Ch : S IN LI

Bi ton 1: CHT IN LI MNH 1. Vit phng trnh cht in li mnh Axit : HCl, H2SO4 , HNO3 ... HCl H+ + Cl H2SO4 2H+ + SO42 Bazo : NaOH, Ca(OH)2 ... NaOH Na+ + OH Ca(OH)2 Ca2+ + 2OH Mui : NaCl, CaCl2 , Al2(SO4)3 NaCl Na+ + Cl CaCl2 Ca2+ + 2Cl Al2(SO4)3 2Al3+ + 3SO422. XC NH NNG MOL ION B1 : Tnh s mol cht in li B2 : Vit phng trnh in li, biu din s mol ln phng trnh in li n B3 : Tnh nng mol ion : CM = V Bi 1.1. Vit phng trnh in li cc cht sau y (nu c ) : 1. HClO4 2. Sr(OH)2 3. K3PO4 4. BaCl2 5. AgCl 6. Fe(OH)3 7. Al2(SO4)3 8. KMnO4 9. KOH 10. HNO3 11. BaSO4 12. Bi 1.2. Vit cng thc ca cht m khi in li to ra cc ion : a. K+ v CrO42- b. Fe3+ v NO3-c. Mg2+ v MnO4d. Al3+ v SO42Bi 1.3. Tnh nng mol cc ion trong dung dch sau : a. 100 ml dung dch cha 4,26 gam Al(NO3)3 b. 0,2 lt dung dch c cha 11,7 gam NaCl a. nAl(NO)3 = 0,02 (mol) Al(NO3)3 Al3+ + 3NO30,02 0,02 0,06 (mol) [Al3+] = 0,02/0,1 = 0,2(M) ; [NO3-] = 0,06/0,1 = 0,6 (M) Bi 1.4. Tnh nng mol cc ion trong dung dch thu c khi : a. Trn 200 ml dung dch NaCl 2M vi 200 ml dung dch CaCl2 0,5M b. Trn 400 ml dung dch Fe2(SO4)3 0,2M vi 100 ml dung dch FeCl3 0,3M c. Trn 200 ml dung dch cha 12 gam MgSO4 v 300 ml dung dch cha 34,2 gam Al2(SO4)3 a).[ Na+] = 0,4/0,4 = 1(M) ; [Ca2+] = 0,1/0,4 = 0,25 (M) ; [Cl-] = (0,4 + 0,2)/0,4 = 1,5 (M) b). [Fe3+] = 0,38 (M) ; [SO42-] = 0,48 (M) ; [Cl- ] = 0,18 (M) c). [ Mg2+] = 0,2(M) ; [Al3+] = 0,4 (M) ; [SO42-] = 0,8 (M) Bi 1.5. a). Ha tan 12,5 gam tinh th CuSO4.5H2O trong nc thnh 200 ml dung dch . Tnh nng mol cc ion trong dung dch thu c . b). Ha tan 8,08 gam Fe(NO3)3 .9H2O trong nc thnh 500 ml dung dch . Tnh nng mol cc ion trong dung dch thu c . . a.) nCuSO4. 5H2O = 12,5/250 = 0,05 (mol) CuSO4.5H2O Cu2+ + SO42- + 5H2O 0,05 0,05 0,05 (mol) [ Cu2+] = [SO42-] = 0,05/0,2 = 0,25 (M) b.) 0,04 (M) ; 0,12 (M) Bi 1.6. a). Tnh th tch dung dch HCl 0,5M c cha s mol H+ bng s mol H+ c trong 0,3 lt dung dch HNO3 0,2M . b). Cn bao nhiu ml dung dch HCl 2M trn vi 180 ml dung dch H2SO4 3M c mt dung dch c nng mol ca H+ l 4,5M . Cho bit H2SO4 in li hon ton. . a). VHCl = 0,12 (lit) b). VHCl = 108 (ml) -1-

Trng THPT Trn ng Ninh Bi ton 2: NH LUT BO TON IN TCH B1 : Pht biu nh lut - Trong dung dch cha cc cht in li, tng s mol in tch dng v m lun lun bng nhau. B2 : p dng gii ton Cng thc chung :

GV: H Vn Qun

Mol dt (+) = Mol dt ()

Cch tnh mol in tch : ndt = sochi dt. nion Khi lng cht tan trong dung dch mmuoi = mcation + manion Bi 2.1. Dung dch A cha Al3+ 0,1 mol, Mg2+ 0,15 mol, NO3- 0,3 mol v Cl- a mol . Tnh a . a = 0,3 mol . Bi 2.2. Dung dch A cha Na+ 0,1 mol , Mg2+ 0,05 mol , SO42- 0,04 mol cn li l Cl- . Tnh khi lng mui trong dung dch . m = 11,6 gam. Bi 2.3. Trong mt dung dch c cha a mol Ca2+ , b mol Mg2+ , c mol Cl v d mol NO3a. Lp biu thc lin h gia a, b, c, v d b. Nu a = 0,01 ; c = 0,01 v d = 0,03 th b bng bao nhiu ? a. p dung nh lut BTT : 2a + 2b = c + d b. b =c + d 2a 0,01 + 0,03 2.0,01 = = 0,01 2 2

Bi 2.4. Mt dung dch c cha hai loi cation l Fe2+ (0,1 mol) v Al3+ (0,2 mol) cng hai loi anion l Cl- (x mol) v SO42- (y mol) . Tnh x v y bit rng khi c cn dung dch v lm khan thu c 46,9 gam cht rn khan . x = 0,2 (mol) v y = 0,3 (mol) Bi 2.5. (C 07) Dung dch A cha cc ion Al3+ = 0,6 mol, Fe2+ = 0,3 mol , Cl - = a mol, SO42- = b mol . C cn dung dch A thu c 140,7 gam . Gi tr ca a v b ln lt l : A. 0,6 ; 0,9 B. 0.9 ; 0,6 C. 0,5 ; 0,3 D. 0,2 ; 0,3 Bi ton 3. CHT IN LI YU 1. Vit phng trnh in li Axit : CH3COOH, H2S , H3PO4 * CH3COOH H+ + CH3COO * H2S H+ + HS- ; HS- H+ + S2 + * H3PO4 H + H2PO4 ; H2PO4- H+ + HPO42- ; HPO22- H+ + PO43 Hirxit lng tnh : Al(OH)3 , Zn(OH)2 ... Tnh bazo : * Al(OH)3 Al3+ + 3OH * Zn(OH)2 Zn2+ + 2OH Tnh axit : * Al(OH)3 H3O+ + AlO2 * Zn(OH)2 2H+ + ZnO22 2. Xc nh in li . B1 : p dng CT tnh in li

-2-

Trng THPT Trn ng Ninh

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=

n C so phantu dienli = dien li = M dien li so phantu hoa tan nhoa tan CM hoa tan

B2 : S dng phng php ba dng . AB A+ + B Ban u : a in li : x Cn bng : a x 0 x x

0 x x (M) .

x a * = 1 : cht in li mnh * 0 < < 1 : cht in li yu * = 0 : cht khng in li Bi 3.1. Vit phng trnh in li ca cc cht sau trong dung dch : 1. Axit mnh H2SeO4 (nc th nht in li mnh) 2. Axit yu ba nc H3PO4 . 3. Hi r xit lng tnh Pb(OH)2 . 4. Na2HPO4 . 5. NaH2PO4 6. Axt mnh HMnO4 7. Bazo mnh RbOH. Bi 3.2. in li dung dch CH3COOH 0,1M c dung dch c [H+] = 1,32.10-3 M . Tnh in li ca axit CH3COOH . iu cn nh : bi ton ny cho nng in li ca cht in li CH3COOH H+ + CH3COO 1,32.10-3 1,32.10-3 (M) in li ca axit CH3COOH 1.32.103 .100 = 1,32% = 0,1 Bi 3.3. Tnh nng mol cc ion H+ v CH3COO- c trong dung dch axit CH3COOH 0,1M . Bit phng trnh in li : CH3COOH CH3COO- + H+ v in li = 4% in li : =

C = C0 = 0,1.4% = 0,004 M T phng trnh in li :[CH3COO-] = [H+] = 0,004 M Bi 3.4. Cho dung dch HClO c nng mol 0,01M, nng ny HClO c in li l = 0,172% . a). Tnh nng cc ion H+ v ClO- . b). Tnh nng mol HClO sau in li . . a). [H+] = [ClO-] = 1,72.10-5 (M) . b). [HClO] = 9,9828.10-3 (M) . Bi 3.5. Ha tan 3 gam CH3COOH v nc c 250 ml dung dch, bit in li = 0,12 . Tnh nng mol ca cc phn t v ion trong dung dch . . [CH3COOH] = 0,176 (M) ; [H+] = [CH3COO-] = 0,024 (M) . Bi 3.5. Mt lt dung dch CH3COOH 0,01 M c cha tng s 6,28.1021 ion v phn t CH3COOH . Tnh in li ca axit ny . iu cn nh : - S phn t N = n . 6,02.1023 - cho lng ban u v lng cn li, nn s dng pp ba dng : Ban u in li Khi cn bng CH3COOH H+ + CH3COO Ban u: 0,01 in li: x x x Khi cn bng 0,01 x x x mol -3-

Trng THPT Trn ng Ninh 6,28.1021 = 1,043.102 x = 0,043.10-2 mol Theo : 0,01 x + x + x = 23 6,02.10

GV: H Vn Qun

0,043.102 = 4,3.102 = 4,3% in li : = 0,01 Bi 3.6. Trong 100 ml dung dch axit nitr nhit nht nh c 5,64.1021 phn t HNO2 v 3,6.1020 ion NO2-. a. Tnh in li ca axit nitr trong dung dch nhit . b. Tnh nng mol ca dung dch ni trn . HNO2 H+ + NO2 Ban u n0 in li 3,6.1020 3,6.1020 21 Khi cn bng 5,64.10 3,6.1020 S phn t ha tan trong dung dch l : n0 = 3,6.1020 + 5,64.1021 = 6.1021 3,6.1020 = = 0,06 = 6% 6.1021 6.1021 = 0,1( M ) . b. Nng dung dch l: 6,02.1023.0,1 Bi 3.7. Tnh nng mol ca cc ion H+ v CH3COO- trong 2 lit dung dch c cha 24 gam CH3COOH ha tan . Bit in li ca axit l = 1,2% C0 = 0,2 M C = 1,2% x 0,2 = 0,0024 M T phng trnh in li : [ CH3COO-] = [H+] = 0,0024 M Bi 3.8. Dung dch axit CH3COOH 0,6% c khi lng ring xp x 1 g/ml . in li ca axit = 1% . Tnh nng mol ca ion H+ trong 1 lt dung dch . m = V.D = 1000 gam maxit = 0,6% x 1000 = 6 gam naxit = 0,1 mol [CH3COOH] = 0,1 M V = 1% C = 0,1x 1% = 0,001 M [ H+ ] = 0,001 M. Bi 3.9. Ha tan 3 gam CH3COOH vo nc c 250 ml dung dch, bit in li = 0,12 . Tnh nng mol ca cc phn t v ion trong dung dch . 3 = 0,05(mol ) S mol ban u ca CH3COOH : nCH3COOH = 60 3 S mol in li ca CH3COOH : nCH3COOH = 0,05.0,12 = 6.10 (mol ) Ban u : in li : Cn bng : CH3COOH H+ + CH3COO 0,05 0 0 -3 -3 6.10 6.10 6.10-3 -3 -3 0,05 6.10 6.10 6.10-3 (mol). [CH3COOH] = 0,176 (M) ; [H+] = [CH3COO-] = 0,024 (M).

Bi ton 4. XC NH HNG S IN LI B1 : Xc nh hng s in li ca axit. HA H+ + A ka = [ H + ].[ A ] [ HA]

- [H+] , [A-], [HA] trng thi cn bng. -4-

Trng THPT Trn ng Ninh - ka : cng ln th tnh axit cng mnh. B2 : Xc nh hng s in li ca bazo. BOH OH- + B+ . [OH ].[B ] kb = [BOH ]

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- [OH-], [B+], [BOH] trng thi cn bng . - kb : cng ln th tnh bazo cng mnh. Bi 4.1. C mt dung dch axit CH3COOH (cht in li yu) . Nu ha tan vo dung dch mt t tinh th CH3COONa (Cht in li mnh), th nng H+ c thay i khng , nu c th thay i th no ? Gii thch . iu cn nh. - S phn li ca cht in li yu l mt qu trnh thun nghch dn n cn bng ng (cn bng in li) Cn bng in li cng c hng s cn bng K v tun theo nguyn l L Sa t li . - in li ch ph thuc vo nhit khi in li . . CH3COOH CH3COO- + H+ [ H + ][CH 3COO ] [CH 3COOH ] Khi ha tan cht in li CH3COONa vo dung dch th nng CH3COO- tng ln do s phn li : CH3COONa Na+ + CH3COOV Ka khng i [H+] gim xung Bi 4.2. Cn bng sau tn ti trong dung dch : CH3COOH H+ + CH3COO- . in li ca CH3COOH bin i nh th no ? a. Khi nh vo vi git dung dch HCl b. Khi pha long dung dch c. Khi nh vo vi git dung dch NaOH d. Nh vo dung dch vi git dung dch CH3COONa . CH3COOH H + + CH 3COO k= [H + ] [CH 3COO ] = in li : = [CH 3COOH ] [CH 3COOH ] a. Khi nh HCl vo lng H+ tng ln cn bng dch chuyn sang phi lng CH3COOH tng ln gim b. Khi pha long dung dch CH3COOH in li nhiu tng c. Khi nh vo dd NaOH cn bng dch chuyn sang phi, (v H+ + OH-) tng d. CH3COO- tng ln cn bng dch chuyn chiu nghch (lm gim nng CH3COO- ) gim. Bi 4.3. Tnh nng mol ion H+ ca dung dch CH3COOH 0,1M , bit hng s phn li ca axit Ka = 1,75.10-5 . . CH3COOH H+ + CH3COO B : 0,1 0 0 li : x x x Cb : 0,1 x x x (M) [H + ][CH 3COO ] x2 5 1,75.10 = Hng s in li ca axit : ka = [CH 3COOH ] 0,1 x V : x