http://d.violet.vn/uploads/resources/436/48442/preview.swfng lit) C%=m *100/m = *M*100)/ *1000 )=C/ ---> vi n/V=C(M), mct=n*M, d=mdd/1000V (do V tinh b ng l ma d lai co n vi la g/ml) => C(M)=(C%*10)/M Cm C(M) = ( 10.C%.d ) / M Tro g C M): g mol (n/l) C%: n g khi lng phn trm d: khi lng ring c a dung dch M: phn t khi c a cht tan

p

u g

g g ro g p g

g m

Trang 1 / 2Trong lnh vc Ha Hc, t cc th nghim n gin trong phng th nghim cho n cc ti nghin cu, u yu cu chun b cc dung dch c bn cn thit. Mi nghe th bt c sinh vin Ha Hc no cng cho l nhng vic n gin, thm ch nhm mt cng c th lm c, nhng chun b mt dung dch no theo yu cu ca cng vic i hi s tinh khit v tnh chnh xc cao th chng ta phi tin hnh chun b chng rt cn thn. Di y l mt vi thut ng thng dng. Cht tan: Cht c kh nng ha tan trong dung dch. Dung mi: Cht dng ha tan mt cht khc thnh dung dch. ng n ha tan trong nc th nc l dung mi dng ha tan ng, cn ng l cht tan. Dung dch: Mt hn hp ca 2 hay nhiu cht tinh khit. Trong mt dung dch mt cht tinh khit c ha tan ng ly bi mt cht tinh khit khc. V d, trong dung dch nc ng th hm lng ng c phn b ng u trong dung dch, v l mt dung dch ng ly. Mol: Mt n v khi lng c dng trong Ha Hc. Thut ng ny cp n mt s lng rt ln cc ht c bn (nguyn t, phn t, ion, electron...vv..) ca mt cht bt k. Mt mol c khong 6.02 x 23 10 ht c bn ca mt cht no (s Avogadro). Dung dch chun: Nhiu dung dch c dng trong lnh vc phn tch m nng ca chng cn c xc nh mt cch chnh xc (n cc ch s c ngha thch hp). Chng c th c chun b bng cch cn mt lng xc nh cht rn tinh khit ri pha thnh dung dch vi th tch xc nh hoc bng cch chun bng mt dung dch chun khc c nng xc nh khng i. Tm li dung dch chun l mt dung dch c nng xc nh v khng b bin tnh theo thi gian. chnh xc v cc ch s c ngha: Nu mt dung dch c pha t 2.5 g cht rn (+/- 0.1g) v ta bit rng con s ny c 2 ch s c ngha v nng ca n cng c th l s c hai ch s c ngha chng hn nh 1.5 M hay l 0.15 M... Nu cht rn c cn l 2.50 g (+/- 0.01g) ta bit rng nng ca dung dch cng c th cha n 3 ch s c ngha chng hn nh: 1.50 M hay l 0.150 M. Nu khi lng cht tan l 2.500 g (+/- 0.001g) th nng ca dung dch c th c 4 ch s c ngha chng hn 1.500 M hoc l 0.1500 M. Ti sao chng ta li quan tm n cc ch s c ngha. Vic xc nh cc ch s c ngha gip bn xc nh chnh xc ca mu cn. Trong cc lp hc ph thng th vic xc nh cc ch s c ngha th khng my ai nhng trong cc phng th nghim phn tch vic xc nh mc chnh xc ca mu l rt quan trng. cn mu chnh xc bn nn s dng cc loi cn in t ph hp cho cng vic

ca bn i hi. Cn c chnh xc cng sao th gi thnh s cng cao. Khi bn cn vi hm lng mu ln th s c t sai s hn. Pha mt lng ln dung dch mu, ct vo kho ri dng t t, nh vy s tit kim c cng sc ca bn va li trnh c cc sai st. Cc ha cht thng dng chun b dung dch mu c lit k di y: Ha cht Phn t khi tinh khit ng dng

Natri Cacbonat

105.99

99.9%

Axt baz Tc nhn kh

Kali HydroPhtalat

204.23

99.9

Kali Dicromat

294.22

99.9

Kali Iotdat

214.01

99.9

Natri thiosulfat Bc nitrat Mui kim loi

NaCl

58.45

99.9

EDTA dinatri, dihydrat

372.25

99.0

Cc ha cht thng thng : Cc ha cht thng thng l cc ha cht hay c dng trong cc trng trung hc, m chnh xc ca cc cht ny khng c quan tm v t nhiu n cng khng lm nh hng n cc th nghim n gin, nh ch th mu axt baz. Yu cu duy nht ca chng l lm sao hc sinh thy c s i mu, si bt khi phn ng xy ra...vv... Cc dung dch Stock: Mt dung dch stock l mt dung dch c nng tng i cao (dung dch chun hay khng chun) m n c th ct tr c trong mt thi gian tng i lu, ch pha long khi no dng n. Vic chun b cc dung dch stock c th tit kim c rt nhiu thi gian ca bn. Chun b cc dung dch. Nhiu phn ng ha hc xy ra trong dung dch m trong chng ta bit hm lng ca cc cht phn ng. iu ny c th lin quan n vic cn mt hm lng chnh xc ca mt cht iu kin kh ro hoc l ong mt lng cht lng chnh xc. Chun b cc dung dch mt cch chnh xc s an ton hn v c hi thnh cng cng cao hn. Pha dung dch theo t l phn trm v khi lng. Cng thc: Phn trm khi lng C% = [Khi lng cht tan(g) / Th tch ca dung dch (ml)] x 100 V d 1: Mt dung dch 10% NaCl c 10g NaCl ha tan trong 100ml dung dch. Cch lm: Cn 10g NaCl vo ng ong hay bnh nh mc c cha khong 50ml nc. Lc u cho NaCl tan

ht sau thm nc n vch 100ml. Ch : ng bao gi ly rt y nc vo ng ong n vch 100ml ri sau mi cho NaCl vo. V nu lm nh th th tch dung dch s b thay i do thnh phn phn trm khi lng cht tan trong dung dch c th nh hn so vi yu cu. (V mt vt l th qu trnh ha tan mui vo nc khng lm thay i th tch cui ca dung dch nhng cng c th xy ra vi mt s cht khc, do bn nn lm ng cch vn hn.) Pha dung dch theo t l phn trm v th tch Khi cht tan ca bn l cht lng, i khi chng ta phi pha ch cc dung dch theo t l phn trm v th tch. Cng thc T l theo phn trm th tch V% = [Th tch cht tan (ml) / Th tch ca dung dch (ml)] x 100 V d 2: Pha 100ml c cha 5% etylen glycol v th tch. Cch lm Trc ht, biu din thnh phn phn trm ca cht tan di dng mt s thp phn 5% = 0.05. Sau nhn s ny vi tng th tch ca dung dch chng ta s thu c th tch ca cht tan cn thit: 0.05 x 100 = 5ml (etylen glycol). ong ra 5ml etylen glycol ri vo bnh nh mc sau thm nc cho n vch 100ml. Pha dung dch theo s mol cho trc Nng mol l mt n v rt hu hiu dng nh gi hm lng cht tan trong dung dch v ng thi n l mt yu t dng tnh th tch (mt tnh cht d o lng ca dung dch) cn thit ca cht tan qua vic chuyn i s mol ca cht tan. Nng mol M = [S mol cht tan / Mt lt dung dch] By gi chng ta xem th, tin hnh chun b dung dch vi nng cho trc nh th no nh. V d 3: Gi s rng bn ang lm mt th nghim v cn dng n cc dung dch sau: 0.5M Na 2CO3 , 1.5M HCl. Dung dch Na2CO3 s c pha t Na2CO3 tinh khit dng rn v dung dch HCl c pha long t HCl m c. i vi Na2CO3 : Trc ht bn phi cn khi lng ca Na2CO3 Pha cht rn vi mt th tch nc nh hn th tch dung dch mong mun. Pha long dung dch n th tch cn dng. Gi s bn cn 250ml dung dch Na2CO3 0.5M. Trc ht cn tnh khi lng cn thit ca Na2CO3 cn cho 250ml dung dch: S gam Na2CO3 cn = [(250ml/1000ml) x 0.5 mol Na2CO3 x 105.989 g Na2CO3] = 13.2 g Na2CO3 Trong 105.989 g l khi lng phn t ca Na2CO3 Cn 13.2 g Na2CO3 ri vo bnh nh mc.

Thm t nc ha tan cht rn. Thm nc cho n vch 250ml. Nh vy t v d trn chng ta thu c cng thc tng qut cho bi ton pha dung dch vi nng v th tch cho trc nh sau: m(g) cht tan = [th tch dung dch(ml) / 1000ml] x [Nng mol ca dung dch] x [Khi lng phn t ca cht tan]

i vi HCl: Tht kh m pha dung dch HCl t kh Hydrochlorua, cc k thut vin phng th nghim thng mua dung dch axt HCl m c v pha long. Axt HCl dng trong cc phn ng thng c nng t 1 n 6M. HCl m c thng cha khong 35-37% khi lng Hydroclorua ha tan vo nc v c khi lng ring l 1.18g/ml. Nh vy 1000ml HCl m c c khi lng ring l 1.18g/ml th s c khi lng l 1000 x 1.18 = 1,180g. 1000ml HCl m c c cha [35/100] x 1,180 = 413g HCl; Phn t lng ca HCl = 1.008 + 35.5 = 36.508 ~ 36.5g/mol vy nng ca n s l: 413/36.5 = 11.3 M Bng cch tng t chng ta c th tnh ton c cho mt s axt thng dng nh di y: Phng php chung l: T khi lng ring tm ra khi lng ca mt 1000ml dung dch axt. T hm lng % c th tnh c khi lng ca axt/1000ml. Tm nng mol ca axt bng cch chia cho phn t khi ca n.

Axt HCl HNO3 H2SO4 CH3COOH NH3

Phn t khi 3 g/cm 36.5 63.0 98.1 60 17

Khi lng ring (g/ml) 1.18 1.42 1.84 1.05 0.89

Hm lng % 35 70 98 100 35

Nng 11.3 15.8 18.4 17.5 18.3

Cc phng thc pha long c hi khc nhau mt cht l cc axt hon ton tinh khit, ging nh sulfuric axt m c c cha khong 95 98% H2SO4. y chng ta ngm hiu rng cc axt m c cng cha mt lng nc ng k no . Tr li vi v d 3 trn: Gi s chng ta mun pha 500ml dung dch HCl 1.5M t HCl m c. Chng ta bit l HCl m c c nng l 11.3M, nh vy trc ht chng ta lp hai t s: 500ml HCl 1.5 M c] [1.5 mol HCl long / 1000 ml HCL long] [12 mol HCl m c / 1000ml HCl m

Mi nng c vit thnh mt t s v c n v r rng xc nh u l dung dch tinh khit u l dung dch pha long. S ml HCl m c cn pha long = 500ml HCL long x [1.5 mol HCl long / 1000 ml HCL long] x [1000ml HCl m c /12 mol HCl m c /] = 66.4 ml HCl m c ong ra 66.4 ml HCl m c vo bnh nh mc 500ml c cha sn mt lng nc ng k pha long axt.

35172-Sao Lai Nhan Nham May Anh-Lam Chan Huy Thm nc vo n vch nh mc 500ml Ngoi ra cn c mt cch gii quyt khc cho bi ton pha long ny. Khi mt lng axt m c c pha long bng nc th hm lng axt s khng thay i trc v sau khi pha long. Do s mol ca axt trong dung dch m c v dung dch sau khi pha long phi bng nhau. Do chng ta c phng trnh. [S mol ca cht tan trong dung m c] = [S mol ca cht tan trong dung dch pha long] S mol ca cht tan c th tnh bng tch ca nng mol v th tch ca dung dch: Nh vy chng ta s c: [Th tch dung m c] x [Nng ca dung dch m c] = [Th tch ca dung dch pha long] x [Nng ca dung dch pha long] Gi [Th tch dung m c] l VC ; [Nng ca dung dch m c] l MC ; [Th tch ca dung dch pha long] l VD ; [Nng ca dung dch pha long] l MD : Chng ta thu c cng thc di y: VC MC = VD MD Tt nhin l bn bit nn lm g vi phng trnh ny tm ra th tch ca cht tan cn pha dung dch. Lu tr cc dung dch Cc l nha c np rt thch hp cho vic ct tr cc dung dch stock, cc l nha cng an ton hn thy tinh khi chng va chm vo nhau. Bn nn dng l bng nha ng cc dung dch kim, v kim n mn thy tinh. Vi cc dung dch m bn ct tr mt th tch tng i nhiu th nn dng bnh nc khong 20L dng ng chng v sau ny khi dng n, bn s rt ra d dng hn v cc bnh nc khong ny c vi. Cc l hay bnh cha dung dch nn c gin nhn v ghi ch cn thn. Cnh bo Axt Cc axt m c u t nhiu c tnh n mn do khi thao tc bn nn thn trng. Nn mang bao tay bng nylon thay v dng bao tay y t bn ngoi ca hng thit b y t. Sau khi rt axt xong nn dng khn t lau sch ming chai. ong axt bng ng ong hoc pipet (c qu bp), sau khi dng xong nhng ng ong hoc pipet vo chu ra cha y nc, khng nn rt nc trc tip vo ng ong ra. Dung dch NaOH, KOH NaOH dng rn hay dng lng m c u c tnh n mn kh cao, thm ch NaOH long cng c tc dng rt xu n mt. Nn khi pha cc dung dch kim nh NaOH, KOH bn nn tin hnh trong t ht, nh eo gng tay v mt n bo v, bt qut t ht khong 10-15 pht sau mi ly ra rt vo bnh cha. Mt iu ng ch l bn nn cc cht rn ca kim vo nc mt cch t t v chng rt

hu nc. Dung dch Hydroperoxit H2O2 bn trn th trng hin nay thng cha khong 30% hoc l '20 volume' thng c xut x t Trung Quc. Nh vy nhng thng s ny l g? v lm sao chng ta c th c c 20 phn th tch t 3 dung dch H2O2 m c 30%. Chng ta cn bit '20 volume' l g? N c ngha l 1cm dung dch s cho 3 3 ra 20cm kh Oxy. Nh vy c lng kh l bao nhiu trong mi cm H2O2 30%. Khi lng ring ca H2O2 l 1.10g/cm nh vy 1000ml dung dch c khi lng l 1,100g v do n cha [30/100]x 1,100g = 330g H2O2. Phn t khi ca H2O2 l 34g/mol. Vy nng mol ca dung dch 30% s l: 330/34 = 9.7M. Phng trnh phn hy:3

H2O2 1mol 9.7mol 1000cm3

=

H2O

+

1/2 O2 mol 4.85mol ? cm kh3

Mt mol kh Oxy iu kin tiu chun (standard temperature and pressure) chim 22.4 dm = 22400 3 3 3 3 cm . Do 4.85mol s chim th tch l 108,640cm . V 1cm H2O2 30% cho ra 108,640/1000 cm kh 3 Oxy =108 cm = '100 phn th tch' hay 100 volume. Nn mun c dung dch H 2O2 '20 volume' chng ta 3 cn pha long 5 ln bnh ng 1000 cm H2O2 30%.

3