problema analítica
TRANSCRIPT
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BOLETIN 1
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2.42. Se determina el N de una muestra de leche en polvo quepesa 3.000 g mediante el mtodo de Kjeldahl. Para ello se hace
destilar el amoniaco sobre 0.0 ml de una disoluci!n de ".
clorh#drico 0.$2$0 %& valor"ndose el e'ceso de "cido con otra
disoluci!n de Na() 0.$40 %& de la que se gastaron 32.2 ml
*alcular a+ Porcentaje de Nitr!geno en la muestra.
b+ Si el ,actor de conversi!n de Nitr!geno en prote#na es -.3&/qu cantidad de prote#nas contiene la masa de muestra
analiada1.
a+ as trans,ormaciones que tienen lugar son como las de
2.4$& con la nica di,erencia de que se recoge el amon#aco
destilado en "cido ,uerte en e'ceso para valorar ste con
sosa.
n la valoraci!n por retroceso del e'ceso de "cido con
sosa& tenemos5
mmol )*l 6 mmol N)3 7 mmol Na()
0.0 8 0.$2$0 6 mg N9$4 7 32.2 8 0.$40
mg N 6 $.3 mg
: N 6 0.0$3 8 $0093.0000
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b+ : Prote#nas en la leche
0.$ 8 -.3 6 3.2:
n la muestra de 3.0000 g habr" pues5
g prot;ot 6 0.0
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A 50.00 mL sample of a white dinner wine required 21.48 mL of
0.03776 M NaoH to achieve a phenolphthalein end point. Express
the acidity of the wine in terms of grams of tartaric acid (H2C4H4O6;
150.09 g/mol) per 100 mL. (Assume that two hydrogens of the acid
are titrated.)
21.48 mL * (0.03776 mol OH-/1000mL) = 0.000811085 mol OH-
Since it takes twice the moles of NaOH to neutralize each mole of
tartaric acid (since there are 2 moles of H+ in each tartaric acid
molecule), 0.0008110085 moles of Base will neutralize half or
0.000405504 moles of Acid.
Convert this equivalence of acid you have found into grams using
the MW:
0.000405504 moles of H+ * (150.09 g/mol) = 0.06087 grams of H+
Now convert grams to concentration of grams/100 mL using the total
volume of all the wine that you started with:
0.06087 grams/50 mL = 0.1217 grams tartaric acid/100mL
Basically you are using the act of adding a Base painstakingly
measuring every single drop until you reach endpoint (neutral pH),
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then using that amount to go backward and calculate how much acid
was in there to start with.
BOLETIN 2
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5. EDTA Titrations
>a+ ? 0.00 m sample containing Ni27@as treated @ith 2.0 m o,
0.000 % A;? to comple' all the Ni27and leave e'cess A;? in solution.
;he e'cess A;? @as then bacBCtitrated& requiring .00 m o, 0.000 % Dn27.
Ehat @as the concentration o, Ni27in the original solution1
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36. The amount of calcium in physiologic uidscan be determined by a complexometrictitration with EDTA. In one such analysisa
!."!!#m$ sample of blood serum was madebasic by adding % drops of &a'( and titratedwith !.!!"") * EDTA re+uiring !.%6, m$ toreach the end point. -eport the concentrationof calcium in the sample as miligrams of aper "!! m$./olution
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