Stopnja protolize(disociacije) - merilo za jakost elektrolita. = c d /c
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1 Stopnja protolize(disociacije) - merilo za jakost elektrolita = N/N 0 = n/n 0 = c d /c = stopnja protolize (disociacije) N = število disociiranih molekul (HCl) oz. formulskih enot (NaCl) N 0 = število vseh molekul v raztopini c d =konc. disoc. molekul c= konc. vseh molekul(analitska konc.) močni elektroliti: 0,8 šibki elektroliti : 0,3 neelektroliti : = 0 1
2 Naloge: 1. Izračunaj masno koncentracijo kloridnih ionov v 0,0512 M MgCl 2! Predpostavi popolno disociacijo! Popolna disociacija = 1 (močan elektrolit) H 2 O(l) MgCl 2 (s) Mg 2+ (aq) + 2Cl - (aq) Lahko pišemo tudi: MgCl 2 (aq) Mg 2+ (aq) + 2Cl - (aq) ali enostavno: MgCl 2 Mg Cl - 2
3 2. V 1,00 ml 0,030 M CH 3 COOH je 4, acetatnih ionov CH 3 COO -. Izračunaj stopnjo ionizacije v 0,030 M CH 3 COOH! CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) H 2 O(l) CH 3 COOH(l) CH 3 COOH(aq) Čista ocetna kislina je molekulsko zgrajena in je pri sobni temperaturi tekoča. Če nekaj ocetne kisline raztopimo v vodi, dobimo vodno raztopino ocetne kisline! 3
4 3. 28,1 ml NH 3 plina pri 20 C in 86,65 kpa uvedeš v 100 ml vode. Koncentracija hidroksidnih ionov v dobljeni raztopini je 1, moll -1. Izračunaj stopnjo protolize raztopljenega NH 3! Spremembo prostornine pri uvajanju NH 3 zanemari! 4. 8,00g NaOH raztopiš v vodi in razredčiš na 250 ml. Izračunaj molarnost natrijevih ionov v raztopini! Predpostavi popolno disociacijo! 4
5 5. V enem litru vode si raztopil 30 g KNO 3 in 40 g NaNO 3. Gostota dobljene raztopine pri 20 C je 1,07 g/ml. Izračunaj molarnost nitratnih ionov v raztopini. Predpostavi popolno disociacijo! 6. Koliko gramov Cu(OH) 2 se izloči iz raztopine, če zmešamo 100 ml 0,10 M KOH in 100 ml 0,10 M Cu(NO 3 ) 2? 7. Če v 100 ml raztopine AgNO 3 uvedeš 2,50 L plinastega HCl, merjenega pri 27 C in tlaku 99,8 kpa, oboriš vse srebro v obliki trdnega AgCl. Izračunaj masno koncentracijo raztopine AgNO 3! 5
6 Opredelitev jakosti kislin in baz s stopnjo protolize(disociacije) Ostwald (1953): za ravnotežje med ioni in nedisociiranimi molekulami je uporabil zakon o vplivu mas: CH 3 COOH H + + CH 3 COO c ( 1 ) c + c K dis. = H +. CH 3 COO / CH 3 COOH K dis. = c c / c(1 ) = c 2 /(1 ) 2 + (K dis / c) K dis /c = 0 Ta zakon ravnotežja velja le za razredčene vodne raztopine šibkih elektrolitov (K dis ) 6
7 K disoc. = c 2 / 1- pribl. formula za << 1 : K disoc. c 2 /1 K disoc. = c 2 = K/c 7
8 Kadar ni zanemarljiv uporabljamo kvadratno enačbo: K disoc.= c 2 / 1- koreni kvadratne enačbe 2 + (K d / c) - (K d / c) = 0 a b c 1 2 = ( -b + b 2-4ac )/ 2a 1 2 = ( -K d /c + (K d /c ) 2 4(K d /c) ) 2 8
9 Vodne raztopine šibkih kisli in baz Zelo malo K in B močno disociira ali sprejema protone. Večina jih je šibkih. Relativno moč K in B lahko izrazimo s KONSTANTO RAVNOTEŽJA, ki jo za šibko kislino zapišemo: HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) K a [H O 3 + [HA] - ][A ] 9
10 Ravnotežje ocetne kisline v vodni raztopini je podano z naslednjo zvezo: CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) K a [H 3 + O ][CH3COO [CH3COOH] - ] Kadar je Ka velika, pomeni, da je ravnotežje reakcije pomaknjeno v desno. Kislina je močna. Nizka vrednost Ka pomeni, da je kislina šibka. 10
11 šibka baza: B(aq) + H 2 O(l) BH + (aq) + OH - (aq) BH + OH - K b [ ][ ] [ B] MOČNE KISLINE, MOČNE BAZE K 1 ŠIBKE KISLINE, ŠIBKE BAZE K 1 11
12 Disociacijske konstante nekaterih kislin v vodi (25 ºC) : Disociacijske konstante nekaterih baz v vodi (25 ºC) : Kislina K a pk a CH 3 COOH 1, ,75 C 3 H 6 O 3 1,4 x ,85 HF 5, ,25 HClO 3, ,43 B(OH) 3.H 2 O 7,3 x ,14 Baza K b pk b NH 3 1, ,75 C 2 H 5 NH 2 5, ,25 (C 2 H 5 ) 2 NH 9, ,02 H 2 SO HSO 4-1, ,92 C 6 H 5 NH 2 3, ,42 HClO HNO ,3 12
13 Aplikacija ZDM na disociacijo vode- ph Kw = [H 3 O + ][OH - ] = 1.0 x (pri 22 C) ionski produkt vode V čisti vodi velja [H 3 O + ] = [OH ] nevtralna raztopina pri 22 o C [ H 3 O + ] = Kw 1/2 = 1x10 7 mol/l [ H 3 O + ] > [ OH ].kisla raztopina [ H 3 O + ] < [ OH ] bazična raztopina 13
14 Sørensen (1909): ph(raztopin) = log[h 3 O + ] do H 3 O + = 0,1 mol/l poh(raztopin) = log[oh - ] ph + poh = 14 sicer ph = log a(h 3 O + ) za celotno konc. območje 14
15 Šibka kislina [H + ] = K k.c k Šibka baza [OH - ] = K b.c b Močna kislina [ H + ] = c k Močna baza [OH - ] = c b 15
16 9. Kemijsko ravnotežje. Ravnotežna konstanta kemijske reakcije. 1. Računske naloge: a) Konstanta K a (CH 3 COOH) = 1, Izračunajte stopnjo disociacije ocetne kisline s koncentracijo 0,01 mol/l! 1) po približni formuli 2) po kvadratni enačbi b) Izračunajte ph 0,1 M baze s stopnjo disociacije...! c) Izračunajte množinsko koncentracijo (molarnost)..., ki ima ph =... in Kd =... Delež disociiranih molekul je zanemarljiv. 16
17 Naloge: 1. 28,1 ml NH 3 plina pri 20 C in 86,65 kpa uvedeš v 100 ml vode. Koncentracija hidroksidnih ionov v dobljeni raztopini je 1, mol L -1. Izračunaj stopnjo protolize raztopljenega NH 3! Spremembo prostornine pri uvajanju NH 3 zanemari! Rešitev: ( =0,013) 2. 15,0 ml 10% H 2 SO 4 z gostoto 1,070 g ml -1 (20 C) razredčiš na 250 ml. Pri titraciji 20 ml razredčene raztopine porabiš 10,0 ml raztopine NaOH. Izračunaj molarnost raztopine NaOH! Rešitev: (0,262 M) 3. Izračunaj koncentracijo H 3 O + ionov v raztopini HNO 2 s c (HNO 2 ) = 0,08 mol/l, če je Ka= 4,5.10-4! a) po približni formuli b) po kvadratni enačbi Rešitev : a) 0,006 mol/l b) 0,0058 mol/l 17
18 4. Raztopina, ki vsebuje 8,08g NaOH v 2L ima ph=12,9. Izračunaj stopnjo protolize! Rešitev : 0,79 5. HNO 2 + H 2 O H 3 O + + NO 2 - Koncentracija c k = 0,08 mol/l, K k = 4, Izračunaj H 3 O +! Rešitev: H 3 O + = 5, mol/l 6. 10,0 L zmesi O 2 in HCl(g) pri T=27 C in tlaku 1, Pa v mnžinskem razmerju 1:1, uvedeš v 1,0 L 0,50 M NaOH. Koliko gramov NaOH še ostane v raztopini po uvajanju plinske zmesi? R: 12g 18
19 10. Kemijsko ravnotežje. Protolitska ravnotežja v vodnih raztopinah.a)titracija močne kisline z močno bazo: 10,0 ml 0,1M HCl dodajamo po 1,0 ml 0,1M NaOH tako, da dodamo skupno 20,0 ml raztopine NaOH! K disoc.=1.74x10-5 HCl + NaOH NaCl + H 2 O.b) Titracija šibke kisline z močno bazo: 10 ml 0.1M CH 3 COOH dodajamo po 1,0 ml 0.1M NaOH tako, da dodamo skupno 20,0 ml raztopine NaOH! K disoc.=1.8x10-5 CH 3 COOH + NaOH CH 3 COONa + H 2 O 19
20 1.a) Titracija močne kisline z močno bazo: 10,0 ml xxx M HCl dodajamo po 1,0 ml xxx M NaOH tako, da dodamo skupno 20,0 ml raztopine NaOH! HCl + NaOH NaCl + H 2 O V(NaOH)v ml [H + ] ali [OH - ] ph 0 [H + ]= c k 1 [H + ]= (n k -n b )/(V k +V b ) računamo do e.t.! Ekvivalentna točka: n k = n b [H + ]= [OH - ] = x K w = [H + ].[OH - ]= x 2 =10-14 ck.vk =cb.vb x=10-7 ph=7 Vb= Po ekvivalentni točki: [OH - ] =( n b -n k )/(V k +V b ) računamp po e.t! 20
21 1.b) Titracija šibke kisline z močno bazo: 10 ml 0.1M CH 3 COOH dodajamo po 1,0 ml 0.1M NaOH tako, da dodamo skupno 20,0 ml raztopine NaOH! K disoc. =1.8x10-5 CH3 COOH + NaOH CH3COONa + H2O V(NaOH) v ml [H + ] ali [OH - ] ph 0 [ H + ] = Kk.ck = 1 [ H + ] = Kk. c k /c s = Kk. (n k -n b )/n b Ekv.točka [OH - ] = (Kw/Kk).cs 11 [OH - ] =( n b -n k )/(V k +V b ) Ekvivalentna točka: n k = n b [OH - ] = (Kw/Kk).cs ck.vk =cb.vb cs= n k / (Vk +Vb) = n b /( Vk +Vb) 21
22 1. Koliko g NH 4 Cl moramo dodati k 500 ml raztopine s c (NH 3 ) = 0,1 mol/l, da dobimo raztopino s ph =9? (Spremembo volumna zanemarimo!) K b = 1, Rešitev : 4,81g soli 2. Izračunaj ph in H 3 O +, če zmešamo 22mL raztopine HAc s c (HAc) = 0,1 mol/l in 22 ml raztopine NaOH s c (NaOH) = 0,1 mol/l! K a = 1, Rešitev : 8,72 3. Pri kateri ph vrednosti dosežemo ekvivalentno točko, če 10 ml raztopine NH 4 OH s c (NH 4 OH) = 0,3 mol/l titriramo z raztopino HCl s c(hcl) =0,3 mol/l? Kb= 1, Rešitev : 5,04 22
23 4. Koliko ml NaOH s c (NaOH) = 1mol/L moramo dodati k 100mL CH 3 COOH s c (CH 3 COOH) = 1mol/L, da bo ph=5? Kk= 1, Rešitev : 64,3 ml 5. Raztopina, ki vsebuje 8,08g NaOH v 2L ima ph=12,9. Izračunaj stopnjo protolize! Rešitev : 0,79 6. HNO 2 + H 2 O H 3 O + + NO 2 - Koncentracija c k = 0,08 mol/l, K k = 4, Izračunaj H 3 O +! Rešitev: H 3 O + = 5, mol/l 23
24 24
25 Kislinsko bazne titracije Postopno dodajanje neke baze h kislini (alkalimetrija) oz. neke kisline k bazi (acidimetrija) imenujemo titracija. Primer: Titracija močne kisline z močno bazo: 10,0 ml 0,1M HCl (začetni ph = 1) dodajamo po 1,0 ml 0,1M NaOH tako, da dodamo skupno 15,0 ml raztopine NaOH! Grafično nam prikazuje spremembo ph v odvisnosti od V (ml) dodanega NaOH spodnja titracijska krivulja. Na krivulji opazimo, da se ph na začetku, ko je v prebitku močna kislina le malo spreminja. Ko je porabljene 90% kisline ph naraste od 1 na 2, pri 99% porabi kisline pa od 1 na 3. Sledi območje, ko ph močno skoči pri le malem dodatku baze (npr. 1 kapljica). Pri točno 10 ml dodatka baze dosežemo t.i. ekvivalentno točko, ki leži v našem primeru pri ph=7 in sovpada z nevtralno točko (po def. ph=7) Po tej točki, raste ph, saj ga določa le še NaOH. 25
26 Kislinsko bazne titracije ph Ekvivalentna in nevtralna točka Začetni ph Titracijska krivulja za HCl (0,1MHCl in 0,1M NaOH) 26
27 Kadar titriramo npr. šibko kislino z močno bazo ima titracijska krivulja nekoliko drugačen potek: Primer: Titracija šibke kisline z močno bazo: 10 ml 0.1M CH3COOH (začetni ph=2,9) dodajamo po 1,0 ml 0.1M NaOH tako, da dodamo skupno 15,0 ml raztopine NaOH! Grafično nam prikazuje spremembo ph v odvisnosti od V (ml) dodanega NaOH spodnja titracijska krivulja. Ekvivalentna točka leži pri ph=9,2, torej se ne ujema z nevtralno točko! Značilno za šibke kisline je da skok ph ni tako močan, kot v prejšnjem primeru. Podobno lahko sklepamo pri titraciji šibke baze z močno kislino. Titracijska krivulja se začne v alkalnem območju, ekvivalentna točka pa je pri ph 7, torej v kislem ph območju. 27
28 Kislinsko bazne titracije ph Ekvivalentna točka Začetni ph Nevtralna točka Titracijska krivulja za CH 3 COOH (0,1M CH 3 COOH in 0,1M NaOH) 28
29 Pufri in njihovo delovanje Puferske raztopine vsebujejo substance, ki omogočajo da se pri dodatku kisline ali baze, ph raztopine neznatno spremeni. Pufri so: Mešanica šibke kisline in njene konjugirane baze npr. ocetna kislina/natrijev acetat Mešanica šibke baze in njene konjugirane kisline npr. amoniak/amonijev klorid Kadar govorimo o 0,2M acetatnem pufru, pomeni to da imamo v 1L vodne raztopine pufra: 0,1mol ocetne kisline in 0,1 mol natrijevega acetata. Kaj se zgodi, ko dodamo H 3 O + oz. OH - ione? 29
30 Ko dodamo H 3 O +, prevzame acetat protone in tvori nedisociirano ocetno kislino. H 3 O + + CH 3 COO - CH 3 COOH + H 2 O (1) Ko dodamo OH - ione, odvzamejo le ti ocetni kislini protone, tvorijo pa se acetatni ioni: OH - + CH 3 COOH CH 3 COO - + H 2 O (2) V obeh primerih nastane nevtralna voda, ki je že prisotna v raztopini. Povišanje koncentracije ene ali druge vrste ionov ne vpliva bistveno na ph! To kar velja za acetatni pufer lahko razširimo na vse puferske raztopine. 30
31 HIDROLIZA h = n h /n 0 stopnja hidrolize je odvisna od temperature [OH - ] = K h.c s [OH - ] = (K w /K k ).c s ali [H + ] = (K w /K b ).c s PUFRI [H + ] = K k. c k /c s [OH - ] = K b. c b /c s 31
32 11. Topnostni produkt Velja za slabo topne a dobro disociirane elektrolite, npr: AgCl. Ravnotežje v heterogenem sistemu: AgCl (s) AgCl (aq) Če je snov slabo topna, je koncentracija nasičene raztopine majhna. 32
33 Lahko predpostavimo, da raztopljena snov popolnoma disociira: AgCl (aq) Ag + (aq) + Cl - (aq) AgCl (s) Ag + (aq) + Cl - (aq) 33
34 K = Ag +. Cl - AgCl Koncentracija AgCl v trdni fazi je konstantna oz. se zelo malo spreminja Ag + Cl - = K AgCl = L AgCl L AgCl = (topnostni produkt AgCl ) = 1, s = Ag + = Cl - mol ion/l Ag +. Cl - = s 2 s = Lp = (1, ) 1/2 = 1, mol ion/l 34
35 PbCl 2 Pb Cl - L PbCl2,25 C = Pb 2+ Cl - 2 = 1, Lp (PbCl2) = s. (2s) 2 = 4s 3 s = Lp(PbCl 2 ) As 2 S 3 2As 3+ +3S 2- L As2S3,25 C = As 3+ 2 S 2-3 =
36 Med topnostnim produktom L, K in topnostjo s (konc. nasičene raztopine v mol/l) velja za poljubni elektrolit zveza: A m B n na +m + mb -n s ns ms K= A +m n B -n m = (ns) n (ms) m = n n m m s n+m s = (L/ n n m m ) 1/(n+m) As 2 S 3 2 As S 2- L As2S3 = As 3+ 2 S 2-3 = = (2.s) 2.(3.s) 3 =108s 5 s = (L As2S3 /108) 1/5 = ( /108) 1/5 = 5, mol ion /L 36
37 Vpliv skupnega iona Na izločanje ali raztapljanje trdnih snovi lahko v skladu z zakonom o vplivu koncentracij vplivamo s spreminjanjem koncentracije enega od ionov. Enačba Ag + Cl - = L AgCl kaže, da se iz nasičene raztopine izloči trdna snov, če je produkt koncentracije obeh vrst ionov - Q večji od topnostnega produkta. 37
38 Q = L..sistem je v ravnotežju Q L..razt. ni nasičena, oborina se ne izloči, ampak se sol še raztopi dokler ni Q = L Q L.izloči se oborina 38
39 L AgCl = Ag + Cl - = 1, NaCl = Cl - = 0,1mol/L s = Ag + = L AgCl / Cl - = 1, /10-1 = 1, mol ion/l V čisti vodi je topnost AgCl: Ag + = Cl - s = ( L AgCl ) 1/2 = 1, mol ion/l 39
40 Naloge: 1. Računski nalogi: a) Topnostni produkt... je.... Koliko mg... se lahko raztopi v 1 L vode? Koliko mg pa se raztopi v 1 L... M...? b) Topnostni produkt... je.... Izračunaj množinsko koncentracijo (molarnost) nasičene raztopine! Koliko mg... se lahko raztopi v 150 L vode? 40
41 30 ml nasičene raztopine PbCl2 Pb2+ + 2Clfiltrat I filtrat II Pb Cl - Pb Cl - dodatek : Na + + Cl - dodatek : 2H + + SO
42 PbCl 2 + Pb Cl - + Na + PbSO 4 + 2H + + 2Cl - filtrat III več PbSO 4 oborine dodatek : 2H + + SO 4 2- PbSO 4 + 2Cl - + Na + +2H + manj PbSO 4 oborine 42
43 12. Reakcije oksidacije in redukcije 43
44 1. Elektrokemijska napetostna vrsta za kovine K Ca Mg Zn Fe Sn Pb H Cu Ag Au Elementi levo od vodika izpodrivajo vodik. Elementi desno od vodika ne izpodrivajo vodika iz kislin. Vsak element izpodriva iz spojin vse elemente, ki so desno od njega. Redoksi reakcije potečejo zato, ker elementi, ioni in spojine različno močno vežejo elektrone npr : Zn + 2HCl = H 2 + ZnCl 2 Vodikov ion močneje privlači elktron, kot cinkov atom, zato pride do reakcije! Zn - e - = Zn 2+ delne reakcije H + + e - = H o 44
45 1. Elektrokemijska napetostna vrsta za kovine Baker močneje veže elektrone kot vodik, zato ga ne izpodriva iz kislin: Cu + 2HCl = ne poteče Cu + CuCl 2 = ne poteče Cink slabše veže elektrone kot baker, zato ga izpodriva iz spojin: Zn+ CuCl 2 = Cu + ZnCl 2 Zn o - 2e - = Zn 2+ delne reakcije Cu e - = Cu o Cu + ZnCl 2 = ne poteče Zn + ZnCl 2 = ne poteče 45
46 Figure 21.2 An oxidation-reduction reaction. A strip of zinc metal was placed in a solution of copper(ii) sulfate (left), and the zinc reacts with the copper(ii) ions to give copper metal and zinc ions in solution. Zn+ CuCl 2 = Cu + ZnCl 2 Zn(s) + Cu 2+( aq) Zn 2+( aq) + Cu(s) Copper metal accumulates on the zinc strip, and the blue color of aqueous 46 copper(ii) ions fades as copper(ii) ions disappear from solution (middle and right).
47 2.Napetostna vrsta za nekovine Elementi I.skupine P.S. so močni reducenti. Halogeni pa so močni oksidanti. NaBr + Cl 2 + H 2 O NaCl + Br 2 + H 2 O Br 2 se raztaplja v CCl 4 (rjava barva dokaz Br 2 ) KJ + Cl 2 + H 2 O KCl + J 2 + H 2 O J 2 se raztaplja v CCl 4 (vijolična barva dokaz za J 2 ) Napetostna vrsta za nekovine: F Cl Br J S 47
48 3. Kalijev manganat (VII) kot oksidant v različnih medijih: KMnO 4 + H 2 SO 4 + Na 2 SO 3 Na 2 SO 4 + K 2 SO 4 + H 2 O + MnSO 4 Mn e - = Mn 2+ x2 S e - = S 6+ x5 KMnO 4 + H 2 O + Na 2 SO 3 MnO 2 + KOH + Na 2 SO 4 Mn e - = Mn 4+ x2 S 4+ -2e - = S 6+ x3 KMnO 4 + NaOH + Na 2 SO 3 K 2 SO 4 + H 2 O + Na 2 MnO 4 Mn e - = Mn 6+ x2 S e - = S 6+ x1 48
49 4. Določevanje Fe 2+ s titracijo z 0,02M KMnO 4 KMnO 4 + H 2 SO 4 + FeSO 4 Fe 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O + MnSO 4 Mn e - = Mn 2+ / x2 2Fe e - = 2Fe 3+ / x5 49
50 Reakcije HNO 3 in H 2 SO 4 z nekaterimi polžlahtnimi kovinami: HNO 3 (razr.) + Cu Cu(NO 3 ) 2 + NO + H 2 O HNO 3 (konc.) + Cu Cu(NO 3 ) 2 + NO 2 + H 2 O Enako velja za: Ag, Hg, Cr, Co in Ni! H 2 SO 4 + Cu CuSO 4 + SO 2 + H 2 O Aktivne kovine in HNO 3 Zn + HNO 3 Zn(NO 3 ) 2 + H 2 50
51 Žlahtne kovine in HNO 3 Au + HNO 3 + 3HCl AuCl 3 + NO + H 2 O 51
52 Naloge: 1. Računska naloga Izračnajte, koliko mililitrov... HNO 3 z gostoto... g/ml potrebujete za raztapljanje... g...! Izračunajte tudi, koliko litrov in kateri plin pri reakciji nastane pri temperaturi... C in tlaku... kpa! 52
b) Računske naloge (z osnovami): 1. Izračunaj in nariši tiracijsko krivuljo, če k 10,0mL 0,126M HCl dodajaš deleže (glej tabelo) 0,126M NaOH!
11. Vaja: Kemijsko ravnotežje II a) Naloga: 1. Izmeri ph destilirane in vodovodne vode, ter razloži njegovo vrednost s pomočjo eksperimentov!. Opazuj vpliv temperature na kemijsko ravnotežje!. Določi karbonatno
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