: theo th d 1, n = 7, Xtb = 3,117

2.5 Gii hn tin cy v khong tin cy

S ln o n cng ln th gi tr Xtb cng gn vi gi tr thc M, nhng trong thc t ta ch thc hin n ln o v trong thng k th xc sut bt gp thng n nh trc l 95%, nn gi tr Xtb s nm trong gii hn c th chp nhn c gi l gii hn tin cy.

- Gii hn tin cy e (Confidence Limit). Hay ni cch khc e l gii hn hai bn ca Xtb c cha gi tr M mc tin cy 95%.

Trong t c tra t bng Student (3.1). Gi tr t ph thuc vo s bc t do k = n 1 v vo mc xc sut c n nh P. S ln thc nghim cng nh, xc sut P cng ln th gi tr t cng cao.

Bng 3.1. Bng gi tr ca chun Student t

(3.4)

(3.5)

(3.6)

t do 90% 0,10 95% 0,05

98% 0,02

99% 0,01

1 6,31 12,71 31,82 63,66



Nh vy gi tr Xtb cn cch gi tr thc M mt gii hn e hay cn gi l khong tin cy (Confidence Intervals):

Th d: vi th d 1, gi tr e c tnh:

ng vi P = 0,95 v n = 7 th tlt = 2,45 nn hm lng KMnO4 nm trong khong 3,117 0,047 (g/l) tc gi tr thc M nm trong gii hn 3,070 3,164 (g/l).

3. SAI S TRONG PHN TCH Vi mt mu c gi tr thc l M, v cc nguyn nhn khc nhau ta khng th xc nh ng gi tr

M m ch xc nh c gi tr trung bnh gia cc ln o lp li. Thut ng sai s c dng din t mc sai lch ca php o, n th hin lch gia cc gi tr thc nghim xi vi nhau, vi gi tr trung bnh v vi gi tr thc ca mu. Trong phn tch ta ch c gng thc hin sao cho sai s l nh nht, cc gi tr thu c t thc nghim nm trong khong chp nhn ch khng th loi tr c hon

2 2,92 4,30 6,96 9,92

3 2,35 3,18 4,54 5,84

4 2,13 2,78 3,75 4,60

5 2,02 2,57 3,36 4,03

6 1,94 2,45 3,14 3,71

7 1,89 2,36 3,00 3,50

8 1,86 2,31 2,90 3,36

9 1,83 2,26 2,82 3,25

10 1,81 2,23 2,76 3,17

12 1,78 2,18 2,68 3,05

14 1,76 2,14 2,62 2,98

16 1,75 2,12 2,58 2,92

18 1,73 2,10 2,55 2,88

20 1,72 2,09 2,53 2,85

30 1,70 2,04 2,46 2,75

50 1,68 2,01 2,40 2,68

1,64 1,96 2,33 2,58

(3.7)



ton sai s.

Khi trnh by v kt qu bng s liu, cc ti liu c th cp n 2 loi l sai s tuyt i v sai s tng i.

3.1. Sai s tuyt i

Vi M l gi tr thc ca mu th cn xc nh, xi l gi tr o c t thc nghim:

Gi tr Xtb thng khc vi gi tr thc M ca mu o. Khi n th Xtb M.

Sai s tuyt i l hiu gia gi tr trung bnh Xtb v gi tr thc M. Thng thng khng th xc nh c gi tr thc m ch chp nhn gi tr ng tin cy nht trong dy s liu l Xtb hay mt cch gn ng l hiu ca gi tr o c v Xtb.

M: gi tr thc ca mu o : sai s tuyt i

Sai s tuyt i c th l s m cng c th l s dng.

3.2. Sai s tng i S

Sai s tng i S l t s gia sai s tuyt i v gi tr thc M hoc gi tr trung bnh Xtb. Thng thng sai s tng i c biu th di dng phn trm.

Th d: hm lng thc ca Paracetamol trong vin nn l 500,2 mg; Codein l 30 mg. Sau khi tin hnh xc nh theo phng php sc k lng hiu nng cao cho thy hm lng Paracetamol l 500,9 mg v Codein l 30,7 mg.

Sai s tuyt i khi xc nh hai hot cht trn l nh nhau: + 0,7 mg.

Sai s tng i th khc nhau:

Nh vy sai s tng i th hin r hn ng ca phng php, vi th d trn cho thy phng php ny xc nh hm lng Paracetamol ng hn hm lng ca Codein phosphat.

Khi trnh by v nguyn nhn dn n cc kt qu thu c trong qu trnh thc nghim b lch nhau. Cc ti liu thng cp n 3 loi sai s: sai s ngu nhin, sai s h thng, sai s th.

3.3. Sai s ngu nhin (random error)

Sai s ngu nhin hay cn gi l sai s khng xc nh l sai s gy nn bi nhng nguyn nhn khng xc nh c, khng bit trc, khng c nh, thay i khng theo quy lut nn khng th hiu chnh hay loi tr m ch c th hn ch bng cch tng s ln phn tch v thao tc mt cch cn thn, ng thi nh gi cc s liu thc nghim bng ton hc thng k.

Cng thc tnh sai s tuyt i: = Xtb - M hay = xi - Xtb (3.8)

(3.9)



3.4. Sai s h thng (systematic error)

Sai s h thng hay cn gi l sai s xc nh l sai s bit r nguyn nhn v c th hiu chnh c. Sai s ny thng do cc nguyn nhn sau:

- Sai s do mu o: gy ra khi mu phn tch khng i din. Vic ly mu l tp hp cc thao tc nhm ly mt lng mu i din tin hnh phn tch, ng thi vic ly mu cn tun theo cc th tc quy nh.

- Sai s do dng c: d t hay nhiu th tt c cc dng c o lng lun c sai s h thng. Sai s ny l do qu trnh ch to v chun ho dng c v.v

Th d: - Cc dng c o lng thng thng nh pipet chnh xc, buret, bnh nh mc l cc dng c dng ly th tch chnh xc nhng do trong qu trnh ch to hoc s dng khng ng quy nh c th gy ra mt sai s: th tch ca mt buret 25 ml c sai s l 0,03 ml c ngha l th tch tht s ca buret l nm trong khong [24,97 25,03] ml; bnh nh mc c th tch l 100ml nhng nu thc t ch c 99,90 ml th sai s l 99,90 100,00 = - 0,10 ml.

Sai s dng c thng d pht hin v hiu chnh c bng cch nh k chun ho cc dng c trong phng th nghim.

- Sai s do phng php o: phng php o lng cng gy ra sai s h thng. V vy, khi p dng mt phng php mi phn tch lun lun phi xy dng v thm nh quy trnh chng minh mt cch khoa hc rng sai s ca phng php l rt thp v c th chp nhn c. Sai s do phng php thng kh pht hin v l nguyn nhn chnh gy ra sai s h thng. Thng thng pht hin sai s phng php c th tin hnh theo cc cch sau:

- Thc hin song song mu trng loi cc p ng gy ra do cc cht khng cn phn tch. - Phn tch mu chun kim tra ng ca phng php. - Phn tch cng mt mu nhng bng phng php d kin v thc hin song song vi t nht mt

phng php khc v so snh hai kt qu. - Sai s do ngi lm cng tc phn tch: ngi lm cng tc phn tch i hi phi c k nng

ngh nghip v kinh nghim phn tch. Sai s do tng c nhn l iu khng trnh khi, mi c nhn c mt kh nng ring v quan st s thay i mu ca dung dch; quan st im tng ng trong chun ; c v tr mc nc gia hai vch; c tn hiu o c t my mc; hoc do cu th; do nh kin u dn n sai s.

Sai s do c nhn c th khc phc c khi thao tc ng theo quy nh v nhiu kim nghim vin thc hin trn cng mt mu th.

XC NH SAI S H THNG Bng cch xc nh mu chun bit hm lng, gi tr M coi nh bit trc, Xtb thu c t thc

nghim. so snh s khc bit gia gi tr M v Xtb (trong trng hp thm nh phng php mi) thng s dng thng k tnh gi tr ttn ri so snh vi tlt theo bng (3.1) cn c vo bc t do (n 1) v xc xut P (95%).

Nu ttn < tlt, kt lun khng tm thy sai s h thng

Nu ttn > tlt, kt lun phng php c sai s h thng



Th d: gi s mu th KMnO4 trong th d 1 c hm lng tht M l 3,110 mg. Hy xt xem phng php c sai s h thng khng?

ttn = 0,3786 < tlt = 2,45 nn kt lun phng php khng c sai s h thng (P = 95%).

Th d: trc khi phn tch xc nh hm lng ca Na2CO3 c trong mu th, mt sinh vin c

tng kim tra li quy trnh phn tch xem c sai s h thng hay khng bng cch phn tch mt mu bit hm lng M = 98,76%. Cc kt qu thu c sau 5 ln thc hin:

98,71(%) 98,59(%) 98,62(%) 98,44(%) 98,58(%)

Nh vy tlt = 2,78 < ttn = 3,91, kt lun phng php c sai s h thng (P=95%).

3.5. Sai s th (gross error)

Ngoi ra cn c sai s th l sai s khi kt qu gia cc ln o lp li khc hn so vi cc gi tr trung bnh hay gi tr thc ca mu. Sai s th do nhiu nguyn nhn khc nhau: do c kt qu o sai, ly nhm qu cn, v.v pht hin v loi tr sai s th cn phi tin hnh phn tch nhiu ln trn mt mu o (n > 6) v loi i nhng gi tr bt thng theo nhng quy tc nht nh.

Loi tr sai s th Thc t sau khi tin hnh phn tch thng thu c mt dy cc s liu, i khi c mt vi s liu

c gi tr khc hn: hoc cao qu hoc thp qu, nhng s liu ny c gi l sai s th hay s liu xu. C hai cch gip chng ta kim tra xem nn gi li hay loi b chng: dng chun Dixon v dng bng kim nh T.

3.5.1. Phng php dng chun Dixon (chun Q) S dng khi n < 10, gm cc bc nh sau: - Bc 1: sp xp cc s liu theo th t tng dn hoc gim dn: x1, x2, x3,..., xn trong x1 l s

liu nghi ng ln nht (max) hay nh nht (min), x2 l s liu ln cn ca s liu nghi ng.

- Bc 3: so snh gi tr Qtn vi Qlt: Qlt > Qtn th x1 c gi li

Qlt < Qtn th x1 b loi b

(3.10)

- Bc 2: dng test Q tnh gi tr Qtn:

(3.11)



Qlt c tra t bng (3.2) vi s ln o n v xc sut bt gp P thng thng l 95%.

Bng 3.2. Bng kim nh Q chun DIXON

Th d: ta c dy s liu sau 17,61 16,86 16,93 16,84 16,95 16,91 Bc 1: sp xp theo th t 16,84 16,86 16,91 16,93 16,95 17,61; ta thy s liu nh nht l

16,84 (ln cn l 16,86) v s liu ln nht l 17,61 (ln cn l 16,95) thng l hai s liu ng nghi ng.

Bc 2: tnh gi tr Qtn ;

Bc 3: tra bng (3.2) vi n = 6; P = 0,95 Qlt (= 0,625) < Qtn (= 0,86) nn gi tr 17,61 b loi b

Qlt (= 0,625) > Qtn (= 0,03) nn gi tr 16,84 c gi li

3.5.2. Phng php dng bng kim nh T S dng vi n bt k. Sai s th thng ri vo cc gi tr cc i hay cc tiu ca dy s liu, ta

tnh Ttn theo cng thc sau:

Cn c vo s ln o n v vo xc sut n nh trc P, tra Tlt theo bng (3.3).

Tlt < Ttn th s liu nghi ng l sai s th b loi b

Tlt > Ttn th s liu nghi ng l sai s th c gi li

Bng 3.3. Gi tr t ca bng kim nh T vi P = 95%

n P

0,1 (90%) 0,05 (95%) 0,01 (99%) 3 0,941 0,970 0,994

4 0,765 0,829 0,926

5 0,642 0,710 0,821

6 0,560 0,625 0,740

7 0,507 0,568 0,680

8 0,468 0,526 0,634

9 0,437 0,498 0,598

10 0,412 0,466 0,568

(3.12)

(3.13)



Th d: vi dy s liu 16,84 16,86 16,91 16,93 16,95 17,61 Vi n = 6; Xtb = 17,02; SD = 0,292; P = 0,95 ta c Tlt = 1,996

Tlt (= 1,996) < Ttn (= 2,02) nn gi tr 17,61 b loi b.

Tlt (= 1,996) > Ttn (= 0,61) nn gi tr 16,84 c gi li.

4. TM TT CC BC TIN HNH KHI X L S LIU Bc 1: sp xp dy s liu thu c theo th t tng hoc gim dn. Bc 2: loi cc gi tr khng ph hp, loi sai s th.

Bc 3: tnh gi tr Xtb; lch chun SD, RSD%, xt sai s h thng.

Bc 4: xc nh gii hn tin cy e v khong tin cy.

Bc 5: bo co kt qu. Th d: theo s liu ca th d 1, cc bc tin hnh nh sau:

Bc 1: 3,056; 3,080; 3,094; 3,107; 3,112; 3,174; 3,198 (g/l). Bc 2: loi sai s th theo chun Q hoc T: khng loi b gi tr no. Bc 3: Xtb = 3,117; SD = 0,05; RSD = 1,63%

Bc 4:

ng vi P = 95% v n = 7 tlt = 2,45 tnh e = 0,047.

Bc 5: hm lng KMnO4 trong khong 3,117 0,047 (g/l) vi tin cy P = 95%.

n Tlt n Tlt

3 1,412 11 2,343

4 1,689 12 2,387

5 1,869 13 2,426

6 1,996 14 2,462

7 2,093 15 2,493

8 2,172 16 2,523

9 2,237 17 2,551

10 2,294 18 2,623



5. SO SNH CC DY S LIU Kt qu thu c t o lng thc nghim thng mc phi cc sai s hoc ngu nhin hoc h

thng. Gi tr trung bnh X thng ch st gn vi gi tr thc M, gi tr ny li nm trong khong tin cy no vi xc sut bt gp mong mun.

Vic so snh gia cc dy s liu c thc hin nh cc bng chun xem cc dy s liu ny l ng nht hay khc nhau c ngha gip cho vic so snh gia cc phng php, gia cc ngi phn tch v gia cc l sn xut.

Cc vn thng gp trong thc nghim l :

- C hai gi tr trung bnh XnA v XnB thu c t 2 dy o c lp nhau, mt dy c nA kt qu, dy kia c nB kt qu. Hai gi tr trung bnh sai khc khng nhiu. Phi kt lun sai khc ny c ng k khng?

- C mt gi tr phn tch X ca sn phm c bit trc qua nhiu ln lm thc nghim. Ngi thc hin phn tch lp li phng php v cn so snh kt qu o c Xi vi gi tr bit trc X xem khc nhau c ngha hay khng?

- C mt l thuc c thc hin nh lng vi 2 phng php khc nhau v cho 2 dy kt qu khc nhau. Cc nh phn tch cn phi so snh kt qu o c trn cng mu bng 2 phng php khc nhau xem c khc nhau hay khng?

- Trong trng hp so snh hai dy kt qu:

S dng th nghim F (F-test) kim tra chnh xc hay lp li ca hai dy kt qu xem c ng nht hay khng.

S dng th nghim t ( t-test) so snh hai gi tr trung bnh xem s khc nhau gia hai gi tr ny c ngha hay khng.

5.1. So snh hai phng sai

S dng chun F so snh chnh xc ( lp li) ca hai dy th nghim A v B vi nA, nB l s ln thc hin ca A v B. Phng sai tng ng v . Quy c :

So snh Ftn vi Flt c tra t bng (3.4). Flt ph thuc vo bc t do ca hai th nghim A, B v ph thuc vo mc xc sut cho trc thng thng l P = 95%.

Bng 3.4. Gi tr F ca chun F (P = 95%)

(3.14)



Bc t do ca A: nA -1; bc t do ca B: nB 1

Nu Flt > Ftn: khng c s khc bit v lp li ca A v B.

Nu Flt < Ftn: lp li ca hai th nghim A v B khc nhau.

Th d: kt qu 7 ln phn tch hm lng aspirin bng phng php A c = 0,00259. Theo 6 ln phn tch bng phng php B l = 0,00138. Hy xt xem chnh xc ca hai phng php c ging nhau khng?

Do Flt = 6,978 > Ftn = 1,88: chnh xc ca hai phng php l nh nhau (P = 95%).

5.2. So snh hai s trung bnh

Trong phn sai s h thng chng ta bit kt qu phn tch b nh hng chnh bi ba nguyn nhn chnh: do mu, do phng php, do kim nghim vin. Thc hin nhng cp th nghim song song tm ra nguyn nhn gy sai s trong ch mt yu t thay i. Th d nh cng thc hin song song hai phng php khc nhau trn cng mt mu th, cng mt kim nghim vin ri so snh c hay khng c s khc bit gia hai gi tr trung bnh ca hai phng php ny. Tng t nh vy so snh hai gi tr trung bnh ca hai kim nghim vin hoc ca hai mu o



S dng php th t (t test) thc hin cng vic ny nh sau: Kt qu phn tch bng phng php A c: XtbA; SDA; ; nA (s ln thc hin ca A).

Kt qu phn tch bng phng php B c: XtbB; SDB; ; nB (s ln thc hin ca B)

Bc 1: so snh hai phng sai (php th F; F test) Bc 2: c hai trng hp

Trng hp 1: phng sai v khng khc nhau, ttn c tnh:

tlt c tra t bng (3.1) vi t do (nA + nB -2), P = 95%.

Trng hp 2: phng sai v khc nhau, ttn c tnh:

tlt c tra t bng (3.1) vi P = 95% v t do c tnh theo cng thc sau:

t do lun l s nguyn nn gi tr tnh c t cng thc (3.18) c gi tr gn s nguyn k cn nht. Th d: = 5,4 th t do l 5; = 5,9 t do l 6.

Bc 3: kt lun c hay khng c s khc nhau ca hai phng php o.

Cho d ttn c tnh theo cng thc (3.15) hay (3.17):

Nu ttn < tlt kt lun hai phng php cho kt qu khng khc nhau.

Nu ttn > tlt kt lun hai phng php cho kt qu khc nhau.

Th d: hm lng ca dung dch KMnO4 c xc nh bng hai phng php A v B, kt qu:

A: 3,080; 3,094; 3,107; 3,056; 3,112; 3,174; 3,198 (g/l) B: 3,052; 3,141; 3,083; 3,083; 3,048 (g/l)

So snh hai phng sai v (chun F) khng khc nhau.

Tnh Schung theo cng thc (3.16):

(3.15)

(3.16)

(3.17)

(3.18)



tlt c tra t bng (3.1), t do n = (7 + 5 - 2) = 10; ttn = 1,34 < tlt = 2,23 nn kt lun hai phng php A v B cho kt qu khng khc nhau (P = 95%).

Th d: hm lng ca Na2CO3 (%kl/kl) c hai kim nghim vin C v D chun bng phng php acid-base cho ln lt kt qu nh sau:

Do Ftn = 45,6 > Flt = 7,15 nn hai phng sai C v D khc nhau.

t do :

Do ttn = 4,62 > tlt = 2,57, nn kt lun kt qu phn tch ca C v D khc nhau (P= 95%).

6. CH S C NGHA S liu thu uc t thc nghim c th l s o trc tip khi c trn thang o ca cc dng c o

lng hoc s o gin tip l s o tnh c t mt hoc vi s o trc tip thng qua cng thc ton hc.

C D

86,82 81,01

87,04 86,15

86,93 81,73

97,01 83,19

86,20 80,27

87,00 83,94

XtbC = 86,83% XtbD = 82,71%

SC = 0,32 SD = 2,16



Cc s liu ny phi c th hin sao cho ngi c hiu c mc chnh xc ca chng, v vy khi ghi kt qu phi tun theo quy tc v ch s c ngha.

6.1. Ch s c ngha (CSCN) trong s o trc tip

Mt s o trc tip c th c nhiu CSCN tin cy (ch s chc chn) nhng duy nht ch c mt CSCN khng tin cy (ch s nghi ng) ng sau cng k t tri sang phi. CSCN phn nh mc chnh xc ca dng c o lng.

Th d:

- Trn thang o ca buret c th tch l 25ml, vch chia nh nht l 0,1ml. Gi s th tch ti im tng ng khi chun l V = 18,75 ml, nh vy c 4 ch s c ngha l 1; 8; 7 v 5. Trong 1; 8; 7 l cc ch s chc chn v ch s 5 l ch s nghi ng hay cn gi l ch s khng tin cy. Ch s 5 cn ty thuc vo kh nng c lng ca ngi quan st: c th c l 2 hoc 4 hoc 7 v.v 9.

- Khi cn trn cn phn tch c nhy l 0,1 mg, kt qu c c l 1,2326 g, nh vy c 5 ch s c ngha trong ch s 6 cui cng l ch s nghi ng.

- S lng CSCN ca mt s o c tnh t ch s u tin khc 0 k t tri sang phi. Mi ch s 0 sau CSCN u tin bt k v tr no u l CSCN. Hay ni cch khc, ch s 0 ng trc ch s khc 0 u tin ca mt s o khng phi l CSCN. Ch s 0 ng gia hai ch s khc 0hoc ng sau cng u l CSCN.

Th d: V = 18,75 ml : c 4 CSCN, 5 l CSCN khng tin cy. M = 0,0020 g : c 2 CSCN, 0 l CSCN khng tin cy.

M = 5,06 g : c 3 CSCN, 6 l CSCN khng tin cy. M = 0,15 (M/l): c 2 CSCN, 5 l CSCN khng tin cy. Khi mun chuyn i n v o lng thun li cho vic tnh ton kt qu o cui cng, s lng

CSCN ca s o phi c gi nguyn.

Th d: chuyn 0,28 g ra n v mg. Ta thy 0,28 g c 2 CSCN nn khng th chuyn thnh 280 mg (3 CSCN) m phi l 0,28.103 mg hoc bng 2,8.102 mg.

6.2. Ch s c ngha trong s o gin tip

Nh nh ngha, s o gin tip l s o tnh c t cc s o trc tip thng qua mt hoc vi biu thc ton hc no . Sai s ca s o trc tip c nh hng n kt qu ca s o gin tip nn s o gin tip cng phi c ghi chp theo nguyn tc ca CSCN.

- i vi php cng v tr: CSCN ca kt qu cui c cng s ch s bn phi ca du thp phn c s CSCN t nht.

Th d: tnh phn t lng ca phn t BaO2 vi Ba = 137,34; O = 15,9994

137,34 + 15,9994 + 15,9994 = 169,3388 c lm trn 169,34 - i vi php nhn v chia: s lng CSCN ca kt qu cui cng bng s lng CSCN ca s o

c CSCN t nht.

Th d: - ly chnh xc 10,00 ml dung dch acid hydroclorid cha bit nng em chun bng dung dch chun natri hydroxid 0,09876 M. Th tch NaOH cn dng l 9,58 ml. Tnh MHCl.



- i vi kt qu phi tnh ton qua nhiu bc: th ch c lm trn kt qu cui cng m khng lm trn cc bc trung gian.

Th d: - pha dung dch chun HCl c nng xp x 0,1N dng chun dung dch NH4OH cn c cc bc nh sau: pha dung dch chun HCl c nng xp x 0,1 N, sau dng dung dch gc Na2B4O7 0,1N xc nh li nng HCl. Ly chnh xc 10ml dung dch Na2B4O7 em chun bng dung dch chun HCl trn vi ch th Helianthin n khi mu chuyn t vng sang cam. Th tch HCl cn dng l 9,75 ml, nng chnh xc ca dung dch acid c tnh nh sau:

Dng acid HCl ny nh lng dung dch NH4OH cha bit nng : ly chnh xc 10 ml NH4OH v vi git ch th Tashri, nh HCl n khi mu chuyn t xanh l sang tm. Th tch HCl cn dng l 9,8 ml, nng dung dch NH4OH c tnh nh sau:

6.3. Cch lm trn s o gin tip

Cch lm trn s phi tun theo quy tc ch s c ngha. Cch thc lm trn CSCN cui cng ca php tnh ty thuc vo gi tr ca ch s nghi ng ng sau n.

- CSCN sau cng ca s o gin tip c tng mt n v nu ng sau n l ch s nghi ngln hn 5; nu nh hn 5 th vn gi nguyn.

Th d: 71,56 lm trn thnh 71,6 71,54 lm trn thnh 71,5 - Khi ch s nghi ng ny bng 5 th CSCN sau cng ca s o gin tip c tng mt n v

nu n l mt ch s l; nu n l mt ch s chn (k c s 0) th vn gi nguyn. Th d: 71,55 lm trn thnh 71,6 71,05 lm trn thnh 71,0 71,25 lm trn thnh 71,2

T KHO

CU HI LNG GI

Sai s h thng, ngu nhin Sai s tng i, sai s tuyt i

Sai s th lch chun, lch chun tng i

Gi tr trung bnh Gii hn tin cy, khong tin cy



1. Gii thch s khc nhau gia: sai s h thng v sai s ngu nhin; sai s tuyt i v sai s tng i.

2. Trnh by cc khi nim v cng thc tnh cc i lng trong ton thng k: gi tr trung bnh, lch chun, gii hn tin cy v khong tin cy v.v

3. Nu cc bc x l kt qu phn tch; ng dng vo bi tp sau: Sau 6 ln xc nh hm lng (mg) ca mt hot cht trong ch phm thu c kt qu nh sau:

116,0 ; 97,9 ; 114,2 ; 100,6 ; 108,3 ; 104,5. Hm lng thc ca mu phn tch nm trong gii hn no? Gi s hm lng thc ca mu M = 100,0; hy tm xem phng php c sai s h thng hay khng?

CU HI TRC NGHIM 1. Khi tin hnh phn tch 1 mu bt k thng mc phi cc loi sai s:

2. Sai s do phng php o dn n:

3. Mt kim nghim vin c nhm th tch ti im tng ung khi nh lng, vy trong phn tnh ton kt qu ngi ny mc phi:

4. Loi sai s no c th hiu chnh v loi tr khi tin hnh phn tch mu?

5. Loi sai s no th hin ng ca phng php phn tch?

6. Loi sai s no th hin chnh xc ca phng php phn tch?

a. Sai s h thng d. C a, b, c u ng b. Sai s ngu nhin e. Ch b, c ng

c. Sai s th

a. Sai s th d. Sai s tuyt i b. Sai s ngu nhin e. Sai s tng i c. Sai s h thng

a. Sai s tng i d. Sai s h thng b. Sai s tuyt i e. Khng cu no ng

c. Sai s th

a. Sai s th d. Sai s tuyt i b. Sai s ngu nhin e. Sai s tng i c. Sai s h thng

a. Sai s th d. Sai s h thng b. Sai s ngu nhin e. Cu b v d ng

c. Sai s tuyt i



7. Loi tr sai s th bng cch:

8. Ch s c ngha (CSCN) trong s o trc tip bao gm:

9. Kt qu nh lng sau cng l M = 0,0020 g, s o ny bao gm ch s c ngha:

10. Kt qu nh lng sau cng l M = 0,0025 g, ch s 5 l ch s:

Chng 4 NNG DUNG DCH

a. Sai s th d. Sai s h thng

b. Sai s ngu nhin e. Sai s tng i c. Sai s tuyt i

a. Tra bng Student tm Ttn v Tlt d. Cu a v c ng

b. Dng phng php chun Dixon (test Q) e. Cu b v c ng c. Dng phng php kim nh T (test T)

a. Nhiu ch s tin cy v nhiu ch s nghi ng d. Cu a v c ng b. Ch c ch s tin cy e. Khng cu no ng

c. Nhiu ch s tin cy v duy nht mt ch s nghi ng

a. 5 ch s c ngha d. 1 ch s c ngha b. 4 ch s c ngha e. Khng cu no ng c. 2 ch s c ngha

a. Ch s c ngha tin cy c. Cu a v b u ng b. Ch s c ngha khng tin cy d. Khng cu no ng

Phn II

PHN TCH KHI LNG V PHN TCH TH TCH



NI DUNG

1. M U

Dung dch l mt h ng th gm hai hay nhiu cu t (phn t hay ion). Mt dung dch bt k u gm cht tan v dung mi. Thnh phn ca cht tan trong dung dch c biu din bng nng .

Nng l mt c tnh nh lng c bn ca dung dch. Ngi ta biu th nng ca dung dch bng lng cht tan c ho tan trong mt lng xc nh

dung dch hay dung mi.

2. CC CCH BIU TH NNG 2.1. Nng phn trm (nng bch phn)

2.1.1. Nng phn trm theo khi lng: C%(kl/ kl) Biu th bng s gam cht tan c trong 100 gam dung dch K hiu: C% (kl/kl) Cng thc

m: khi lng ca cht tan (gam). m1: khi lng ca dung dch (gam).

V: th tch dung dch (ml). d: khi lng ring ca dung dch. Th d: Dung dch amoniac 25% (kl/kl) c ngha l trong 100g dung dch c cha 25g amoniac T cng thc (4.1), suy ra lng cht tan cn thit

Trng hp pha mt dung dch trong ho tan m gam cht tan vo b gam dung mi th nng phn trm ca dung dch l:

MC TIU

Trnh by c cc cch biu th nng dung dch.

Tnh c ng lng ca mt cht trong phn ng.

Gii c cc bi ton v nng dung dch.

(4.1)

(4.2)



Th d 1: Tnh nng phn trm C% (kl/kl) ca dung dch natri carbonat nu cn 25 g Na2CO3 pha trong 250 ml nc

Gii:

Trong bi ton ny, m: khi lng cht tan = 25 g b: khi lng dung mi = V.d = 250 1 = 250 (g) m + b: khi lng dung dch = 25 g + 250 (g)

Nng phn trm ca natri cacbonat:

Th d 2: Alizarin c dng lm thuc th. Khi pha ngi ta ho tan 0,25 g trong 100 ml nc, nh vy nng phn trm C%(kl/kl) ca alizarin s l:

Trong ho phn tch, nng phn trm c coi l gn ng, ho cht c cn trn cn k thut.

2.1.2. Nng phn trm khi lng theo th tch: C%(kl/ tt)

Biu th s gam cht tan trong 100 ml dung dch K hiu: C% (kl/tt) Cng thc

Khi lng cht tan c tnh

Th d 1: Khi pha mt dung dch glucose u trng, nu s dng 200 g glucose pha thnh 1000 ml, nng dung dch glucose tnh theo nng phn trm:

Th d 2: pha 1 lt dung dch tim truyn NaCl 0,9%, lng NaCl c tnh theo cng thc:

(4.3)

(4.4)

(4.5)



2.1.3. Nng phn trm theo th tch: C% (tt/ tt)

Biu th s mililt (ml) cht tan c trong 100 ml dung dch K hiu: C% (tt/tt) Cng thc

V: th tch ca cht tan.

V1: th tch ca dung dch.

Th d: Ngi ta ly 960 ml ethanol tuyt i pha thnh 1000 ml dung dch. Nh vy nng ca dung dch cn l:

3. NNG PHN T (NNG MOL) CM

Biu th s mol ca cht tan c trong 1 lt (1000 ml) dung dch K hiu: CM

Cng thc

m: khi lng ca cht tan (gam) M: khi lng mol (khi lng phn t) ca cht tan. V: th tch ca dung dch cn pha.

S mol ca dung dch c tnh:

(4.6)

(4.7)

(4.8)

(4.9)



Th d 1: Tnh nng mol ca dung dch H2SO4, bit rng pha mt dung dch c th tch l 500 ml, lng H2SO4 m c cn dng l 49 gam. Khi lng mol ca H2SO4 (M = 98 g)

Gii

Th d 2: Nc bin tiu chun cha 2,7 g mui NaCl trong mi 100 ml dung dch. Xc nh nng mol ca NaCl trong nc bin.

Gii:

Khi lng NaCl: m = 2,7 (g) Khi lng mol ca NaCl M = 58,5 g Th tch ca dung dch V = 100 ml

Mt dung dch cha mt phn t gam cht tan trong mt lt dung dch c gi l dung dch phn t. T dung dch phn t c ly lm n v, ngi ta so snh dung dch m c hn: 2M, 3M... hay dung dch long hn: 0,1M (decimol), 0,01M (centimol), 0,001M (milimol)...

3.1. Nng ng lng

Cch biu th nng ng lng do nh ho hc ngi c Karl Friedrich Mohr pht biu mc ch n gin ho cc php tnh.

Cch biu th nng ny da trn khi nim khi lng ng lng (ng lng gam) ca mt cht.

3.1.1. ng lng gam

K hiu E gam ng lng gam ca mt cht c tnh tu thuc vo bn cht ca phn ng ho hc.

3.1.1.1. Trng hp phn ng acid base

i vi mt acid: ng lng gam l phn t mol ca acid chia cho s proton hot tnh (H+)

Th d 1: ng lng gam ca acid hydrochloric trong phn ng trung ho l:

HCl + NaOH NaCl + H2O

Th d 2: ng lng gam ca acid sulfuric c tnh trong trng hp phn ng vi NaOH



H2SO4 + 2NaOH Na2SO4 + 2 H2O

i vi mt base, ng lng gam l phn t gam ca base chia cho s proton hot tnh (H+) cn thit trung ho n.

Th d: ng lng gam ca natri hydroxyd c tnh khi phn ng vi HCl

3.1.1.2. Trng hp phn ng oxy ho kh

i vi phn ng oxy ho ng lng gam ca bt k cht no a vo phn ng cng l khi lng mol ca cht chia cho s electron (n) cho hay nhn trong qu trnh phn ng

Th d 1: Trong phn ng oxy ho gia st (II) v kali permanganat

Th d 2: Phn ng oxy ho gia Cr3+ vi Sn2+

3.1.1.3. Trng hp phn ng ta v phc cht

ng lng gam ca mt cation c tnh bng cch ly khi lng gam ca cation chia cho in tch ca cation

ng lng gam ca mt anion l khi lng gam ca anion chia cho s ng lng ca ion kim loi tng ng to ta hoc phc cht.

Th d: - Trng hp 1. Cation v anion c cng s in tch



- Trng hp 2. Cation v anion c s in tch khc nhau

- Trng hp 3. Trng hp to phc gia cation v anion

3.1.2. S ng lng gam

S ng lng gam ca cht A bng s gam cht A chia cho ng lng gam ca cht

3.1.3. Nng ng lng

Nng ng lng c biu din bng s ng lng gam ca cht tan c trong 1 lt (1000 ml) dung dch.

K hiu: CN

Cng thc:

(4.10)

(4.11)

(4.12) (4.13)



Th d: Xc nh nng ng lng ca dung dch nitrat bc khi ho tan 1,35 g AgNO3 thnh 250 ml dung dch.

Bit khi lng mol AgNO3 = 108, ng lng gam AgNO3 = 108 (s in tch Ag+ = 1).

Nng ng lng ca dung dch AgNO3:

Mt dung dch cha mt khi lng ng lng ca cht tan c go l dung dch c nng nguyn chun N. Bn cnh nng nguyn chun, ngi ta cn s dng bi s ca N nh dung dch 2N, 4N..., c s ca N nh dung dch 0,1N, dung dch 0,01N...

tnh ton lng cht tan A cn thit pha dung dch ng lng cho bit trc, ngi ta suy lng m t cng thc (4.13)

3.1.4. Cng thc tnh nng ng lng

Cc tnh ton trong cch chun theo nng ng lng gn ging vi cch tnh ton trong chun theo nng phn t. C mt s khc bit l ti im tng ng tch ca cc th tch (tnh bng ml) ca cc dung dch cc cht phn ng vi nng ng lng th bng nhau

ml A. NA = ml B. NB

VA.NA = VB.NB

Hay ti im tng ng cc th tch dung dch thuc th v ca cc cht cn xc nh phn ng vi nhau t l nghch vi nng ng lng ca chng.

Quy tc ny gi l quy tc hp thc (stochiomtrie stoichiometry) Th d: Dung dch acid hydrochloric c xc nh nng bng cch em chun vi dung dch

chun natri hydroxyd. Nu sau khi phn ng kt thc, 25,00 ml dung dch HCl c trung ho cn mt th tch NaOH 0,0950 N l 32,20 ml. Tnh nng ng lng ca dung dch acid hydrochloric.

Gii:

p dng phng trnh trn, ta c: NAVA = NBVB

Vi VA: th tch ca dung dch HCl cn chun , VA = 25,00 ml.

(4.14)



NA: nng ca HCl.

VB: th tch ca dung dch chun NaOH, VB = 32,20 ml.

NB: nng ca NaOH NB = 0,0950 N

Nng ca dung dch acid hydrocloric:

4. MT S CCH BIU TH NNG KHC 4.1. Nng gam

4.1.1. Nng g/l

Biu th bng s gam cht tan c trong 1 lt dung dch.

K hiu: g/l Cng thc:

m: khi lng ca cht tan (gam). V: th tch dung dch cho tnh bng ml.

Loi nng ny thng dng trong mt s thuc th acid hay base.

Th d: Nc thch cao hay dung dch CaSO4 bo ho c nng 2 g/l.

4.1.2. chun

Biu th s gam hay miligam cht tan c trong 1 mililt dung dch

K hiu: T

Th d: Acid nitric m c l acid c chun T = 1,40 g/ml. Trong thc t, ngi ta thng dng nhng dung dch chun c nng chn sao cho c 1 ml

dung dch chun tng ng vi mt lng cht cn nh lng tnh ton c d dng. Th d: xc nh cng ca nc, ngi ta nh lng nc bng dung dch complexon III

chun 0,01M vi ch th en Eriocrom T. Khi phn ng kt thc, c s th tch ml dung dch complexon III s dng.

Tnh kt qu: 1 ml complexon III ng vi 0,561 mg CaO, nh vy ng vi V ml ngi ta tnh c

(4.15)

(4.16)

(4.16a)



hm lng Ca2+ c trong nc.

Ngi ta cn dng chun trong chuyn i t nng ny sang nng khc. Th d: Tnh nng phn trm C%(kl/tt) ca dung dch H2SO4 2M.

T dung dch H2SO4 2M, p dng cng thc tnh CM suy ra khi lng

Nng phn trm C% (kl/tt) ca H2SO4 = 0,196 100 = 19,6%

4.2. Nng ion gam

Biu th s ion gam cht tan c trong 1 lt dung dch. K hiu: ion gam/l. Th d: Dung dch FeSO4 c cha 5,6 g Fe

2+ nng ion gam ca Fe2+ l 0,1 ion g/l v 1 ion

gam Fe2+ = 56.

4.3. Nng phn triu (ppm: parts per million)

i vi nhng dung dch c nng rt long, ngi ta thng s dng nng phn triu (ppm). Nng phn triu biu th s gam cht tan c trong 106

gam dung dch hay hn hp. Ngi ta cn biu th s microgam ca cht tan c trong 1 ml dung dch hay s miligam cht tan c trong 1 lt dung dch.

K hiu: ppm Cng thc:

V khi lng ring ca nc l 1, khi mt dung dch c lng cht tan qu t so vi lng dung mi, ngi ta quy c 1 ml dung dch = 1 g, v l , dung dch c nng 1 ppm tng ng vi 1g cht tan c trong 1 ml dung dch hay 1,00 mg/1 lt dung dch

Th d: TCVN 5501 - 1991 quy nh gii hn nng nitrat c trong nc ung l 5ppm.

(4.17)

(4.18)

(4.19)



4.4. Nng phn t (ppb: parts per billion)

Nng ppb hay nng phn t biu th s gam ca cht tan c trong 109 gam ca dung dch hay hn hp.

K hiu: ppb Cng thc:

Dung dch 1 ppb ca mt cht tng ng vi 1 mg cht trong 1 ml dung dch hay 1 g trong 1 lt dung dch.

Trng hp biu th nng ca cht kh, ngi ta thng dng n v cht kh trong mt th tch hn l trong khi lng.

Th d: 8ppb monoxyd carbon trong khng kh c ngha l 8l CO c trong 1 lt khng kh.

5. H THC LIN LC GIA CC NNG 5.1. Nng phn trm (C%) v nng mol (CM)

Th d 1: Tnh nng phn trm ca dung dch NH4OH 14,8M, bit rng khi lng ring l 0,899 g/ml, khi lng mol phn t l 17,03.

Gii

p dng cng thc (4.22)

Th d 2: Tnh nng mol ca dung dch H2SO4 98% (d = 1,84 g/ml), phn t gam = 98.

Gii

Nng mol v nng ng lng

(4.20)

(4.21)

(4.22)

(4.23)

(4.24)



n c tnh tu theo bn cht

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