Laboratorio de huevo flotante

Happiness by Alexander J Krzyston

Laboratorio de huevo flotante

Introducción:

En este experimento intentamos encontrar la densidad de dos óvulos diferentes, un huevo fresco (huevo A) y un huevo cocido (huevo B). Las densidades fueron determinadas usando dos técnicas diferentes. Primero nos coloca un huevo en el agua y agregar sal hasta el huevo apenas toca la superficie del agua, en este punto el huevo y el agua tenían la misma densidad. Por saber cuánta sal agregamos al agua que podríamos determinar la densidad del agua y, por tanto, la densidad del huevo. El segundo método que utilizamos para encontrar la densidad del huevo fue primero concentrando el huevo y luego colocándola en un vaso de agua. Observando el desplazamiento de las aguas como resultado de agregar el huevo podemos determinar el volumen del huevo y luego dividir la masa del huevo por este volumen para calcular la densidad.

Procedimiento experimental:

El procedimiento fue el mismo en el manual de laboratorio. La única excepción a esto fue en lugar de utilizar un cilindro aforado de 1000 mL y agregar 500 mL de agua en él, se utilizó un vaso de precipitados de 400 mL y agrega 200 mL de agua.

Análisis de datos:

Datos:

Sal agregando método #1 al agua

Huevo A

Ensayo 1 ensayo 2 ensayo 3

masa inicial de sal 92,9 g g 68,2 79,8 g

masa final de sal 44,6 g 21,2 g g 32,2

Diferencia en la final e inicial calculado masa

Ensayo 1 = g 76.6

Ensayo 2 = 47,1 g

Ensayo 3 = 47,5 g

Huevo B

Ensayo 1 ensayo 2 ensayo 3

masa inicial de sal 68,3 g g 72,9 86,5 g

masa final de sal 47,3 g 52,5 g g 66,3

Diferencia en la final e inicial calculado masa

Trail 1 = 21.0 g

Ensayo 2 = 20,4 g

Trail 3 = 20,2 g

Desplazamiento de agua #2 método para encontrar el volumen

Huevo A

Ensayo 1 ensayo 2 ensayo 3

masa de huevo g 50,9 51,1 g g 51,2

nivel del agua antes de 200,0 mL 200,0 mL 200,0 mL

nivel de agua después de mL 240,0 240,0 mL mL 240,0

Diferencias de nivel del agua antes y después de calculado

Ensayo 1 = 40,0 mL

Ensayo 2 = 40,0 mL

Ensayo 3 = 40,0 mL

Huevo B

Ensayo 1 ensayo 2 ensayo 3

masa de huevo g 52,4 52,2 g g 52,7

nivel del agua antes de 200,0 mL 200,0 mL 200,0 mL

nivel de agua después de 250,0 mL 250,0 mL 250,0 mL

Diferencias de nivel del agua antes y después de calculado

Ensayo 1 = 50,0 mL

Ensayo 2 = 50,0 mL

Ensayo 3 = 50,0 mL

Cálculos:

Ver documento adjunto para cálculo e impresión preguntas final del laboratorio.

Discusión:

Tenemos algunas fuentes de error en nuestro experimento. Notamos como que fuimos agregando sal a nuestro vaso que nuestro huevo B tenía una grieta que continuaron aumentando mientras pasaba el experimento. Sin embargo, la cáscara de huevo B nunca se cayó y se mantuvo en el huevo. Además, en cada uno de nuestros tres ensayos para cada huevo había un poco de sal sin disolver en el fondo del vaso. No podríamos conseguir la sal a disolver lo suficientemente rápido antes de que nuestro huevo comenzó a flotar. Como resultado, podría ser que nuestros cálculos para la densidad del agua salada para esta parte del laboratorio son ligeramente más altos porque fue vertido más sal en el vaso que en realidad fue disuelto en el agua. Otra fuente de error surge del hecho de que mientras concentrando nuestro huevo en la segunda parte del laboratorio, nuestro huevo estaba mojada que agrega una pequeña cantidad de su masa, así nuestro cálculo para la segunda parte del laboratorio podría dar un huevo con una densidad ligeramente superior a lo que se esperaría debido a la masa adicional.

Acid Titration posted by Alexander James Krzyston

Happiness by Alexander J Krzyston

Introduction:
    In this experiment, we titrated HCl against NaOH in order to find the initial concentration of HCl.  We first titrated HCl without the use of a pH meter and then with a pH meter.
Alexander Krzyston Acid Titration| Alexander J Krzyston Acid Titration| Alex James Krzyston
Alex Krzyston Acid Titration| Alex J Krzyston Acid Titration| Alexander James Krzyston
NORTHWESTERN UNIVERSITY Acid Titration| EVANSTON Acid Titration| BURR RIDGE
In the titration without a pH meter, we used a phenolphthalein indicator to determine when the solution has reached the equivalence point, when the solution turned bright pink.  For the titration with a pH meter, we recorded the change in volume with every change of 0.3 in pH. We then plotted ΔpH /ΔV vs. V, at the peak of the graph is the equivalence point and used this point to calculate the initial concentration of HCl. 
     We also titrated Acetic Acid against NaOH to find its concentration and pKa. 

OVERVIEW:
1. Sometimes chemists use indicators that change color at a pH of 9 to 10 instead of a pH of 7 because not all acid-base titrations reach their equivalence pints at a pH of 7 when they involve a weak acid or a weak base.  For example in the titration of a weak acid with a strong base, like the one we just did of HOAc against NaOH, the equivalence point was at a pH around 9 which is why we used phenolphthalein as an indicator because it changes from colorless to pink over a pH of 8.2 to 10.
2. In the titration of HCl against NaOH we used pH= 7 as the equivalence point because it was the titration of a strong acid against a strong base so when the two were at the equivalence point, all of the OH- from the NaOH neutralized with all of the H+ from the HCl and the concentrations are equal resulting in H2O.  All of the HCl went into forming H+ and all of the NaOH went into forming OH- to react with the H+.  In contrast, with the Acetic Acid, not all of the HOAc went into forming H+ ions.

This lab called for very careful measurements which was difficult to do and resulted in data that was not as accurate as it could be.  Going one drop over the equivalence point had a huge impact on the titration especially since the pH changes so radically near the equivalence point.  This made tracking the pH with the pH meter near the equivalence point difficult.  These drastic pH changes are clearly visible on our graphs.  We had some difficulty in trying to track the change in volume with each change of 0.3 in pH because the pH meter slow in reaction time in adjusting to the pH.  Often times we would add NaOH until there was a 0.3 change in pH and when we then stopped to record the change in volume the pH would continue to rise.  Also, because of the drastic change in pH near the equivalence point, there were huge fluctuations in the pH meter near this point (after adding only 0.10 the change in pH was 1.39 and after adding 0.02mL of NaOH, the change in pH was 1.22).  Initially, there would be a huge change in pH and then as the added NaOH dispersed throughout the solution, the pH would drop.  For this reason there are some inconsistencies in our data, especially with the titration of HCl and NaOH with the pH meter.  The most noticeable is the change in pH from 3.43 to 6.07 in which the volume changed from 22.60 to 22.85.  This had a minimal impact on the shape of the pH vs. volume graph, but did affect the shape of the ΔpH/ΔV vs. volume graph.  Instead of a slow steady rise followed by a large, drastic rise and fall, the graph had a slight peak followed by a slight drop followed by the huge rise.  I think we should have allowed more time for the pH meter to stabilize before continuing with the titration.  However, our data for the titration of NaOH and Acetic Acid was more consistent which shows that as the experiment when on, we became better at working with the pH meter. In addition, because we were working with a weak acid and strong base which created a buffer solution, the pH did not change as rapidly as with the strong acid.
According to the ΔpH/ΔV vs. volume graph our equivalence point was at 32.02mL which is at a pH of 9.10.  The equivalence pH was supposed to be at 7, when the added OH- from the NaOH neutralized all of the H+ from the HCl. This difference could result from the fact that we did not wait long enough for the pH meter to stabilize to get the final reading.  Had our ability to control the addition of NaOH been better, I do not think we would have had such issues. However, the shape of the graph was correct and shows a steep rise in the region from a pH of about 6.00 to 10.00.

Procedure:
Data:
Titration of HCl against NaOH
without pH meter

Run    Initial Buret Reading (mL)    Final Buret Reading (mL)     ΔV (mL)
1    0.00    23.25    23.25
2    0.00    23.00    23.00
3    0.00    23.00    23.00
Mean     0.00    23.08    23.08

Titration of HCl against NaOH
with pH meter

 Vbase (mL)    pH of Solution    ΔV (mL)    ΔpH      ΔpH
  ΔV         (mL)-1

0.00    1.50             
17.32    1.87    17.23    0.37    0.02
19.71    2.12    2.39    0.25    0.10
20.60    2.31    0.89    0.29    0.21
21.71    2.63    1.11    0.32    0.29
22.60    3.43    0.89    0.80    0.90
22.85    6.07    0.25    2.73    12.92
22.90    6.49    0.15    0.42    2.80
23.00    7.88    0.10    1.39    12.90
23.02    9.10    0.02    1.22    61.00
23.05    9.49    0.03    0.39    13.00
23.41    10.46    0.36    0.07    2.70
23.72    10.77    0.31    0.31    1.00
24.10    11.00    0.38    0.27    0.71

Titration of Acetic Acid against NaOH
 Vbase (mL)    pH of Solution    ΔV (mL)    ΔpH     ΔpH
  ΔV        (mL)-1
0.00    3.08             
0.91    3.37    0.91    0.29    0.32
2.00    3.67    1.09    0.30    0.28
3.95    3.99    1.95    0.32    0.11
6.83    4.30    2.88    0.31    0.11
10.28    4.59    3.45    0.29    0.08
14.20    4.90    3.92    0.31    0.08
17.85    5.26    3.65    0.36    0.90
19.32    5.48    1.47    0.22    0.15
20.90    5.80    0.77    0.32    0.42
21.53    6.07    0.63    0.27    0.43
22.00    6.40    0.47    0.33    0.70
22.21    6.67    0.21    0.27    1.30
22.60    9.10    0.39    0.33    0.85
23.70    9.75    1.10    0.65    0.59
25.00    10.37    1.30    0.62    0.48
26.20    11.00    1.20    0.63    0.53

Analysis:
pH vs. VNaOH

pH vs. VNaOH


Initial concentration of HCl based on titration with pH meter
[NaOH] = 0.2200M
Volume of NaOH to reach equivalence = 32.08mL
32.08mL ×      L     × 0.2200mol = 0.007058mol of NaOH
              1000mL         L 

[OH-] = [H+] = [HCl]

Volume of HCl solution = 100mL

 0.007058mol = 0.07058M
     0.100 L

Initial concentration of HCl based on titration with pH meter
[NaOH] = 0.2200M
Volume of NaOH to reach equivalence point = 32.02mL
32.02mL ×      L     × 0.2200mol = 0.007044mol of NaOH
              1000mL         L 

[OH-] = [H+] = [HCl]

Volume of HCl solution = 100mL

 0.007044mol = 0.07044M
     0.100 L

Vbase (mL)
2.7
ΔpH
ΔV         (mL)-1

pH vs. VNaOH

 OVERVIEW:
1. Sometimes chemists use indicators that change color at a pH of 9 to 10 instead of a pH of 7 because not all acid-base titrations reach their equivalence pints at a pH of 7 when they involve a weak acid or a weak base.  For example in the titration of a weak acid with a strong base, like the one we just did of HOAc against NaOH, the equivalence point was at a pH around 9 which is why we used phenolphthalein as an indicator because it changes from colorless to pink over a pH of 8.2 to 10.
2. In the titration of HCl against NaOH we used pH= 7 as the equivalence point because it was the titration of a strong acid against a strong base so when the two were at the equivalence point, all of the OH- from the NaOH neutralized with all of the H+ from the HCl and the concentrations are equal resulting in H2O.  All of the HCl went into forming H+ and all of the NaOH went into forming OH- to react with the H+.  In contrast, with the Acetic Acid, not all of the HOAc went into forming H+ ions.

This lab called for very careful measurements which was difficult to do and resulted in data that was not as accurate as it could be.  Going one drop over the equivalence point had a huge impact on the titration especially since the pH changes so radically near the equivalence point.  This made tracking the pH with the pH meter near the equivalence point difficult.  These drastic pH changes are clearly visible on our graphs.  We had some difficulty in trying to track the change in volume with each change of 0.3 in pH because the pH meter slow in reaction time in adjusting to the pH.  Often times we would add NaOH until there was a 0.3 change in pH and when we then stopped to record the change in volume the pH would continue to rise.  Also, because of the drastic change in pH near the equivalence point, there were huge fluctuations in the pH meter near this point (after adding only 0.10 the change in pH was 1.39 and after adding 0.02mL of NaOH, the change in pH was 1.22).  Initially, there would be a huge change in pH and then as the added NaOH dispersed throughout the solution, the pH would drop.  For this reason there are some inconsistencies in our data, especially with the titration of HCl and NaOH with the pH meter.  The most noticeable is the change in pH from 3.43 to 6.07 in which the volume changed from 22.60 to 22.85.  This had a minimal impact on the shape of the pH vs. volume graph, but did affect the shape of the ΔpH/ΔV vs. volume graph.  Instead of a slow steady rise followed by a large, drastic rise and fall, the graph had a slight peak followed by a slight drop followed by the huge rise.  I think we should have allowed more time for the pH meter to stabilize before continuing with the titration.  However, our data for the titration of NaOH and Acetic Acid was more consistent which shows that as the experiment when on, we became better at working with the pH meter. In addition, because we were working with a weak acid and strong base which created a buffer solution, the pH did not change as rapidly as with the strong acid.
According to the ΔpH/ΔV vs. volume graph our equivalence point was at 32.02mL which is at a pH of 9.10.  The equivalence pH was supposed to be at 7, when the added OH- from the NaOH neutralized all of the H+ from the HCl. This difference could result from the fact that we did not wait long enough for the pH meter to stabilize to get the final reading.  Had our ability to control the addition of NaOH been better, I do not think we would have had such issues. However, the shape of the graph was correct and shows a steep rise in the region from a pH of about 6.00 to 10.00.

Vitamin C posted by Alexander James Krzyston

Happiness by Alexander J Krzyston

Introduction:

The purpose of the this experiment was to determine the vitamin C concentration of vitamin c in various foods, a vitamin C tablet, orange juice, and fresh lime juice.  We did this by first making an ascorbic acid solution consisting of ascorbic acid (from the food), 5mL of HCl, 1.0g of KI, 2-3mL of a 0.5% starch solution, and 50mL of water.

Alexander Krzyston |Vitamin C Alexander J Krzyston |Vitamin C Alex James Krzyston
Alex Krzyston |Vitamin C Alex J Krzyston |Vitamin C Alexander James Krzyston
NORTHWESTERN UNIVERSITY |Vitamin C EVANSTON |Vitamin C BURR RIDGE

We titrated the ascorbic acid solution against 0.01M KIO3-.  Vitamin C is easily oxidized in acidic conditions, which is why we added the HCl.  The IO3- ions will react with the ascorbic acid, oxidizing it, to form dehydro-L-ascorbic acid and I3-.  When all of the ascorbic acid is used up, I2 will build up and react with I- to form I3-, which produced a purple color in the presence of starch.  When the solution turned purple, we had reached the end point.  My knowing the mass of the vitamin c table and the volume of orange juice and fresh lime juice in the ascorbic acid solution, we can determine the amount of vitamin C in each of the three.

Overview

In this experiment we found that the bottled orange juice had a higher concentration of vitamin c than the fresh lime juice.  This surprised me at first because I would think that the fresh fruit juice would have more vitamin c than bottled juice.  However, according to http://www.naturalhub.com/natural_food_guide_fruit_vitamin_c.htm
Lime juice has 29mg of vitamin C per 100mg of fruit, which it approximately 1mg per average slice, where as an orange has 53mg of vitamin C per 100mg of fruit, which is approximately 70mg per orange.  Thus, the differences in vitamin C concentration do not depend on whether the juice is bottled or not, but on the fruit.  To determine whether or not bottled versus fresh juice had an impact f vitamin C concentration, we should have tested the same fruit juice both bottled and fresh. By doing so we could determine if processing has any impact on vitamin C concentration. 

Procedure:

Procedure was same as in lab manual. 
*For the titration of fruit juices and powered drink mixes, we titrated orange juice (no pulp)
*For the titration of fresh fruits we titrated lime juice, we had to use more than one lime to get 50.0mL of juice so in the final calculations we could not calculate the amount of vitamin C per fruit so value so given in mg/mL.

Data:
Part 1: Vitamin C Tablet
Trial 1:                            Trial 2:
mass of 1/2 vitamin C tablet    0.1624g        mass of 1/2 vitamin C tablet    0.1208g
initial volume of KIO3    0.00mL        initial volume of KIO3    0.00mL
final volume of KIO3    24.00mL        final volume of KIO3    18.15mL


Part 2: Orange Juice
volume of orange juice    20.0mL
initial volume of KIO3    0.00mL
final volume of KIO3    1.26mL
   
Part 3: Fresh Lime Juice
   
volume of fresh lime juice    50.0mL
initial volume of KIO3    0.00mL
final volume of KIO3    2.71mL

Calculations:
    See attached sheet for calculations.
Discussion:
1.    3C6H8O6 + IO3- → 3C6H6O6 + I- + 3H2O
2.    Two electrons are involved in the redox reaction.  Ascorbic acid is oxidized (loses 2 electrons) and iodine is reduced.
3.    Vitamin C will react with dissolved oxygen under acidic condition because vitamin C is easily oxidized in acidic conditions: 3C6H8O6 - → 3C6H6O6 + 2H+ + 2e-
4.    Pasteurization would decrease the concentration of vitamin C in juices.  Pasteurization involves heating of the juices to kill bacteria, this in turn also destroys vitamin C.  The concentration of vitamin C decreases as pasteurization temperature and time increases this was proven in a study of the Effect of Pasteurization on Vitamin C in Guava Juice in January 2008 at King Mongkut’s University of Technology Thronburi in Thailand (http://pdfcast.org/pdf/effect-of-pasteurization-on-vitamin-c-content-of-guava-juice) and has alos been rproven in various experiments involving raw and pasteurized milk.  “Eighty percent of a food's vitamin C content is lost merely by boiling it for twenty minutes and at 120°C” (http://www.daviddarling.info/encyclopedia/P/pasteurization.html).

5.    Describing vitamin C as an antioxidant would be a correct description.  Vitamin C is an electron donor which makes means that it is a reducing agent. By donating electrons, it prevents other compounds from being oxidized, in humans this would be radicals such as oxygen related radicals (superoxide, hydroxyl radical, peroxyl radical), sulfur radicals, and nitrogen-oxygen radicals or compounds that are reactive, but not radicals like hypochlorous acid, nitrosamines, and other nitrosating compounds, nitrous acid related compounds and ozone.  “all known physiological and biochemical actions of vitamin C are due to its action as an electron donor” (http://www.jacn.org/cgi/reprint/22/1/18.pdf).

*Other source also used: http://www.vitaminstuff.com/antioxidants.html

In this experiment we found that the bottled orange juice had a higher concentration of vitamin c than the fresh lime juice.  This surprised me at first because I would think that the fresh fruit juice would have more vitamin c than bottled juice.  However, according to http://www.naturalhub.com/natural_food_guide_fruit_vitamin_c.htm
Lime juice has 29mg of vitamin C per 100mg of fruit, which it approximately 1mg per average slice, where as an orange has 53mg of vitamin C per 100mg of fruit, which is approximately 70mg per orange.  Thus, the differences in vitamin C concentration do not depend on whether the juice is bottled or not, but on the fruit.  To determine whether or not bottled versus fresh juice had an impact f vitamin C concentration, we should have tested the same fruit juice both bottled and fresh. By doing so we could determine if processing has any impact on vitamin C concentration. 

Calcium Benzoate Solubility posted by Alexander James Krzyston

Happiness by Alexander J Krzyston

Introduction:
    In this lab we sought to determine the solubility product of Calcium Benzoate.  We did this by titrating calcium benzoate against EDTA standard solution.


When the EDTA is added to the calcium benzoate solution, it will react with Ca2+.  Thus by knowing the endpoint of the solution, we can calculate the final concentration of Ca2+ in solution (it will be equal to the volume of EDTA used in titration). Since calcium and benzoate are in a 1:2 ratio, the final concentration of Bz- will be double the change in [Ca2+] concentration.  By knowing the final concentration of both Ca2+ and Bz-, we can calculate the solubility product of CaBz2:
[Ca2+][Bz-]2 = Ksp

Alexander Krzyston Calcium Benzoate| Alexander J Krzyston Calcium Benzoate| Alex James Krzyston
Alex Krzyston Calcium Benzoate| Alex J Krzyston Calcium Benzoate| Alexander James Krzyston
NORTHWESTERN UNIVERSITY Calcium Benzoate| EVANSTON Calcium Benzoate| BURR RIDGE
Overview:
In this lab source of error could be due to the fact that the specific end point of the solution was rather difficult to determine.  This is because the color change was from red to purple to blue, and the difference between purple and blue can be rather objective and difficult to differentiate purple from blue when the two are mixed in solution.  As a result, it would be easy to stop before the end as the solution begins to turn bluish, but is not really totally blue yet.  A way to avoid this problem would be to have an over titrated solution as reference to compare the two solutions to ensure that you do not stop titrating too early.  Although we did have the first practice titration to do this, if you under titrated that solution you would not know what the actual color would be and all the other titration would also be under titrated.  This would result in a lower final concentration of Ca2+ ions and a higher final concentration of Bz- ions.  Also, because the blue was difficult to see when mixed with the purple color during titration, it would also be easy to over titrated. 
Another source of error in this lab would come from the EDTA solution.  If you messed up on the preparation of the EDTA solution initially, then all of the titrations would be wring because that was the solution we were using to titrated. 
Perhaps a way to improve this lab or an interesting variation would be to determine the solubility product of Calcium Benzoate by filtering out the precipitate from the over saturated solution and then measuring that value and use it to determine the solubility product and the final concentration of the two ions.   

Procedure:
    Procedure was same as given in lab manual
*as oppose to adding the MgCl2(6H2O) salt to each CaBz2 solution to sharpen the end point, we added the MgCl2(6H2O) to the standardized Na2H2E solution.  By doing this the slat only had to be added once as oppose to six separate times.

Data:
Weight of Na2H2E powder: 8.8986g
Volume of Na2H2E solution: 250mL
Solution #    T(oC)     [Ca2+]i     [Bz-]i    Vi(mL)    Vf (mL)    ΔV (mL)    Ksp
1     19.00     0.070     0.320      0.47     12.10     11.63   
1     19.00     0.070     0.320     12.10     23.70     11.60   
2     19.00     0.200     0.200      3.89     41.75     37.77    
2     19.00     0.200     0.200      0.22     37.94     37.72   
3     19.00     0.125     0.250      0.60     21.89     21.29   
3     19.00     0.125      0.250     21.89     43.20     21.31   
    *all concentrations in molarity (mol/L)
Analysis:
1. concentration of [Na2H2E]:
    0.09562 M

2. final concentration of [Ca2+]f
    See table for values
    See attached sheet for calculations

3. final concentration of [Bz-]f
See table for values
    See attached sheet for calculations

4. Ksp
See table for values
    See attached sheet for calculations

Solution #    [Ca2+]i     [Bz-]i    [Ca2+]f     [Bz-]f    Ksp
1     0.070     0.320     0.0556     0.0555     0.00474
1     0.070     0.320     0.292     0.290     0.00467
2     0.200     0.200     0.181     0.162     0.00475
2     0.200     0.200     0.180     0.160     0.00461
3     0.125     0.250     0.102     0.204     0.00424
3     0.125      0.250     0.102     0.204      0.00424
    *all concentrations in molarity (mol/L)
5. Mean Value of  Ksp:
    0.00424
        *See attached sheet for calculations

6. Recalculation of Ksp at 20oC:
    (Starting at Ksp value at 19oC and based on 2.5% increase per degree)

    0.00465
        *See attached sheet for calculations

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Free Energy of Cobalt posted by Alexander James Krzyston

Happiness by Alexander J Krzyston

Introduction:
In this experiment we used colorimetric analysis to study the free energy and solubility of a cobalt salt.  We did this by measuring the transmittance of four solution of the cobalt salt at different concentrations.
Alexander Krzyston Cobalt Alexander J Krzyston Cobalt| Alex James Krzyston
Alex Krzyston Cobalt| Alex J Krzyston Cobalt| Alexander James Krzyston
NORTHWESTERN UNIVERSITY Cobalt| EVANSTON Cobalt| BURR RIDGE
We then applied Beer’s Law to the data to show that A/r of the solution was equal.  We then measured the absorbance of four other solutions of cobalt that were prepared at four different temperatures.  A/r = kd we used this equation (value obtained from previous four solutions) and A = kCd to calculate the concentration of each of the solutions, A, B, C, and D.  Because A, B, C, and D are saturated solution, the concentrations for them are also the solubilities, thus s = c and we can use the equation Ksp = (s)(2s)2 = 4s3 to find Ksp We can then make a graph of lnKsp vs. 1/T, which is a linear relationship.  From the graph we can find the slope and intercept.  From the slope we can calculate ΔHo and from the intercept we can calculate ΔSo. 


Overview:
Sources of error in this experiment are that it was clear that some of the cobalt solution remained on the paper we used to weigh it out with, this result in a higher cobalt concentration than there actually was.  In addition, not all of the cobalt may have been dissolved into the solution; this too would produce a solution with a lower concentration. Another source of error is that when we were using the 10mL pipettes to measure out the solutions it was difficult to get all of the solution out of the pipette, often times so would remain at the very tip.  In addition, although the Solutions A, B, C, and D were prepared at different temperatures, they were all kept at room temperature so this may or may not have had an effect of their absorbance. 
An interesting observation we found was that when we were measuring out 10ml of water to fill each of the 50mL beakers, we used the 10mL pipette and filled it to the line but when we out to distilled water in the beaker it measured more than 10mL.  In this case, we went by the measurements of the pipette and not the beakers.
One way to improve this experiment would be to obtain data using both the UV-vis spectrophotometer and the Spec-20.  This would allow students to compare data between the two machines and determine which one is more accurate. Also I think keeping the four solution that were saturated at different temperatures at the temperatures they were saturated at would improve the accuracy of the experiment.
*In calculating ΔHo and ΔSo from slope and intercept I had to use the average slope because my graph was not linear.  To do this I calculated the slope from (0.0034, -12) to (0.0035, -13) which was -38000, the slope from (0.0035, -13) to (0.0036, -13) which was zero, and the slope from (0.0036, -13) to (0.0037, -12) which was 10000.  I then found the average slope which was -12000.  I then used the equation:  m = -ΔHo / R to find the change in enthalpy.  To find the change in entropy I used the average slope (-12000) that I had already calculated in the equation y = mx+b and applied each of the four data points to find an intercept.  Using (0.0037, -12) I got an intercept of 360, using (0.0036, -13) I got an intercept of 350, using (0.0035, -13) I got an intercept of 340, and using (0.0034, -9.2) I got an intercept of 330.  I then took the average of the four intercepts and got and average intercept of 340.  I then used the equation b = ΔSo / R to find the change in entropy.

Experimental Procedure:   
The procedure was the same as that given in the lab manual.  We used the UV-vis spectrophotometer and the Spec-20, thus we did not need to calculate absorbance, we could directly read off the UV-vis spectrophotometer (but I did calculate transmittance in the lab report).

Data Analysis:

1. see attached sheet for calculation of cobalt concentration
2. see attached sheet for calculation of transmittance and mean absorbance
3. see attached sheet for calculation of molar absorptivity (k)
4. see attached sheet for Beer’s Law and calculation of A/r

Solution #    Transmittance    Absorbance    A/r
0    0.13    0.072    0.072
1    0.13    0.027    0.079
2    0.13    0.036    0.072
3    0.13    0.048    0.073

5. see attached sheet for calculation of concentrations and Ksp

solution    Saturated at (oC)    Absorbance    Concentration    Ksp
A    0.0    0.185    0.015    1.5 x 10-5
B    6.0    0.098    0.0082    2.2 x 10-6
C    12.0    0.101    0.0084    2.4 x 10-6
D    18.0    0.356    0.030    1.0 x 10-4

6.
solution    oC    T (in K)    1/T (in K-1)    Ksp    ln Ksp
A    0.0    273    0.0037    1.5 x 10-5    -12
B    6.0    279    0.0036    2.2 x 10-6    -13
C    12.0    285    0.0035    2.4 x 10-6    -13
D    18.0    291    0.0034    1.0 x 10-4    -9.2

7.


8. see attached sheet for calculation of ΔHo and ΔSo from slope and intercept
*slope and intercept values are averages (see discussion of explanation)
Slope(m) = -12000
Intercept(b) = 350

ΔHo = 1.0 x 103kJ
ΔSo = 2910 J



In this experiment, we took a solid cobalt salt and dissolved it into water.  The cobalt salt went from a solid to a solution; it was ionized which contributed to the entropy change, increasing it.  In the previous experiment involving Sn and Fe, Sn and Fe were already in solution form.

Overview:
Sources of error in this experiment are that it was clear that some of the cobalt solution remained on the paper we used to weigh it out with, this result in a higher cobalt concentration than there actually was.  In addition, not all of the cobalt may have been dissolved into the solution; this too would produce a solution with a lower concentration. Another source of error is that when we were using the 10mL pipettes to measure out the solutions it was difficult to get all of the solution out of the pipette, often times so would remain at the very tip.  In addition, although the Solutions A, B, C, and D were prepared at different temperatures, they were all kept at room temperature so this may or may not have had an effect of their absorbance. 
An interesting observation we found was that when we were measuring out 10ml of water to fill each of the 50mL beakers, we used the 10mL pipette and filled it to the line but when we out to distilled water in the beaker it measured more than 10mL.  In this case, we went by the measurements of the pipette and not the beakers.
One way to improve this experiment would be to obtain data using both the UV-vis spectrophotometer and the Spec-20.  This would allow students to compare data between the two machines and determine which one is more accurate. Also I think keeping the four solution that were saturated at different temperatures at the temperatures they were saturated at would improve the accuracy of the experiment.
*In calculating ΔHo and ΔSo from slope and intercept I had to use the average slope because my graph was not linear.  To do this I calculated the slope from (0.0034, -12) to (0.0035, -13) which was -38000, the slope from (0.0035, -13) to (0.0036, -13) which was zero, and the slope from (0.0036, -13) to (0.0037, -12) which was 10000.  I then found the average slope which was -12000.  I then used the equation:  m = -ΔHo / R to find the change in enthalpy.  To find the change in entropy I used the average slope (-12000) that I had already calculated in the equation y = mx+b and applied each of the four data points to find an intercept.  Using (0.0037, -12) I got an intercept of 360, using (0.0036, -13) I got an intercept of 350, using (0.0035, -13) I got an intercept of 340, and using (0.0034, -9.2) I got an intercept of 330.  I then took the average of the four intercepts and got and average intercept of 340.  I then used the equation b = ΔSo / R to find the change in entropy.

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Floating Egg posted by Alexander James Krzyston

Happiness by Alexander J Krzyston

Introduction:
In this experiment we sought to find the density of two different eggs, a fresh egg (Egg A) and a boiled egg (Egg B).  The densities were determined using two different techniques.
Alexander Krzyston egg Alexander J Krzyston egg Alex James Krzyston
Alex Krzyston egg Alex J Krzyston egg Alexander James Krzyston
NORTHWESTERN UNIVERSITY egg EVANSTON egg BURR RIDGE
First we placed an egg in water and added salt until the egg just barely touched the surface of the water, at this point the egg and water had the same density.  By knowing how much salt we added to the water we could determine the density of the water and thus the density of the egg.  The second method we used to find the density of the egg was by first massing the egg and then placing it in a beaker of water.  By looking at the displacement of the water as a result of adding the egg we could determine the volume of the egg and then divide the mass of the egg by this volume to calculate the density.


Overview:
We did have some sources of error in our experiment. We notice as we were adding salt to our beaker that our Egg B had a crack in it which continued to increase as the experiment went on.  However, the shell of Egg B never fell off and remained on the egg.   In addition, in the each of our three trials for each egg there was some undissolved salt on the bottom of the beaker.  We could not get the salt to dissolve fast enough before our egg started to float.  As a result, it could be that our calculations for the density of the salt water for this part of the lab are slightly higher because more salt was poured into the beaker than was actually dissolved in the water.  Another source of error arises from the fact that while massing our egg in the second part of the lab, our egg was wet which added a small amount to its mass, thus our calculation for the second part of the lab could give an egg with a density slightly greater than what would be expected due to the added mass.

Experimental Procedure:
The procedure was the same given in the lab manual.  The only exception to this was that instead of using a 1000mL volumetric cylinder and adding 500mL of water into it, we used a 400mL beaker and added 200mL of water to it.
Data Analysis:
Data:
Method #1 adding salt to water
Egg A
     Trial 1    Trial 2    Trial 3
initial mass of salt     92.9g    68.2g    79.8g
final mass of salt    44.6g    21.2g    32.2g

Difference in final and initial mass calculated
Trial 1 =47.6g
Trial 2 = 47.1g
Trial 3 = 47.5g

Egg B
     Trial 1    Trial 2    Trial 3
initial mass of salt     68.3g    72.9g    86.5g
final mass of salt    47.3g    52.5g    66.3g

Difference in final and initial mass calculated
Trail 1 =21.0g
Trial 2 = 20.4g
Trail 3 = 20.2g

Method #2 water displacement to find volume
Egg A
     Trial 1    Trial 2    Trial 3
mass of egg    50.9g    51.1g    51.2g
water level before    200.0mL    200.0mL    200.0mL
water level after    240.0mL    240.0mL    240.0mL

Differences of water level before and after calculated
Trial 1 = 40.0mL
Trial 2 = 40.0mL
Trial 3 = 40.0mL

Egg B
     Trial 1    Trial 2    Trial 3
mass of egg    52.4g    52.2g    52.7g
water level before    200.0mL    200.0mL    200.0mL
water level after    250.0mL    250.0mL    250.0mL

Differences of water level before and after calculated
Trial 1 = 50.0mL
Trial 2 = 50.0mL
Trial 3 = 50.0mL

Calculations:
See attached paper for calculation and questions at end of lab print out.

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Flame Test posted by Alexander James Krzyston

Happiness by Alexander J Krzyston

Introduction:

     In this lab, we sought to determine the identity of 10 unknown solutions by using the flame test.  We burned the 10 unknown solutions using a Bunsen burner and an iron wire and observed the various colors of light produced.
Alexander Krzyston FT Alexander J Krzyston FT Alex James Krzyston
Alex Krzyston FT Alex J Krzyston FT Alexander James Krzyston
NORTHWESTERN UNIVERSITY FT EVANSTON FT BURR RIDGE
Different chemical produce different colors of light when burned because they have different energy levels (as a result of having different amounts of electrons).  By known which chemical produce which colors of light we could determine the identities of the 10 solutions.
 
Overview:

    The biggest challenge in this lab was differentiating become the colors.  Deciding upon what color flame was produced by the different ions was very subject especially when trying to differentiate between the various colors of red and orange, between blue and green, and between pale blue and bluish white.  Some of the solutions produced colors so similar we had to guess which ions were present.  This was especially problematic with solution #1 and solution #8, both of which produced an orange yellow flame. Another issue we encountered with trying to determine the flame color was the fact that the iron wire loop we were using to burn the solution would start to turn orange when heated and alter the color of the flame.  In addition, because the iron wire loops were had been frequently used before, there was residue left behind which would also alter the flame color.   As a result, we had to guess which ions were present in a few of the solution. 
    An alteration to this lab which I think would be helpful in determine which ions were present in which solutions would be if when were given 10 solution in which the ions present were given. This way would could compare between the known and unknown solutions which would be help to determine what the colors produced by the ions looked like we burned. (This would prevent error due to the subjecting of differentiating between red and red orange and yellow is very subjective.)

Data/Observations:
Solution    Flame Color and Intensity
1     Yellow, fairly intense
2     Pale blue/green intense
3     White  Fairly intense
4     Red-orange Intense and persistent
5     Red Very intense and bright
6     Green Very intense
7     Yellow/green Pale
8     Yellow Intense
9     Pale purple almost white Not intense
10    Red intense



Results:
solution    Identity of Ions
1     Na
2     Pb
3     Sb or As
4     Ca
5     Li
6     Cu
7     Ba
8     Na
9     K
10     Sr


A.    Colored light is emitted as a result of burning the solutions because the electrons in the solution absorb energy in the form of heat from the flame and they are they are excited causing them to jump to the next energy level.  When the electrons fall back to their ground state they emit energy in the form of light.  Because different chemicals have different energy levels, they require different amounts of energy to be excited.  As a result, they emit different colors of light when the electrons fall back to their ground state. (the different energy levels are a result of chemicals having different amounts of electrons)
The purpose of the cobalt blue glass is to eliminate the sodium effect.  Sodium in flame produces a strong yellow light that empowers the weaker light produced by other metal ions.  The cobalt blue glass absorbs the yellow light but allows other colors to pass through.

B.    The flame test cannot be used to determine the identity of unknown mixtures of solution because the color produced by heating the solution would be a mixture/combination of the two colors.  For instance if a solution contained Na and Pb, the resulting color of light would be green and one might confuse the solution of Na and Pb for one of Cu.  In addition, some chemicals produce the same color light such as Sb and As which makes in difficult to differentiate between the two.  Other chemicals produce weak colors that are difficult to see. 

C.    The electrons are excited by the heat from the flame, meaning that they are absorbing energy in the form of heat which raises them to the next energy level. 

D.    The chemicals need to be burned to get color in the flame because if they were just heated and not burned, they would not absorb enough energy to jump to the next energy level. 


Additional Analysis:
i.    Ions highest to lowest energy

K, As, Sb, Pb, Cu, Ba, Na, Ca, Sr, Li

ii.    Ions highest to lowest frequenc

K, As, Sb, Pb, Cu, Ba, Na, Ca, Sr, Li

iii.    Ions shortest to longest wavelength

K, As, Sb, Pb, Cu, Ba, Na, Ca, Sr, Li

iv.    The relationship between energy, wavelength, and frequency is that as energy increase frequency increases and wavelength decreases.  This is seen in our results; the elements with the highest energy also had the highest frequencies, but had the shortest wavelengths. 


Overview:

    The biggest challenge in this lab was differentiating become the colors.  Deciding upon what color flame was produced by the different ions was very subject especially when trying to differentiate between the various colors of red and orange, between blue and green, and between pale blue and bluish white.  Some of the solutions produced colors so similar we had to guess which ions were present.  This was especially problematic with solution #1 and solution #8, both of which produced an orange yellow flame. Another issue we encountered with trying to determine the flame color was the fact that the iron wire loop we were using to burn the solution would start to turn orange when heated and alter the color of the flame.  In addition, because the iron wire loops were had been frequently used before, there was residue left behind which would also alter the flame color.   As a result, we had to guess which ions were present in a few of the solution. 
    An alteration to this lab which I think would be helpful in determine which ions were present in which solutions would be if when were given 10 solution in which the ions present were given. This way would could compare between the known and unknown solutions which would be help to determine what the colors produced by the ions looked like we burned. (This would prevent error due to the subjecting of differentiating between red and red orange and yellow is very subjective.)