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

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P /IJyfD *>6 XUAN HUNG •— ^ Gi&o vifen chuySn luy§n thi OH-CD H6a Pfian ddng& pnifong phapgidk HOA HOC^ PHAN V6 coll 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 Cd D6 XUAN Hl/NG , Chju trach nhi^m xud't ban NGUYEN TH| THANH HlTCfNG Bien lap : HAI AU Sura ban in : H6NG HAI Trinh bhy : C6ng ty KHANG VI^T Bia : C6ng ty KHANG VI|;T NHA XUAT BAN idNG HOP TP. H6 CHI MINH NHA SACH TdNG H0P 62 Nguyen Thj Minh Khal, Q.l DT: 38225340 - 38296764 - 38247225 Fax: 84.8.38222726 Email: tonghop®nxbhcni.com.vn Website: www.nxbhcm.com.vn/ www.fiditour.com Tong phdt hank C6NG TY TNHH MTV D|CH Vg VAN H6A KHANG VI|T 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 Website: www.nhasachkhangviet.vn In Ian thur 1. So lUcfng 2.000 cuon, kho 16x24cm. T^i: Cong ty TNHH MTV in an MAI THjNH DLfC. Dja chl: 71, Kha Van Can, P.Hi$p Binh Chanh, Q.Thu Dure, Tp.HCM. So DKKHXB: 1482-12/CXB/12-1SIAHTPHCM ngay 06/12/2012. Quyet djnh xuat b5n s6': 1771/QD-THTPHCM-2012 do NXB T6ng hap Thanh pho Ho Chi Minh cS'p ngdy 27/12/2012. In xong va nop IIAJ chieu qwy I nSm 2013. A. T6M TAT Lf THUY€T I. SI/DI|NU * SI/ di^n li 1^ qui trlnh phan li cac chat trong nifdc ra ion. Vf du : NaOH > Na* + OH" * ChS't di$n li Ih nhifng chat tan trong nurdc phan li ra ion. C6 hai loai: Chat di0n li manh va chatdi^n li yeu. * DO diOn li (a): Dp diOn li (a) cua chat dipn li la ti so giffa s6' phan tur phan li ra ion (n) va tong so phan tur hoa tan (no). a=JL O^a^l «() Neu bieu di^n diTdi dang phan trSm : 0% < a ^ 100% u a = 0 : chat kh6ng di$n li u 0 < a < 1: cha't di^n li yeu u a = 1 : cha't diOn li manh. II. AXIT-BAZCJ-MU6'l 1. Axit * Theo thuye't A-r6-ni-ut: axit Ift cha't khi tan trong nrfdc phan li ra cation H*. Vi du : HCl > H* + CI" * Theo thuyd't Bron-stet: axit Ih chat nhtfcJng proton (H*). Vi du : CH3COOH + H2O U CH3COO- + H3O* Nhffng axit khi tan trong nurdc ma phan tur phan li nhieu na'c ra ion H* la cac axit nhieu nac. 2. Bazcf: * Theo thuye't A-re-ni-ut: bazd la chat khi tan trong nirdc phan li ra anion OH'. Vidu: KOH J-K^ + OH" * Theo thuyet Bron-stet: bazd la cha't nhan proton. Vidu: NH3 + H20-> NH;+0H- Nhffng bazd khi tan trong nxSdc mi phSn tur phan li nhieu nS'c ra ion OH" la cdc bazd nhieu nSfc. 3. Mu6'i Muoi la h0p chSft khi tan trong nurdc phan li ra cation kim loai (ho&c cation NH4) va anion g6c axit. C6 hai loai: + Muoi axit: NaHCOj, KHSO4 + Muoi trung hoa : KCl, CaC03, Phan dgng phi^ng phap giai H6a tipc 11 V6 CO - D5 XuSn Hang _______ 4. Hidroxit li^ng tinh La hidroxit khi tan trong nifdc vita c6 the phan li nhiT axit vifa c6 the phan 11 nhuf bazd. Vi du : Zn(OH)2, A1(0H)3. III. Si; DI5N LI CUA NLTdC. pH. CHAT CHI TH! AXIT - BAZd 1. Nvldc la cha't di$n li rat yg'u H2O H* + OH" * Tich so' ion cua nuTdc (kn^o) la h^ng so cl nhi$t do xdc dinh kH20 =[H^]-[OH'] = 1-010''^ ([Hi = [0H-] = I.O.IQ-' (mol//) d 25"C) * Moi trurdng axit la moi trifdng trong d6 : [H1>[0H'] hay [H^ > 1,0.10"^M * Moi triTcJng kiem 1^ m6i triTdng trong do : [H1<[0H-] hay [Hi < 1,0.1 Q-'M * Moi trirdng trung tinh Ih moi triT&ng trong do: [H^] = [OH"] = 1,0.10"'M 2. Khai ni^m vl pH. ChS't chi thj axit - bazof [Hi = 1,0.10"''" (M) Ve mat toan hoc : pH = -lg[Hl + Khi pH < 7 : moi triTdng axit + 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. IV. PHAN CfNG TRAO DOI ION TRONG DUNG DjCH CAC CHAT DIEN LI * PhSn tfng xay ra trong dung dich cdc chat dien li la phan ti'ng giffa cdc ion. * Phan tfttg trao doi trong dung dich cac cha't dien li chi xay ra khi cac ion ke't hdp dtfdc vdi nhau tao thinh it nhS't mot trong cdc cha't sau : + Cha't ket tua; + Chat dien li y6'u; + Chat khi; K2SO4 + BaClz > 2KC1 + BaS04>t PhiTdng trinh ion thu gon : Ba^* + SO^" > BaS04. NaOH + HCl > NaCl + H2O PhiTdng trinh ion thu gon : H"" + OH" > H2O K2CO3 + 2HC1 > 2KC1 + C02t + H2O PhiTdng trinh ion rut gon : 2H* + CO^" > COzt + H2O. B. PHAN LOAI VA PHlTdNG PHAP GIAI CAC DANG BAI TAP Dang 1. - Viet phi/cfng trinh di#n li cac chdt - Tinh nong dp mol cua tCfng Ion BAI TAP MAU VA BAI TAP NANG CAO Bai 1. Viet phifdng trinh dien li cua nhiTng chat sau : a) Cac chat dien li manh : NaCl, HNO3, KOH, Na2S04, Ba(N03)2, H2SO4, [Ag(NH3)2]Cl, CUSO4.5H2O. b) Cac cha't dien li yeu : CH3COOH, Mg(0H)2, H2S, HCIO, HCN, Bi(0H)3. y Gidi a) NaCl > Na* + Cl" b) CH3COOH ^ CH3COO" + H^ HNO3 > H* + NO3 Mg(0H)2 ^ Mg'^ + 20H" KOH > K* + OH" H2S ^ 2H^ + S^" Na2S04 >2Na^+SO^~ HCIO-> H* + CIO" Ba(N03)2 > Ba^* + 2N0J HCN -> H^ + CN" H2SO4 > 2H^ + SO^- Bi(0H)3 ^ Bi^* + SOH". [Ag(NH3)2]Cl > [Ag(NH3)2]" + Cr CUSO4.5H2O > Cu^* + S04^' + 5H2O Bai 2. Viet phi/dng trinh c&c chat dien li manh va tinh nong do mol cua tilfng ion trong cac dung dich sau : Ba(N03)2:0,10M; HNO3: 0,020M; KOH : 0,01 OM HBr04 : 0,025M; HMn04 : 0,030M; NaC104 : 0,040M. Gidi • Ba(N03)2 > Ba^* + 2NO3 0,1 OM 0,1 OM 0,20M =>[Ba^1 = 0,10M [NO3] = 0,20M • HNO3 > H* + NO3 0,020M 0,020M 0,020M =>[H1= [NOJ] =0,020M • KOH > K* + OH" 0,0 lOM 0,0 lOM 0,0 lOM =>[K1 = [OH"] = 0,010M Phan dang va pniiono ph^p giai H6a hpc 11 VP cc - Bi XuSn Hung HBr04 0.Q25M + BrO^ 0.025M 0,025M r:>[Hl= {BrOJ] =0,025M • HMn04 f W * Mn04 0,030M 0,030M 0.030M r^iHi = IMnO^J =0.030M • NaClOa > Na* + ClO^ (J.040M 0.040M ();040M ==>lNan= lC10;j =0,()40M. Biii 3. Vic'l phifdng irinh di^n li thco lifng na'c ciia ctic axit saii: H2SO4, H3PO4, H2S. H2SO3. H2SCO4. Giai • H2SO4 > H* + HSOi HSO4 > H* + SOi' • H3PO4 -> H* + H2FO4 H2PO; -> + HPO;' H2S -> + HS- Hs->ir + s^- H:S03 -> ir + USO^ H.S05; -> + so;" H2SCO4 —> H' + HSco;; HSc(J4 -> + ScOj-, Hsii 4. Vic't phiftJng trinh di^n li 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. cm a) Nong do mol dung djch H2SO4: 2H* + SO-V H2SO4 0,15M n.3M 0,I5M =>|H*1 = 0.3M; [S05-1 = 0.15M. h) So mol NaC!: HN,,;, = ~^ = 0.1 (mo!) NaCl > Na* + Cf 0,5M 0,5M 0,5M => [Nal = [Cn = 0,5M. c) NaOH > Na* + OH" 0,250M 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 nong dp mol nhiT sau : [Kl = O,O50M;[Mg^*] = O,20M; [Cr] = 0,050M; lSOf]=0.20M. Vie't cong Ihtfc phan tijr cua X, Y V£k vie't phiTdng trinh di^n li cua chung tronj dung djch. Giai Theo de tW hai hdp chat X va Y li : KCl va MgS04. Kci —> K* + cr 0,050M 0.050M 0,050M MgS04 > Mg^* + SOj- 0,20M 0,20M 0,20M. Dqnq 2. - Tinh d$ di^n H a. h^ng s6 di^n li - Tinh (H*), (OHl. pH cua dung djch BAI TAP MAU Bai 1. Tinh nong dp H\, pH ciia dung dich HCl 0.1 OM va dung dich NaOF COIOM. Giai • HCl —• H* + cr 0,10M 0,10M =>IH*] = 0,10M; pH = -lg[H*J = -lgO,l = l =>[OH-]= =1.0.10"'^M. 1.0.10"' • NaOH > Na* + OH" O.OIOM O.OIOM =>[OH-] = 0.010M =>[H1= i:^:H!!l = 1.0.I0"'2M 1,0.10"^ =>pH = -lglH*] = -lgl0-'^=12. 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 li cua axit axetic trong dung dich tren. Neu hoa tan thtm 0,01 mol HCl vao 1 lit dung dich tren thi do dien li cua mit axetic tftng hay giam? GiU thich? Giai a) PhiTctng trinh di6n li: CH3COOH ^ CHiCOO" + H* Trong 1 lit dung dich c6 0,25 mol CH3COOH c6 pH = 2 => [Hi = 10"''" = 10"^ = 0,01M V|y troog 1 lit dung dich c6 0,01 mol CH3COOH phan li ra ion. Bo di^n li cua CH3COOH la: a = —x 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 li chuydn dich sang trai, do do do diSn U giam. Bai 3. Mot dung dich c6 [H^ = 0,010M. Tinh [OH^ va pH cua dung dich. Moi tri/cfng cua dung dich nay la axit, kiem hay trung tinh? Hay cho bie't mau cua quy tim trong dung dich nay? Gidi [HI =0,010M => [0H-] = ' , =1,0.10-'^M 1,0.10"^ pH = -lg[Hl = -lgI0-' = 2 Moi 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. b) Dung djch H2SO4 0,02M. c) Dung dich Ca(0H)2 0,004M (a = 0,8). d) Dung djch CH3COOH 0,004M (a = 0,7). Gidi a) KOH > K* + OH- 0,003M 0,003M [0H-] = 0.003M • I n 10"'"* ^ [Hi = ii^ii^ = 0,33.10-'' :^ pH = -lg[Hl = 11,48 3,0.10"^ b) H2SO4 > 2H^ + SO^- 0,02M 0,04M [HI = 0,04M => pH = -lg[Hl = 1,4. KHAWTfi Vim c) Ca(OH)2 > Ca'^ + 20H- 0,004M 0,008M hay [OH-] = 0,008M N6ng do OH" cua dung dich: [OH'] = 0,008.a = 0,008.0,8 = 0,0064M => [Hi = ^'"'^^ = l,5625.10-'2 =:> pH = -lg[Hl = 11,81. 6,4.10 ^ d) CH3COOH -> CH3COO- + H^ 0,004M 0,004M hay [H1 = 0,004M => Nong dp H* cua dung dich: [Hi = 0,004.a = 0,004.0,7 = 0,0028M pH = -lg[Hl = 2,55. Bai 5. Mot dung dich c6 pH = 9,0. Tinh nong do mol cua cac ion H"" va OH" trong dung dich. Hay cho bid't mau cua phenolphetalein trong dung djch nay. Gidi lQ-14 pH = 9,0 [Hi = lO ^" = lO-^'M =>[0H1 = = lO'^M VI pH = 9,0 nen dung dich c6 moi tru'dng kiem => phenolphetalein doi sang 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 IM c6 a = 0,43%. Tinh hang so Kb va pH cua dung djch do. Gidi a) Vi phan tijT anilin c6 goc C6H5- la goc hut electron nen lam giam mat do electron tren nguyen tijf N, do do c6 tinh bazcJ yeu hdn phan tiif NH3. • (NH3)>^b (CfiHs-NHa)- b) Phan tfng : NH3 + H2O NH^ + OH" 1 mol 0 0 (l-x)mol xmol x mol Ma a = 0,43% = 0,0043 = y =:> x = 0,0043 = 4,3.10"^ Taco: Kb = \^KllO^Kl± jL= ^8,57.10-^- [NH3] 1-x 1-x 1-4,3.10"^ Vay Kb = 1,857.10"'. Phan dsino va phiiong phip giai H6a hpc 11 VP co- D5 XuSn Hung IH*] = = JO:!!. = 0,23.10-'' = 2.3.10-'^ [OH"] 4,3.10"^ pH = -lg[Hl = -lg(2,3.10-'')= 11,64. Bai 7. Tinh nong dp mol ciSa cdc ion H* va OH' trong dung djch NaN02 1,0M. Biet n^ng h^ng so phan li bazd cda NOj la Kh = 2,5.10'". Giai Phuang trinh di§n li: NaNO: > Na* + NOj IM IM Phan tfng thuy phan : NOj + H2O -> HNO2 + OH" Bandau: IM 0 0 Canbhng: (1-x) x x [HNO,J.[OH-l^^^ [NO2] (1-x) „2 hay — = 2,5.10-" 1-x Giai ra ta c6: x = 5.10''^ hay [OH'] = 5.10-''M [H*]= -l^i = -^^^ = 0,2.10-''(M). 10-'^ _ 10-"* [OH-] 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-^ (Dieuki|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 Phtfdng irinh diOn li: HCl > H* + Cl" Goi V, la the lich dung djch HCl ban dau c6 pH = 3. . pH = 3 IH*] = lO'^M mh [H^] = -Jl => n^^ ^ = lO-^V, (mol) Goi V2 la the lich dung djch HCl sau khi pha loang c6 pH = 4. n + . pH = 4 => [H*] = lO-'^M ma [H*l = -j^- => n^+ = 10-^V2 (mol) Khi pha loang dung djch so mol H* khong thay doi: n + =n ^. H d H s =>10-^V, = 10-^V2 :^ = ^ = 10 =:>V2=10V, V$y pha loang dung djch HCl 10 Ian nghla la phai pha loang 1 the tich dung djch HCl vdi 9 the lich niTdc nguycn cha't. Bai 10. Dung djch X gom CH3COOH IM (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 Ta c6: HCl -> H* + Cr 10-^ 10-^ CH3COOH < > CHjCOO + H* Bandau 1 0 lO' Phanli x x x Can bhng 1-x x x +10"^ Taco: iLi^tlE!! = 1,75.10-^ (*) 1-x VI X < < 1 => 1 - X => 1, do do ttr (*) => x^ +10-\-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) Phan d^ng phipng ph&p giii H6a hoc 11 VP co - D5 XuSn Hung Gidi Ta c6: n = 5.10'Wl; HHCI = 0,01 mol => 2 n + = 0,02 mol H2SO^ H "Ba(OH)2 = 0.01 mol; HNaOH = 0,02 mol I n^^. = 0,04 mol Phan ung trung h6a: H* + OH" -* H2O 0,02 0,02 n . d„ = 0,04 - 0,02 = 0,02 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) Gidi Ta CO : nNaOH = 0,0l.V mol nHci = 0,03.V mol n =0,01.Vmol OH n ,-0,03.Vmol Phurong trinh ion: H^ + OH' H.O 0,01V 0,01V =^ = 0,03V - 0,01V = 0,02V (mol) H H^ 0,02 V 2V = 0,01 = 10"2M =>pH = 2 ^ DapanC. 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^ = IQ-'M =^ n ^ = 0,1.0,1 = 0,01 mol H Va: VNaOH = 200 - 100 = 100 ml fiNaOH = 0,1a mol n . = 0,1a mol OH Phifdng irinh phdn tfng: 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. I • Theo phiTdng irlnh phan tfog: n phaming = n . 0,01 mol OH H => n _j^ = (0,1a-0,01) mol OH Mat kh^c, ta co: pH = 12 => pOH = 2 0,1a-0,01 0,2 = 0,01 a = 0,12 => [OUT ] = 10-^ = O.OIM Dap an D. 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 Tacd: nH2S04 = 0,025 mol "HCI - 0,02 mol nBa(0H)2 = 0'25a mol n ,=0,07 mol H+ n 2 =0,025 mol •n = 0,5a mol OH" = 0,25a mol Ba Dung djch sau khi tron co pH = 12 (moi trifdng bazd) => 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 OH 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. m BAITAPAPDyNG Mi 1. Tinh nong dp mol cua ion H* trong dung dich HNO2 0,10M bie't ring hlng so phan li axit cua HNO2 la Ka = 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 13 Phan djinfl vi phuang pMp giii H6a hqc 11 VP cO - Dg Xuin Hung [HNO2] (0,1-X) 0,1-X Vi HNO2 la axit ycu nen x « 0,1 nen — = 4.10"* => x = 6,3.10"^ 0,1 Vay [H1 = 6,3.10-^M. Bai 2. Co hai dung dich sau : 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". Giai a) CH3COOH CH,COO- + H* Bandau: 0,1M 0 0 Canb^ng: 0.1-x x x ^ _ [CH3COO-].[H-] X.X _ x^ 5 [CH3COOH] 0,1-X 0,1-X ' • Vi CH3COOH la mOt axit yeu nen X « 0,1 nen — = 1,75.10"^ 0.1 Giai ratac6x= 1,32.10-^M hay [H*] = 1,32.10'^M. b) NH3 + H2O -> NH; + OH- Bandau: 0.1 0 0 (M) Canbkng: 0,1-x x x (M) [NH;].[0H-]_ X.X _ x^ 3 [NH3] -0,l-x-0,l-x-^'^°-^° VI NH3 la mpt bazd yeu nen x « 0,1 — = 1,80.10"^ =>x= 1,34.10"^ Vay [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 Giai a) Khoi liTdng chat tan H2SO4 : C%.m,,d 98%. 100 , 98 , . ,. '"^••= -^ = -lB5^ = ''^^^'"^=^ "" ^ =^-Un.ol) H2SO4 > 2H* + SO^ 1 mol 2 mol =>[H1= Y = 2(M). b) Goi V la the tich dung djch NaOH c6 nong do 1,8M =>nNaOH= 1.8.V(mol) 2NaOH + H2SO4 )• Na2S04 + 2H2O 1,8V 0,9V =>nH2S04d^=(l-0.9V) mol Sau phdn i?ng : Vjj = 1 + V . pH=l =>[H1 = 10-'=0,1(M) H2SO4 > 2H* + SO^ (1-0,9V) (2-1,8 V) =>[H1=^-^ = 0.1 1 +V 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 (1,8V-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 •Dung djch hon hcJp gom HNO3 v^ HCl c6 : lpH= 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 co- D5 Xuan Hi/ng nBa(OH)2 = 0,025.V (mol) Ba(OH)2 > Ba^*+ lOW 0,025V 0,025V (mol) Phifdng trlnh phin tfng : H* + OH" —» H^O 0,05V 0,05V Dung dich tao thanh c6 pH = 2 => [Hi dir = 10"''" = 10-' = 0,01M Vjj sau phan iJng = V + 0,1 ^ Vd>/=^M-V = 0,01.(V4-0,l) oO,01 - 0,05V = 0,01V + 0,001 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. Gidi 1 46 a) nHci= —- = 0,04 (mol) 36,5 HCl —> H* + cr 0,04 0,04 (mol) ^ [HI =^ = ^ = 0,im=>pli = -IgO, 1 = 1. V 0,4 b) Ta CO : nNaOH = 0,4.0,375 = 0,15 (mol) NaOH > Na* + OH" 0,15 mol 0,15 mol nHci = 0,1.1 =0,1 (mol) HCl —> H^ + cr 0,1 mol 0,1 mol Vdj sau phan i?ng = 0,4 + 0,1 = 0,5 (lit). PhiTcJng trmh phan tfng : H^ + OH" —-» H2O 0,1 mol 0,1 mol n^ . =0,15-0,1 = 0,05(mol) "OH dU [OH-] = -^ = 0,l(M> [Hi = h2:lI^=x(r^^.(Pi^-^ 0^110^^ = [§, 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. Gidi Phircfng trinh dien li: CH3COOH -> CH^COO" + H* (1) Bandau: 2 0 0 (M) Can bang: 2-x x x (M) K.= [CH3C00-j.[H"]^ x.x ^ ^2 10-5 [CH3COOH] 2-x 2-x Vi la axit yeu nen x « 2. 2 =^^ = 2.10-5 2 X = 6,324.10" => [Hi = 6,324.10'^ {M)=> [0H-] = 10 -14 = 1,58.10"'^ (M). (2) 6,324.10" Neu them vao moi lit dung djch axit 0,2 mol muoi CHsCOONa : Phi/dng trinh dien li: CHjCOONa > CHjCOO" + Na* 0,2 mol// 0,2 mol// nen lam cho nong do CH3COO- trong dung djch tang len do do can bang d phdn tfng (i) dich chuyen theo chieu tOf phai sang irai, nen can b^ng duftJc thanh lap. CH3COOH -> CH3COO- + H* Bandau: 2 0 0 Canb^ng: (2-y) (0,2+ y) y Vi Ih axit yeu ndn y « 2 => [CH3COOH] = 2 - y = 2 (M) [CH3COO-] = 0,2 + y = 0,2 (M) tCH,COO-l.[H'1^0^^^,^ [CH3COOH] 2 => y = 2.10"* hay [Hi = 2. lO"" (M) (M) (M) [OH1 = 10 -14 = 0,5.10"'" =5.10-" (M) 2.10 -4 So vdi dung dich CH3COOH ban diu: [0H1 tang len = 5.10 -3rWe-4i^4ft-v4-ttI1 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. Bie't rang hang so axit (hhng so ion hoa) cua HF la = 6,8.10""*. Cho log6,8 = 0.83. Gidi a) Phtfdng trinh dien 11: HF -> H^ + F Trong dung dich c6 F" nen lam cho can bang it bi chuyen dich nen c6 the coi [HF] = 0,1M=>[F] = 0,1M k^=mlLlE:i.6,8.10-^M) [HF] [H1=^'^-^Q"'-^'^=6.8.10-"(M) 0,1 pH = -lg[Hl = -lg(6,8.10"') = 3,17. b) * Khi them 0,01 mol HCl vao thitac6:H^+ F > HF [HFJ = 0,1 + 0,01 = 0,11 (mol) [F] = 0,1-0,01 =0,09 (mol) [HI = 6,8.10-r = 6.8.10"^ — - 8,31.10-^ [F-] , 0,09 => pH =-lg(8,31.10"") = 3,08. * Khi them 0,01 mol NaOH vao ta c6 : HF + OH" > F + H2O =:>[F] = 0,1+0,01 =0,11 (mol) [HF]= 0,1-0,01 =0,09 (mol) [H^] = 6.8.10-*. = 5,56.10"^(M) pH = -lg[Hl = 3,25. Mi 8. Tron ba dung dich H2SO4 0,05M; HNO3 0,1M va HCl 0,15M vdi nhffng the tich bang nhau, thu dtfcfc dung dich A. Lay 600ml dung djch A cho tac dung vcti dung dich B gom NaOH 0,3M va KOH 0,15M. Tinh the tich dung dich B can dijng de sau khi tac dung vdi 600ml dung dich A thu difdc dung dich CO pH = 3. Gidi Vi the tich blng nhau nen the tich moi dung dich axit trtfdc khi trpn Ian la: — = 200 ml = 0,2 (lit) So'mol m6i axit trong 600ml dung dich A : nH2S04 =0,05.0,2 = 0.01 (mol) nHN03 =0,1.0,2 = 0,02 (mol) n HCl =0.15.0.2 = 0.03 (mol) • Phifdng trinh dien li cdc axit: H2SO4 —^ 2H* + SO^~ • 0.01 mol 0,02 mol . . HNO3 > H^ + NO3 . 0,02 mal 0,02 mol HCl —> H^ + cr . 0,03 mol 0,03 mol => I = 0.02 + 0.02 + 0.03 = 0.07 (mol) Goi the tich dung djch B can diing la V. HNaOH = 0,3.V (mol) nKOH = 0.l'5.V(mol) NaOH > Na* + OH" 0.3V (mol) 0,3V (mol) KOH > K* + OH" 0,15V (mol) 0,15V (mol) => Z n_. = 0,3V + 0,15V = 0,45 V (mol) OH Phan tfng trung hoa dung dich A va dung dich B : H* + OH- > H2O . . 0,45V 0.45V (mol) Dung dich thu dufdc cd pH = 3 => moi trtfcfng axit ^^H"" di/. The tich dung dich sau phan tfng : Vjj = 0,6 + V pH = 3 [Hi = 10-'= 0.001 (M) [HI = ^ = "'Q^-Q''^^^ = 0.001 V = 0.154 lit V 0.6 + V Vay the tich dung dich B la 0,154 lit. 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, tichso [Hl.[OH-] = 10-^ 19 [...]... dung dich c6 chtfa cac ion: x mol 0,6 mol S04^", 0,4 mol NOf Co can dung dich nay thu du'rtc 116 ,8 gam hon = 2M 0,05 hcJp cac muo'i khan Tim M Gidi Bai 7: Dung djch A chiJa HCl I M va H2SO4 0,6M Cho 100ml dung djch B gom KOH I M va NaOH 0,8M vao 100ml dung djch A, c6 ciin dung djch sau n^^^o^ = 0,06 mol • => 116 ,8 = 0 , 2 M M + 0,2.24 + 0,3.64 + 0,6.96 + 0,4.62 n^, = 0,22 mol "Hci^O.lmol Dung dich... C.(2),(4),(5),(7) C (2), (4), (6) (7) B Cdc dung dich NaHC03, KHS NaHS04 d6u c6 pH < 7 C.5% D 6% dung dich c6 pH = 9 A 100 Ian B 110 Ian C 99 Ian D 80 Ian Cfiu 20 De thu diTdc 1 lit dung dich HCl c6 pH = 5 tir dung dich HCl c6 pH = 3 thi the tich nufdcca't can diingia: Cfiu 11 Phdt bieu nao sau day kh6ng chinh xdc? CuCb, NaHS04 deu c6 pH < 7 B 3% Cfiu 19 Phai pha loang dung dich KOH 0,00 I M vcti niTctc... 0,25 X 0,08 = 0,02 mol "Ba(OHi2 v,+v, O Cau 25 "H^SO, =0,0025 mol => Chon B pH = 4 1,8 V - 2 = 0,1 V +0,1 o 1,7 V = 2,1 11 9 => 0,5 a - 0,025 = 0,005 => 0,5a = 0,03 => a = 0,06M nBaso4 = " s o i = 0 ' " " 2 5 m o l niaaso, = 0,5825 g =>Chon C 55 Phan d?ng va phuong pMp g'tii H6a hpc 11 VP co - D5 Xuan Hang C f l u 26 HHCI = 0,25 X 0,08 = 0,02 N 6 n g dp NH4CI = N 6 n g dp N H 4 " = 53 5^250 mol A r>r>ic... 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... ion rial gon: 2 r + H2O2 ^ GiaM > (CH3COO)2Ca + C O t + H2O B o t nS ( N H 4 H C O 3 ) ^ > 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... 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 f) K N O , + NaCl g) S n ( O H ) 2 + N... 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... 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... 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+... 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
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