Laboratory reports

Beoordeling 2.5
Foto van een scholier
  • Proef door een scholier
  • 5e klas vwo | 1672 woorden
  • 31 augustus 2009
  • 9 keer beoordeeld
Cijfer 2.5
9 keer beoordeeld

Taal
Engels
Vak
Laboratory reports of the following experiments:

· Acid rain stimulation
· Determination of Ca in shells using two different analysis method

Acid rain stimulation

1. Make a curve of pH versus time. What is the maximum and what the minimum pH?
The initial pH of the water is measured, which is 6.75.
After the vaporization of SO2 gas to the water in an enclosed space, its pH is measured continuously every 30 seconds until no change is observed.
Table1. Shows the change of pH during 6 minutes.
Time (sec) 0 30 60 90 120 150 180 210 240 270 300 330 360
pH 6.75 6.48 6.45 6.00 5.34 3.65 3.48 3.12 2.86 2.46 2.08 2.03 2.03
Graph 1. PH change measurements of water with SO2 during 6 minutes
Maximum of the curve is 6.75
Minimum of the curve is 2.03

2.
Also, pieces of magnesium, paper and shell was added to the made “rain water” and the changes were observed:
Material Magnesium Paper shell
pH 2.31 2.62 2.49

Observation Oxygen bubbles surrounded the magnesium metal No change, least acidity compared to other materials No change
Table 2. Different material is added to the made rainwater and the results are observed.

Conclusion:
To conclude, SO2 gas that is mostly released from industries cause neutral water to become acidic. It does have an impact on different materials, but because of shortage of time, we couldn’t get the proper result of the made acid rain effects on paper, magnesium and shells.

Determination of Ca in shells using two different analysis methods

During which step of your experimental procedure can sampling errors occur?
In order to asses during which step of the experimental procedure a sampling error could have occured sampling needs to be defined. According to Pierre (1998) “Sampling in a broad sense covers all the operations which, beginning with the object to be sampled, the ‘lot’, end with the fraction of pulp that is finally analysed in its entirety.” The aim of sampling is to obtain a representative sample from a defined population. In order to avoid sampling errors it first needs to be specified for which population of shells the calcium concentration should be measured. Therefore, it needs to be decided from which type of shells and from which geographical area our sample should be derived. A sampling error could therefore occur if the population to which the sample is supposed to be generalized is not clearly defined.
The sample of shell s is unbiased if all shells from the defined population have the same probability of being selected (Piere, 1998). Therefore a random sample of shells should be taken and each shells is supposed to be taken one by one. However, in this experiment no random sample of shells might have been taken.
The fundamental error, which is defined as the variance in the sample which is only caused by the heterogeneity of the sample, can be minimized by “grinding and milling”. Therefore, an error can occur in the experiment if the shells are not milled enough. Moreover, also the right concentration of the sample based on the “typ of heterogeneity and expected concentration” should be taken. Consistently, a sampling error could occur if not the proper concentration of shells is taken.
2. CaCO3 + 2HCl → CaCl (aq) + H2CO3 (aq)→Ca2+ + 2Cl- + H2O + CO2

3. Supposing the shells are 100% pure CaCO3, what is the Ca molarity (mol/L) of the 100 mL solution? What is the concentration in ppm?

4) M (Ca)=40 g/mol
M(C)=12 g/mol
M (O)= 16 g /mol
The total molecular mass of CaCO3t is 100g per mol. The atomic mass of calcium is 40% of the total molecular mass of CaCo3. In the experiment 5 g CaCO3 were used. In order to calculate the proportion of calcium of 5 gram CaC03 one has to multiply 5g (CaCO3) with 0,4.
0,4 X 0,5g= 0,2 g
To calculate the concentration in ppm one has to multiply the mass of Calcium in this solution by 1000.
1000 X 0,2 g=200 ppm
In 1 Liter the solution contains would contain 200 ppm Calcium. However, the calcium is dissolved in 100 ml.
Calculation of ppm in 100 ml
200 ppm/ 0,1= 2000 ppm
The concentration is 2000 ppm.
Calculating the Ca molarity of the 100 ml solution
n=m/M
=> n= 2g/40 g mol^-1
=> n= 0,05 mol
C=n/V
=> C=0,05 mol/0,1 L
=>c=0,5 mol/ L
The concentration of the solution is 0,5 mol per liter.

4. How many moles of Ca are present in the 20 mL solution that you titrate, supposing 100% pure calcium carbonate shells?
Calculation of the amount of substances of Calcium
C=n/V
=>n= c X V
=>n= 0,5 mol/L X 0,02l
n= 0,001 mol
20 ml solution contain 0,001 mol calcium.

5.

How many mL of 0.1M EDTA would you theoretically have to use, supposing 100% pure calciumcarbonate shells.
Determining the mass concentration of EDTA
We would need to use theoretically the same mass concentration of EDTA as we Calcium in the solution since one calcium ion reacts with one EDTA molecule to form a EDTA complex. For the reason that we have 0,001 mol Calcium we would need 0,001 mol EDTA if calcium reacts completely.

Determining the how many ml of the 0,1mol EDTA solution is needed to have 0,01 mol

C=0,1 mol/ l , n=0,001 mol
C=n/V
=>v=n/C
=>v= 0,01 / 0,1
=>v =0,01 L
Theoretical are 10 ml of the 0,1 molar EDTA solution needed.

6. How many mL EDTA did you actually use, if it deviates from the theoretical pure calcite titration value, explain why.

We used actually 33,63 ml of the EDTA solution. This are 22,63 ml too much than the theoretical titration value. This big difference can only be explained by a mistake during the experiment. One explanation could be that too much CaCO3 was added. Moreover, the volumetric flask might have not been clean well enough. Therefore some metals from earlier experiments might have still been in the volumetric flask and might have blocked the EDTA. The high calcium concentration could have also been caused by the fact that we did not homogenize the dilutions sufficiently.
Moreover, it was unclear whether the concentration of EDTA and Ca are equal if the solution is light or dark blue. At 33,63 ml the solution was dark blue. That means that we could have actually used too much EDTA.
The unexpected EDTA concentration could have also been caused by experimental mistakes during the titration itself. We did not clean the titration apparatus. Moreover, the morality of the EDTA solution might have been differed from 0,1 mol per liter.

7. Evaluate the precision of your titration.

As already said we did not clean the titration apparatus therefore might our titration have been not precise enough.

8. What is the Ca concentration in the shells based on the titration results.

1. Calculating the amount of substances of EDTA
C=n/V
n=c X V
n= 0,1 mol /L X 0,033L
n=0,0033 mol
Therefore the amount of substance of EDTA is O, OO33 mol the amount of substance of EDTA is equal to the amount of substance of calcium since EDTA and Calcium react with each other in a ratio of 1:1.

2. Calculating the concentration of Ca in the solution

C=n/V
=>c=0,0033 mol/ 0,1L
=> c= 0,033 mol L
In the concentration of calcium in the 100 ml solution is 0,033 mol/ L.

9. Draw the calibration curve for Ca measured on the AAS.
Ppm of Ca AAS
0 0
2 0.092
4 0.177
6 0,244
8 0.313
unknown 0,140

10. How did you dilute your sample for AAS measurement?
The AAS measurement measures the atomic absorption spectromy. According to Lambert-beer the more calcium is present in a solution the more radiation is absorpt by the calcium and the less light can be measured. The physical absorption of a given Ca stock solution (c=1000ppm) was measured and compared against the value of a solution with unknown Ca (the solution with the shells) concentration. However, only calcium concentrations ranging between 0 till 8 ppm can be measured. Therefore it needs to be ensured that the sample with the unknown Ca concentration (from the shell) has a calcium concentration which lies on the range of the Lambert-Beer. With the formula C1 X V1=C2 X V2 it can be determined how many ml should be taken from 100 ml dilutions (the solution with the shell) to receive a sample of 50 ml containing at least 4 ppm calcium.
C1 X V1=C2 X V2
=>2000 X V1=4 X 500 => V1= 1ml
1 ml of 100 ml dilutions should be taken to 49 ml distilled water to receive a solution which includes as the highest concentration 4 ppm. The solution should include a maximum value of 4 ppm if the shell consists of 100% pure carbon.
Calculation of the sample containing a given ca solution:
C1 (concentration of the stock solution (1000ppm Ca)) X V1=C2 (concentrations of Ca lying within the range of Lambert-Beer) X V2
=>1000 ppm X V1=C2( 0 ppm; 2ppm; 4ppm; 6ppm; 8ppm Ca) X 50 ml
To obtain a sample which contains 0 ppm (Ca) 50 ml of distilled water were used and o,1 ml stock (c(Ca)= 1000 ppm) were used. For a sample of 2 ppm Calcium concentration 49,9ml distilled water (DW) were used an d o,1 ml stock. For a sample with a 4 ppm Ca concentration 49,8 ml DW were used and 0,2 ml stock. For a sample with a 6 ppm Ca concentration 49,7 ml DW and 0,3 ml stock solution were used. For a sample containing 8 ppm Ca 0,4 ml stock solution and 49,6 ml distilled water were used.

11. What is the concentration in the shells based on the AAS measurement?
To obtain the concentration of the unknown sample one needs to locate the value obtained by the ASS of the unknown sample on the graph plotting the ASS results of known Ca concentrations. The regression curve for the sample is:
Y=0,0383X+0,0096 To calculate the x of the unknown concentration the ASS result is put into the equation.
=>0,140= 0,038 X + 0,0096
=> 0,140- 0,0096=0,038 X
=> X=3,43 ppm
To calculate the Ca concentration of the 500 ml dissolution we need to multiply 3,43 with 500. The 500 ml dissolution has a concentration of 1715 ppm Ca. The Ca concentration is as the result of the titration already indicated to high. Therefore, we can conclude that a mistake must have occurred during the experiment itself. As already said it is likely that the solutions were not homogenized sufficient or the experimenter must have added too much CaCO3.

12. Assuming the value given by AAS is correct, what is the average error (%) of the titrations?
The average error of the titration is: (1500 / 1715) x 100 = 87.5 %

References:

Pierre, G. (1998). Sampling for Analytical Purpose. United States: Libri

REACTIES

Log in om een reactie te plaatsen of maak een profiel aan.