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!
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1 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 trdoto vodovodne vode! b) Računske naloge (z osnovami): 1. Izračunaj in nariši tiracijsko krivuljo, če k 10,0mL 0,16M HCl dodajaš deleže (glej tabelo) 0,16M NaOH! V(NaOH) [ml] ph V(NaOH) [ml] ph 0 0,90 9,5,9 1 0,99 9,8,90 1,08 9,97,7 1, ,00 1,7 10,0 10,8 5 1,8 10, 11,10 6 1,50 10, 11,9 7 1, ,78 8 1, ,51 9,18 Začetni ph: To je ph raztopine 0,16M HCl, ko ji baze še nismo pričeli dodajati. Ker HCl popolnoma disociira je koncentracija vodikovih in kloridnih ionov, na katere kislina razpade, kar enaka koncentraciji kisline (). 0,16/L + HCl H + Cl [ ] ph log H + log 0,16 0,90 ph pred ekvivalentno točko: To je ph 10mL raztopine 0,16M HCl v katero dodamo največ tolikšno množino NaOH, da še ne doseže množine HCl (n A ). V našem primeru bomo dodali npr. 1mL NaOH s koncentracijo 0,16M (). Ves NaOH tako zreagira s HCl, njena množina pa se zmanjša za množino dodanega NaOH (n B ). Množina kisline, ki ostane (n A1 ) je enaka množini vodikovih ionov, ki ob disociaciji le te nastanejo. Koncentracijo vodikovih ionov pa enostavno izračunamo, če poznamo nov volumen raztopine, ki je kar vsota volumna kisline in baze. 7
2 V(HCl) 10mL 0,16/L V(NaOH) 1mL HCl + NaOH NaCl + H O + ( HCl) na nana1 1 na1 na + n HCl n NaOH V ( HCl) V ( NaOH ) 1 H VR V ( HCl) + V ( NaOH ) 0,16 0, 01 0,16 L L L 0, 001L 0,10 L 0,01L+ 0,001L [ ] ph log H + log 0,10 0,99 ph v ekvivalentni točki: To je ph 10mL raztopine 0,16M HCl, ki ji dodamo tolikšno množino NaOH, da je enaka množini HCl. Ker imata v našem primeru kislina in baza enaki koncentraciji, je tudi volumen dodanega NaOH enak volumnu kisline. HCl in NaOH torej popolnoma zreagirata v NaCl in H O. Voda v raztopini pa disociira na oksonijeve in hidroksidne ione, kateri ji določajo ph in katerih koncentracijo enostavno izračunamo s pomočjo ionskega produkta vode. Kw 10-1 /L HCl + NaOH NaCl + HO / / HO HO + OH + l aq () ( ) ( aq) Kw H O OH + 10 HO OH Kw Kw Kw 7 10 L 1 ph H O + 7 log log 10 7,00 ph po ekvivalentni točki: To je ph 10mL raztopine 0,16M HCl v katero dodamo najmanj tolikšno množino NaOH (n B ), da še preseže množino HCl (n A ). V našem primeru bomo dodali npr. 10,0mL NaOH s koncentracijo 0,16M (). Tako vsa HCl zreagira z NaOH, množina baze pa se zmanjša za množino kisline. Množina baze, ki ostane (n A1 ) je enaka množini hidroksidnih ionov, ki ob disociaciji le te nastanejo. Koncentracijo hidroksidnih ionov pa enostavno izračunamo, če poznamo nov volumen raztopine, ki je kar vsota volumna kisline in baze. 8
3 V(NaOH) 10,0mL HCl + NaOH NaCl + H O + ( NaOH ) na na 1 + NaOH Na + OH R 1 na n NaOH n HCl V ( NaOH ) V ( HCl) OH V V ( HCl) + V ( NaOH ) 0,16 0, 0100L0,16 0, 001L 0, 0100L+ 0, 001L L L 1,89 10 L ph + OH + 1 log 1 log 1, ,8 Titracijska krivulja: Ko po zgornjih postopkih izračunamo ph za vse prostornine NaOH v tabeli, lahko narišemo tiracijsko krivuljo, titracije močne kisline z močno bazo, ki nam prikazuje naraščanje ph vrednosti ob dodajanju NaOH. 1 TITRACIJSKA KRVULJA MOČNE KISLINE Z MOČNO BAZO: HCl z NaOH ph V [ml] - NaOH 18 9
4 . Izračunaj in nariši tiracijsko krivuljo, če k 10,0mL 0,08M CH COOH dodajaš po 1mL (glej tabelo) 0,08M NaOH tako, da dodaš skupno 0,0mL raztopine NaOH! V(NaOH) [ml] ph V(NaOH) [ml] ph 0, ,85 1, ,86,1 1 1,0,8 1 1,1, ,0 5,7 16 1,7 6,9 17 1, 7 5, ,6 8 5,5 19 1,9 9 5,70 0 1, 10 8,67 Začetni ph: To je ph raztopine 0,16M CH COOH, ko ji baze še nismo pričeli dodajati. Ker CH COOH ne disociira popolno, je koncentracija vodikovih (oksonijevih) in acetatnih ionov enaka samo koncentraciji disociiranega dela kisline (α). Da bi lahko določili koncentracijo oksonijevih ionov, moramo poznati stopnjo disociacije te kisline (α) oz. konstanto kemijskega ravnotežja za ocetno kislino (Ka), ki nam podaja konstantno razmerje med produkti in reaktanti. Če je stopnja disociacije zelo majhna lahko pri pisanju izraza za konstanto kemijskega ravnotežja, disociiran del kisline zanemarimo. Ka 1, ,08/L + CHCOOH ( ) + HO aq ( l) CHCOO( aq) + HO( aq) α α α + + CHCOO HO HO Ka CH COOH α α << [ ] HO Ka HO Ka 1, 8 10 [ 0, 08] 1, 10 L ph H O + log log 1, 10,9 0
5 ph pred ekvivalentno točko: To je ph 10mL raztopine 0,08M CH COOH v katero dodamo največ tolikšno množino NaOH, da še ne doseže množine CH COOH (n A ). V našem primeru bomo dodali npr. 1mL NaOH s koncentracijo 0,08M (). Ves NaOH tako zreagira s CH COOH v CH COONa, množina kisline pa se zmanjša za množino dodanega NaOH (n B ). Množina kisline, ki ostane (n A1 ) delno disociira na oksonijeve in acetatne ione. Da bi izračunali koncentracijo oksonijevih ionov moramo zapisati izraz za konstanto kemijskega ravnotežja disociiranega dela kisline ki ostane, ob čemer pa ne smemo pozabiti da nastali CH COONa disociira tudi na acetatne ione, katerih koncentracija () vpliva na ravnotežje, zato jo ob pisanju izraza ne smemo izpustiti. Če je stopnja disociacije zelo majhna, lahko pri pisanju izraza, disociiran del kisline zanemarimo, prav tako lahko zanemarimo tudi acetatne ione, ki so nastali pri disociaciji kisline, če je koncentracija acetatnih ionov nastali pri disociaciji CH COONa toliko večja. V(CH COOH) 10mL 0,08/L V(NaOH) 1mL ( ) CH COOH + NaOH CH COONa + H O + CHCOOH 1 na na1 na nana1 + CHCOONa CHCOO + Na + CHCOOH( ) + H O aq ( l) CHCOO( aq) + H O( aq) 1 1 1α 1α+ 1 ( ) ( ) ( ) ( V V ( CH COOH ) + V ( NaOH ) n CH COOH n NaOH V CH COOH V NaOH R 0,08 0,01L0,08 0,001L 0, ,01L+ 0,001L L L L n NaOH V NaOH 0,08 L 0,001L 7,7 10 V V CH COOH + V NaOH 0,01L+ 0,001L R α ) [ ] ( α + ) + + CHCOO HO 1 HO Ka CH COOH α 1 1 α <<, 1 1 H O Ka ,8 10 0, Ka L 1,6 10 H O L 1 7,7 10 L ph H O + log log 1,6 10,79 1
6 ph v ekvivalentni točki: To je ph 10mL raztopine 0,08M CH COOH, ki ji dodamo tolikšno množino NaOH, da je enaka množini CH COOH. Ker imata v našem primeru kislina in baza enaki koncentraciji ( ), je tudi volumen dodanega NaOH enak volumnu kisline. HCl in NaOH torej popolnoma zreagirata v CH COONa in H O. Acetatni ioni, ki nastanejo pri disociaciji CH COONa pa delno reagirajo z vodo pri čemer nastaneta ocetna kislin in hidroksidni ioni. Slednji določajo ph raztopini, da pa bi določili njihovo koncentracijo moramo zapisati izraz za konstanto kemijskega ravnotežja konjugirane baze (acetatnih ionov) (Kb'). Del acetatnih ionov, ki zreagirajo, lahko pri pisanju izraza zanemarimo, če je del nezreagiranih acetatnih ionov toliko večji. Kw 10-1 CH COOH + NaOH CH COONa + H O K B / / + CHCOONa CHCOO + Na B' CHCOO( ) + H O aq ( l) CHCOOH( aq) + OH( aq) α α α ( ) V NaOH 0, 08 L 0,1L 0,0 L V CH COOH + V NaOH 0,1 L [ CH COOH ] + 1 OH HO Kw 10 Kb' Kb' 5, CH Ka COO HO 1, [ ] CHCOOH OH OH Kb' CH COO α α << OH OH Kb' 10 ' 5, ,08 Kb OH L, L ph + OH log 1 log,7 10 8,67
7 ph po ekvivalentni točki: To je ph 10mL raztopine 0,08M CH COOH v katero dodamo najmanj tolikšno množino NaOH (n B ), da še preseže množino CH COOH (n A ). V našem primeru bomo dodali npr. 11mL NaOH s koncentracijo 0,08M (). Tako vsa CH COOH zreagira z NaOH v CH COONa, množina baze pa se zmanjša za množino kisline. Množina baze, ki ostane (n A1 ) je enaka množini hidroksidnih ionov, ki ob disociaciji le te nastanejo. Vedeti pa moramo, da hidroksidni ioni nastanejo tudi ob reakciji CH COONa z vodo, zato moramo kljub vsemu zapisati izraz za konstanto kemijskega ravnotežja konjugirane baze (acetatnih ionov) (Kb') in izračunati koncentracijo hidroksidnih ionov, ki nastanejo ob reakciji acetatnih ionov z vodo ([OH - ] ). Če je le ta dovolj velika, jo moramo nato prišteti h koncentraciji hidroksidnih ionov nastalih ob disociaciji ostanka baze ([OH - ] 1 ). Del acetatnih ionov, ki zreagirajo z vodo, lahko pri pisanju izraza zanemarimo, če je del nezreagiranih acetatnih ionov toliko večji. ( ) CH COOH + NaOH CH COONa + H O + NaOH na na 1 na 1 + NaOH Na + OH + CHCOONa CHCOO + Na CHCOO( ) + H O aq ( l) CHCOOH( aq) + OH( aq) α α α+ 1 ( ) n( CHCOOH ) V ( NaOH ) V ( CH COOH ) V V ( CH COOH ) + V ( NaOH ) n NaOH OH R 0,08 0,011L0,08 0,01L 0,01L+ 0,011L V ( CHCOOH ) L L,81 10 L 0,08 L 0,01L 0,081 V CH COOH + V NaOH 0,01L + 0,011L L [ CH COOH ] OH ( α)( α + ) Kb' CHCOO 1 α α <<, 1 Kb OH Kb 10 1 ' 5, ,081 9 ' L OH 6,81 10 L,7 10 L
8 OH << OH Koncentracije ne prištevamo! 1 ph + OH + 1 log 1 log, ,58 1 Titracijska krivulja: Ko po zgornjih postopkih izračunamo ph za vse prostornine NaOH v tabeli, lahko narišemo tiracijsko krivuljo, titracije šibke kisline z močno bazo, ki nam prikazuje naraščanje ph vrednosti ob dodajanju NaOH. 1 TITRACIJSKA KRVULJA ŠIBKE KISLINE Z MOČNO BAZO: CH COOH z NaOH ph V [ml] - NaOH c) Izvedba vaje: 1. V 50mL odmerimo 0mL destilirane vode in ji izmerimo ph. Destilirani vodi nato dodamo 0,5g Na CO 10H O, ko se sol raztopi, ponovno izmerimo ph, ter razložimo zakaj pride do spremembe! Podobno izmerimo ph 0mL vodovodne vode. Nato ji dodamo 0,5g (NH ) SO, ponovno izmerimo ph, ter razložimo zakaj pride do spremembe!. V epruveto nalijemo ml M raztopine natrijevega acetata in dodamo kapljice raztopine fenolftaleina. Segrevamo do vrenja in pustimo, da se ponovno ohladi. Opišemo in razložimo opažene spremembe!
9 . V erlenmajerico odmerimo z merilnim valjem 100mL vodovodne vode in ji dodamo nekaj kapljic indikatorja metiloranža. Titriramo jo z 0,1M HCl do preskoka barve. Izračunamo, koliko mg O je v 100mL vode, če predpostavimo samo prisotnost O v vodovodni vodi. Titracijo ponovimo vsaj dvakrat! Trdoto vode podamo v nemških stopinjah ( N), 1 N predstavlja 1mg O v 100mL vode. d) Meritve z diskusijo: 1. ph destilirane in vodovodne vode:: Izmerjeni ph destilirane vode znaša 6,5. Do kislosti pride, ker se zračni ogljikov dioksid raztaplja v destilirani vodi pri čemer nastaja ogljikova(iv) kislina, ki kislo disociira. CO + H O H CO q ( g ) ( l) ( a ) HCO + HO HCO + HO + ( aq) ( l) ( aq) ( aq) Po dodatku natrijevega karbonata dekahidrata, ph naraste na 11,61 saj le ta disociira na karbonatne ione, ki z vodo bazično reagirajo. Na CO 10H O Na + CO + 10H O + CO + H O HCO + OH ( aq) ( aq) ( aq) Izmerjeni ph vodovodne vode znaša 7,0. Do bazičnosti pride, ker so v vodovodni vodi raztopljene karbonatne soli, katere z njo bazično reagirajo. Mg CO Mg + CO + + CO + H O HCO + OH aq aq aq ( ) Po dodatku amonijevega sulfata(vi), ph rahlo pade na 6,97 saj le ta disociira na amonijeve ione, ki z vodo kislo reagirajo in na sulfatne ione, ki z vodo bazično reagirajo, a ker je amonijevih ionov za več, se ph nekoliko zniža. ( NH ) SO NH SO + + NH + H O NH + H O + + ( aq) ( aq) ( aq) ( aq) SO + H O HSO + OH aq l aq aq 5
10 . Vplivi temperature na kemijsko ravnotežje: Ko k natrijevemu acetatu dodamo indikator fenolftalein, se le ta rahlo vijolično obarva, saj ob disociaciji soli nastali acetatni ioni, bazično reagira z vodo. CH COONa CH COO + Na + CH COO CH COOH + OH ( aq) ( aq) ( aq) Ob segrevanju raztopine se le ta močneje vijolično obarva, kar pomeni, da se ravnotežje pomakne v smer produktov (v desno) in zato nastane več hidroksidnih ionov. Če ogreto raztopino ponovno ohladimo, pa nastopi rahlo razbarvanje (ravnotežje se zopet premakne bolj v levo). CH COO CH COOH + OH SEGREVANJE ( aq) ( aq) ( aq) CH COO H COOH + OH OHLAJANJE ( aq) ( aq) ( aq). Trdota vode: Karbonatne soli, raztopljene v vodovodni vodi, ob titraciji le te s klorovodikovo kislino, reagirajo, pri čemer nastaneta kalcijev klorid in voda. O + HCl Cl + H O V(HCl) 1,6mL V(HCl),6mL V(HCl),5mL c(hcl) 0,1/L e) Izračun:. V(HCl),6mL V(H O) 100mL M(O) 56,06g/ ( ) 0,01 0,006,6 10 L n HCl c HCl V HCl L ( ) ( ),6 10 n HCl n HCl n O n O, 10 ( ) m O n O M O, 10 56, 06 1,9mg g 1 N 1mg O 100mL H O 1,9 N 6
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