Creating Solutions of Standard Molarity

Transcription

1 Roberts 1 Zoë Roberts Chemistry Lab Creating Solutions of Standard Molarity I. Purpose When dealing with high concentrations it is known that there is a large amount of solute or substances dissolved in a unit of solvent, usually a liquid. In this lab the focus was on creating a solution of the correct molarity, which is found by dividing the number of moles of solute by the number of liters in the solution. Once the correct molarity was calculated of each solution used, the next step was to make a correlation chart that compared concentration and absorbance. Because the class was working with homogeneous mixture there was not a clear way to visually detect the differences in concentration. Using a Spec -20 made it capable to see how much light is absorbed, the particular wavelengths of light absorbed are found to be characteristics of the species. II. Procedure The first thing to do was measure the absorbance of standards. After calibrating the Spec-20 next was to fill a cuvette with the.200 M standard solution of CuSO4-5H2O leaving at least 1 cm unfilled at the top of the cuvette. Finding the maximum absorbance wavelengths for wavelengths-580, 590, and 600nm. After finding the maximum wavelength it was important to set the Spec-20 to the wavelength for maximum absorbance. Measuring the percent transmittance for the.500m,.200m and.100 M and.050m CuSO4-5H2O standard solutions. After this we had to measure the concentration of solutions. To do this one would start by first creating a 20.0ml of a.500m solution from the solid CuSO4-5H2O. After mixing the water and the solid and making sure the crystals had dissolved the percent transmittance of

2 Roberts 2 our solution in the Spec-20 was checked. From this solution a different solution was made of exactly 20.0 ml of a.200 M CuSO4-5H2O. It was important to figure out how much water was needed to use to dilute our sample. Next the percent transmission of the solution in the Spec-20 was checked to make sure the concentration was correct. Repeat this process for 20.0ml of.100m CuSO4-5H2O from our.200m solution and then 20.0ml of.050m CuSO4-5H2O from our.100m solution. Lastly fill a clean cuvette with the solution of CuSO4-5H2O not knowing the concentration. We measured out the percent transmittance of this solution and recorded the value. III. Results M= The molarity equation used to create 20ml of.5m solution from the solid CuSO4-5H2O is shown above where M is the molarity, which is the calculated concentration of the solvent. Moles is the amount of moles in the solution which is divided by the volume in liters of solvent. Mi x Vi = Mf x Vf In the above equation is an example of a dilution equation used in the lab. This was used to create the.200m,.100m, and.050m solutions of CuSO4-5H2O. The Mi and Vi are the initial Molarity and initial volume and the Mf and Vf are the final molarity and the final volume. A= - log (10) The above equation is used to calculate the absorbance of the solutions using the respective percent transmittance. Where A is the absorbance which is the product when log and the percent transmittance (T) is multiplied together. This equation was used in the below table.

3 Roberts 3 Table 1: Absorbance of Solutions Concentration %T Absorbance.5M 9% M 39%.409.1M 50% M 63%.201 Unknown 49%.310 Data from table 1 is used in the graph shown in figure Absorbance vs Concentration Series M.1M Unknown.2M.5M Absorbance is shown in the Y value, Concentration is showed as the x value. IV. Discussion

4 Roberts 4 The standard value of the.5m sample of CuSO4 was 32 percent transmission and the absorbance value was.495. The team made data differed. For the.5m sample of CuSO4 was 9 percent transmission and the absorbance value was For the calculations to find the dilution the equation should be set up using Mi x Vi = Mf x Vf. The initial volume was the value that was unknown. In the 20ml of the.2m sample the calculations of the initial volume of the.5m sample was founded to be 8ml. Comparing then the initial values of the.2m sample and.1m sample and then again comparing the.1m sample with the.05m sample. The values of the absorbance from the standard and the team made values differed but still maintained the same concept that the lower the percent transmission the higher the absorbance value. The correlation between the absorbance and the concentration was a positive correlation. As the concentration was at an increased value the absorbance was also a larger value. For example.5m had a 1.05 absorbance value where.1m had a.301 absorbance value. V. Conclusion Looking at the data there is a correlation between the concentration and the percent transmission. The higher the concentration the lower the percent transmission is. For example.500m of CuSO4-5H2O has a 9 percent transmission while.05m of the solution has a 63 percent transmission. The correlation between percent transmission and absorbance though had a different relationship. The lower the percent transmission the higher the absorbance was of the solution. When the percent transmission is 9 the absorbance measure was 1.05 and when the percent transmission was at 63 percent the absorbance measured.901.

5 Roberts 5 VI. Sample Calculations: Dilution: Mi x Vi = Mf x Vf.200 x (Vi) =.100 x (20ml).200(Vi)/.200= 2/.200 Vi = 10 ML