Tài liệu Phân dạng và phương pháp giải hóa học 11-phần vô cơ

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P •— /IJyfD ^ *>6 X U A N HUNG Gi&o vifen c h u y S n l u y § n t h i OH-CD H6a Pfian ddng& pnifong phapgidk HOA HOC^ PHAN V 6 c o l l D3nh cho hoc sinh lop 11 on tap va nang cao la nang lam bai. i THi; VIEW TINHBINH IHUAN MXUJTBANldNGHQPliNHPHdHdCHlMMI PHAN DANG VA PHUdNG PHAP GlAi HOA HQC 11 Vd C d A. T 6 M T A T L f T H U Y € T D6 X U A N H l / N G I. , Chju trach nhi^m xud't ban Vf du : N G U Y E N T H | T H A N H HlTCfNG Bien lap C6 hai loai: Chat di0n l i manh va chatdi^n li yeu. Sura ban in : H6NG HAI : C6ng ty K H A N G V I ^ T Bia : C6ng ty K H A N G V I | ; T > Na* + O H " NaOH * ChS't di$n li Ih nhifng chat tan trong nurdc phan li ra ion. : HAI A U Trinh bhy SI/DI|NU * S I / di^n l i 1^ qui trlnh phan li cac chat trong nifdc ra ion. * DO diOn l i (a): Dp diOn l i (a) cua chat dipn l i la ti so giffa s6' phan tur phan l i ra ion (n) va tong so phan tur hoa tan (no). a=JL «() NHA XUAT BAN idNG HOP TP. H6 CHI MINH NHA SACH TdNG H0P Neu bieu di^n diTdi dang phan trSm : 0% < a ^ 100% u a = 0 : chat kh6ng di$n l i 62 Nguyen Thj Minh Khal, Q . l u 0 < a < 1 : cha't di^n li yeu D T : 38225340 - 38296764 - 38247225 u a = 1 : cha't diOn li manh. Fax: 84.8.38222726 II. A X I T - B A Z C J - M U 6 ' l Email: tonghop®nxbhcni.com.vn 1. Axit Website: www.nxbhcm.com.vn/ www.fiditour.com * Theo thuye't A-r6-ni-ut: axit Ift cha't khi tan trong nrfdc phan li ra cation H*. V i du : Tong phdt hank C6NG TY TNHH MTV C H 3 C O O H + H 2 O U C H 3 C O O - + H3O* axit nhieu nac. 2. Bazcf: Djachf: 71 Dinh Ti§n Ho^ng - P.Da Kao - Q.1 - TP.HCM Dientho^i: 08. 39115694 - 39105797 - 39111969 - 39111968 Fax: 08: 3911 0880 Email: khangvietbookstore©yahoo.com.vri * Theo thuye't A-re-ni-ut: bazd la chat khi tan trong nirdc phan li ra anion O H ' . Vidu: KOH J-K^ + OH" * Theo thuyet Bron-stet: bazd la cha't nhan proton. Website: www.nhasachkhangviet.vn Vidu: NH3 + H 2 0 - > N H ; + 0 H - Nhffng bazd khi tan trong nxSdc m i phSn tur phan l i nhieu nS'c ra ion OH" la In Ian thur 1 . So lUcfng 2 . 0 0 0 c u o n , kho 1 6 x 2 4 c m . cdc bazd nhieu nSfc. T ^ i : C o n g ty T N H H M T V in a n M A I T H j N H DLfC. 3. Mu6'i Dja c h l : 7 1 , Kha V a n C a n , P.Hi$p Binh C h a n h , Q . T h u Dure, T p . H C M . Muoi la h0p chSft khi tan trong nurdc phan l i ra cation kim loai (ho&c cation So D K K H X B : 1 4 8 2 - 1 2 / C X B / 1 2 - 1 S I A H T P H C M ngay 0 6 / 1 2 / 2 0 1 2 . In xong va n o p IIAJ c h i e u qwy I nSm 2 0 1 3 . > H* + CI" Nhffng axit khi tan trong nurdc ma phan tur phan l i nhieu na'c ra ion H* la cac D|CH Vg V A N H 6 A K H A N G V I | T T h a n h p h o H o C h i M i n h cS'p ngdy 2 7 / 1 2 / 2 0 1 2 . HCl * Theo thuyd't Bron-stet: axit Ih chat nhtfcJng proton (H*). V i du : Q u y e t d j n h xuat b 5 n s6': 1 7 7 1 / Q D - T H T P H C M - 2 0 1 2 do N X B T 6 n g O^a^l hap N H 4 ) v a anion g6c axit. C6 hai l o a i : + Muoi axit: NaHCOj, K H S O 4 . . . . + Muoi trung hoa : KCl, CaC03,... phi^ng phap giai H6a tipc 11 V6 CO - D5 XuSn Hang Phan dgng _______ 4. Hidroxit li^ng tinh La hidroxit khi tan trong nifdc vita c6 the phan li nhiT axit vifa c6 the phan 11 B . P H A N L O A I V A PHlTdNG P H A P G I A I C A C D A N G B A I T A P Dang 1. nhuf bazd. V i du : - Viet phi/cfng trinh di#n li c a c chdt Zn(OH)2, A1(0H)3. - Tinh nong d p mol c u a tCfng Ion III. Si; DI5N LI CUA NLTdC. pH. CHAT CHI TH! AXIT - BAZd 1. Nvldc la cha't di$n li rat yg'u H2O BAI TAP M A U V A BAI TAP NANG C A O H * + OH" * Tich so' ion cua nuTdc (kn^o) la h^ng so cl nhi$t do xdc dinh kH20 * =[H^]-[OH'] = 1-010''^ M o i trurdng axit la moi trifdng trong d6 : [H1>[0H'] * hay hay a) Cac chat dien l i manh : NaCl, H N O 3 , KOH, Na2S04, Ba(N03)2, H2SO4, [Ag(NH3)2]Cl, CUSO4.5H2O. b) Cac cha't dien li yeu : C H 3 C O O H , Mg(0H)2, H2S, HCIO, HCN, B i ( 0 H ) 3 . [ H ^ > 1,0.10"^M a) NaCl [ H i < 1,0.1 Q - ' M > Na* + C l " KOH 2. Khai ni^m v l pH. ChS't chi thj axit - bazof Ve mat toan hoc : pH = - l g [ H l Ba(N03)2 + Khi pH < 7 : moi triTdng axit H2SO4 + Khi pH = 7 : moi trifcfng trung tinh + Khi pH > 7 : moi tru-dng kiSm * Thang pH thuTdng dilng c6 gid tri ttt" I -> 14. Mau ciia hai cha't chi thi axit - bazd la quy tim va phenolphetalein. * PhSn tfng xay ra trong dung dich cdc chat dien l i la phan ti'ng giffa cdc ion. * Phan tfttg trao doi trong dung dich cac cha't dien l i chi xay ra khi cac ion ke't hdp dtfdc vdi nhau tao thinh it nhS't mot trong cdc cha't sau : H2S ^ >2Na^+SO^~ > Ba^* + 2N0J Bi(0H)3 ^ [Ag(NH3)2]Cl > [Ag(NH3)2]" + C r CUSO4.5H2O > Cu^* + S04^' • > H2O > 2KC1 + C 0 2 t + H 2 O PhiTdng trinh ion rut gon : 2H* + CO^" > COzt + H2O. 5H2O KOH : 0,01 OM HBr04 : 0,025M; HMn04 : 0,030M; NaC104 : 0,040M. Ba(N03)2 HNO3 0,020M > NaCl + H2O PhiTdng trinh ion thu gon : H"" + OH" SOH". H N O 3 : 0,020M; > Ba^* + 0,1 OM 2NO3 0,20M [NO3] = 0 , 2 0 M > BaS04. + Bi^* + Ba(N03)2:0,10M; =>[Ba^1 = 0,10M + Chat khi; PhiTdng trinh ion thu gon : Ba^* + SO^" 2H^ + S^" HCN -> H^ + CN" > 2H^ + SO^- 0,1 OM > 2KC1 + BaS04>t 20H" H C I O - > H * + CIO" Gidi • + Chat dien l i y6'u; K 2 C O 3 + 2HC1 K* + OH" Mg'^ + trong cac dung dich sau : + Cha't ket tua; NaOH + HCl Mg(0H)2 ^ Bai 2. Viet phi/dng trinh c&c chat dien l i manh va tinh nong do mol cua tilfng ion IV. PHAN CfNG TRAO DOI ION TRONG DUNG DjCH C A C CHAT DIEN LI K2SO4 + BaClz > Na2S04 [ H i = 1,0.10"''" (M) b) C H 3 C O O H ^ CH3COO" + H^ > H* + NO3 HNO3 M o i trirdng trung tinh Ih moi triT&ng trong do: [H^] = [OH"] = 1,0.10"'M * y Gidi M o i triTcJng kiem 1^ m6i triTdng trong do : [H1<[0H-] * ( [ H i = [ 0 H - ] = I.O.IQ-' (mol//) d 25"C) Bai 1. Viet phifdng trinh dien l i cua nhiTng chat sau : > H* + NO3 0,020M 0,020M =>[H1= [NOJ] =0,020M • KOH 0,0 lOM > K* 0,0 lOM + OH" 0,0 lOM =>[K1 = [OH"] = 0,010M Phan dang va pniiono ph^p giai H6a hpc 11 VP cc - Bi XuSn Hung HBr04 + BrO^ 0.Q25M 0.025M r:>[Hl= • • 0,025M f W 0,030M 0,030M * > Na* (J.040M 0.040M 0,5M > Na* 0,250M 0.030M + OH" 0.250M 0.250M => [Na*] = [OH-] = 0,250M. Bai 5. Hai hdp cha't X va Y khi tan vao nirdc moi chat di^n li ra hai loai ion vdi + ClO^ nong dp mol nhiT sau : ();040M [ K l = O,O50M;[Mg^*] = O,20M; [Cr] = 0,050M; dung djch. H2S. H2SO3. H2SCO4. Giai Giai Theo de tW hai hdp chat X va Y l i : KCl va MgS04. H2SO4 > H* + HSOi H2S HSO4 > H* + SOi' H s - > i r + s^- 0,050M 0.050M H3PO4 - > H* + H2FO4 H:S03 -> i r + USO^ MgS04 > Mg^* + H2PO; -> H.S05; -> 0,20M 0,20M + HPO;' lSOf]=0.20M. Vie't cong Ihtfc phan tijr cua X , Y V£k vie't phiTdng trinh di^n li cua chung tronj Biii 3. Vic'l phifdng irinh di^n l i thco lifng na'c ciia ctic axit saii: H2SO4, H3PO4, • 0,5M c) NaOH Mn04 ==>lNan= lC10;j =0,()40M. • + Cf => [ N a l = [Cn = 0,5M. = IMnO^J =0.030M NaClOa > Na* 0,5M {BrOJ] =0,025M HMn04 r^iHi NaCl -> + HS- Kci + so;" —> K* + cr 0,050M SOj0,20M. H2SCO4 — > H ' + HSco;; HSc(J4 -> Dqnq 2. - Tinh d$ di^n H a. h^ng s6 di^n li - Tinh (H*), (OHl. pH cua dung djch + ScOj-, Hsii 4. Vic't phiftJng trinh di^n l i va linh nong do mol cua cac ion trong cac dung djch sau: a) 0,3 mol dung djch axit sunfuric vdi the tich 2 lit. b) 5,85g dung dich natri clorua vtJi the tich 200nil. c) Dung dich NaOH 0.25M. BAI TAP MAU Bai 1. Tinh nong dp H \, pH ciia dung dich HCl 0.1 OM va dung dich NaOF COIOM. cm a) Nong do mol dung djch H 2 S O 4 : Giai • HCl — • H* + 0,10M H2SO4 2H* 0,15M =>|H*1 = 0.3M; 0,I5M =>[OH-]= [S05-1 = 0.15M. • h) So mol NaC!: HN,,;, = ~ ^ = 0.1 0,10M =>IH*] = 0,10M; pH = -lg[H*J = -lgO,l = l + SO-V n.3M cr (mo!) NaOH 1.0.10"' =1.0.10"'^M. > Na* + OH" O.OIOM O.OIOM =>[OH-] = 0.010M =>[H1= i:^:H!!l = 1.0.I0"'2M 1,0.10"^ =>pH = - l g l H * ] = - l g l 0 - ' ^ = 1 2 . KHAWTfi jPijilff 4$fi9 ¥^ phuong pWp 9i5i Hda hoc 11 VP CO - B5 Xuan Hung Uhi I. Dung dich axit axetic nong do 0,25M co pH = 2. g) TInh do dien l i cua axit axetic trong dung dich tren. Neu hoa tan thtm 0,01 mol HCl vao 1 lit dung dich tren thi do dien l i cua mit axetic tftng hay giam? GiU thich? Giai a) PhiTctng trinh di6n l i : CH3COOH ^ CHiCOO" + H* Trong 1 lit dung dich c6 0,25 mol CH3COOH c6 pH = 2 c) Ca(OH)2 > Ca'^ + 2 0 H - 0,004M hay 0,008M [OH-] = 0,008M N6ng do OH" cua dung dich: [OH'] = 0,008.a = 0,008.0,8 = 0,0064M => [ H i = ^'"'^^ = l,5625.10-'2 =:> pH = - l g [ H l = 11,81. 6,4.10--^ d) C H 3 C O O H -> C H 3 C O O - + H^ => [ H i = 10"''" = 10"^ = 0,01M 0,004M V|y troog 1 lit dung dich c6 0,01 mol CH3COOH phan l i ra ion. hay Bo di^n l i cua C H 3 C O O H la: a = — x => Nong dp H * cua dung dich: 100% = 4%. 0,25 b) Khi them 0,01 mol HCl vao 1 lit dung dich tren thi nong do H"" tang len, do d6 cSn bhng dien l i chuydn dich sang trai, do do do diSn U giam. 0,004M [H1 = 0,004M [ H i = 0,004.a = 0,004.0,7 = 0,0028M quy tim trong dung dich nay? Gidi [HI =0,010M => [0H-] = ' , =1,0.10-'^M 1,0.10"^ pH = - l g [ H l = - l g I 0 - ' = 2 M o i irirftng cija dung djch nay lii axil. Quy tim doi sang milu do. Bai 4. Tinh pH cOa cac dung dich sau : a) Dung djch KOH 0,003M. trong dung dich. Hay cho bid't mau cua phenolphetalein trong dung djch nay. Gidi lQ-14 pH = 9,0 c) Dung dich Ca(0H)2 0,004M (a = 0,8). Gidi m^u hong. Bai 6. a) Cho hai chat NH3 va C6H5NH2 (anilin) chat nao c6 hang so bazd (kb) Idn hdn? Giai thich? b) Dung djch NH3 I M c6 a = 0,43%. Tinh hang so Kb va pH cua dung djch do. Gidi electron tren nguyen tijf N , do do c6 tinh bazcJ yeu hdn phan tiif NH3. • (NH3)>^b 1 mol 0,003M 0,02M Vay 0,04M [ H I = 0,04M => pH = - l g [ H l = 1,4. 0 xmol x mol Ma a = 0,43% = 0,0043 = y =:> x = 0,0043 = 4,3.10"^ Taco: SO^- N H ^ + OH" 0 (l-x)mol • I n 10"'"* ^ [ H i = i i ^ i i ^ = 0,33.10-'' : ^ pH = - l g [ H l = 11,48 3,0.10"^ > 2H^ + (CfiHs-NHa)- b) Phan tfng : NH3 + H2O > K* + OH- [ 0 H - ] = 0.003M b) H2SO4 = lO'^M VI pH = 9,0 nen dung dich c6 moi tru'dng kiem => phenolphetalein doi sang d) Dung djch C H 3 C O O H 0,004M (a = 0,7). 0,003M [ H i = lO--^" = lO-^'M = > [ 0 H 1 = a) V i phan tijT anilin c6 goc C6H5- la goc hut electron nen lam giam mat do b) Dung djch H2SO4 0,02M. a) KOH pH = - l g [ H l = 2,55. Bai 5. Mot dung dich c6 pH = 9,0. Tinh nong do mol cua cac ion H"" va OH" Bai 3. Mot dung dich c6 [ H ^ = 0,010M. Tinh [ O H ^ va pH cua dung dich. Moi tri/cfng cua dung dich nay la axit, kiem hay trung tinh? Hay cho bie't mau cua Vim Kb = \^KllO^Kl±..jL= [NH3] Kb = 1,857.10"'. 1-x 1-x 1-4,3.10"^ ^8,57.10-^- Phan dsino va phiiong p h i p IH*] = giai H6a hpc 11 VP c o - D5 XuSn Hung Giai = J O : ! ! . = 0,23.10-'' = 2.3.10-'^ [OH"] 4,3.10"^ Phtfdng irinh diOn l i : HCl > H* + Cl" Goi V, la the lich dung djch HCl ban dau c6 pH = 3. pH = - l g [ H l = -lg(2,3.10-'')= 11,64. Bai 7. Tinh nong dp mol ciSa cdc ion H* va O H ' trong dung djch NaN02 1,0M. . pH = 3 IH*] = lO'^M mh [H^] = - J l => n^^ ^ = lO-^V, (mol) Biet n^ng h^ng so phan li bazd cda NOj la Kh = 2,5.10'". Goi V2 la the lich dung djch HCl sau khi pha loang c6 pH = 4. Giai Phuang trinh di§n l i : NaNO: > Na* IM . IM Phan tfng thuy phan : N O j + H2O -> HNO2 + OH" Bandau: IM 0 0 Canbhng: (1-x) x x [HNO,J.[OH-l^^^ [NO2] hay n+ + NOj pH = 4 => [H*] = lO-'^M ma [H*l = - j ^ - => n^+ = 10-^V2 (mol) Khi pha loang dung djch so mol H* khong thay d o i : n + = n ^. H =>10-^V, = 10-^V2 : ^ =^ = 10 d H s =:>V2=10V, V$y pha loang dung djch HCl 10 Ian nghla la phai pha loang 1 the tich dung (1-x) djch HCl vdi 9 the lich niTdc nguycn cha't. „2 — = 2,5.10-" 1-x Bai 10. Dung djch X gom CH3COOH I M (Ka = 1,75.lO') va HCl 0.001 M . Tinh pH cAa dung djch X . (Tn'ch de thi tuyen sink Dai hoc khoi A nam 2011) Giai ra ta c6: x = 5.10''^ hay [OH'] = 5.10-''M 10-'^ _ 10-"* [ H * ] = -l^i = - ^ ^ ^ = 0,2.10-''(M). [ O H - ] 5.10"^ Bui 8. Cho dung djch X chua hon hpp gom CH3COOH 0,1M va CHjCOONa 0,lM. Biet a 25"C cua CH3COOH la 1,75.10-^ va bo qua sy phan li cua nuac. Tinh pH cua dung djch X. '' Trich de thi tuyen sink Dgi hgc khoi B'' Giai Phuang trinh dien li: CH.iCOONa CH3COO + Na^ 0,1 0,1 Can bang: CH3COOH -> CH3COO + Bandau: 0,1 0,1 0 Phan li: X x x Can bang: 0,1-x x + 0,1 x =:»Kc='^'^'^"^"'^^=: 1,75.10-^ ( D i e u k i | n : 0 < x < 0 , l ) 0,1-x => X ,= 1,75.10'(nh^n);X2 = -0,1 (lo^i) => pH = - Ig [ H = - log (1,75.10') = 4,76 Bai 9. Cho dung djch HCl c6 pH = 3. Ctin pha loang dung dich axit nay bhng nurdc bao nhieu Ian dc thu diTtJc dung djch HCl c6 pH = 4? Giai HCl - > H* + Cr Ta c6: 10-^ 10-^ CH3COOH < > CHjCOO + H* Bandau 1 0 lO' Phanli x x x 1-x x x +10"^ Can bhng Taco: iLi^tlE!! = 1,75.10-^ (*) 1-x VI X < < 1 => 1 - X => 1, do do ttr (*) => x^ + 1 0 - \-5 X, =-4,71.10-'(loai) X2 = 3.71.10' (nhan) => pH = -IglH*] = -lg(3.71.10-' + 10-') = 2,33 Bai 11. Trpn 100 ml dung djch hon hrtp gom H2SO4 0,05M va HCl O.IM vdi 100 ml dung djch hon hdp gom NaOH 0.2M va Ba(OH)2 O.IM ihu diTcJc dung djch X. Tinh pH cua dung djch X. {Trich de thi tuyen sinh Dai hoc khoi B) phipng ph&p giii H6a hoc 11 VP co - D5 XuSn Hung Phan d^ng Gidi Ta c6: n = 5.10'Wl; HHCI = 0,01 mol => 2 n + = 0,02 mol H2SO^ H = 0.01 mol; HNaOH = 0,02 mol Phan ung trung h6a: H* + OH" -* H2O 0,02 0,02 n . d„ = 0,04 - 0,02 = 0,02 mol "Ba(OH)2 I n^^. = 0,04 mol OH => [OH] = 0,02 : 0,2 = 0,1M => pOH = 1 ^ pH = 13 Bai 12. Trpn Ian V ml dung djch NaOH 0,01M vdi V ml dung dich HCl 0,03M difcJc 2V ml dung dich Y. Dung dich Y co pH la: A. 4 B. 3 C. 2 D. 1 (Trich de thi tuyen sinh Ccio ddn^ khoi A, B) Ta C O : n N a O H = 0,0l.V mol nGidi =0,01.Vmol OH nHci = 0,03.V mol n ,-0,03.Vmol Phurong trinh ion: H^ + OH' H.O 0,01V 0,01V =^ = 0,03V - 0,01V = 0,02V (mol) I Va: VNaOH = fiNaOH = 0,1a 200 - 100 mol = 100 ml n . = 0,1a mol Phifdng irinh phdn tfng: OH H* + OH" -> H2O 0,01 0,01 Dung djch sau phan tfng co pH = 12 (moi triTdng bazcJ) => sau phan \ing tren OH' dir, H* het. 0,01 mol = n . H => [OUT ] = 10-^ = O.OIM Mat kh^c, ta co: pH = 12 => pOH = 2 = 0,01 a = 0,12 Dap an D. 0,1a-0,01 0,2 Bai 14. Tron 250ml dung djch hon hdp HCl 0,08M va H2SO4 0,1M vdi 250ml dung dich Ba(0H)2 aM thi thu du'dc m gam ke'l lua va 500ml dung dich co pH =12. Tinh a va m. Gidi n , = 0 , 0 7 mol Tacd: nH2S04 = 0,025 mol "HCI H+ - 0,02 mol n 2 =0,025 mol nBa(0H)2 = 0'25a mol •n Ba OH" = 0,5a mol Dung djch sau khi tron co pH = 12 (moi trifdng = 0,25a bazd)mol => Sau phan iJng OH" dif. H* + OH" ^ H2O 0,07 0,07 n,^„_ = (0,5a - 0,07) mol pH = 12 => pOH = 2 => [OH"] = 10"'M n . = 0,01.0,5 = 0,005 mol H^ H phaming OH OH H 0,02 V = 0,01 = 10"2M =>pH = 2 ^ D a p a n C . 2V Bai 13. Tron 100 ml dung djch c6 pH = 1 gom HCl va HNO3 vdi 100 ml dung dich NaOH nong do a (mol/1) thu di/dc 200 ml dung dich co pH = 12. Gia Iri cua a la: (biel trong moi dung djch [H* ][0H""] = 10"'^) A. 0,15 B.0,30 C. 0,03 D. 0,12. (Trich de thi tuyen sinh Dai hoc khoi B) Gidi Ta co: pH = 1 :^ [ H ^ = I Q - ' M =^ n ^ = 0,1.0,1 = 0,01 mol • Theo phiTdng irlnh phan tfog: n => n _ j ^ = (0,1a-0,01) mol OH m Ta co: 0,5a - 0,07 = 0,005 ^ a = 0,15 Ba^* + S04^BaS04 i 0,025 0,025 mi = 0,025.233 = 5,825 (g) Dap an B. BAITAPAPDyNG Mi 1. Tinh nong dp mol cua ion H* trong dung dich hlng so phan li axit cua HNO2 la K a = 4,0.10-^ Gidi I PhiTdng trinh di^n li: HNO2 ^ H^ + NO2 Bandau: 0,1 OM 0 0 Canbkng: (0,1-x) x x HNO2 0,10M bie't ring 13 Phan djinfl v i phuang p M p giii H6a hqc 11 VP cO - Dg Xuin Hung Giai (0,1-X) [HNO2] Vi HNO2 la axit ycu nen x « nen Vay a) Khoi liTdng chat tan H2SO4 : C%.m,,d 98%. 100 0,1-X H2SO4 — = 4.10"* => x = 6,3.10"^ 0,1 a) CHjCOOH O.IOM (K, = 1,75.10"*). Tinh nong do mol cija ion H\ b) NH3 0,1 OM (Kh = 1,80. lO-'). Tinh nong do mol cua ion OH". 0,1M 0 0.1-x 0,1-X [CH3COOH] 1,8V Vi CH3COOH la mOt axit yeu nen X « x^ 5 0,1-X ' • Bandau: Canbkng: hay 0,1 N H ; + OH- 0.1 0 0 (M) 0,1-x x x (M) [NH3] X.X _ x^ 3 -0,l-x-0,l-x-^'^°-^° VI NH3 la mpt bazd yeu nen x « 0,1 — = 1,80.10"^ =>x= 1,34.10"^ [0H-] = 1,34.10-^M. Bai 3. a) Them tir tir lOOg dung dich H2SO4 98% v^o niTdc va dieu chinh de diTcJc 1 lit dung dich A. Tinh nong do mol cua ion H* trong dung dich A. b) Phai th6m v^o 1 lit dung dich A tren bao nhieu lit dung dich NaOH 1,8M de thu dU'dc: - Dung djch c6 pH = 1; - Dung dich c6 pH = 13. 14 pH=l =>[H1 H2SO4 > (1-0,9V) = 10-'=0,1(M) 2H* + SO^ ( 2 - 1 , 8 V) 1 +V [H*] = 1,32.10'^M. H2O -> [NH;].[0H-]_ Vay . = > [ H 1 = ^ - ^ = 0.1 Giai r a t a c 6 x = 1,32.10-^M + 0,9V Sau phdn i?ng : Vjj = 1 + V — = 1,75.10"^ 0.1 NH3 )• Na2S04 + 2H2O = > n H 2 S 0 4 d ^ = ( l - 0 . 9 V ) mol x X.X _ 1.8.V(mol) 2NaOH + H2SO4 0 x ^ _ [CH3COO-].[H-] b) 2 mol =>nNaOH= CH,COO- + H* CH3COOH nen SO^ b) Goi V la the tich dung djch NaOH c6 nong do 1,8M Giai Canb^ng: + 98 , . ,. ""--^ =^-Un.ol) =>[H1= Y = 2(M). Bai 2. Co hai dung dich sau : Bandau: > 2H* 1 mol [H1 = 6,3.10-^M. a) , '"^••= - ^ = -lB5^ = ''^^^'"^=^ 0,1 Giai ra ta CO V = 1 (lit). • pH= 13 => dung dich c6 tinh bazd => [H*] = lO-'^M => [0H-] = 10-'M = 0,1M H2SO4 + 2NaOH > Na2S04 + 2H2O I mol 2 mol =>nNaOHdif=(1.8V-2)moI NaOH > Na* + OH" (l,8V-2)mol ( 1 , 8 V - 2 ) mol 1 8V-2 => [0H-] = ' = 0.1 V = 1.2353 (lit). 1 +V Bai 4. Tinh the tich dung dich Ba(0H)2 0,025M can cho vao 100ml dung djch hSn hdp gom HNO3 v^ HCl c6 pH = 1 de tao th^nh dung dich c6 pH = 2. Giai • D u n g djch hon hcJp gom HNO3 v^ HCl c6 : l p H = 1 =>[H1 = 10"' =0.1Mii> n , =0,1.0,1 = 0,01 (mol) [Goi V \h. th6 tich dung dich Ba(OH)2 0,025M ^ 15 Phati djing va phuang phap glSi H6a hge 11 VP c o - D5 Xuan Hi/ng nBa(OH)2 = 0,025.V (mol) Ba(OH)2 Bai 6. Tinh nong do [H*] va [OH"] trong dung dich CH3COOH 2M. Neu sau do them v.^o moi lit dung djch axit tren 0,2 mol muo'i CHjCOONa thi [H*] va [OH"] cua dung dich tang hay giam bao nhieu Ian. Bic't hang so' phan 11 axit la 2.10"'* va the tich dUng dich thay do'i khong dang ke. > Ba^*+ lOW 0,025V 0,025V (mol) Phifdng trlnh p h i n tfng : H * + 0,05V OH" — » H^O 0,05V Gidi Dung dich tao thanh c6 pH = 2 Phircfng trinh dien l i : CH3COOH -> CH^COO" + H * => [ H i dir = 10"''" = 10-' = 0,01M Bandau: 2 0 0 (M) V j j sau phan iJng = V + 0,1 Can bang: 2-x x x (M) ^ V d > / = ^ M - V = 0,01.(V4-0,l) o O , 0 1 - 0,05V = 0,01V + 0,001 K.= o V = 0,15 (lit). Bai 5. a) Tinh pH cua dung dich chtfa l,46g HCl trong 400ml. b) Tinh pH cua dung djch tao thanh sau khi tron 100ml dung dich HCl 1,00M vdi 400ml dung dich NaOH 0,375M. 1 46 a) nHci= — - = 0,04 (mol) 36,5 HCl CO : 0,4 nNaOH NaOH n H c i = 0,1.1 =0,1 (2) 0,2 mol// nen lam cho nong do CH3COO- trong dung djch tang len do do can bang d (mol) phdn tfng ( i ) dich chuyen theo chieu tOf phai sang irai, nen can b^ng duftJc thanh lap. CH3COOH -> CH3COO- + H * + OH" Bandau: 2 0 0 (M) 0,15 mol Canb^ng: (2-y) (0,2+ y) y (M) V i Ih axit yeu ndn y « 2 (mol) => [CH3COOH] = 2 - y = 2 (M) HCl — > H^ + c r 0,1 mol = 1,58.10"'^ (M). > CHjCOO" + Na* 0,2 mol// = 0,4.0,375 = 0,15 (mol) 0,15 mol -14 6,324.10" Phi/dng trinh dien l i : CHjCOONa = 0,im=>pli = -IgO, 1 = 1. > Na* 10 Neu them vao moi lit dung djch axit 0,2 mol muoi CHsCOONa : 0,04 ^ [ H I =^ = ^ b) Ta Vi la axit yeu nen x « 2. 2 = ^ ^ = 2.10-5 X = 6,324.10" 2 — > H* + c r 0,04 V [ C H 3 C 0 0 - j . [ H " ] ^ x.x ^ ^2 10-5 [CH3COOH] 2-x 2-x => [ H i = 6,324.10'^ {M)=> [ 0 H - ] = Gidi (1) [CH3COO-] = 0,2 + y = 0,2 (M) 0,1 mol Vdj sau phan i?ng = 0,4 + 0,1 = 0,5 (lit). PhiTcJng trmh phan tfng : H^ tCH,COO-l.[H'1^0^^^,^.. + OH" — - » H2O 0,1 mol [CH3COOH] 0,1 mol => y = 2.10"* n ^ . . - . = 0 , 1 5 - 0 , 1 = 0,05(mol) hay [ H i = 2. lO"" (M) 10 -14 = 0,5.10"'" = 5 . 1 0 - " (M) -4 2.10 So vdi dung dich CH3COOH ban d i u : [OH1 = "OH dU [OH-] = - ^ = 0,l(M> [ H i = h2:lI^=x(r^^.(Pi^-^ 0^110^^ 2 = [§, [ 0 H 1 tang len = 5.10 - 3 r W e - 4 i ^ 4 f t - v 4 - t t I 1 gi^ni 31,64 Ian. l.:;aH).7'ViEN TINH BIN.H THUAN ' JO 17 PhSn dgng va phuong phap giii H6a hgc 11 Vfl cO - B5 XuSn Hang Bai 7. a) Tinh pH ciia dung dich A la hon hcJp gom HF 0,1M va NaF = 0,1M. b) Tinh pH cua 1 lit dung djch A d tren trong hai trUdng hcTp sau : - Them 0,01 mol HCl vao; - Them 0,01 mol NaOH vao. So'mol m6i axit trong 600ml dung dich A : . nHN03 =0,1.0,2 = 0,02 (mol) = 6,8.10""*. Gidi — ^ 2H* + 0.01 mol H^ + F > 0,02 mal [HF] = 0 , 1 M = > [ F ] = 0,1M HCl . [HF] [H1=^'^-^Q"'-^'^=6.8.10-"(M) 0,1 0,03 mol nKOH F > HF [HFJ = 0,1 + 0,01 = 0,11 (mol) H^ + NaOH =:>[F] = 0,1+0,01 =0,11 (mol) 0,3V (mol) > K* = 0,3V + 0,15V = 0,45 V (mol) OH Phan tfng trung hoa dung dich A va dung dich B : . 0,45V = 5,56.10"^(M) + OH" 0,15V (mol) H* + OH- [ H F ] = 0 , 1 - 0 , 0 1 =0,09 (mol) pH = - l g [ H l = 3,25. M i 8. Tron ba dung dich H2SO4 0,05M; HNO3 0,1M va HCl 0,15M vdi nhffng la V. > Na* + OH" => Z n _ . > F + H2O diing (mol) 0,15V (mol) => pH =-lg(8,31.10"") = 3,08. * Khi them 0,01 mol NaOH vao ta c6 : HF + OH" cr = 0.l'5.V(mol) KOH = 6.8.10"^ — - 8,31.10-^ , 0,09 . = 0.02 + 0.02 + 0.03 = 0.07 (mol) 0.3V (mol) [ F ] = 0 , 1 - 0 , 0 1 =0,09 (mol) > H2O . 0.45V (mol) Dung dich thu dufdc cd pH = 3 => moi trtfcfng axit ^^H"" di/. The tich dung dich sau phan tfng : V j j = 0,6 + V the tich bang nhau, thu dtfcfc dung dich A. Lay 600ml dung djch A cho tac pH = 3 dung vcti dung dich B gom NaOH 0,3M va KOH 0,15M. Tinh the tich dung [HI = ^ = "'Q^-Q''^^^ = 0.001 V 0.6 + V dich B can dijng de sau khi tac dung vdi 600ml dung dich A thu difdc dung dich CO pH = 3. . 0,03 mol HNaOH = 0,3.V b) * Khi them 0,01 mol HCl vao thitac6:H^+ NO3 + Goi the tich dung djch B can pH = - l g [ H l = -lg(6,8.10"') = 3,17. [H^] = 6.8.10-*. H^ . 0,02 mol — > => I SO^~ • 0,02 mol HNO3 Trong dung dich c6 F" nen lam cho can bang it bi chuyen dich nen c6 the coi k^=mlLlE:i.6,8.10-^M) • Phifdng trinh dien l i cdc axit: H2SO4 [F-] . nH2S04 =0,05.0,2 = 0.01 (mol) Cho log6,8 = 0.83. [ H I = 6,8.10-r . n HCl =0.15.0.2 = 0.03 (mol) Bie't rang hang so axit (hhng so ion hoa) cua HF la a) Phtfdng trinh dien 11: HF -> . [ H i = 10-'= 0.001 (M) V = 0.154 lit Vay the tich dung dich B la 0,154 lit. Gidi V i the tich blng nhau nen the tich moi dung dich axit trtfdc khi trpn Ian la: Bai 9. Trpn 200ml dung dich g6m HCl O.IM va H2SO4 0.05M vdi 300ml dung dich Ba(0H)2 cd nong dp a mol// thu di/dc m gam ket tua va 500ml dung dich I cd pH = 13. Tinh a va m. Cho biet trong cac dung dich vdi dung moi la niTdc, — = 200 ml = 0,2 (lit) tichso [ H l . [ O H - ] = 1 0 - ^ 19 Phan dgng va phuong ph^p g i i i H6a hpc 11 VP CO - 05 Xuan Hi/ng Gidi Ta CO: n H c i = 0,1.0.2 = 0,02 (mol); nH2S04 -0,05.0,2 = 0,01 (mol) HCl — > H^ + 0,02 mol 0,02 mol H2SO4 > 2H^ + cr SO4" 0,01 mol 0,02 mol 0,01 mol = > I n + =0,02+ 0,02 = 0,04 (mol) V a t a c o nB^(OH)2 =0'3.a(mol) Ba(0H)2 > Ba^* + 2 0 H - Gidi a) PhiTdng trinh dien li: C H 3 C O O H - » C H 3 C O O - + BandSu: 0,1 0 0 Can bang: 0.1-x x x -.100% = 1% =^ X = 10-' hay [H^ = IQ-' (M) a = (M) (M) 0,1 =>pH = - l g [ H l = - l g l 0 - ' = 3. b) The tich dung djch sau khi trpn : Vjj = VA + VB = 2.V (lit) OHci = 0,2.V (mol) HCl — > H^ + cr 0,2V 0,2V nH2S04 =0,l.V(mol) 0,3a mol 0,3a mol 0,6a mol Khi trpn dung dich gom HCl va H2SO4 vdi dung dich Ba(0H)2 ta c6 cac H2SO4 > 2H* + SO^" phu'dng trinh sau : + OH" > H 2 O (1) 0,1.V 0,2.V 0,04 0,04 =>In„. = 0,2V + 0,2V = 0,4V (mol) Ba^* + SO^- > BaS04i (2) 0,4V = 0,2(M) [ H iH = V 2V 0,01 0,01 0,01 Vay pH = - l g [ H l = -lgO,2 = 0,7. Dung dich CO p H = 13 [ H i = lO^'^M Bai 11. Trpn dung dich X chuTa NaOH 0,1M, Ba(0H)2 0,2M vdi dung dich Y =>[0H-]= - i — = 10"' =0,1M (HCl 0,2M; H2SO4 0,1M) theo ti Ip nao ve the tich de dung dich ihu diTpc c6 p H = 13? io~'-^ Gidi => n _ =0,1.0,5 = 0,05 (mol) Taco: nB,(OH)2 = 0,2Vx m o l j OH M Tac6: nOH _dir = 0,6a-0,04 " N a O H = 0,lVx mol J~"^'0HnH2S04 = 0,1VY mol T o 0,6a - 0,04 = 0,05 o a = 0,15 (M) = 0,4VY mol n„ 2+ = 0,3a = 0,3.0,15 = 0.045 (mol) "HCI - 0,2VY mol Bu Sau khi trpn dung dich c6 pH = 13 moi triTclng bazd => sau phan ifng giiTa Ttr(2)=:>n 2+ = 0.045-0.01 = 0,035(mol) => ng^so. =0.01 mol axit va bazd thi OH" diT: H^ + OH" -> H 2 O => m = mB„s04 = 233.0,01 = 2.33 (gam). B^i 10. a) Dung dich CH3COOH O.IM c6 dp di^n li a = 1%. Vifi't phUdng trinh dien li CHiCOOH va tinh pH cua dung dich nhy. b) A Ih dung djch HCl 0.2M; B la dung dich H2SO4 O.IM. Trpn cac the tlch b<^ng nhau Cho lg4 = 0,6; lg2 = cua 0.3. A va B diTpc dung dich X. Tinh pH cua dung dich X. (DHQG.d0l) 0,4VY 0,4VY (0,5VX-0,4VY) mol n^,^.u. = T a c 6 : p H = 13 => [ H i = 1 1-13 0 0,5VY - 0 , 4 V X Vx + VY [ 0 H 1 = 10,-1 VX _ 4 91 KifAwrfiinET Phfln-djino va phuong ph^p gSi HdaJipc 11 VP co - D5 Xuan Hifng Hkil2. D a n a 3. Tron iOOmI dung^ljch g6m Ba(OH)2 O.IM va NaOH 0,1M vdi 400 ml dung dich gom H2SO4 0/)375M va HCl 0,0125M thu di/dc dung djch X. Gia - tri pH cua dung dlchX 1^: - X a c dinh vol tro axit, bazd, trung tinti hay li/dng tinh A. 2 B. 1 C.6 D. 7. (Trich de thi tuyen sink Dai hoc khoi B) Gidi Ta c6: nBa(OH)2 = O'O^ mol Z ^=> "NaOH = 0,01 > In. mol - NhOn bid't dung djch v a ion BAI T A P M A U Bai 1. Viet " H C I = 0,005 mol Khi tron Ian hon hdp 2 axit 0,03 . = 0,035 mol = 0.035 - 0,03 = 0,005 mol 0,5 => pH = 2 => D a p a n A. axit H2SO4 0.5M thu difdc 5,32 lit Hj (dktc) va dung dich Y (coi the tich dung dich khong doi). Dung dich Y c6 pH la: A. 7 HHCI = 0,25 5,32 22,4 2H* 0,475 , => > 2Na^ + 2Cr > 2Na" + SO^" + CO2T + H2O • COzt + H2O. >-AgCli + KN03 K* + Cr + Ag* + NO3 Ag* + cr + H.St > 2Na2S04 + C O j t + H2O 2Na* + C 0 | - + 2H* + S O ^ > AgCli + + NO3 > AgCli. Bai 2. Chi dung mot thuoc thijr h\, hay trinh b a y e a c h nhan b i c l ba dung djch co cClng nong do mol sau : NaOH, H3PO4 va H2SO4. 0,2375 Gidi = 0,5 - 0,475 = 0,025 mol n + -1 H^ >• khong xay ra. j-HzSt. 0 KCl + AgNOj -> H2 > BaS04i + Cu'^ + 2NO3 > 2NaCl + H2St 2H* + C O | - Khi cho Mg, Al tac dung vdi hon hdp 2 axit HCl va H2SO4, ta c6 sd do phan «?ng: > BaS04i + Cu(N0,)2 2Na* + S^- + 2H* + 2Cr S^- + 2H* ~ Cho dung djch phenolphetalcin tifng giot Ian 0,025 0,25 Dap an B. 2Cr > BaS044'. e) NazCOj + H2SO4 = 0,2375 mol > FeCOH).! + 2Na* + ).Fe(0H)2i. d) Na2S + 2HC1 mol mol sau : > Fe(OH)24' + 2NaCl + 2Na* + 20H- c) KzCOs + NaCl mol " H 2 S 0 4 = 0,125 nH2 = 0 KCl + AgNOj. Ba^* + SO4- D. 6. (Trich de tlii tuyen sinh Bai hoc khoi A) cm Ta c6: e) NazCOj + H2SO4 Cu^* + SO4" + Ba^* + 2NO3 C. 2 B. 1 cha't d) NajS + HCl Fe^* + 20H- B a i 13. Cho m gam hon hdp Mg va Al vio 250ml dung djch X chiJa hon help axit HCl I M cac cap c) K2CO3 + NaCl b) CUSO4 + Ba(N03)2 = 0,01 = 10-2 dung dich giffa b ) CUSO4 + Ba(N03)2 Fe^* + 2Cr 0.005 r a trong a) FeCb + NaOH a) FeCb + 2NaOH H •HI (neu c6) xay Gidi 0.03 n^^. uTng 2 bazd xay ra phan xing trung hoa: H2O H* + OH" phiTdng trinh p h a n tuf, phifdng trinh ion, phiTdng irinh ion thu g o n ciia cdc p h a n = 0,03 mol OH- nH2S04 = 0.015 mol -j Tn. Phan a n g trao ddi ion = 0.1 = 10"' M liTdt vao ba mau thi'r. mau nao c6 mau hong h\g dich NaOH. con lai la H3PO4 va H2SO4. pH = 1 - Lay ba dung dich vdi V b^ng nhau sau do them viio hai dung dich axit Ian iiTdt vai giot phcnolphctalcin. Do tiep V ml dung djch NaOH vao tiTng axit. phuong ph^p giai H6a hgc 11 V6 cO - D8 Xuan Htflig Phan dgng t h e m m o t i t dung djch N a O H ntJa neu thay dung djch c6 m ^ u hong Ih H3PO4, con h i i la dung djch H2SO4 B a i 5. V i c t cdc phuTcfng trinh phan lit \k ion riit gon cua cac phan iJng (neu c6) khong hiOn tiTcfng. xay ra trong dung djch giiJa cdc cap chat sau : B a i 3. Rau qua kho dU'cJc bao quan bhng k h i SO2 thi/dng chtfa mot liTtJng nho a) Fe2(S04)3 + N a O H b) N H 4 C I + A g N 0 3 hdp cha't CO goc S O 3 " . D e xac dinh suT c6 mat cua cac ion SO3" trong rau c) NaF + H C l d) MgCl2 + K N O 3 qua, m o t hoc sinh ngam mot it qua dau trong niTdfc. Sau mot thcli gian Ipc lay e) FeS i") K O H + H C I O . + HCl Gidi dung dich r o i cho tac dung v d i dung djch H2O2 (cha't oxi hoa), sau do cho tac diing t i c p v d i dung djch B a C l 2 . V i e t cac phiTcfng trinh ion rilt gon da xay ra. a) Fe2(S04)3 + 6 N a O H Fe^* + 3 0 H - Gidi SO^ + H2O2 Ba^* + S O 4 " > SOJ- Cr + A g * + H2O F + H* g i a m , bot n d ( N H 4 H C O 3 ) , phen chua (KAI(S04)2.12H20), m u o i iot ( N a C l + K I ) . Hay dilng cac phan iJng hoa hoc de phan biet chiing. V i e t phiTtfng Irinh ion riit gon cua cac phan tfng. Dung djch m u o i an (NaCl) Cho dung dich AgNOs vao m u o i an -> hien ti/cfng A g C l i trang. A g * + Cr * > AgCli e) FeS (r, + 2HC1 FeS + 2 H * 0 > FeCh + H2St >Fe^^ + H2St HCIO + K O H HCIO + O H " > K C I O + H2O > CIO" + H2O. B a i 6. V i e t phiTdng trinh ion rut gon cua cac phiin tfug hoa hoc sau: Giam ( C H 3 C O O H ) (3) Na2S04 + BaCl2 - > (4) H2SO4 + BaS03 - > Cho CaC03 vao c6 hipn tifdng sui bot k h i . (5) (NH4)2S04 + B a ( 0 H ) 2 -> (6) Fc2(S04)3 + Ba(N03)2 > N H j t + H2O Phen chua (KAI(S04)2.12H2O) Khi hoa tan phen chua v i o niTdc c6 ket tua keo trttng xusi't hien. PhiTcJng trinh ion rut gon: A l ' * + 3H2O ^ Al(0H)3i + 3H* M u o i iof ( N a C l + K I ) Cho chat o x i hoa H2O2 vao - > c6 hi pt ion thu gon: Ba^"" + S04^' -» B a S 0 4 i (2) C U S O 4 + Ba(N03)2 ^ Cu(N03)2 + BaS04 i - » pt ion thu g o n : Ba^* + S04^" -> BaS04 i (3) Na2S04+ BaCl2 - * 2 N a C l + -> pt ion thu gon: BaS04i Ba^"^ + S04^" BaS04 i (4) H2SO4 + BaS03 -> SO2 + H2O + BaS04 i -> pt ion thu gon: 2 H * + S04^" + BaS03 ^ SO2 + H2O + BaS04 i (5) (NH4)2S04 + Ba(OH)2 -> 2 N H , + B a S 0 4 i + 2H20 -> pt ion thu gon: 2NH4* + S04^"+ Ba'^+ 2 0 H - ^ 2NH3 + BaS04 i + 2H2O mau xanh. Phi/cJng trinh ion rial gon: 2 r + H2O2 ^ GiaM > (CH3COO)2Ca + C O . t + H2O B o t nS ( N H 4 H C O 3 ) N H ; + OH- * > khong c6 phan tfng xay ra (2) C U S O 4 + Ba(N03)2 Cho dung djch N a O H vao - > k h i , m u i khai. * >HFt (1) (NH4)2S04 + BaCl2 2 C H 3 C O O H + CaCOj * > NaCl + H F t d) MgCl2 + K N O 3 Giai Cho cAc chat tren vao nxidc de dufcJc cac dung dich : > A g C l i + NH4NO3 ).AgCl4^ c) N a F + H C l > BaS04>l' B a i 4. NhuTng hoa chat sau thi/clng di/dc dung trong cong viec noi trcJ : m u o i an, * >Fe(OH)3i b) N H 4 C I + A g N 0 3 PhiTdng trinh ion r u t gon : > 2 F e ( O H ) 3 i + 3Na2S04 > h + 20H-. (6) Fe2(S04)3 + 3Ba(N03)2 pt ion thu gon: -> 2Fc(N03)3 + 3BaS04 i Ba^"" + S04^" -> BaS04 i H6a hQC 11 V6 CO - D S XuSn Hung Bai 7. Trong cac ion sau: C H j C O O " , CO^", HCO^, HSO4, CP, N H 4 , B a i 9. a) Theo dinh nghTa m d i cua Bronstet. Cho quy t i m vao cac dung djch sau se c6 AKHzO)^", S^', Q H 5 O " , K"" la axit, bazd, trung tinh hay lU3ng tinh? T a i sao? mau gi? C H j C O O N a , Ba(N03)2. N H 4 C I , K 2 C O 3 ? Gifii thich? Gidi * b) D i / doan dung dich cho difdi day c6 gia t r i p H Idn hOn hay nho hcfn 7? Cac ion la axit : H S O 4 , N H 4 , A K H j O ) ^ * v i chung c6 kha ming cho proton NazCOj, K C l . C H 3 C O O K , NaHS04. Al2(S04)3. Na2S, CfiHsONa? (theo Bronstet). H S O ; + H2O ^ I HjO^ + Gidi SO^- a) Theo dinh nghTa m d i cua Bronstet: A x i t la cha't nhi/dng proton ( H * ) , bazO Iti * NH^ + chat nhan proton. H j O ^ + NH3 H2O * A 1 ( H 2 0 ) ' V H2O - > H3O* + A l ( O H ) ' ^ * Vi: Cac i o n la bazd : C H j C O O " , C O ^ , S^", CfH^Q- v i chung c6 kha nang nhan * S^' + * HCO3 + OH" C O 3 " + H2O Cho quy t i m vao dung djch N H 4 C I - quy t i m hoa do. VT : * * Cac i o n la irung tinh : C r , K"^ v l chung khong cho cung khong nhan proton. * Ion la i m l n g tinh : H C O J v i viTa cho, viTa nhan proton H C O 3 + H j O ^ - > C O 2 + 2H2O HCO3 + H2O > C a C O j i +? c) BaC03+? e) FeCl3+? >Ba(N03)2+? d) H P O ^ " +? a) CaCl2 + Na2C03 ^CaCOji b) Pb(N03)2 + 2KC1 > * Dung dich c6 p H < 7 : NaHS04, Al2(S04)3 * Dung djch c6 p H > 7 : C H 3 C O O K . Na2C03, NiXjS, > CaSOA + H3PO4 > Fe(0H)3i + KCl > Fe(0H)3i. CfiHsONa. B a i 10. C h i dung quj' t i m hay phan bi?t niim lo mat nhan sau: H C l , H2SO4, K C l , N a O H , BaCl2. ~ Laymoilomotitdelammliuthiif. - K h i cho q u j ' t i m vao tifng mSu thuf: >PbCl24' d) CaHP04 + H2SO4 Fe^^ + 3 0 H - +? PbCl2>t + 2 K N O 3 BaC03 + 2HNO3 > Ba(N03)2 + CO.T + BaC03 + 2 H * > Ba^* + C02t + H2O -> H C O 3 + OH" Gidi > C a C O j i + 2NaCl Ca^*+CO|- e) FeCb + 3 K 0 H >-H3P04+? +H2O COJ- b) * Dung djch c6 p H = 7 : K C l >KC1 +? Gidi Pb^* + 2Cr > PbCbi NH3 + HjO^ > IK" + K2CO3 CO^ C 0 | - + HiO\ b) Pb(N03)2 +? H2O + Cr Cho quy t i m vao dung dich K 2 C O 3 - quy t i m hoa xanh. Vl: B a i 8. V i e t phiTcJng trinh phan tuT, phU'cJng trinh ion rut gon cua ctic phan tfng sau: a) CaCl2 +? > NH|+ NH4CI NH;^ H 2 O - > CfiHjOH + O H " (phenol) c) Cho quy t i m vao dung dich Ba(N03)2 - > quy tim khong d o i mau do khong bi H2O - > HS" + O H " CfiHjO" + )• C H j C O O " + N a * + H 2 O C H 3 C O O H + OH" thfiy phan. CH3COOH + O H " + H2O CHjCOONa CH3COO proton (theo Bronstet). CH3C06- Cho quy t i m vao dung dich CHsCOONa - > quy t i m hoa xanh. H2O - + M a u l a m quy t i m hoa do la H C l va H2SO4. + Mixu Ihm quy t i m hoa xanh la N a O H . + H a i m a u k h o n g Ic^m d o i mau quy t i m la K C l vii B a C b . Cho mot trong hai mau l a m quy t i m hoa do vao hin liTtft hai dung djch m u o i : + D u n g djch H C l khong tac dung difdc \6i hai mu6'i. + Dung dich H2SO4 tao ket tua triCng vdi B a C l j con v d i K C l t h i khong hien tircJng. BaCl2 + H2SO4 > BaS04>l + 2HC1. : Phan dgng va phiiong ph^p oi5i H6a hpc 11 VP cO - D5 XuSn Hang * BAITAPAPDgNG B a i 1. V i e t phufdng trlnh ion rut gon cua cac phan iJng (neu c6) xay ra trong b) C u ( 0 H ) 2 + HCl c) NajHPOa + H C l d) Sn(0H)2 ,r, + H2SO4 e) M g S 0 4 + N a N O , f) K N O , + NaCl g) S n ( O H ) 2 + N a O H h) Cu(N03)2 + H2O. NH4C1—> NH^ NH;+H20 HSO3 + OH- SO^~ + H2O b) C u ( 0 H ) 2 + 2 H ^ - > Cu^* + 2H2O c) H P O 4 " + 2H* ^ H3PO4 d) Sn(0H)2 + 2 H " -> Sn'^ + 2H2O e) khong xiiy ra 1) khong g) S n ( 0 H ) 2 + 2 0 H - -> SnO^" + 2H2O h) C u ' * + H . O -> Cu(OH)^ + H ^ Xiiy ra B a i 2. V i c t phU'cfng trinh dang phan tur ufng v d i phu'dng trinh ion rut gon sau > CaCOsi b) Pb^^ + SO^" > HjSt > N H j t + H2O a) C a C l . + N a . C O j d) K O H + H C l c) N H 4 C I + N a O H 1) C O 2 + 2 K O H cm > PbSOai H C O O " + H2O N H ; + H2O K2SO4 - > NH3 + H3O* > 2K* + S O ^ - NU'dc khong thuy phan nen H2O trung tinh. cap dung djch nko c6 the ph5n uTng v d i nhau? V i sao? Vie't phifdng trinh hoa hoc cua cac phan lirng xay ra du'di dang phan tuf va ion riit gon. Gidi * Dung dich H C l phan tfug difdc v d i dung dich K O H HCl + K O H > 2NaCl + H.St H* > K C l + H2O * > N a C l + NH3T + H2O + OH- > K C l + H2O > H2O Dung djch B a C b phan uTng diTdc v d i dung dich Fe2(S04)3 3 B a C l 2 + Fe2(S04)3 v K i C O j + HjO. Ba^* + SO^~ a) The nao Ih muo'i trung h6a, muoi axit? Cho v i d i i . A x i t pholphord H3PO3 la axit hai lan axit, vay hdp chat Na2HP03 la muo'i axit hay m u o i trung hoa? b) Vic't phuTdng trlnh ion rut gon cua cac phan iJng thuy phan cac > HCOOH + OH" _ H2O dong vai tro krSng tinh. > P b S 0 4 i + 2NaN03 muoi : N a H C O j , N H 4 C I , H C O O N H 4 , K2SO4. Trong cac phan itng nay nu'dc dong vai tro axit hay bazd? Gidi a ) * M u o i trung hoa \h. muo'i ma anion goc axit khong con hidro c6 kha nang phan l i ra i o n H"^ (hidro co ifnh axit). V i du : K C l , N a j C O j , . . . Muo'i axit la muo'i mh anion goc axit con hidro co kha niing phan l i ra ion H * . V i du : NaHS04, KH2PO4,... H30" > N H : + HCOO" HCOONH4 1) C O 2 + 2 0 H - - — > C O ^ + H2O. B a i 3. * >NH3 + B a i 4. Cho b o n lo di/ng bon dung djch : H C l , B a C b , K O H , Fe2(S04)3. Nhffng > H2O > CaCO.ii + 2NaCl b) Pb(N0.,)2 + Na2S04 c) Na.S + 2HC1 d) H ^ + O H " +cr _ H2O dong vai tr6 bazd. a) c) N H ; + O H " > H2CO3 + O H " _ H2O dong vai lr5 axit. Gidi c) S^- + 2 H ^ > Na* + H C O 3 H C O 3 + H2O a) N a H S O , + N a O H a) Ca^* + COJ~ Na2HP03 la muo'i trung hoa. b) N a H C 0 3 dung djch giiJa cac c$p chai't sau : KHAlgfiVngr * > 3BaS044 + 2FeCl3 > BaS044' Dung djch K O H phan uTng diTdc v d i dung djch Fe2(S04)3 6 K 0 H + Fe2(S04)3 Fe^* + 3 0 H - > 3K2SO4 + 2Fe(OH)3>l' > Fe(0H)3>l'. B a i 5. a) L a n liTc^t nhung giay qu>' tim v^o cAc dung djch m u o i sau : N H 4 C I , K N O 3 , Na2S thl gia'y quy dd co mau gl? G i a i thich ng^n gon vil vie't phiTdng trinh phAn tJng m i n h hoa. (DH Buu chlnh Vien thon^ TP.HCM) b) T r i n h b ^ y phiTdng phdp h6a hoc nhan bi6't suf c6 m&t cua cac ion trong dung dich chtfa h5n h(?p ba mu6'i: F e C b , C u C h , A I C I 3 . (DHAit ninit) 29 PhSn djng va phuonq phap giai H6a hpc 11 VP cO - D8 Xuan Hung Gidi a) Nhung giS'y quy tim v^o Ian liTcft cac muo'i: * NH4CI Ihm cho quy tim h6a hong. Vi dung djch NH4CI c6 moi triTdng axit, pH < 7. NH4C1 — > NH^ cr + C2H5NH2 + H2Q * Dung dich K N O 3 : quy tim khong doi mau. Vi dung dich KNO3 c6 moi triT^ng trung tinh, pH = 7. KNO3 >K^+ NOJ Ion K*, NO3 khong c6 kha nang cho vtl nhan proton. * Dung dich Na2S : quy tim hoa xanh. Vi dung dich NajS c6 moi triTcfng kiem, pH > 7. NajS > 2Na* + S^" S^- + H2O -> HS- + OH" + H 2 O - > H2S + OH". b) Cho dung dich AgN03 v^o hon hcfp gom cac dung dich thay xua't hien ket tua tring AgCl chiJng to c6 ion CI". Ag^ + cr > AgCli * Cho dung dich NH3 diT vao hon hcJp dung dich, ta thay xuat hien dung dieh mau xanh tham va ke't tua -> chtfug to c6 ion Cu^""; cho dung dich NaOH du" vao ket tua thu dtfdc thay xua't hien ke't tua nau do chiJng to c6 ion Fe^"", sue khi CO2 dir vao dung djch con lai, ta tha'y xua't hien ke't tua keo trfng A1(0H)3 chu-ng to c6 ion AI^^ FeCb + 3NH3 + 3H2O > Fe(OH)34' + 3NH4CI (nau do) AICI3 + 3NH3 + 3H2O > Al(OH)34'+ 3NH4CI (keo tr^ng) CUCI2 + 2NH3 + 2H2O > Cu(0H)2>l' + 2NH4CI Cu(0H)2 + 4NH3 > A1(0H)3 + NaOH > [Cu(NH3)4](OH)2 (dung dich xanh tham) NaA102 + 2H2O NaA102 + CO2 + 2H2O > Al(0H)3>t + NaHC03 NaOH + CO2 > NaHC03. B a i 6. a) Cho a moi NO2 hap thu ho^n toan vao dung dich chiJa a moi NaOH. Dung dich thu diTdc c6 gii tri pH Idn hdn hay nho hdn 7? Tai sao? C6H5OH + H2O Gidi C H 3 C O O - + H2O CH3O- + H2O. a) Phifdngtrinh phan tfng : 2NO2 + 2N aOH > NaNQz + NaNOj + H2O NaN02 — ^ + H2O -> NH3 + H3O* NH^ HS" b) Hoan thanh cac phU'ctng trinh sau vh. cho biet cha't nao la axit, bazcJ? Na* + NO2 NO2 + H2O -> HNO2 + . OH"" Do do NO2 la mot bazd p pH > 7. b) . C2H5NH2 + H2O . CzHjNH^ + OH- • bazd axit . C H 3 C O O - + H2O ^ bazd CH3COOH + axit . • C6H5OH+ H2O -> C6H5O- + H3O* • CH3O- H2O axit + OH- bazcf -> CH3OH + OH- bazd axit. Bai 7. Vie't phiTdng trinh phan tijT va phiTdng trinh ion rut gon cua cac phan vCng sau : a) FeS +? > FeClz +? b) HNO3 +? > C H 3 C O O H +? c) Ba(HC03)2 +? > BaC03 +? d) AlBr3 +? > A1(0H)3 +? e) K3PO4 +? > Ag3P04 +? f) Ca(0H)2 + NaHC03 )-? + CaC03+? Gidi a) FeS + 2HC1 > FeCh + H2St FeS + 2H* > Fe^* + HjSt b) HNO3 + CH3C00Na > C H 3 C O O H + NaN03 H* + C H 3 C O O - > CH3COOH c) Ba(HC03)2 + Ba(0H)2 > 2BaC03i + 2H2O Ba^* + HCO3 + OH" > BaC03i + H2O d) AlBrj + 3NaOH > Al(OH)34' + 3NaBr Al'* + 30H> Al(0H)3i e) K3PO4 + 3AgN03 > Ag3P04^ + 3KNO3 3Ag* + P 0 | > Ag3P04i f) Ca(OH)2 + 2NaHC03 > Na2C03 + CaC03^ + 2H2O Ca'* + 20H- + 2HCO3 > COJ- + CaC034< + 2H2O. Phan d?ng va phtfOng phap g\i\a hQC 11 V6 c d - D 5 Xuan Hang B a i 8. Cho mot i t phenolphetalein vao dung djch amoniac loang chiJa a m o l N H , dxidc dung dich A c6 mau. H o i m ^ u cua dung dich bie'n d d i nhU' the nao trong til'ng trUdng hdp. b) T h e m ^ mol AICI3 v ^ o dung djch A . V i e t cac phuTcfng trinh phan uTng xdy ra (TS DH Van Lana, khoi B) Gidi a) Phifdng t n n h phan ufng : NH3 + H C l > NHj Do do dung djch A chuy6'n sang khong mau v i muoi NH4CI la mot axit c6 p H < 7. > d) A g ^ H\ vh CV e) N a * . N H ^ SO^" va C 0 | - . Gidi Al(0H)3i Al(0H)3i Mg^* + 2 0 H - >Mg(OH)2i b) T 6 n t a i difcfc cdc ion tren U-ong mpt dung djch. > NH4CI NHj + H^ A l ^ ^ + 3NH3 + 3 H 2 O b) N a ^ Fe'*, C I " va SO^" a) K h o n g ton t ^ i di/dc v i c6 sU" tao thanh ke't tua M g ( 0 H ) 2 . (dang ion thu gon)? G i a i thich? > a) K.\ O H - va C r c) B a ^ N a ^ C I " va SO^" a) T h e m a m o l H C l vi\ dung djch A . b) A I C I 3 + 3 N H 3 + 3 H 2 O B a i 2. Trong dung dich c6 the ton t a i dong thcJi cac ion sau day dufc^c khong? + 3NH4CI + 3NH^ D o do dung djch A chuyen sang khong mau. V i N H 4 C I va A I C I 3 d c u la nhufng axit c6 p H < 7. c) K h d n g ton t a i diTcJc \\ s\i tao thanh ke't tua BaS04. Ba^*+ SO^" )-BaS04>l' d) Kh6ng t o n t a i di/dc v i c6 sif tao thanh kd't tua A g C l trong dung dich va siji bpt k h i . Ag^ + C r 2H* + CO^- >AgCl4' > C O z t + H2O e) T 6 n tai dufdc cac ion tren trong mot dung dich. B a i 3. C6 ba ong nghiem, m o i ong chtfa hai cation va hai anion (khong triJng lap Dang 4. giffa cdc 6ng nghiem) trong so cdc cation va anion sau : - SU ton tqi dong thdi ion trong dung djch N a \;, A g ^ Ba^^ M g ^ ^ A 1 ^ cr, Br". NO3 va SO^", PO^-, C O ^ " . - Xic dinh cic cation va anion trong tuTng 6'ng nghiem. Bai t6p dUa vdo ptiUdng trinh Ion thu gon. Gidi BAI TAP MkU M u o n tao thanh trong dung dich thl cic cap cation va anion phai khong tao Bail. thanh ke't tua, cha't k h i , chat di?n l i y e u trong til'ng dung dich. D o d6 cac ong a) Trong dung djch A c6 cac i o n K"", M g ^ " , Fe^* va C I " . N e u c6 can dung dich nghi^ra Ian li/dt la : thu duTdc hon hop g o m nhurng m u o i n^o? V i e t cong thtfc h6a hoc v^ ten cua Ong 1 g6m : N a ^ N H ^ , P O ^ " , C 0 | " . muoi. 6ng 2 g6m : A g ^ M g ^ NO3 . S O ^ . b) Can lay nhi?ng muoi nao de pha chd' duTcfc dung dich c6 cic ion N a * , C u ^ \ SO^" va NO3. B a i 4. Gidi a) N e u CO can dung djch A thu di/dc m u o i : K C l : k a l i clorua M g C I j : magie clorua F e C l j : s^t ( I I I ) clorua. b) Can l a y cdc muo'i tfng v d i cdc ion trSn : Na2S04 va Cu(N03)2 ho^c hai m u o i C U S O 4 va N a N O j . 32 6ng 3 g6m : B a ^ A l ^ cr. Br". a) M p t dung dich chtfa a m o l Na*, b m o l Ca^\ m o l H C O i va d m o l C r . L a p bieu thiJc l i 6 n he giffa a, b , c, d va c6ng thuTc tinh t6ng khS'i luTcJng m u o i trong dung dich. ^) K6't q u ^ xic dinh nong dp m o l ciia c^c ion trong mOt dung dich nhU" sau Na* : 0,05; Ca^^ : 0 . 0 1 ; NO3 : 0 , 0 1 ; C r : 0,04 va HCO3 : 0,025. H 6 i ke't qua do dung hay sai? T a i sao? PhSn d?ng va phuang phAp g\i\a hqc 11 Va cO - D5 Xuan Hung Gidi Ba'" a) V i irong mgt dung djch Irung hoa vc d i c n nen theo djnh luat biio totin dicn 0,025 tich thi tong dien tich dU'dng bang tdng d i c n tich am. T a CO : m= m , +m Na+ S O ^ > BaS04i (mol) 0,025 0,025 a + 2b = c + d Vdd BaCl2 ~ Kho'i lifdng muoi trong dung dich chinh la tong k h o i lufdng cac ion. T a CO : + 0 2 = 0,125 (lit). B a i 6. M o t dung djch co chiJa hai cation Fe^"" (0,1 mol) va Ar^"" (0,2 mol) va hai 7+ + m _ +m _ Ca'^* HCO3 CI anion : C r (x mol) va SO4' (y mol). T i n h x va y, biet rang khi c6 can dung dich thu du-dc 46,9 gam chaft ran khan. m = 23a + 40b + 61c + 35,5d. (TSDHQGTP.HCM,c1(/t2) b) Tong d i c n tich dU'dng : S d t ( + ) = 0 , 0 5 + 2.0,01 =0,07 Gidi (1) Tong dien tich am : Fe^^ :0,1 m o l I d t ( - ) = 0,01+0,04 + 0,025 = 0,075 Tir (1) va (2) ta c6 : I d t ( + ) ^ Z d t ( - ) (2) Dung djch c6 hay 0,07 ^ 0,075 Nen ket qua phan tich sai. 0,1.2+ 0,2.3 = x + 2 y i i > x + 2y = 0,8 b) Tinh the tich dung djch NaOH 0,25M du de lam ket tua het ion difdng trong m = 56.0,1 + 27.0,2 + 35,5x + 96y = 46,9 c) Tinh the tich dung dich BaCl^ 0,2M du de lam ket tua het ion 804". => 35,5x + 96y = 35,9 Gidi 0,025 0,025 = 0,l(M) )• Cu(OH)24' 0,05 b) Chia dung dich A l^m hai phan b^ng nhau : - - Phan 2 : tiic dijng vcKi dung dich HCl diT thu diTdc 235,2ml k h i d 13,5"C va T i n h tdng k h o i lifcing cac muoi trong ^ dung dich A . a) Dung dich A dMc dieu che tif hai m u o i : 0,25 Na2S04 va ( N H 4 ) 2 C 0 3 >Ba^-^ + 2Cr 0,025 Phan 1 : tac dung \6i dung djch Ba(0H)2 duT dun ndng, thu di/dc 4,3g k c l tua Gidi V=JL.M5.o,2(lit) 34 a) Dung dich tren diTcJc d i c u che tuf hai muoi trung hoa nao? 1 atm. (mol) The tich dung dich N a O H 0,25M: 0,025 [y = 0 , 3 m o l . X va 470,4ml k h i Y d 13,5"C va 1 atm. 0,05 mol Cu^^ + 2 0 H - BaCl2 ' 35,5x + 96y = 35,9 — - > Na* + O H " 0,05 mol fx = 0,2 mol B a i 7. Dung dich A chiJa cac ion : N a \, SO4" va CO]~. a) Nong do mol cac ion trong dung dich : [Cu^1=[SOr] = ^ x + 2y = 0,8 (mol) _4_ "CUS04 = T 7 r = 0.025 (mol) 160 (2) Ta CO he phiTdng trinh : S04~ > Cu^^ + 0.025 CM (1) K h o i lU'dng chat ran : dung dich. 0,025 : X mol D j n h luat biio toan dien tich : a) Tinh nong do mol cua cac ion trong dung dich. b) N a O H cr SO4" : y m o l B a i 5. Hoa tan 4 gam CUSO4 vao mot liTcfng niTdc viTa du 250ml dung dich. CUSO4 Al''^ :0,2"mol (mol) hoSc Na2C03 va (NH4)2S04. b) Phan 1 : tac dung v d i B a ( 0 H ) 2 dM 35 Phan dgng va phuong phAp giai H6a hpc 11 VP co - D5 Xuan Hiing Ba(OH)2 > Ba'* + 20HBa'^ + COJ> BaCOji 0,01 molj 0,01 mol Ba'"+ SO^-J -> BaS044' 0,01 moli 0,01 mol NH; + OH- H NaOH 0,7 mol HNaOH = 0,7 0,02 mol I 0,02 mol PV 1.0,4704 Somolkhi Y: i ., „ „ NH3 = — = 0,082.(13,5 + 273) - 0 , 0 2 mol m=rn BaC03 + m BaS04 = 4,3 (g) (I) Phan 2 : tac dung vdi dung dich HCl d\i Hci — > + cr 2H^ + CO^" > CO2T + H2O 0,01 mol 0,01 mol ''''' PV _ 1.0,2352 nco2 =• = 0,01 (mol) RT 0.082.(13,5 + 273) Kh6i hfang BaCOj : m^^coj = 197.0,01 = 1,97 (g) = 4,3-1.97 = 2,33(g) =^ 0 3 , 5 0 4 = ^2,33 = 0,01 (mol) Theo djnh luat Ho tokn dien tich : n ^ 2- + 2.n so 2mBaS04 n^^, = 2.0,01 + 2.0,01 - 0,02 = 0,02 (mol) CO: Tong khoi liTdng cdc muoi trong dung dich A : m = 0,02.23 + 0,02.18 + 0,01.60 + 0,01.96 = 2,38 (g). Bai 8. Mot dung dich X c6 chura cdc ion Zn^^, Fe'* \i S0^~. Biet r^ng diing het 350ml dung dich NaOH 2M thi lam k6't tua het ion Zn^* va ion Fe^* trong 100ml dung dich X, neu do tiep 200ml dung dich NaOH thi mot chat ket tua vCra tan het, con lai mpt chat ke't tua m^u nau do. Tinh nong dp mol cua moi muo'i trong dung djch X. Gidi Zn^^: a mol 100ml dung dich X Fe^^: b mol 36 > Na*+ OH" 0,7 mol (mol) PhiTcfng trinh ion rut gpn : -> NHjt + H2O RHAJPG VIET Zn^* + 20H- > Zn(OH)2i (1) a mol 2a mol a mol Fe^* + 30H- > Fe(0H)3i b mol 3b mol b mol Neu do' tiep dung dich NaOH Zn(OH)2 + 20H> ZnOJ' + 2H2O a mol 2a mol Theo(l)va(2)tac6: 2a + 3b = 0,7 (I) So mol NaOH trong 200ml dung dich NaOH 2M HNaOH = 2.0,2 = 0,4 (mol) => n_= 0,4 (2) (3) mol OH Theo (3): 2a = 0,4 i=> a = 0,2 (mol) The a = 0,2 vao (I) =>b = 0,l(mol) So' mol muo'i : ZnS04 > Zn^^ + S04~ 0,2 mol 0,2 mol Fe2(S04)3 > 2Fe'* + 3S0^~ 0,05 mol 0,1 mol Nong dp mol moi muoi trong dung dich X [ZnS04]= — —-2M V = 0,1 [Fe2(S04)3]= ^ = 0,5M. Bai 9. Cho dung dich A chiJa cac ion AP^ Fe^*, SO4- va Br'. - Trpn Ian 100ml dung dich A vdi dung djch AgNOs d\i thu diTpc 75,2g ke't tua. - Trpn Ian 100ml dung dich A vdi dung djch KOH dU" thu diTpc m gam ke't tua X, nung nong X den khoi li/png khong doi thu di/pc 16g chat ran Y. - Trpn Ian 100ml dung djch A vao dung dich BaClj diT thu du-pc 93,2g ke't tua. a) Tinh nong dp mol moi ion trong dung dich A. b) Neu thoi mpt luTdng khi NH3 vao 100ml dung dich A. Sau do Ipc, tach lay ket tua. Dem nung ke't tua den kho'i liTpng khong doi thi thu diTpc bao nhieu gam 37 cha't r^n? Phan dgng phuong phip gi3i H6a hpc 11 VP c o - D5 XuSn Hung Gidi a) 100ml dung dich A t,) Thdi khi NHj v^o dung djch A : mol AP^:X Al'^ + 0,2 mol SOJ-.z Fe^* + 3 N H 3 + 3H2O mol Al(OH)3>l' + 3NH* 0,2 mol > Fe(OH)3i + 3NH^ 0,2 mol B r " : t mol 0,2 mol Dem kct tua nung : I * Tron dung dich A vdi dung djch AgNOi dif: Ag* + Br" > AgBri 0,4 mol "AsBri 0,4 mol = - ^ = 0,4(mol) 1 OO f => n„ , = t = 0,4 mol i * Tron dung dich A vdi dung djch KOH dU": Al^* + 30H~ > Al(OH)3i A1(0H)3 + OH" > AIO2 + 2H2O i| ' I , Fc^* + 3 0 H 0,2 mol >Fc(0H)3i 0,2 mol 2Fe(OH)3 -A 0,2 mol FejOj + 3H2O 0,1 mol nFc203 =1^ = 0,1 (mol) => Ba'* + SO^- I I 0,4 mol =^ = 0,4(mol) "BaS04i CO : 2z + t - 3 y 2Fe(OH)3 Fe203 + 3H2O 0,2 mol 0,1 mol Khoi liTcJng A I 2 O 3 : m^i,^o.^ = 102.0,1 = 10,2 (g) Khoi li/dng Fe203: mFe^Os = 160.0,1 = 16(g) (chat nln Y la FcjOj) Gidi PhuTdng trinh phan u^ng : np^3^ =y MCO3 = 0,2mol 0,000518 ' NaOH nHci n^^2- = z = 0,4 mol , 0,001036 + HCl 0,000564 . > MCI2 + C O . t + H2O + 2HC1 (1) (mol) > NaCl + H20 0,000564 (2) (mol) = 0,08.0,02 = 0,0016 (mol) HNaOH = 2.0,4 — + 3H2O djch NaOH 0,10M. Xac dinh kim loai M . HHCKD = 0,1.0,00564 = 0,000564 (mol) = 0,0016 - 0,000564 = 0,001036 (mol) 0.1022 0,000518 3x + 3y = 2z + t => X = 0,1 mol 200ml dung djch HCl 0,080M. De trung hoa liWng HCl diTcan 5,64ml dung I * Djnh luat bao toan dien tich : Ta AI2O3 0,2 mol Bai 10. Hoa tan hoiin toan 0,1022g mot muoi kim loai hoa trj (11) M C O 3 trong (ketluaX) > BaS04i 0,4 mol 93 2 2Al(OH)3 Khoi liTrtng cha't ntn : m = 10,2 + 16 = 26,2 (g). * Tron dung djch A vdi dung dich BaCh di/ ' > Fe''+ : y mol v 5. + 3H2O 3NH3 + 0,4-3.0,2 '- '„ — = 0,2 mol) ' M = 60 = 197 => M = 137 (Bari) Vay kim loai M la Ba. I Ndng do mol cac ion trong dung djch A : I I [ A P 1 = ^ =2(M) 0,1 (FC'1=^-2(M) 0,1 ; i i •> 0 4 lSOl-J--^-4(M) 0 4 [ B r - J = - ^ = 4(M). ^ ,j BAITAPAPOgNG Bai 1: Dung dich X c6 chiJa: 0,07 mol Na"^; 0,02 mol SO4'" va x mol OH'. Dung dich Y c6 chiJa CIO4"; NO3" va y mol H ^ Tong so mol C104' va NO, la 0,04. Tron X va Y diTdc 100 ml dung dich Z. Tinh pH cua dung dich Z (bo qua sif dicn li cua H2O). 39
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