Question

Many metals in seawater can be preconcentrated for analysis by coprecipitation with\({\rm{Ga}}{({\rm{OH}})_3}\). A \(200 - \mu {\rm{L}}\)HCl solution containing \(50\mu {\rm{g}}\)of \({\rm{G}}{{\rm{a}}^{3 + }}\)is added to \(10.00\;{\rm{mL}}\) of the seawater. When the \({\rm{pH}}\)is brought to \(9.1\)with\({\rm{NaOH}}\), a jellylike precipitate forms. After centrifugation to pack the precipitate, the water is removed and the gel is washed with water. Then the gel is dissolved in \(50\mu {\rm{L}}\) of \({\rm{HN}}{{\rm{O}}_3}\)and aspirated into an inductively coupled plasma for atomic emission analysis. The preconcentration factor is\(10\;{\rm{mL}}/50\mu {\rm{L}} = 200\). The figure shows elemental concentrations in filtered and unfiltered seawater as a function of depth near hydrothermal vents.

(a) What is the atomic ratio (Ga added): (Ni in seawater) for the sample with the highest concentration of\({\rm{Ni}}\)?

(b) The results given by gray lines were obtained with seawater samples that were not filtered prior to coprecipitation. Colored lines are from filtered samples. Results for Ni do not vary between the two procedures, but results for Fe vary. Explain what this means.

Short answer

(a)The \({\rm{Ga}}/{\rm{Ni}}\)atomic ratio is \(53\).

(b)Results for Ni do not vary between the two procedures, but results for Fe vary are explained clearly.

Step-by-step solution

Derivation of atomic ratio.

The atomic ratio is a measure of the proportion of one type of atom I to another type of atom (ii) (j). The atomic percent (or at. percent) is a closely related concept that expresses the percentage of one type of atom relative to the total number of atoms. The molar fraction or molar percent are the molecular equivalents of these concepts.

Determining the amount of moles.

In this task we have a figure that shows elemental concentrations in filtered and unfiltered seawater as a function of depth near hydrothermal vents.

a) Determine the atomic ratio (Ga added) : (Ni in seawater) for a sample with the greatest concentration of Ni

As seen from the figure the highest concentration of Ni would be \(80{\rm{ng}}/{\rm{mL}}\)In \(10\;{\rm{mL}}\) of the sample there would be \(800{\rm{ng}}\)of \({\rm{Ni}}\)

800 ng of Ni would give us the following amount of moles:

\(n({\rm{Ni}}) = \frac{m}{M}\)

\(n({\rm{Ni}}) = \frac{{800 \times {{10}^{ - 9}}\;{\rm{g}}}}{{58.6934\;{\rm{g}}/{\rm{mol}}}}\)

\(n({\rm{Ni}}) = 1.36 \times {10^{ - 8}}\;{\rm{mol}}\)

\(50\mu {\rm{g}}\)of Ga are added to 800 ng of Ni which would give us the following amount of Ga:

\(n({\rm{Ga}}) = \frac{m}{M}\)

\(n({\rm{Ga}}) = \frac{{50 \times {{10}^{ - 6}}\;{\rm{g}}}}{{69.723\;{\rm{g}}/{\rm{mol}}}}\)

\(n({\rm{Ga}}) = 7.17 \times {10^{ - 7}}\;{\rm{mol}}\)

Determining the atomic ratio.

Next calculating the \({\rm{Ga}}/{\rm{Ni}}\)atomic ratio

\({\rm{Ga}}/{\rm{Ni}} = \frac{{n({\rm{Ga}})}}{{n({\rm{Ni}})}}\)

\({\rm{Ga}}/{\rm{Ni}} = \frac{{7.17 \times {{10}^{ - 7}}\;{\rm{mol}}}}{{1.36 \times {{10}^{ - 8}}\;{\rm{mol}}}}\)

\({\rm{Ga}}/{\rm{Ni}} = 53\)

Therefore the \({\rm{Ga}}/{\rm{Ni}}\)atomic ratio is \(53\).

Analysing the results.

b) Explaining the reason why do results for Ni don 't vary between the two procedures, but results for Fe vary

• In these procedures all of the Ni is in the solution because its concentration didn't decrease with filtration.
• Considering that filtration did remove most of the Fe it would be present as a suspension of solid particles.