Exercícios Resolvidos Envolvendo Volumetria de Precipitacao
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Volumetria de precipitação . #QuímicaAnalíticaQuantitativa...
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Uma moeda de prata pesando 0,5000g é dissolvida com acido nítrico e a prata é titulada com 38,22ml de uma solução de KSCN que contém 4,5000 g/L de solução. Calcular a porcentagem de prata na moed moeda a anal analis isad ada. a. Ag°
HNO3
Ag+(aq) + SCN-(aq)
AgSCN(aq)
KSCN KS CN ------- 4, 4,5 5 g/L 97 ----------1,0L -------1,0mol 4,5000----1,0L --------X X= 0,046 mol/L
Nº de de mmol mmolss de Ag+= Nº de mmol mmolss de SCN-(aq) = V(ml)x[SCN -] = 38,22 x 0,046mol/L Nº de de mmol mmolss de Ag+= 1,758 mmols 1,0 mmol de Ag+----108/1000 1,758 ------------------y y = 0,1869 g de prata
0,1869g-----K 0,5000g-----100% K=38% de prata
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Qual deve ser a molaridade de uma solução de AgNO 3 para que sendo usada na determinação do grau de pureza de amostras de NaCl, tornando-se o-se mass assas de amostras de 0,5000g, cada 1,00ml de solução corresponda a 1,0% de NaCl na amos mostra tra anal analiisada? ada? 1,0% -----X 100%----0,5 X=0,005g de NaCl
1 mm mmol ol de NaC NaCll ---------- 58, 58,45 45/10 /1000 00 Y------0,005g Y=0,0855 Y=0,08 55 mmols mmols de NaCl NaCl
Nº de mmols de AgNO3 = Nº de mols de NaCl V(mL V(mL)) x[AgNO3] = 0,0855 1,0mLx[AgNO3] = 0,0855mol/L [AgNO3] = 0,0855mol/L
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Amostra de 20 tabletes de sacarin rina solúvel foi tratada com 20,0mL de solução 0,08181mol/L de AgNO3. Após remoção do sólido, a titulação do filtrado necessitou de 2,81 mL de solução 0,04124mol/L de KSCN. Determine a massa em mg de sacarina em cada ada table ablette. Segu Segun ndo a Reação ação:: NaC6H6CONSO2(aq) + Ag+(aq) AgC6H6CONSO2(s) + Na+(aq) Nº de mmols mmols de Sacar Sacarina ina = Nº de mmol mmolss de Ag+(adc) – Nº de mmo mmols ls de Ag+ Ag+(exc) = 20 ml x 0, 0,08 08181 181mo mol/ l/LL - 0, 0,115 1159 9 =1,6362 – 0,1159 Nº de mmols mmols de Sacarin Sacarina= a= 1,503 1,503 mmols mmols de Sacarina Sacarina
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Uma solução de FeCl3 . 6H2O contem em cada mL, uma massa de ferro que produziria através de um processo químico adequado 0,300 mg de Fe2O3. Quantos mL de AgNO3 0,05 0,050 0 N ser serão neces ecessá sári rios os para titular 50,0 mL da solução de FeCl3 . 6H2O? C = 1,015 g/L de FeCl3 . 6H2O
N=
m = 1,015 N = 0,0112 N eq g . V(L) 90,06 . 1,0 L PF .
.
3
FeCl3 . 6H2O
Fe2O3
540,40g 159,70g X 0,300 g X= 1,015m 1,015mg g ou 1,015 1,015 . 10-3 = 1,015g/L
Nº de meq FeCl3 = Nº de meq de AgNO3
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O cloreto contido em uma amostra de salmoura foi determinado pelo Método de Volhard. Volhard. Uma alíquota de 10, 10,00 mL da salmoura foi tratada com 15, 15,00 mL de solução 0,1182 mol/L de AgNO3 padronizada. padronizada. O excesso de prata foi titulada com solução 0,101 mol/L de KSCN padronizada, requerendo 2,35 mL até a formação do complexo vermelho Fe(SCN)2+. Calcule a concentração de NaCl na salmoura, expressando o resultado em g/L. g/L . mmols ls de Ag Ag+ exc Nº de de mmol mmolss de NaC NaCll = Nº de de mmol mmolss de Ag Ag + adic – Nº de mmo = 15,00 x 0,1182 molL -1 - 0,2374 mmols = 1,773 – 0,2374 Nº de mmo mmols ls de NaCl NaCl = 1,5356 1,5356 mmol mmolss Nº de mmo mmols ls de Ag Ag+ exc = Nº de mmol mmolss de KSCN KSCN
= 2,35 2,35 mL x 0,101 0,101 mol mol L-1
1,0 mmol 1,0 mmolss Na NaCl Cl ---------- 58 58,4 ,45 5 / 1000 1000 1,5356mmol NaCl----W
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O método de Fajans usa um indicador fluorescente, que se adsorve eletrostaticamente sobre a superfície de um precipitado coloidal imed imedia iata tame ment ntee após após o pont pontoo este estequ quio iomé métr tric ico. o. Tais ais indi indica cado dorres poss possue uem m cor cores diferentes nos estados livre (em solução) e adsorvido. O indicador mais comumente usado é a eosina (tetrabromofluoresceína). O diretor de um laboratório de análises químicas fecha acordo com um cliente para determinação de brometo num grande número de amostras. Para evitar perda de tempo calculando o resultado final, o químico responsável pelas análises quer que o volume de AgNO 3 usado na titulação de Fajans seja igual, numericamente, à percentagem de bromo na amostra. Se de cada amostra serão pesados 500,0 mg, qual a concentração molar de AgNO 3 que deve deverá rá ser ser usad usada? a? Br –
V AgNO3 = % 1,0 mL = 1% de Br-
Nº de mmol mmolss de Br Br - = Nº Nº de mm mmol olss de Ag + 0,062257 mmol = 1,0 ml x [ AgNO3]
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Se 25,0 mL de uma solução de NaCl foi requerido para precipi ipitar tar quantitativamente a prata de uma solução obtida pela solubilização de 0,2365 g de prata metálica com pureza de 98%. Qual a concentração da solução de NaCl em g/mL e normalidade. Pureza Pure za do Ag ----- 0,2365 g----100% g----100%
NaCl + Ag°
AgCl (s) + Na+
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Uma solução de AgNO3 0,1000 mol L-1 foi padronizada com o sal KCl de 99,60% de pureza. Uma amostra de 0,3254g do sal é titulada de acordo com o método Mohr gastando 44,20mL da solução de AgNO3. Calcule o fator de correção da solução de AgNO3. M.M =KCl 75,436
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Uma amostra impura de KI pesando 2,145g foi dissolvida em agua, acidificada e tratada com 50,0 mL de AgNO3 0,0243mol/L. Após a precipitação do AgI, a mistura resultante foi titulada com KCSN 0,121 mol/L , observa rvando-se um consumo de 3,32 mL para titular o excesso de AgNO3 . Calcular a porcentagem de KI na amostra.
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No método de Liebig, o cianeto é determinado por titulação com AgNO3 padr padrão ão,, quan quando do as segu seguin inte tess reaçõ eações es acon aconte tece cem: m: 2CN- + Ag+ [Ag(CN)2][Ag(CN)2]- + Ag+ Ag[Ag(CN)2] A segunda reação indica o ponto final da titulação. Uma amostra de NaCN, pesando 0,4029 g, é dissolvida em água e titulada com 40 25 mL de AgNO 0 1012 mol L-1 Calcule o grau
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Em uma titulação de 50,0 mL de NaCl 0,1000 mol L -1 com AgNO3 0,1000 mol L-1, determinar as concen centrações dos íons clor loreto e íons prata na solução, em cada uma das seguintes situações: Kps AgCl= 1,56 ,56x10 -10 Após a adição adição de 20,0 mL de AgNO AgNO 3; [Cl-] = (VCl- x CCl-) – (VAg+ x CAg+) = (0,05 x 0,1000) – (0,02 x 0,1000) = 4,28 x 10 (VCl- +VAg+) 0,05 +0,02
-2
mol /L
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Uma amostra de NaCl e NaI pesando 0,4000 g produziu através de um tratamento químico adequado um precipitado de AgCl e AgI
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Uma amostra impura de KBr pesando 0,4100 g foi dissolvida em
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25,00 mL de água potável potável foram diluídos para para 250,0 mL. Uma Uma
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É sabido que o método de Volhard pode ser empregado para a determinação de haletos. Considere uma amostra cuja massa em
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0,2142 g do sal MCl2 são dissolvidos em água destilada e titulados
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Um químico deseja determinar o grau de pureza de um lote de ferrocianeto de potássio, cujo rótulo rótulo diz estar trihidratado {K4[Fe(CN)6] . 3H2O}. 3H2O}. Ele pesa, então, 1,6801 g do sal como amostra, transfere para um erlenmeyer, erlenmeyer, dissolve em água e titula com uma solução de Ce(IV) em H2SO4, gastando 39,63 mL. Esta mesma solução solução gastou na padronização padronização 49,06 mL para titular 50,00 mL de uma solução contendo 4,848 g/L de As2O3. Qual é o grau de pureza do ferrocianeto de potássio? As2O3 + 6OH- ↔ 2AsO33- + 3H2O AsO33- + 3H+ ↔ H3AsO3 H3AsO3 + 2Ce4+ + H2O ↔ HAsO42- + 2Ce3+ + 4H+ As2O3 -----------2 -2 H3AsO3 Ce4+ + Fe2+ ↔ Ce3+ + Fe3+ 197,8g -------251,84g -------251,84g Nº de de mmol mmolss de Ce4+ = (nº (nº de mmols mmols deH3AsO3 )x 2
49,06[Ce4+] = (50ml (50ml x 0,0489)x2 [Ce4+] =(50 x 0,0489)x2 = 0,0996mol/L
4,848---------W W= 6,170 g
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Uma amostra de minério de ferro, pesando 750,0 mg, foi dissolvida em ácido e tratada para oxidar todo o ferro ao íon férrico. Após eliminar todo o excesso de agente oxidante, um excesso de KI foi adicionado. O I2 liberado requereu 18,50 mL de Na2S2O3 0,0750 mol L-1 para titulação. Qual é a percentagem de ferro na amos amostr tra? a? Reaç Reaçõe õess quím químic icas as:: 2Fe 2Fe3+ + 2I- ↔ 2Fe2+ + I2 I2 + 2S2O32- ↔ 2I- + S4O622Fe3+ + 2S2O32- ↔ 2Fe2+ + S4O62Nº de mmol mmolss de Fe3+ =
nº nº de de mmo mmo,s ,s de S2O32= 18,50ml x 0,0750 mol/L Nº de mmol mmolss de Fe3+ =1,3875 mmols 750mg ------ 0,75g ------100% 0,0774-----Y
1,0 1,0 mmol mmol de Fe3+ ------0,05585 g 1,3875mmols-------------X X= 0,07749g
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A prata contida em 24,00 mL de uma solução 0,1000 mol L-1 é precipitada completamente como Ag3AsO4. O precipitado isolado é dissolvido em ácido forte e o arseniato arseniato determinad determinadoo iodometrica iodometricament mentee com Na2S2O3 Na2S2O3 requere requerendo ndo 20,00 mL desta solução. Qual é a concentração da solução solução de tiossulfato em mol L-1? Resposta: 0,0800 mol L-1. As4O6(s) + 6H2O 4H3AsO3 - Ag3AsO 3Ag+ + AsO33 3 ↔
↔
AsO33- + I3- + H2O ↔ AsO43- + 3I- + 2H+ IO3- + 8I- + 6H+ ↔ 3I3- + 3H2O 2S2O32- + I3- ↔ S4O62- + 3I-
AsO33- + IO3- + 2I- + 2S2O32- + 4H+ ↔ AsO43- + I3- + S4O62- +2H2O Nº de de mmol mmolss de Ag+ = nº nº de mmols mmols de AsO43-
3
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Uma amostra de minério de ferro, pesando 750,0 mg, foi dissolvida em ácido e tratada para oxidar todo o ferro ao íon férrico. Após eliminar todo o excesso de agente oxidante, um excesso de KI foi adicionado. O I 2 liberado liberado requereu requereu 18,50 mL mL de Na2S2O3 0,0750 mol L-1 para titulação. Qual Qual é a percentagem de ferro na amostra? Reações químicas: 2Fe3+ + 2I- ↔ 2Fe2+ + I2
I2 + 2S2O32- ↔ 2I- + S4O62-
m= 750mg= 0,75g
f= 0,0750x10= 0,75
Fe3+ % = 100x0,75 x 0,005585 x 18,5 0,75
Fe 3+ % = 1,7492 0,75
Fe3+ % = 10,33
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Com a finalidade de determinar o teor de cálcio numa solução, tomou-se uma alíquota de 25,02 mL de amostra à qual se adicionou solução contendo 3,5000 g de oxalato de sód sódio dess esseca ecado (pad (padrã rãoo prim primár ário io). ). Apó Após a filt filtra raçã çãoo do prec precip ipit itaado, do, o exce excess ssoo de oxa xala latto foi determinado por permanganometria e foram gastos 23,40 mL de solução de permanganato 0,2050 mol L -1. Quais as reações envolvidas? Qual a concentração de cálcio na amostra? Que massa de óxido de cálcio seria obtida caso se calcinasse o prec precip ipit itad adoo a 90 900° 0°C? C? 1,0mmol Na2C2O42------0,1340g Ca2+ + C2O42- ↔CaC2O4(s) + C2O42- (exc) X-------X------------------------------------------- 3,500g 3,500g C2O42- + MnO4- ↔CO2 + Mn2+ X= 26,12 mmols C2O42- ↔ 2CO2 + 2e (x5) MnO4- + 8H+ + 5e ↔Mn2+ + 4H2O (x2) 5 C2O42- + 2MnO4- + 16H+ ↔ 10CO2 + 2Mn2+ + 8H2O Nº de mmols C2O42-que reagiu de mmols mmols de MnO MnO42 Nº de mmol mmolss de C2O42- = 5 nº de 26,12 – 11,99 = 14,13 mmols = 5 x (23,40x0,2050 (23,40x0,2050)) 2 2-
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Suponha que titulamos 100,0 mL de uma solução de Fe 2+ 0,0500 mol L-1 com uma solução de Ce4+ 0,100 mol L-1. O ponto de equiv quivaalênci ênciaa oco ocorre rre quand uandoo VCe VCe4+ = 50,0 mL, porque o Ce4+ tem o dobro da concentração do Fe2+. Calcule o potencial da pilha em 36,0; 50,0 e 63,0 mL.
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Dez comprimidos de aspirina (AAS, HOOCC 6H4OCOCH3), pesando juntos 6,2550 g, são pulverizados. Uma amostra de 0,1251 g do pó resultante é brom bromin inad adaa por trata ratame men nto com 20 20,0 ,000 mL de KBrO BrO 3 0,0400 mol L-1, contendo 75 g/L de KBr. Após a brominação se completar, a solução é tratada com um excesso de KI e o I2 liberado é titulado com 14,12 mL de Na 2S2O3 0,1039 mol L-1. Quan Quanto toss grama ramass (em médi édia) de aspi aspirrina ina exis existe tem m em cada cada compr omprim imiido? do? BrO3- + 5Br - + 6H+ ↔ 3Br2 + 3H2O HOOCC6H4OCOCH3 + 3Br2 ↔ HOOCC6HBr3OCOCH3 + 3H+ + 3Br -
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