Europium is a lanthanide element found at parts per billion levels in natural wa

Question

Europium is a lanthanide element found at parts per billion levels in natural waters. It can be measured from the intensity of orange light emitted when a solution is illuminated with ultraviolet radiation. Certain organic compounds that bind Eu(III) are required to enhance the emission. The figure below shows standard addition experiments in which 10.00 mL of sample and 20.00 mL containing a large excess of organic additive were placed in 50-mL volumetric flasks. Then Eu(III) standards (0, 5.00, 10.00, or 15.00 mL) were added and the flasks were diluted to 50.0 mL with H, 0. Standards added to tap water contained 0.153 ng/mL (ppb) of Eu(III), but those added to pond water were 100 times more concentrated (15.3 ng/mL) Calculate the concentration of Eu(III) (ng/mL) in pond water and tap water. pond water ng/mL water ng/mL For tap water, emission peak area increases by 4.57 units when 10.00 mL of 0.153 ng/mL standard are added. This response is 4.57 units/1.53 ng = 2.99 units per ng of Eu(III). For pond water, the response is 12.6 units when 10.00 mL of 15.3 ng/mL standard are added, or about 0.0824 units per ng. How would you explain these observations? Why was standard addition necessary for this analysis? There is likely to be a in which something in the water the Eu(III) emission. The standard addition method allows us to accurately determine the analyte concentration in the .

Explanation / Answer

From the the above figure the intercept for tap water is -6.0 mL, corresponding to an addition of

(6.0mL) (0.153 ng/mL) = 0.918 ng Eu(III).

This much of Eu (III) is in 10.00 mL of Tap water, so the concentration is 0.918 ng / 10 mL = 0.0918 ng/mL.

For pond water the intercept of - 14.6 mL corresponds to an addition of

(14.6 mL)(15.3 ng/mL) = 223.38 ng Eu(III)

This much of Eu (III) is in 10.00 mL of pond water, so the concentration is 223.38 ng /10 mL = 22.338 ng/mL.

b)

Added standard Eu (III) gives a response of 2.99 unit / ng for tap water and 0.0824 units/ng for pond water. The relative response is

2.99/0.0824 = 36.29 times greater in tap water than in pond water. This should be probably due to the matrix effect in which something in pond water decreases the Eu(III) emission. By the use of standard addition, we can measure the response in the actual sample matrix. Even though Eu(III) in pond water and tap water do not give equal signals, we cab measure the actual signal in each matric an can therefore carry out the accurate analysis.

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