Trace amounts of rare elements are found within groundwater and are of interest
ID: 982878 • Letter: T
Question
Trace amounts of rare elements are found within groundwater and are of interest to geochemists. Europium and terbium are lanthanide-series elements that can be measured from the intensity of their fluorescence emitted when a solution is illuminated with ultraviolet radiation. Certain organic compounds that bind Eu(lll) and Tb(lll) enhance the emission, and substances found in natural waters can decrease the emission. For that reason it is necessary to use standard additions to the sample to correct for such interference. The graph at the right shows the result of such an experiment in which the concentration of Eu(lll) and Tb(lll) was measured in a sample of groundwater. In each case 10.00 mL of sample solution and 15.00 mL of of organic additive were placed in 50-mL volumetric flasks. Eu(lll) standards (0, 5.00, 10.00, 15.00, and 20.00 mL) were added and the flasks were diluted to 50.0 mL with water. Eu(lll) analysis The Eu(lll) standards added to the groundwater sample contained 0.132 ng/mL (ppb) of Eu(lll). Calculate the concentration of Eu(lll) in the groundwater sample. Tb(lll) analysis The Tb(lll) standards added to the groundwater sample contained 13.2 ng/mL (ppb) of Tb(lll). Calculate the concentration of Tb(lll) in the groundwater sample.Explanation / Answer
In the graph the extrapolation of the x-axis is the concentration of the unknown after it has been diluted to final volume which is equal to -23ml of standard
given dilution factor= 50/10= 5
the original unknown had been diluted by a factor of 5(from 10 up to 50.0 mL) by standard additions, then the original unknown was 115ml of standard sample
1ml of standard contains 0.132ng of Eu(III)
So 115ml of standard contains 15.18ng which is equal to concentration of Eu(III) in the ground water
Conc. of Eu(III) in the ground water is 15.18ng/ml
In the graph the extrapolation of the x-axis is the concentration of the unknown after it has been diluted to final volume which is equal to -9.82ml of standard
given dilution factor= 50/10= 5
the original unknown had been diluted by a factor of 5(from 10 up to 50.0 mL) by standard additions, then the original unknown was 49.1ml of standard sample
1ml of standard contains 13.2ng of Tb(III)
So 49.1ml of standard contains 648.12ng which is equal to concentration of Tb(III) in the ground water
Conc. of Tb(III) in the ground water is 648.12ng/ml
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