In practice, one of the most widely used polymers is PDMS because it is a material that has good partitioning properties such as low mass transfer resistance for hydrophobic and nonpolar compounds (Mayer et al. 6148) and recently there has been many studies investigating PDMS absorptive properties, however, only a few have focused on the ability of PDMS to release absorbed chemicals into water. In general, experiments have proven that PDMS can be used to calculate the equilibrium concentrations of many different compounds dissolved in water once the concentrations of these analytes absorbed in the pellets are determined. Since PDMS is not affected by the nature of the matrices or by exposure to hydrophilic chemicals, it can be used for …show more content…
Assuming that the process exhibit a labguimir equilibrium distribution, the thermodynamic behavior of the partitioning process can be modeled by using a first order one compartment equation (Vrana et al. 845).
C_PDMS (t)=C_w (k_1/k_(2 ) )(1-e^(-tk_2 ))
C_w represents the concentration of analyte in water, C_s(t) the concentration of analyte at a given time on the PDMS, and k1/k2 are the uptake and release rates constants in the kinetic process. The exchange of solute into and out of a passive sampling system usually follows a predictable languimuir isotherm as shown in figure 1 (Seethapathy, Górecki, and Li 234).Labumier isotherms occurs when… Figure 1. Passive sampling devices operate in two main regimes (kinetic and equilibrium).
Therefore, a mathematical equation that relates the concentration of analyte absorbed in the PDMS pellets as a function of uptake or release rate constants k, equilibrium concentration of the compounds in the polymer and the t95% can be used (Smith, Oostingh, and Mayer 55). T95% is the time to reach 95% of the maximum measure concentration or steady state (Smith, Oostingh, and Mayer 55), so at t95% the concentration of analyte at time t is 0.95 of the equilibrium concentration (Ct = 0.95Ceq) (Brown and A. 4097).
C_(PDMS(t))=C_PDMS(eq) (1-e^(-tk_upload )) t_(95%)=3/k Where C_(PDMS(t)) represents the concentration of analyte that is absorbed on
At the end of the 75 minutes, two 1.0 mL samples of the albumin/glucose solution from the beaker were added to two test tubes labeled solution end. Then, the dialysis tube was removed from the beaker and rinsed off with distilled water. Once the tubing was rinsed and blotted dry the final water weight was recorded. After measuring the final water weight, the contents in the tubing was dumped into a beaker and 1.0 mL of starch/sodium sulfate solution was added to two test tubes labeled bag end (Keith et al., 2010).
3ml of sample was taken first flask at 4 minutes and added to the appropriate tube of sodium hydroxide, from the second flask at 4.5 minute and so on, each flask was sampled at 30 second intervals. The sampling was then repeated starting at 8,12,16 minutes. The final sample from the last flask was taken at 18.5 minutes. Once the sampling was completed, measurements of absorbance were obtained for solution in each tube at 405 nm.
10 microliters of the sample is then added and the assay absorption is measured at 340nm. If absorbance was above 1.5, samples were diluted.
In this experiment, there were 6 different test tubes, each containing a different amount of phosphate concentration. Using a balance, 1 g of phosphate in a 400 mL beaker was measured and then inserted into a 100 mL graduated cylinder. 5 mL of ammonium vanadomolybdate (AVM) was added, along with deionized water until the solution reached to 25 mL. Then the solution was poured into one of the six test tubes. Another test tube contained 2 g of phosphate (phosphate was again measured using the balance) and the step described earlier was repeated until five of the six test tubes were filled, each with the same volume of 25 mL. The sixth test tube was filled with only deionized water, with 25 mL as the volume. A spectrophotometer was then used to determine the percent transmittance for each solution. The absorbance was also calculated using the equation A = -log (T). Each given amount of phosphate was converted into moles and then applied the value of .025 L for the volume to calculate the molarity of phosphate in each solution.
There were several steps completed to prepare for the experiment. Three dialysis tubes were filled with approximately the same volume of distilled water and then were tied shut. The initial mass (in grams) of the tubes was taken using a triple beam scale. I then filled three 500 mL beakers with 400 mL of water each and dissolved different masses of solute (table sugar) in each beaker in order to make 5%, 10%, and 20% solutions. The beakers were labeled accordingly, and then 20 g, 40 g, and 80 g (respectively) of table sugar was weighed out using a digital scale and placed into the corresponding beakers. The sugar was stirred in using a stirring rod until all of the solute was completely dissolved.
The independent variable in this lab is the molarity of sucrose each dialysis bag is filled with. The time (30 minutes), the temperature (23C) and the type of dialysis tubing used are all constants.
#2. 50.00 mL of solution #2 was removed from the flask using the volumetric pipet and was placed into a 100 mL volumetric flask. It was diluted to the line, stopped, and mixed gently. The flask was labeled solution #3. 10.00 mL of solution #3 was removed with the volumetric pipet and placed into a 25 mL volumetric flask and was diluted to the line. The flask was stopped and mixed gently.
This allows for the distinction between the DCM layer and water layer, since the DCM layer will collect at the bottom of the test tube. It can then be transferred out of the test tube and collected, thus separating the organic compound from the
Meanwhile, tubes 2, 4,6 & 8 used .5ml of Dopa added instead. Before we began, we calculated the concentration of Dopa first. We measured the increase of the absorbance at 475 nm, which helped to determine the rate of the reactions. As the reaction was initiated, the absorbance values were read out loud while the other timed the reaction and recorded the measurement at 15 seconds, 30 seconds, 1 minute, 2 minutes and 4 minutes
From figure 2, we can observe diffusion taking place.’S’ represent starch, ‘i’ represents iodine and ‘g’ represent glucose. As initially, 1 inch dialysis tube consists of iodine and glucose and 3 inch dialysis tube consists of starch. However, after 1 hour, the iodine molecules got diffused into 3 inch dialysis tube and also the water in the beaker turned slightly cloudy in the final stage.
Tube three contained 1.0 mL of Buffer, pH 6.0 0.1 M NaPO4 including 0.4 mL of 0.006 M catechol, with 600 uM of substrate concentration and 1.6 mL of distilled water. Tube four contained 1.0 mL of Buffer, pH 6.0 0.1 M NaPO4 including 0.8 mL of 0.006 M catechol, with 1200 uM of substrate concentration and 1.2 mL of distilled water. And finally, Tube five contained 1.0 mL of Buffer, pH 6.0 0.1 M NaPO4 including 1.6 mL of 0.006 M catechol, with 2400 uM of substrate concentration and 0.4 mL of distilled water. All five of the test tubes also contained 1.0 mL of enzyme, which would be added right before testing each tube for absorbance. After preparing all of the test tubes, each tube was inserted in the spectrophotometer one-by-one and recorded its change in absorbance for five minutes.
Incorporation of assay controls included setting up a spectrophotomer and running the chart recorder with a full-scale deflection before the start of the assay. The set recorder had a corresponding value of 1 for the change in the absorbance. Therefore, prior testing was done to observe whether a change occurred in the readings. This helped to indicate that the results were valid, as they could have been affected by a fault during the setting up of the spectrophotometer. On the other hand this was considered as one of the controls for the experiment. Nevertheless, a new cuvette had to be used for each assay.
Manipulated with settings on the program to display a graph showing absorbance vs. concentration. Collected absorbance data for the three remaining known solutions to obtain a linear fit1.
Six plastic cups were obtained and labeled 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M. Next, each of the six dialysis tubes were knotted on one end and filled with the sucrose samples, and then tied off. These samples were then dried by patting with a paper towel, weighed and placed into their corresponding cups. The mass of each sample was recorded in table 2 on the data sheet. Each cup was then
where, kd is the intra-particle diffusion rate constant (mg g-1 min-1/2) and C (mg g-1) is the intercept of the vertical axis. The intercept reflects the boundary layer effect and the linearity of the plot of q versus t1/2 denotes the presence of intra-particle diffusion. If C = 0, then intraparticle diffusion is the rate-controlling step, whereas if C ≠ 0, intra-particle diffusion is not the rate-controlling step, but film diffusion plays an important role in the metal sorption process. (Al-Degs et al. 2006). As may be seen in Fig. 4, there is a linear dependence of q on t1/2 at all temperatures