Laboratorio de Análisis Instrumental en ingeniería I ingeniería química DETERMINACIÓN DE AGUA I OBJETIVOS 1.1 Operar adecuadamente la balanza analítica eléctrica y la electrónica 1.2 Aplicación del método gravimétrico por volatilización 1.3 Determinar el contenido de humedad y porcentaje de sólidos en un alimento II PRINCIPIO DEL METODO El método de análisis toma el nombre de método gravimétrico indirecto por volatilización. Consiste en someter una porción pesada de muestra a calentamiento a una temperatura adecuada y a un período de tiempo determinado. El residuo sólido se pesa y el agua se determina en función de la variación de peso que experimenta la sustancia en estudio. Este método es válido sólo cuando en el caso de que la pérdida de peso sea debido exclusivamente a la eliminación de agua. III GENERALIDADES Los métodos gravimétricos son métodos de análisis basados en pesar un analito al estado puro o de compuesto químico de composición química conocida, estos métodos miden la masa de la sustancia. Se clasifican en métodos por. a) Precipitación b) Por desprendimiento o volatilización o desprendimiento c) Electrogravimétricos Los métodos por volatilización consisten en desprender de la sustancia en estudio el analito como un producto gaseoso por tratamiento generalmente por calentamiento a una temperatura adecuada. La determinación de agua en las sustancias sólidas es uno de los casos más importantes de análisis, gravimétrico por volatilización basado en la pérdida de peso de una sustancia después de haber sido sometida a tratamiento debido a que un componente común en los cuerpos sólidos lo constituye el agua. El agua en los cuerpos sólidos se encuentra bajo dos formas: a) AGUA ESENCIAL El agua o los elementos que lo integran existen como parte de la composición estequiométrica de los compuestos. Pueden ser: AGUA DE CRISTALIZACIÓN O HIDRATACIÓN Es aquella que forma parte integrante de la estructura de algunas sustancias denominadas hidratos cristalinos. Durante el calentamiento, estos hidratos se descomponen, desprendiendo agua. El agua se encuentra bajo la forma de “n H2O “.
AGUA DE CONSTITUCIÓN En este tipo de agua los grupos componentes del agua, el hidrógeno y el oxígeno, o el grupo hidroxilo unidos dentro de los compuestos por fuerzas de valencia primaria, se pueden convertir en agua por descomposición descompos ición térmica del producto. 2 Na H C O3 Na2 C O3 + H2O + CO2
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B) AGUA NO ESENCIAL
Es el agua retenida en las sustancias sólidas mediante fuerzas no químicas. Pueden ser: AGUA HIGROSCÓPICAS Es el agua retenida por los sólidos y que son absorbidos por su superficie a partir del aire, o cuando se añade agua a materiales en procesos industriales. Su contenido no es constante, usualmente se determina por calentamiento a unos 100 o C. Es llamada simplemente humedad AGUA OCLUIDA O INCLUIDA Es un componente habitual de los compuestos que se forman por cristalización en disolución acuosa. Forma parte de minerales y productos naturales. Para eliminar este tipo de agua se requiere de elevadas temperaturas. AGUA DE INHIBICIÓN Es el agua retenida por las sustancias que presentan fenómenos gelatinosos o coloides liófilos orgánicos. Este tipo de agua se encuentra en el agar agar, óxidos de silicio y aluminio. MÉTODOS DE DETERMINACIÓN Para analizar agua por los métodos gravimétricos se tiene dos formas: A) Métodos Directos El agua eliminada de la sustancia es recogida y determinada por un aumento de peso que esta va ha provocar en sustancias especiales que actúan como absorbentes de agua. Por el aumento de peso se determina el agua. B) Métodos Indirectos El agua se determina por la pérdida de peso que sufre la muestra cuando se somete a calentamiento. La pérdida de peso se debe solo a la pérdida de agua que experimenta la muestra. Este método para que pueda ser aplicado se requiere que el desprendimiento de agua debe ser completo y que la eliminación del producto volátil corresponda solamente al agua. IV APARATOS 4.1 Balanza analítica con sensibilidad de 0.1 mg 4.2 Crisoles de porcelana 4.3 Pinza para crisoles 4.4 Estufa 4.5 Desecador con agente deshidratante. V REACTIVOS VI PROCEDIMIENTO 6.1 Análisis de la Muestra
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Retirar el crisol de la estufa y colocarlo en un Desecador para que enfríe. Pese el conjunto y nuevamente traslade el crisol a la estufa por espacio de 30 minutos, enfriar y pesar. Repita esta operación hasta conseguir peso constante. Con los datos conseguidos calcule el porcentaje de humedad en la muestra. VII CALCULOS Para calcular el porcentaje de agua se aplica la siguiente relación: %H = G1 - G2 X 100 G1 Dónde: % H = Porcentaje de humedad G1 = Masa en gramos de la muestra original G2 = Masa en gramos de la muestra seca. El porcentaje de sólidos se calcula en base a la siguiente relación: Los resultados obtenidos se expresan en porcentaje de agua y con dos cifras decimales: CUESTIONARIO 1.- ¿En qué consiste la determinación de agua por el método gravimétrico por volatilización indirecto? 2.- En que consiste y ponga ejemplos de agua esencial y no esencial 3.- Que requisitos debe cumplir el método para ser utilizado como una técnica de análisis 4.-Conceptualize los términos de absorción y adsorción 5.-Se tiene una muestra de BaCl2.2H2O que presenta 244 g/mol, si el peso del crisol con la sustancia húmeda pesa 9.6787 g y el peso del crisol con la sustancia seca presenta una masa de 9.3747 g. Calcular el porcentaje de agua teórico y experimental, así como el porcentaje de error que presenta 6.- 250ml de una disolución de ácido clorhídrico se trata con un exceso de nitrato de plata. El precipitado obtenido pesa 0.3931 g ¿Que peso de ácido estaba presente en los 250 ml de la disolución? 7.- Calcular que peso de azufre se puede obtener a partir de 0.2555g de FeS2, así como su porcentaje como azufre INFORME DE LABORATORIO # 1 DETERMINACIÓN DE AGUA Y SOLIDOS ANÁLISIS: determinación de humedad y determinación de porcentaje de solidos MÉTODO: GRAVIMETRIA
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CALCULOS a) Datos: Peso de la muestra Peso crisol + muestra Peso de crisol + desecado Peso de crisol solo Peso del desecado
= 1.0059 =c+1.0059 = c+0.9770 =c =0.9770
b) Solución:
+ − + x100% p h % H2O = p x100% . % H2O = . x100 =2.8730% % H2O =
% sustancia de solidos %solidos = (100-H2O)% %solidos = 97.127%
Peso de muestra seca= Peso de crisol crisol + desecado - Peso de crisol solo =0.9770 TABLA DE RESULTADOS No Muestra % H2O 1 2 3 4 5 6 7
2.85 2.97 3.37 3.18 2.91 3.02 2.87
%SOLIDOS TOTALES SECA) 97.15 97.03 96.63 96.82 97.09 96.98 97.13
PESO DE MUESTRA (BASE SECA
DISCUSIÓN DE RESULTADOS: Media ∑ % Promedio del % H2O = # Mediana Valor que se encuentra en la mitad de la secuencia =2.97
=3.0243
0.9866 0.9824 0.9902 0.9842 0.9714 0.9908 0.9770
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2,97 3,37 3,18 2,91 3,02 2,87
3,0243 3,0243 3,0243 3,0243 3,0243 3,0243
-0,0543 0,3457 0,1557 -0,1143 -0,0043 -0,1543
/ X100 ERRO ERRORR RELAT RELATIVIVOO = /VALOR MEDIDOMEDIA MEDIA
GRUPO1 =5.76 GRUPO2=1.79 GRUPO3=11.43 GRUPO4=5.14 GRUPO5=3.779 GRUPO6=0.1421 GRUPO7=5.1020 OBSERVACIONES
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FIRMA DE ALUMNO: ............................................... CUESTIONARIO 1. En este método se miden los componentes de la muestra que son o pueden ser volátiles. El método será directo si evaporamos el analito y los hacemos pasar a través de una sustancia absorbente que ha sido previamente pesada asi la ganancia de peso corresponderá al analito buscado; el método será indirecto si volatilizamos el analito y pesamos el residuo posterior a la volatilización, asi pues, la perdida de peso sufrida corresponde al analito que ha sido volatilizado el método por volatilización solamente puede utilizarse si el analito es la única sustancia volátil o si el absorbente es selectivo para el a nalito.
2. El agua no esencial es aquella retenida en una molecula por medios físicos, el agua incluida o ocluida es agua no esencial que es retenida al formarse la molecula
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3. Los métodos de análisis han de elegirse teniendo en cuenta su rendimiento en relación con los requisitos convenidos, lo mas importantes de los cuales, desde el punto de vista técnico son los siguientes:
EXACTITUD
PRECISION
LIMITE DE DETECCION
SENSIBLIDAD
ESPECIFIDAD
Falta de error sistematico; grado en que los resultados medios de varias determinación se aproximan al valor verdadero, en particular cuando se aproximaran a la concentración en la que podría adoptarse una medida o, en el caso de una encuesta, por encima del intervalo de las concentraciones previstas Grado de concidencia entre repetidas determinaciones independientes realizadas con el mismo material de ensayo, en particular en concentraciones próximas a aquellas en las que podría adoptarse una medida o, en el caso de una encuesta, por encima del intervalo de las concentraciones previstas Especificado en función del nivel mas bajo en el que pueda cuantificarse la concentración con un grado dado de confianza Cambio en la respuesta por unidad de concentración habitualmente especificada en concentraciones próximas a aquellas en las que podría adoptarse una medida Grado en que otras sustancias puden dar origen a una señar perturbadora
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244g/mol de BaCl2.2H2O y 36g de H2O
244g ----100% 36g ----- x% X = 14.72% teorico Hallando la cantidad de masa seca Cantidad de masa seca = peso de crisol húmedo – peso peso de crisol seco Cantidad de masa seca = 9.6787 – 9.3747 9.3747 Cantidad de masa seca = 0.304g % H2O = 0.304 *100/1 = 30.4% Error = [( Vt-Ve)/ Vt]* 100 Error = [(14.72-30.4)/14.72]*100 [(14.72-30.4)/14.72] *100 = 1.065% 6. HCl + AgNO3 -- HNO3 + ClAg 36.5 x
143.4 0.3931
x = 0.10005g de ClAg *(1mol ClAg/143.4g ClAg)*(1mol HNO3/1mol ClAg)*(63gHNO3/1molHNO3) x = 0.04415 g HNO3 7. 2 Fe + S2 2 FeS2 64 2(119.8) x 0.2555 x = 0.0682g de S2 0.2555 ---- 100% 0.0682 ------- x
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Laboratorio de Análisis Instrumental en ingeniería I ingeniería química DETERMINACIÓN DE CENIZAS I OBJETIVOS 1.1 Aplicar los métodos de destrucción de la materia orgánica por calcinación seca 1.2 Analizar el contenido de cenizas en una muestra de leche en polvo. II PRINCIPIO DEL MÉTODO se basa en la incineración de una porción exactamente pesada de muestra, se destruye la materia orgánica, operando a una temperatura de 500 a 600 o C, quedando como residuo la materia inorgánica o cenizas cuya cantidad exacta se determina por diferencia de masa. Esta porción representa el contenido mineral de la muestra. III GENERALIDADES Las muestras orgánicas, biológicas, alimentos, etc. emplean los métodos de calcinación seca para hacer la determinación de cenizas o de contenido mineral que presenta la muestra. Las cenizas de los productos alimentarios están constituidas por el residuo inorgánico que queda después de que la materia orgánica se ha quemado. Las cenizas obtenidas no tienen necesariamente la misma composición que la materia mineral presente en el alimento original, ya que puede haber habido pérdidas por volatilización o alguna interacción entre los constituyentes. Estos métodos consisten en someter la muestra trabajando a una temperatura de 400 a 700 grados centígrados en una mufla. En estos métodos, el oxígeno atmosférico sirve como oxidante, es decir, la materia orgánica se quema dejando un residuo inorgánico. En el caso de los líquidos y de los tejidos húmedos se secan en un baño de vapor o por calentamiento suave antes de colocarlos en la mufla. El calor de la mufla debe aplicarse gradualmente hasta alcanzar la temperatura final para evitar la combustión rápida y formación de espuma. Una vez terminada la calcinación, por lo general el residuo se solubiliza en el recipiente añadiendo 1 o 2 ml. de ácido clorhídrico 6 M, en caliente y luego se filtra si es necesario, y se transfiere a un vaso de precipitado para continuar el análisis. Se determina cenizas para obtener un índice que se utiliza junto con otros para caracterizar
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909 mg de calcio frente a los 118 mg que se obtienen por la misma cantidad de leche entera. Componentes del polvo de leche entera o polvo de leche descremada contiene agua 3.5 %, grasa 26.10 %, proteínas 27 %, carbohidratos 36.10 %, cenizas 6.90 %. IV APARATOS 4.1 Balanza analítica electrónica con sensibilidad de 0.1 mg. 4.2 crisoles de porcelana o de platino 4.3 Horno - mufla eléctrico, con termorregulador 4.4 Desecadores a base de silicagel, cloruro de calcio u otro deshidratante. V REACTIVOS VI PROCEDIMIENTO a) Medición de la muestra Se limpia un crisol, marcarlo para identificarlo y pesarlo. Se pesa entre 2 a 5 gramos de muestra. b) Calcinación Con la pinza se coloca el crisol que contiene la porción de muestra en el horno mufla. Se regula el horno - mufla para que alcance una temperatura de 550oC. Se quema la porción de muestra con el horno parcialmente cerrado hasta que la combustión sea completa. Se cierra el horno - mufla y se incinera la porción de muestra hasta la obtención de cenizas. El tiempo de incineración debe ser de dos horas mínimo, contado desde el momento que se alcanza los 550oC. c) Pesada del calcinado Transcurrido el tiempo señalado, se extrae el crisol de la mufla utilizando la pinza y se pone a enfriar en el desecador. Una vez enfriada hasta temperatura ambiente, se pesa lo más rápido posible para evitar la absorción de humedad ambiental. Anote el peso obtenido. VII CALCULOS DATOS peso de muestra = ---------------- Peso crisol + cenizas = --------------- Peso crisol solo = --------------- Peso de cenizas = ----------------
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2. 3. 4. 5.
Defina el método por calcinación seca Qué pasaría si no se controla la temperatura de calcinación a 550 o C Qué le indica el contenido de cenizas en una muestra ¿Los contenidos de cenizas en muestras de leche en polvo es igual a la leche fresca?.
INFORME DE LABORATORIO # 2 DETERMINACIÓN DE CENIZAS ANÁLISIS: determinación de humedad y determinación de porcentaje de solidos MÉTODO: GRAVIMETRIA MUESTRA: leche en polvo CONTENIDO TEORICO Datos:
Peso de la muestra =4.0237 Peso del crisol + calcinado = c+0.2039 Peso del crisol solo =c Peso del calcinado o ceniza =0.2039
Solución: CALCULOS % cenizas = (Peso del crisol con cenizas - Peso del crisol solo ) X 100% Peso de muestra % cenizas = cenizas X 100% Peso de muestra % cenizas = 0.2039 X 100% 4.0237
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3 4 5 6 7
4,026 4,0105 4,0162 4,013 4,0237
0,202 0,4033 0,2016 0,1701 0,2039
5,01738698 10,0561027 5,01967034 4,23872415 5,06747521
DISCUSIÓN DE RESULTADOS: Media ∑ %z Promedio del % cenizas = # Mediana Valor que se encuentra en la mitad de la secuencia =5.0196 Desviación absoluta, error absoluto y relativo valor O media M O - M = /D/ D 5 5,63 -0,63 0,63 5,0348 5,63 -0,5952 0,5952 5,0173 5,63 -0,6127 0,6127 10,056 5,63 4,426 4,426 5,0196 5,63 -0,6104 0,6104 4,2387 5,63 -1,3913 1,3913 5,0674 5,63 -0,5626 0,5626
=5.63%
El error medido, en estos casos se conserva el signo del error para indicar si éste es por exceso o por defecto, el signo negativo indica que el resultado experimental es menor que el valor aceptado
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Rechazo de datos
5,15>4*0,2192 La desviación que presenta el valor dudoso es 5.15 y este valor es cuatro veces 2 de la desviación media de los demás valores. Por tanto, se debe rechazar el valor de 10.056.
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Incinerar el papel en el mismo crisol usado antes, enfriar y pesar, anotar como masa de cenizas Insolubles en agua
2. Defina el método por calcinación seca Se refiere a la determinación determinación de las cenizas en una mufla a temperaturas que oscilan entre 500 a 700 ºC el agua y sustancias volátiles son evaporadas, mientras que las sustancias organizas son incineradas en presencia del oxígeno del aire para producir co2 y óxido de nitrógeno, Las cenizas obtenidas no tienen necesariamente n ecesariamente la misma composición que la materia mineral presente en el alimento original, ya que puede haber habido pérdidas por volatilización o alguna interacción entre los constituyentes.
La calcinación tiene como objeto la remoción de agua, co2 y otros gases que se encuentran químicamente unidos en forma de hidratos o carbonatos El proceso de calcinación: durante el proceso de calcinación sale el agua libre, el agua que esta débilmente unida y el agua química o fuertemente unida
En este proceso cambia la naturaleza química del mineral donde: Las sustancias orgánicas se descomponen hasta llegar a co2 y el h y o del aire forman agua. El oxígeno desprende o actúa con otros elementos Los elementos simples de la materia orgánica van sufriendo un proceso de transformación química DISTRIBUCION DE TEMPERATURAS TEMPERATURAS EN EL FORNO DE CALCINACION CALCINACION
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son calentadas a 500-550°C en crisoles de sílica o de platino
3. Qué pasaría si no se controla la temperatura de calcinación a 550 o C La técnica no es adecuada para elementos volátiles (por ejemplo Se, Hg) los cuales se pierden durante la calcinación. Generalmente, la técnica puede utilizarse para el análisis de metales tales como Na, K, Mg, Ca, Fe, Cu y Zn aunque se ha informado ocasionalmente pérdidas de estos elementos. Por lo tanto, es aconsejable verificar los resultados para cada matriz de la muestra analizando un material de referencia o comparando con una técnica establecida de calcinación por vía húmeda.
4. Qué le indica el contenido de cenizas en una muestra La muestra orgánica que queda en cenizas representa la naturaleza química del mineral donde: Las sustancias orgánicas se descomponen hasta llegar a CO2 y el H y O del aire forman agua como también solo se desprende o como también actúa con otros componentes u elementos.
5. ¿Los contenidos de cenizas en muestras de leche en polvo es igual a la leche fresca? La evaluación realizada a tiempo cero, para todas las leches, y después de 90 días ” Student a los promedios de análisis químicos y test de Duncan” de
almacenamiento a temperatura ambiente, para leche UHT y en polvo, no mostró cambios en los macronutrientes para ningún tipo de leche siendo similares al
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1.3 Aplicar los métodos de análisis gravimétrico por precipitación 1.4 Analizar el contenido de sílice impura gravimétricamente en una muestra de cemento II PRINCIPIO DEL MÉTODO se basa en tratar la muestra de cemento con HCl evaporando hasta sequedad para separar la sílice como SiO2XH2O. El residuo se extrae con HCl diluido y se diluye con agua destilada calentando para disolver las sales anhidras. El residuo insoluble se filtra utilizando papel de filtro cuantitativo, se lava, calcina y se pesa como sílice impura. III GENERALIDADES El silicio, segundo elemento en abundancia en la corteza terrestre, no se encuentra jamás libre en la naturaleza. La sílice (dióxido de silicio) se presenta en diferentes grados de pureza, esta sílice o los silicatos se hallan en cantidad considerable en todas las rocas importantes, excepto en los carbonatos. Existen silicatos solubles a la acción de los ácidos, como el HCl, que forma ácido silícico (sílice hidratada), para separar la sílice, se evapora a sequedad la disolución clorhídrica que contiene la sílice hidratada coloidal y el residuo se calienta a 110-120 o C, con lo que se obtiene un producto representado por la fórmula SiO2.XH2O, que se separa por filtración. El cemento Portland resulta de calcinar en hornos rotatorios a temperaturas de 1400-1500 oC, dos tipos de materiales, uno rico en calcio como la roca caliza y otro en sílice, como la arcilla. En ocasiones se emplea arena, bauxita y pequeñas cantidades de hierro para ajustar la composición de la mezcla.
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(a) Disolución de la muestra Se pesan de 0.2000 gramos de la muestra cemento y se coloca en un vaso de 250 ml. Agregar 20 ml de HCl 1:1, agitar para homogenizar y evapore el ácido lentamente hasta sequedad, añadir 1.5 ml de HCl concentrado y 25 ml de agua destilada; calentar suavemente. (b) Filtración Filtrar por decantación la solución y lavar el residuo con agua destilada caliente, dejar escurrir el filtrado (la solución filtrada reservarla para la siguiente practica). Pesar un crisol limpio y seco. Retire el papel de filtro del embudo, doble adecuadamente el papel y su contenido y colocarlo en el crisol (c) Calcinación Colocar el crisol y su contenido en el horno mufla y calcinar a 1000 oC, por espacio de 3 a 4 horas. Retirar el crisol más el calcinado, enfriar y pesar.
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(a) % de sílice impura (b) % de sílice en la sílice impura 6.-0.2540 g de una muestra que contiene bromuro de potasio y bromuro de sodio fueron tratados con nitrato de plata hasta que todo el bromuro como 0.6320 g de bromuro de plata. ¿Cuál es la fracción de bromuro de sodio en peso hay en la muestra original? 7.- Se formula la siguiente reacción: BaCl2 (ac) + Na2SO4 (ac) ------- NaCl(ac) + BaSO4 (s) Calcular: (a) ¿Cuántos gramos de sulfato de sodio reaccionaran con 6 gramos de cloruro de d e bario (b) Si precipita 4 gramos de sulfato de bario ¿Cuántos gramos de cloruro de bario se han
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TABLA DE RESULTADOS
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a) ERROR ABSOLUTO DE LA MUESTRA. E= VALOR MEDIDO - VALOR VERDADERO VERDAD ERO E= 0.1144 – 0.1000 E=0.0144
Es un error por exceso b) PORCENTAJE DE ERROR RELATIVO
− % .−.% %E= . %E=
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0.1120 0.1053 0.1064 0.1132 0.1118 0.1141 ∑= 0.6625
0.1104 0.1104 0.1104 0.1104 0.1104 0.1104
a) ∑IOI – VALOR VALOR SOSPECHOSO = 0.6625 MEDIA:
M=
.
0.0160 0.0510 0.0400 0.0280 0.0140 0.0370 ∑=0.186
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SiF K~ + 3H.O = SiO, H, + 2FK + 4FH 4FH + 4NaOH. 4FNa + 4H~O
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(a) % de sílice impura (b) % de sílice en la sílice impura SOLUCION:
Cemento
⇄
SiO₂
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4g BaSO4 x (1molBaSO4/233.37gBaSO4) x (1molBaCl2/1molBSO4) x ( 208.2g BCl2/1molBCl2) = 3.56g BCl2
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Se realiza la precipitación partiendo de una disolución de ion magnesio que contiene HCl concentrado en unos 150 ml de solución. Se añade exceso de fosfato diamónico y luego hidróxido de amonio con agitación intensa. La mezcla dejar en reposo al menos cuatro horas en frio (mejor toda la noche) antes de la la filtración.
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Se filtra el líquido límpido sobrenadante por papel de filtro de poro fino. Se lava en el vaso el precipitado con pequeños volúmenes de agua fría, se pasa a través del papel de filtro la solución y se ensaya en los filtrados oxalatos y cloruros. Cuando la solución da resultados negativos de oxalatos y cloruros pasar cuantitativamente el precipitado al papel de filtro, separar este precipitado que
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TABLA DE RESULTADOS No muestra
ppm CaCO3 agua de tomilla 44.62 43.68 49.84 46.32 46.18 50.2 54.72
1 2 3 4 5 6 7
Valor teorico de la alcalinidad del agua 25-75 Tomaremos como dato el promedio que es :50 Hallando el error absoluto Error Absoluto = VM-VV Vm= valor medido Vv= valor teorico =50ppm 49,84
50
- 0,16
46,32
50
- 3,68
46,18
50
- 3,82
50,2
50
0,2
54,72
50
4,72
HALLANDO PROMEDIO promedio de ppm
=
47,9371429
HALLANDO HALLANDO LA DESVIACION DESVIACIO N VM
Medi a A
de svi acii on dudoso
44,6 44 ,62 2
47,9 47 ,93 371 714 429 -3, -3,31 3171 714 428 286 6
43,6 43 ,68 8
47,9 47 ,93 371 714 429 -4, -4,25 2571 714 428 286 6
49,8 49 ,84 4
47,9 47 ,93 371 714 429
46,3 46 ,32 2
47,9 47 ,93 371 714 429 -1, -1,61 6171 714 428 286 6
46,1 46 ,18 8
47,9 47 ,93 371 714 429 -1, -1,75 7571 714 428 286 6
1,90 1, 902 285 857 714
ppm CaCO3 agua de acequia 49.08 48.16 48.73 50.95 49.48 51.32 59.29
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DISCUSIÓN DE RESULTADOS: 43,68
46,8 46 ,806 0666 6667 67
3,12 3, 1266 6666 6666 667 7
49,8 ,84 4
46,8 ,80 066667
3,0 ,03 33333333
46,3 ,32 2
46,8 ,80 066667
0,4 ,48 86666667
46,1 ,18 8
46,8 ,80 066667
0,6 ,62 26666667
se rechaza el valor de
50,2
46,8 ,80 066667
3,3 ,39 93333333
54.72
desviacion media
6.783<
4*2.142
2,142222222
hallando el error relativo
=
/− /
Er1
10,76
Er2
12,64
Er3
0,32
Er4
7,36
Er5
7,64
Er6
0,4
x100
OBSERVACIONES:
El valor dudoso fue rechazado gracias a la prueba de 4d El error relativo indica que no ay mas de un 15% de error Medimos el CaCO3 en ppm para comodidad del investigador Utilizamos 2 indicadores con el fin de determinar la acidez y el viraje
FIRMA DEL ALUMNO: ............................................... CUESTIONARIO 1. ¿Cuántos tipos de alcalinidad en el agua conoce? 1. Alcalinidad total (M o T): que representa el total de iones que intervienen, se determina con el indicador anaranjado de metilo:
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2. Definir que es acidimetría es el método que se encarga de determinar la cantidad de ácido que se encuentra de manera libre en una disolución
3. ¿Porque en la determinación de una muestra de agua se usa como indicador la fenolftaleína y luego metil naranja? FENOLATELEINA: FENOLATELEINA: para determinar su carácter si es básico o acido
METIL NARANJA: para determinar su punto de equivalencia 4. ¿Porque se expresa el resultado de la alcalinidad referida a carbonato de calcio en ppm? La gran importancia la medición la medición y control de la alcalinidad del agua en un proceso un proceso de generación de vapor, ya que las calderas las calderas pueden pueden sufrir daños (implosiones o explosiones) por causa de las incrustaciones de los carbonatos y bicarbonatos de calcio y magnesio, por este motivo es que se mide la alcalinidad junto con la dureza del agua ya que están en relación, además para el control de la corrosión la corrosión 5. ¿Cuántos ml de solución 0.1200 N de un ácido se necesitan para neutralizar exactamente 32 ml de una solución 0.1500 N de una base? C1*V1=C2*V2
0.12*V1=0.5*32 V1=133.33ml 6. Se mezclan las siguientes soluciones: 50 ml de NaOH 01450 N, 22 ml de H2SO4 0.1050 N y 40 ml de HCl 0.0950 N ¿Cuál es la normalidad de la solución en ácido o en base? Hallando masas Masa de NaOH = 0.1450*50ml*0.04=0.29g Masa de H2SO4 =22*0.098*(0.105/2)=0.11319 Masa de HCl =40*0.0950*0.0365=0.1387 Base – acido acido = 0.29-(0.11319+0.1387)=0.03811 solucion básica Volumen= v1+ v2 +v3 = 50+22+40 = 112ml Hallando normalidad: 0.03811/(0.04*112ml)=8.51*10´(-3) 7. Que peso de Na2CO3 deberá tomarse para el análisis de modo que cuando se neutralice completamente con ácido 0.145=n, cada ml del ácido utilizado represente el 1% de Na2CO3 en la muestra? Peso de N2CO3=106*0.01*0.145=0.1537g
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. Alcalinidad total (M o T): que representa el total de iones que intervienen, se determina con el indicador anaranjado de metilo:
Laboratorio de Análisis Instrumental en ingeniería I ingeniería química DETERMINACIÓN DE CLORUROS EN EL AGUA POTABLE I OBJETIVOS 1.1 Aplicar los conocimientos de volumetría por precipitación en la preparación y valoración de una solución de Nitrato de plata. 1.2 Analizar el contenido de cloruros que contienen una muestra de agua potable por medio del método Mohr. II PRINCIPIO DEL MÉTODO El método empleado es el de Mohr. Los cloruros presentes en la muestra reaccionan en medio neutro o débilmente alcalino, dando un precipitado blanco de cloruro de plata, que precipita cuantitativamente de acuerdo a la siguiente reacción : C l - + Ag + ----------
Ag Cl
El punto final de la valoración queda establecida cuando aparece un precipitado rojo de cromato de plata formado por adición de un ligero exceso de nitrato de plata después de precipitar todo el ión cloruro. La reacción en el punto final es la siguiente: s iguiente: CrO4 indicador
-2
+
2 Ag + ------------ Ag2CrO4 titulante
rojo
III GENERALIDADES Los métodos argentométricos se basan en las reacciones de precipitación de sales de plata muy pocos solubles y muy aplicables a determinación de haluros. El análisis de aguas, dentro de las relaciones de un moderno estado industrial, se ha desarrollado como un extenso campo de la ciencia en el cual se aplican principios y técnicas que van desde las pruebas sensoriales pasando por las clásicas titrimétricas, espectrofotométrica, etc., hasta las altamente sofisticadas de la cromatografía de gas, cromatografía de capa fina, absorción atómica, etc. El agua de procedencia natural no se encuentra nunca químicamente pura, sino que
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Debido a sus múltiples propiedades químicas y físicas, se utiliza el agua de acuerdo a la correspondiente preparación en numerosos ramos, por ejemplo como agua potable, refrigeración, disolvente en la limpieza, en la energía como transmisor de ésta, en la industria como elemento básico y auxiliar en la obtención de infinidad de productos. Para muchos campos farmacéuticos y químicos se utiliza el agua doblemente destilada, especial importancia presenta el agua en los procesos en el organismo animal y vegetal que sirve como materia prima en la síntesis así como disolvente y elemento de transporte. Una condición básica de un agua potable debe contener en iones cloruros por debajo de 40 ppm ( mg/ L ). Según la norma INDECOPI No 214.003, el agua potable debe cumplir el siguiente requisito en contenido de cloruros:
Compuesto Cloruros ( Cl) mg/L
Valor máximo recomendable 250
valor máximo admisible 600
TEORÍA DEL CÍRCULO DE MOHR PARA DOS DIMENSIONES Considere un cuerpo sobre el cuál actúa un estado plano de cargas. Consideremos al plano de carga para nuestro sistema nuestro sistema al plano xy (ver figura 1), de modo de que no existan esfuerzos en el sentido perpendicular a este (esfuerzos en z nulos). Adoptamos un elemento triangular donde se supone que los ejes x e y son principales, o sea las tensiones de corte en esos planos son nulas. Esta suposición se hace con el fin de no complicar por demás la matemática la matemática siendo el objeto de este desarrollo este desarrollo conocer el desarrollo matemático a fin de ser asociado con el modelo el modelo físico:
Figura 1
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Se puede llegar a calcular la cantidad de concentración del cromato, que debe de encontrarse en la disolución para que la precipitación inicie exactamente en el punto de equivalencia. Cuando la disolución se encuentre saturada de las dos sales, el cloruro y el cromato de plata, se cumplirán de manera simultanea los equilibrios siguiente ( los ( los valores de Kps indicados son aproximados) :
Si despejamos la concentración de catión plata de las dos constantes y pasamos a igualar, obtendremos una relación de donde deduciremos que la relación de las concentraciones de ambos iones es: Como en el punto de equivalencia se cumple que la concentración de iones plata es igual a la concentración de iones cloro y todo ello a su vez es igual a : Con dicho valor dicho valor de concentración de cloro, tendremos la concentración de cromato que debe encontrarse presente en la disolución para que se produzca la precipitación del cromato de plata en el momento, dándonos un resultado de 0.02 M. Por lo cual, teóricamente hablando, se comenzará la formación de Ag2CrO4, en el punto de equivalencia, cando la concentración de cromato en la disolución en concreto en concreto sea, como hemos mencionado, de 0.02 M. Pero, sin embargo, se debe agregar un cierto exceso de plata antes de sea bien visible el color rojo. En la práctica, en los laboratorios, se ha podido comprobar que una concentración de indicador del orden de unos 10^-3 M, es la idónea para poder detectar sin problemas sin problemas ni equívocos el punto final. Como ya se ha mencionado, el método de Mohr se utiliza para la determinación del cloruro y de los bromuros. No es aplicable a la determinación del yoduro o los sulfocianuros, pues en estos casos, el punto final resulta bastante difícil de observar, debido a los fenómenos de adsorción que tiene lugar en dichos procesos. dichos procesos. Otra de las variables las variables a tener en cuenta en el método de Mohr es el pH. el pH. La La valoración debe
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IV APARATOS 4.1 Balanza analítica con sensibilidad de 0.1 mg 4.2 Frascos erlenmeyer de 250 ml 4.3 Bureta por 25 ml. V REACTIVOS 5.1 Solución de nitrato de plata 0.0282 N 5.2 Solución indicadora de cromato de potasio al 5 % Disolver 5 gramos de cromato de potasio en pequeña cantidad de agua destilada, libre de cloruros y se completa hasta 100 ml. VI PROCEDIMIENTO 6.1 Preparación de una solución de Nitrato de Plata 0.0282 N En la reacción siguiente se tiene : Cl -
+
Ag + -----------
Ag Cl
Se observa que un mol de nitrato de plata contiene un equivalente, porque es la cantidad de dicha sustancia que aporta un átomo gramo de ión plata , catión monovalente en el precipitado. El peso equivalente del nitrato de plata es : PE = AgNO3 = 1
169.89
Para preparar una solución standard de nitrato de plata 0.0282 N pesar y disolver 4.79089 gramos de nitrato de plata con agua destilada, libre de cloruros, llevar a un matraz aforado de 1000 ml, enrasar y homogeneizar. 6.2 Valoración
con Cloruro de sodio ( método del pipeteo ) El cloruro de sodio, categoría patrón primario, tiene una pureza de 99.9 - 100 %, por esto es un excelente patrón primario. Pesar 0.2000 gramos de cloruro de sodio y disolverlo con agua destilada en una fiola de 250 ml aforar y homogeneizar. De esta solución, pipetear una
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Agregar 1 ml de solución indicadora de cromato de potasio al 5 % y titular con solución de nitrato de potasio valorado hasta que el color amarillo de la solución se torne rosada amarillenta o naranja, que indica el punto final de la reacción. Determinar luego el gasto del “blanco “ para corregir el volumen gastado en la titulación.
Su técnica es la siguiente : Medir un volumen de agua destilada igual al volumen final de la titulación, agregar 1 ml de solución indicadora de cromato de potasio al 5 % y añada 0,2 g. de carbonato de calcio, libre de cloruros, se titulan con la solución de nitrato de plata hasta que sea visible un cambio definido de color. VII CALCULOS El resultado se suele expresar bajo la forma de cloruros: Solución: Cálculo referido a partes por millón de cloruros: ppm. De Cl - = ( A - B ) x 0.0282 meq/ml x 35.45 mg/meq x 1000 ml ml muestra x L Cálculo como partes por millón referida a cloruro de sodio ppm. NaCl = ( A - B ) x 0.0282 meq/ml x 58.45 mg/meq x 1000 ml ml muestra x L donde : A = ml gastados en la titulación B = ml gastado en el ensayo en blanco VIII EXPRESIÓN DE RESULTADOS El resultado se suele expresar como ppm referida a cloruro y como ppm referida a cloruro de sodio
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DETERMINACIÓN DE CLORUROS FECHA : 18-07-17
GRUPO: 3
APELLIDOS Y NOMBRES: Valeriano Mamani Ivan ANÁLISIS: valoración del nitrato de plata MÉTODO: argentometrico de mohr MUESTRA: agua potable 50ml CALCULOS 1.- Valoración de la solución 0.0282 N de nitrato de plata con cloruro de sodio Datos:
Peso del reactivo patrón = 0.2004 Volumen aforado = 250ml Volumen alícuota = 10ml Gasto de la solución de AgNO3 = 5ml Peso equivalente de NaCl = 0.05845 Miliequivalente de NaCl = 58.45mg-eq
Solución: Normalidad
Molaridad
− ∗1meq 8.016∗10 N = 50.3 ∗0.05845 =0.029179N − ∗1mmol 8.016∗10 M = 50.3 50.3 ∗0.05845 =0.029179M
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b) Solución: Resultado como ppm Cl ppm. De Cl - = ( A - B ) x 0.0282 meq/ml x 35.45 mg/meq x 1000 ml 50ml muestra x L ppm. De Cl - = ( 5 – 4.3 ) x 0.0282 meq/ml x 35.45 mg/meq x 1000 ml 50ml muestra x L ppm. De Cl - =30.10 Resultado como ppm NaCl ppm. NaCl = ( A - B ) x 0.0282 meq/ml x 58.45 mg/meq x 1000 ml ml muestra x L ppm. NaCl = ( 5 – 4.3 ) x 0.0282 meq/ml x 58.45 mg/meq x 1000 ml 50ml muestra x L ppm. NaCl = 46.60 TABLA DE RESULTADOS No Muestra 1 2 3 4 5 6
ppm Cl 28.95 31.08 30.10 30.38 24.47 31.99
ppm Na Cl 47.74 51.25 46.60 50.10 40.34 52.75
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Error Absoluto 2 Error Absoluto 3 Error Absoluto 4 Error Absoluto Error Absoluto Error Absoluto
=31.08-50 = -18.92 =30.10-50 = -19.9 =30.38-50 = -19.62 5=24.47-50 = -25.53 6=31.99-50= -18.01 7=31.84-50 = -18.16
HALLANDO EL ERROR RELATIVO
%E= VMEDIDOVTEORICO VTEORICO X100 %E1= 21.05 50 X100%=42.1% %E2= 18.92 X100=37.84% 50 %E3= 19.9 X100=39.8% 50 %E4= 19.62 X100=39.24% 50 %E5= 25.53 50 X100=51.06%
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N° muestra
valor
media arimetica 29,83
desviacion
1
medido 28,95
2
31,08
29,83
1,25
3
30,1
29,83
0,27
4
30,38
29,83
0,55
5
24,47
29,83
- 5,36
6
31,99
29,83
2,16
7
31,84
29,83
2,01
- 0,88
RECHAZO DE DATOS valor va lor obs observ ervad ado o va valor lor med medido ido media ar arimetic imetica a valor dudoso
promedio
desviac desv iacion ion
24,47
29,83
-5,36
28,95
30,72
1,7733
31,08
30,72
0,3567
30,1
30,72
0,6233
30,38
30,72
0,3433
31,99
30,72
1,2667
31,84
30,72
1,1167
30,72
0,9133
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1.- Porqué el método de Mohr Mohr utiliza una solución valorada de nitrato de plata diluida La solución patrón de AgNO3 se puede preparar por el método directo dado que el nitrato de plata es un reactivo tipo primario; con el objeto de compensar los errores en la precipitación del punto final se prefiere el método indirecto y la solución se valora con NaCl químicamente puro. puro . Cuando la solución tipo se prepara p repara por el método indirecto no es necesario el necesario el ensayo en blanco, porque el exceso empleado en la valoración de la sustancia problema se compensa con el empleado en la valoración del AgNO3. 2.- Señalar razones porque el medio de la solución debe ser neutra y no fuertemente ácida ni básica La valoración de método de Mohr debe llevarse a cabo a valores a valores de pH comprendidos entre 7 y 10, ya que a pH>10, el ión Ag+ precipita como AgOH antes que como Ag2CrO4 no detectándose el color rojo de indicación del punto final. Por otro lado, si pH<7, el Ag2CrO4 se solubiliza al protonarse los iones cromato. Normalmente, se logra un pH adecuado al saturar la disolución del analito con carbonato ácido de sodio. 3.- Porque se hace un blanco con carbonato de calcio El método de Mohr se puede adaptar para determinar el contenido total de cloro de una muestra, al encender la muestra con carbonato con carbonato de calcio, y calcio, y luego carbonato luego carbonato de hierro(III). El carbonato de calcio «fija» el cloro libre, precipita los carbonatos y neutraliza la solución resultante. El carbonato de hierro(III) remueve los fosfatos. Se disuelven todos los cloruros
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Laboratorio de Análisis Instrumental en ingeniería I ingeniería química DETERMINACIÓN DE ACIDEZ I OBJETIVOS 1.1 Aplicar los conocimientos volumétricos en la preparación de soluciones álcalis 1.2 Estandarizar la solución de hidróxido de sodio 1.3 Analizar el contenido de Acido acético en el vinagre y ácido láctico en leche II PRINCIPIO DEL MÉTODO El método utilizado es el Alcalimétrico. Consiste en titular una porción en volumen de la muestra que contiene el analito, ácido Acético, con una solución valorada de una base fuerte y empleando como indicador la fenoltaleína. La reacción principal de titulación es la siguiente: CH3 - COOH Analito
+ NaOH → titulante
NaCH3 - COO
+
H2 O
El ácido predominante en este tipo de muestra, es el ácido acético, aunque están presentes otros ácidos, el resultado se suele expresar como contenido en ácido acético.
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La composición depende de la alimentación del animal así como la especie y del tipo de raza. Los componentes son: Carbohidratos (lactosa), proteínas (caseína, lacto albúmina), grasa (ácido cáprico, palmítico, oleíco), minerales, etc. Interferencias En ciertos tipos de muestras el color proveniente de ella tiende a enmascarar el vire de color del indicador causando problemas en observar con nitidez el punto final de la valoración al diluir el vinagre, se reduce la intensidad de su color de modo que no interfiere en el vinagre el indicador. IV APARATOS 4.1 Balanza Electrónica Denver 4.2 Bureta por 25 ml 4.3 Vasos de precipitados por 250 ml 4.4 Matraces de erlenmeyer por 250 ml. 4.5 Pipeta graduada por 2 ml V REACTIVOS 5.1 Solución valorada de hidróxido de sodio 0.100 N. 5.2 Solución indicadora de fenolftaleína Se pesa 0.1 gramos del reactivo sólido de fenolftaleína, se disuelve y diluye con alcohol al 60% hasta un volumen de 100 ml. VI PROCEDIMIENTO
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2.0 ml x N1 = (ml gastados x factor) x 1 N N1 = Preparación de la solución de hidróxido de sodio N/10 Previamente se calcula la cantidad de hidróxido al 50% requerido para preparar un litro de solución de hidróxido de sodio N/10. V1 x
N1
= 1000 ml x 0.1 N
Medir el volumen (V1) calculado, se lleva a una fiola de un litro y se completa con agua destilada hervida y fría hasta la línea de enrase. Finalmente agitar para homogeneizar. 6.2
Valoración de la solución preparada con biftalato de potasio Se pesa 0.2000 gramos de Ftalato ácido de potasio, calidad reactivo analítico,
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VII CALCULOS De la reacción de titulación se tiene que el peso equivalente del ácido acético (60 g/mol) es igual a su peso molecular por lo que su miliequivalente gramo es 0.060 g / meq. El peso equivalente del ácido láctico (90.08 g/mol) es de 0.0900 g/meq % ácido acético = ml gastados x 0.1000 meq / ml x 0.060g. A. acético/meq ml muestra
x 100
% ácido láctico = ml gastados x 0.1000 meq/ml x 0.009 g Acido láctico/meq x 100% ml de muestra VIII EXRESIÓN DE RESULTADOS El resultado se expresa como porcentaje de ácido acético
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Caso I.Cuando hay presencia solo de iones OH- ; típicos son las soluciones cáusticas NaOH, KOH, etc. Caso II.Cuando hay presencia solo de iones HCO3-, típicos de esto son las aguas naturales, que contienen bicarbonato de calcio y bicarbonato de magnesio, constituyendo la Dureza Temporal del Agua. Caso III.Cuando hay presencia de CO3=, típico de esto son las soluciones decarbonato de sodio y carbonato de potasio, etc; y en aguas tratadas por el método con precipitación con cal 2.- Señale razones porque utiliza el indicador fenoltaleína en los análisis de acidez El cambio del color de la fenolftaleina del rojo al incoloro se produce cuan do el Ph es igual