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 bindEu(III)are required to enhance the emission. The figure shows standard addition experiments in which10.00mLof sample and20.00mLcontaining a large excess of organic additive were placed in 50-mL volumetric flasks. Then Eu(III) standards (0,5.00,10.00,or15.00mL) were added and the flasks were diluted to50.0mLwithH₂O. Standards added to tap water contained0.152ng/mL(ppb) of Eu(III), but those added to pond water were 100 times more concentrated (15.2 ng/mL).

(a) Calculate the concentration of Eu(III)(ng/mL) in pond water and tap water.

(b) For tap water, emission peak area increases by.4.61units when 10.00mL of 0.152ng/mL standard are added. This response is4.61 units/1.52ng = 3.03units per ng ofEu(III). For pond water, the response is12.5units when10.00mLof15.2ng/mLstandard are added, or0.0822units per ng. How would you explain these observations? Why was standard addition necessary for this analysis?

Concentration is defined as the abundance of an ingredient divided by the total volume of a mixture in chemistry.

Standards added to pond water contain 15.2 ng/mL of Eu(III).

Standards added to tap water contain 0.152 ng/mL of Eu(III).

Volume of pond water=10 mL

Volume of tap water =10 mL

Using the given figure, the pond water intercepts thex-axis at -14.6, so the volume of the pond water sample equals 14.6 mL.

Now, we can calculate the mass:

$$\begin{gathered} \mathrm{Mass}=\mathrm{Concentration}\times \mathrm{Volume} \\ =15.2\mathrm{ng}/\mathrm{mL}\times 14.6\mathrm{mL} \\ =221.92\mathrm{ng} \end{gathered}$$

Finally, we can calculate the concentration of Eu(III) in pond water:

$$\begin{gathered} \mathrm{Concentration}\left(\mathrm{w}/\mathrm{v}\right)=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Eu}\left(\mathrm{III}\right)}{\mathrm{Volume}\mathrm{of}\mathrm{pond}\mathrm{water}} \\ =\frac{221.92\mathrm{ng}}{10\mathrm{mL}} \\ =22.192\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Now for the tap water, using the given figure, the tap water intercepts thex-axis at-6.0, so the volume of the tap water sample equals6.0mL.

Now, we can calculate the mass:

$$\begin{gathered} \mathrm{Mass}=\mathrm{Concentration}\times \mathrm{Volume} \\ =0.152\mathrm{ng}/\mathrm{mL}\times 6.0\mathrm{mL} \\ =0.912\mathrm{ng} \end{gathered}$$

Finally, we can calculate the concentration of Eu(III) in tap water:

$$\begin{gathered} \mathrm{Concentration}\left(\mathrm{w}/\mathrm{v}\right)=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Eu}\left(\mathrm{III}\right)}{\mathrm{Volume}\mathrm{of}\mathrm{pond}\mathrm{water}} \\ =\frac{0.912\mathrm{ng}}{10\mathrm{mL}} \\ =0.0912\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Given information:

For tap water:

Peak area increases by 4.61 when 10 mL of 0.152 ng/mL standards are added.

Response =3.03 per ng of Eu(III)

For pond water:

Peak area increases by 12.5 units when 10 mL of 15.2ng/mL standards are added.

First, let's calculate the response of the pond water:

$$\begin{gathered} \mathrm{Response}=\frac{12.5\mathrm{units}}{152\mathrm{ng}} \\ =0.0822 \end{gathered}$$

As we can see the response of tap water (3.03) is approximately 37 times larger than the response of pond water (0.0822), this is due to a matrix effect, something in pond water decreases the Eu(III) emission.

For this study, the standard addition is required since it allows us to determine the real response of the sample's matrix, resulting in a more accurate analysis.