Low iron concentration (as low as 0.02nM) in the open ocean limits phytoplankton growth. Preconcentration is required to determine such low concentrations. Tracefrom a large volume of seawater is concentrated onto achelating resin column. The column is then rinsed with 10mLof 1.5M high-purity water and eluted withofhigh-purity${\mathbf{HNO}}_{\mathbf{3}}$.

(a) For each sample, seawater is passed through the column for 17hours at $\mathbf{10}\mathbf{}\mathbf{mL}\mathbf{/}\mathbf{min}$. How much is the concentration of ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$in the 10mLof ${\mathbf{NHO}}_{\mathbf{3}}$eluate increased by this preconcentration procedure?

(b) What is the concentration of ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$in the seawater when 57 nm ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$is found in the nitric acid eluate?

(c) Reagent-grade concentrated nitric acid is 15.7 M and contains $\mathbf{\le }\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{ppm}$ iron. What would be the apparent concentration of Fe (nM) in a seawater blank if reagent-grade acid were used to prepare the 1.5M ${\mathbf{HNO}}_{\mathbf{3}}$eluent?

Step by step solution

01

a) Calculate the sample volume

$\mathrm{V}=10\mathrm{mL}/\min \times 17\mathrm{h}\times 60\min /\mathrm{h}/1000\mathrm{mL}/\mathrm{L}=10.2\mathrm{L}=10200\mathrm{mL}$

Then calculate increase in ${\mathrm{Fe}}_3$concentration

$$\begin{gathered} \mathrm{concentration}\left(\mathrm{increased}\right)=\frac{\mathrm{total}\mathrm{sample}\mathrm{volume}}{\mathrm{eluted}\mathrm{sample}\mathrm{volume}} \\ \mathrm{concentration}\left(\mathrm{increased}\right)=\frac{10200\mathrm{mL}}{10\mathrm{mL}}=1020\mathrm{times} \end{gathered}$$

02

b) Now calculate the concentration of Fe3+ in the seawater when 57 nm Fe3+ is found in the nitric acid eluate

$$\begin{gathered} \mathrm{c}\left({\mathrm{Fe}}^{3+}\right)={\mathrm{c}}_1/\mathrm{concentration}\left(\mathrm{increased}\right) \\ \mathrm{c}\left({\mathrm{Fe}}^{3+}\right)=57\mathrm{nM}/1020 \\ \mathrm{c}\left({\mathrm{Fe}}^{3+}\right)=0.056\mathrm{nM} \end{gathered}$$

03

c) Then calculate apparent concentration of Fe (nM) in a seawater blank if reagent-grade acid were used to prepare 1.5M HNO3 the eluent

$$\begin{gathered} \le 200\mathrm{ppb}/1020=0.196\mathrm{ppb}=0.196\times {10}^{-9}\mathrm{g}/\mathrm{mL}=0.196\times {10}^{-6}\mathrm{g}/\mathrm{L} \\ 0.196\times {10}^{-6}\mathrm{g}/\mathrm{L}/55.85\mathrm{g}/\mathrm{mol}=0.00351\times {10}^{-6}\mathrm{M}=3.51\mathrm{nM} \end{gathered}$$

Note that 1ppm=1000ppb

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