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Trace amounts of rare elements are found within around water and are of interest

ID: 1065661 • Letter: T

Question

Trace amounts of rare elements are found within around water and are of interest to geochemists Europium and tectum 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(III) 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(III) and Tb(III) was measured in a sample of groundwater In each case 10 00 mL ot sample solution and 15.00 ml of to organic additive wore placed in 50-ml volumetric flasks Eu(III) standards (0. 5 00, 10.00. 15 00. and 20 00 mL) were added and the flasks were drifted to 50 0 ml with water The Eu(III) standards added to tho groundwater sample contained 0 16G ng/mL (ppb) of Eu(III) Calculate the concentration of Eu(III) the groundwater sample The Tb(III) standards added to the groundwater sample contained 16 6 ng/ml (ppto) of Tb(III) Calculate the concentration of Tb(III) In the groundwater sample

Explanation / Answer

From the plot

concentration of Eu(III) standard in sample solution = 0.166 ng/ml x 5 ml/50 = 0.0166 ng/ml when 5 ml std was added

let Cx be the concentration of Eu(III) in ground water

Cx/(Cx + 0.0166) = 14/18

18Cx = 14Cx + 0.2324

Cx = 0.2324/4 = 0.0581 ng/ml in 50 ml solution

So concentration of Eu(III) in original 10 ml sample = 0.0581 x 50/10 = 0.2905 ng/ml

concentration of Tb(III) standard in sample solution = 16.6 ng/ml x 5 ml/50 = 1.66 ng/ml when 5 ml std was added

let Cx be the concentration of Tb(III) in ground water

Cx/(Cx + 1.66) = 2.5/3.5

3.5Cx = 2.5Cx + 4.15

Cx = 4.15/1 = 4.15 ng/ml in 50 ml solution

So concentration of Tb(III) in original 10 ml sample = 4.15 x 50/10 = 20.75 ng/ml

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