IChO 2002 · Theory problem 2

Nitrogen Cycle in Nature

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Ammonia is a toxic substance to marine animals at levels exceeding 1 ppm. Nitrifying bacteria play an important role in the conversion of \(\ce{NH3}\) first to nitrite and then to nitrate, the storage form of nitrogen in the soil.

Figure 1

\(\ce{NADH}\) is the biochemical reducing agent of the coenzyme nicotinamide dinucleotide (NAD), \(\ce{NAD+}\) is the oxidized form of the coenzyme NAD.

Figure 2

2.1 Give the oxidation states of N in the following series: (Use the boxes below the compounds)

Figure 3

The spectrophotometric analysis of nitrite is based on a reaction with an indicator. The coloured product then obtained has an absorbance maximum at \(\lambda = 543\ \mathrm{nm}\). For quantitative analyses a calibration curve has to be made, in which absorbance at the maximum absorbance wavelength \(\lambda = 543\ \mathrm{nm}\) is plotted against nitrite concentration in a series of standards.

2.2 The measurements are performed at the wavelength with the maximum absorbance because:

  • ☐ There is no interference of impurities.
  • ☐ There is no contribution of stray light.
  • ☐ There is optimal accuracy of the measurement.
  • ☐ None of these statements.

Mark the correct answer.

The absorption is measured with a single beam spectrophotometer. However 5 % of the light, the so-called stray light \(I_s\), strikes the detector directly (see Figure 2).

Figure 4

Figure 2

2.3 Calculate the value of the absorbance \(A\) shown by the spectrophotometer if \(\varepsilon = 6\,000\ \mathrm{M^{-1}\,cm^{-1}}\), \(l = 1\ \mathrm{cm}\) and \(c = 1\times10^{-4}\ \mathrm{M}\)

For a nitrite determination in water the following data have been measured.

Table 1

concentration of nitrite nitrogen (ppm) absorbance at 543 nm (1.000 cm cell)
blank 0.003 (due to impurities in the solvent)
0.915 0.167
1.830 0.328

2.4 Determine (show calculation) from the data given above, using the values corrected for the solvent impurities, the slope \(m\) and the intercept \(b\) of the calibration curve \(A = mc + b\).

The duplicate analyses of a water sample are given below. The measurements have been performed at a wavelength of 543 nm and in a 2.000 cm cell.

Table 2

water sample absorbance
analysis 1 0.562
analysis 2 0.554

For the calculation of the concentration of the nitrite nitrogen (\(c\) in ppm) the equation obtained by the method of least squares

\[ \text{corrected absorbance}=0.1769\,c+0.0015 \tag{a} \]

may be applied, using the measurements in a 1.000 cm cell.

2.5 Calculate the average nitrite nitrogen concentration in ppm and \(\mathrm{\mu g\,cm^{-3}}\).

Hint: Take the blank from problem 2.4.

Official English text of the problem.

Answers

  • 2.1: Ammonia, \(\ce{NH3}\), SMILES N: −3; nitrite ion, \(\ce{NO2-}\), SMILES O=N[O-]: +3; nitrate ion, \(\ce{NO3-}\), SMILES [O-][N+](=O)[O-]: +5.
  • 2.2: Select the third statement: “There is optimal accuracy of the measurement.”
  • 2.3: \(\boxed{A=0.540}\).
  • 2.4: \(\boxed{m=0.176\ \mathrm{ppm^{-1}},\quad b=0.003}\).
  • 2.5: \(\boxed{\overline c=1.55\ \mathrm{ppm}=1.55\ \mathrm{\mu g\,cm^{-3}}}\), expressed as nitrogen and scaling the solvent blank with path length. The alternative blank interpretation, subtracting \(0.003\) unchanged from each \(2.000\ \mathrm{cm}\) reading, gives 1.56 in both units.

Solution

2.1

The sum of oxidation states equals the overall charge. Hydrogen has oxidation state \(+1\) and oxygen has oxidation state \(-2\) in these species. Writing \(x\) for the oxidation state of nitrogen:

\[ \begin{aligned} \ce{NH3}:&\quad x+3(+1)=0 &&\Longrightarrow x=-3,\\ \ce{NO2-}:&\quad x+2(-2)=-1 &&\Longrightarrow x=+3,\\ \ce{NO3-}:&\quad x+3(-2)=-1 &&\Longrightarrow x=+5. \end{aligned} \]

Fill the left, middle and right boxes with −3, +3 and +5, respectively.

The ionic SMILES given above represent individual resonance contributors; the actual nitrogen–oxygen bonding is delocalized.

2.2

The Beer–Lambert law is

\[ A=\varepsilon lc. \]

At the absorption maximum, \(\varepsilon\) is greatest, so the concentration sensitivity,

\[ \frac{\partial A}{\partial c}=\varepsilon l, \]

is greatest. For a given small absorbance uncertainty, this reduces the uncertainty in the inferred concentration.

Also, at a smooth absorption maximum,

\[ \frac{d\varepsilon}{d\lambda}=0, \]

so small wavelength-setting errors have a smaller effect on absorbance. These considerations support the third statement. Measuring at the maximum does not eliminate impurity absorption or stray light.

2.3

The absorbance of the light passing through the solution is

\[ A_{\mathrm{true}} =\varepsilon lc =(6000\ \mathrm{M^{-1}\,cm^{-1}}) (1\ \mathrm{cm}) (1\times10^{-4}\ \mathrm{M}) =0.600. \]

Its transmittance is therefore

\[ T_{\mathrm{true}}=10^{-0.600}=0.2511886. \]

As shown by the branching diagram, \(5\%\) of the incident light bypasses the cell, while \(95\%\) passes through it. The total fraction reaching the detector is

\[ T_{\mathrm{measured}} =0.05+0.95T_{\mathrm{true}} =0.05+0.95(0.2511886) =0.2886292. \]

Thus,

\[ \boxed{ A_{\mathrm{measured}} =-\log_{10}(0.2886292) =0.5396597 \approx0.540 }. \]

The stray light increases the detected intensity and consequently lowers the displayed absorbance.

2.4

Subtract the solvent blank from each standard:

\[ A_1=0.167-0.003=0.164, \qquad A_2=0.328-0.003=0.325. \]

The slope of the line through the two corrected standards is

\[ m=\frac{A_2-A_1}{c_2-c_1} =\frac{0.325-0.164}{(1.830-0.915)\ \mathrm{ppm}} =0.175956\ \mathrm{ppm^{-1}}. \]

Its intercept is

\[ b=A_1-mc_1 =0.164-(0.175956\ \mathrm{ppm^{-1}})(0.915\ \mathrm{ppm}) =0.003. \]

Hence,

\[ \boxed{m=0.176\ \mathrm{ppm^{-1}},\qquad b=0.003} \]

and the calibration line is

\[ \boxed{A=(0.175956\ \mathrm{ppm^{-1}})\,c+0.003}. \]

Absorbance and \(b\) are dimensionless. Subtracting the solvent blank does not force the line through the two measured standards to have zero intercept.

2.5

Use the supplied least-squares calibration,

\[ A_{\mathrm{corrected,\;1\,cm}} =(0.1769\ \mathrm{ppm^{-1}})\,c+0.0015. \]

The sample readings were obtained in a \(2.000\ \mathrm{cm}\) cell, whereas both this calibration and the tabulated solvent blank refer to \(1.000\ \mathrm{cm}\).

Under the Beer–Lambert law, the solvent-impurity absorbance scales with path length. Thus the blank in the \(2.000\ \mathrm{cm}\) cell is

\[ A_{\mathrm{blank,\;2\,cm}} =0.003\frac{2.000}{1.000}=0.006. \]

Subtracting this blank and then halving is equivalent to halving the readings first and subtracting \(0.003\):

\[ A_{1,\mathrm{corrected,\;1\,cm}} =\frac{0.562}{2}-0.003=0.278, \]

\[ A_{2,\mathrm{corrected,\;1\,cm}} =\frac{0.554}{2}-0.003=0.274. \]

Therefore,

\[ c_1=\frac{0.278-0.0015}{0.1769\ \mathrm{ppm^{-1}}} =1.56303\ \mathrm{ppm}, \]

\[ c_2=\frac{0.274-0.0015}{0.1769\ \mathrm{ppm^{-1}}} =1.54042\ \mathrm{ppm}. \]

The average is

\[ \boxed{ \overline c=\frac{1.56303+1.54042}{2} =1.55172\ \mathrm{ppm} \approx1.55\ \mathrm{ppm} }. \]

For dilute water, taking \(\rho\approx1.00\ \mathrm{g\,cm^{-3}}\),

\[ 1\ \mathrm{ppm} =1\ \mathrm{mg\,kg^{-1}} \approx1\ \mathrm{mg\,L^{-1}} =1\ \mathrm{\mu g\,cm^{-3}}. \]

Consequently,

\[ \boxed{\overline c=1.55\ \mathrm{\mu g\,cm^{-3}}}. \]

Both results are nitrite nitrogen concentrations, so no conversion to the total mass of the nitrite ion is required.

The solution was checked against the official solution.