IChO 2017 · Theory problem 5

Phosphate and silicate in soil

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Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:

Part A. Determination of total phosphate (\(\ce{PO4^3-}\)) and silicate (\(\ce{SiO4^4-}\))

A 5.00 gram of soil sample is digested to give a final volume of 50.0 \(\mathrm{cm^3}\) digesting solution which dissolves total phosphorus and silicon. The extract is analyzed for the total concentrations of phosphorus and silicon. The concentrations of phosphorus and silicon are found to be 5.16 \(\mathrm{mg\,dm^{-3}}\) and 5.35 \(\mathrm{mg\,dm^{-3}}\), respectively.

5.1.1 Determine the mass of \(\ce{PO4^3-}\) in mg per 1.00 g of soil.

5.1.2 Determine the mass of \(\ce{PO4^3-}\) in mg per 1.00 g of soil.

Part B. Determination of available \(\ce{PO4^3-}\) in acid extract

Phosphate can be analyzed by using molybdenum blue method. One mole of phosphate is converted into one mole of molybdenum blue compound. This method is used for determination of phosphate in the acid extract. Absorbance (\(A\)) and transmittance (\(T\)) are recorded at 800 nm. The molar absorptivity of the molybdenum blue compound is 6720 \(\mathrm{dm^3\,mol^{-1}\,cm^{-1}}\) and all measurement is carried out in a 1.00-cm cuvette.

Transmittance and absorbance are given by the following equations:

\[ T = I/I_{\mathrm{o}} \]

\[ A = \log(I_{\mathrm{o}}/I) \]

where \(I\) is the intensity of the transmitted light and \(I_{\mathrm{o}}\) is the intensity of the incident light.

5.2.1 When the sample containing high concentration of phosphate is analyzed, a reference solution of \(7.5\times10^{-5}\ \mathrm{mol\,dm^{-3}}\) of molybdenum blue compound is used for adjusting zero absorbance. The transmittance of the sample solution is then measured to be 0.55. Calculate the concentration of phosphate (\(\mathrm{mol\,dm^{-3}}\)) in the sample solution.

Part C. Determination of \(\ce{PO4^3-}\) and \(\ce{SiO4^4-}\) in alkaline extract

Both phosphate and silicate ions can react with molybdate in alkaline solution, producing the yellow molybdophosphate and molybdatosilicate. Further reduction with ascorbic acid produces intense color molybdenum blue compounds. Both complexes exhibit maximum absorption at 800 nm. Addition of tartaric acid helps preventing interference from silicate in the determination of phosphate.

Two series of phosphate standard are treated with and without tartaric acid whereas a series of silicate standard is not treated with tartaric acid. Linear equations obtained from those calibration curves are as follows:

Conditions Linear equations
Phosphate with and without tartaric acid \(y = 6720x_1\)
Silicate without tartaric acid \(y = 868x_2\)

\(y\) is absorbance at 800 nm,

\(x_1\) is concentration of phosphate as \(\mathrm{mol\,dm^{-3}}\),

\(x_2\) is concentration of silicate as \(\mathrm{mol\,dm^{-3}}\)

Absorbance at 800 nm of the alkaline fraction of the soil extract after treated with and without tartaric acid are 0.267 and 0.510, respectively.

5.3.1 Calculate the phosphate concentration in the alkaline soil extract in \(\mathrm{mol\,dm^{-3}}\) and calculate the corresponding phosphorus in \(\mathrm{mg\,dm^{-3}}\).

5.3.2 Calculate the silicate concentration from the soil sample in t the alkaline fraction in \(\mathrm{mol\,dm^{-3}}\) and calculate the corresponding silicon in \(\mathrm{mg\,dm^{-3}}\).

Part D. Preconcentration of ammonium phosphomolybdate

A 100 \(\mathrm{cm^3}\) of aqueous sample of ammonium phosphomolybdate (\(\ce{(NH4)3PMo12O40}\)) compound is extracted with 5.0 \(\mathrm{cm^3}\) of an organic solvent. The organic-water partition coefficient (\(K_{\mathrm{ow}}\)) is defined as the ratio of concentration of the compound in the organic phase (\(c_{\mathrm{o}}\)) to that in the water phase (\(c_{\mathrm{w}}\)). \(K_{\mathrm{ow}}\) of the ammonium phosphomolybdate is 5.0.

The molar absorptivity of ammonium phosphomolybdate in the organic phase is 5000 \(\mathrm{dm^3\,mol^{-1}\,cm^{-1}}\).

5.4 If the absorbance in the organic phase is 0.200, calculate the total mass of phosphorus (in mg unit) in the original aqueous sample solution. The optical pathlength of the cuvette is 1.00 cm.

Official English text of the problem.

Answers

  • 5.1.1: \(\boxed{0.158\ \mathrm{mg}}\) of phosphate, \(\ce{PO4^3-}\), per 1.00 g of soil.
  • 5.1.2: As printed, \(\boxed{0.158\ \mathrm{mg}}\) of phosphate per 1.00 g of soil. If the intended species is silicate, \(\ce{SiO4^4-}\), the answer is \(\boxed{0.175\ \mathrm{mg}}\) per 1.00 g of soil.
  • 5.2.1: \(\boxed{c(\ce{PO4^3-})=1.14\times10^{-4}\ \mathrm{mol\,dm^{-3}}}\).
  • 5.3.1: \(\boxed{c(\ce{PO4^3-})=3.97\times10^{-5}\ \mathrm{mol\,dm^{-3}}}\), corresponding to \(\boxed{1.23\ \mathrm{mg\,dm^{-3}}}\) of phosphorus.
  • 5.3.2: \(\boxed{c(\ce{SiO4^4-})=2.80\times10^{-4}\ \mathrm{mol\,dm^{-3}}}\), corresponding to \(\boxed{7.86\ \mathrm{mg\,dm^{-3}}}\) of silicon.
  • 5.4: \(\boxed{m(\ce{P})=0.031\ \mathrm{mg}}\) in the original aqueous sample.

Solution

The ions quantified are phosphate (orthophosphate), \(\ce{PO4^3-}\), SMILES O=P([O-])([O-])[O-], and orthosilicate, \(\ce{SiO4^4-}\), SMILES [O-][Si]([O-])([O-])[O-]. Both have tetrahedral arrangements of oxygen atoms around the central atom. The phosphate SMILES represents one localized Lewis structure; it does not convey resonance delocalization.

For mass conversions, use the standard atomic molar masses \[ M(\ce{P})=30.97,\qquad M(\ce{Si})=28.09,\qquad M(\ce{O})=16.00 \quad \mathrm{g\,mol^{-1}}. \] Thus, \[ M(\ce{PO4^3-})=94.97\ \mathrm{g\,mol^{-1}},\qquad M(\ce{SiO4^4-})=92.09\ \mathrm{g\,mol^{-1}}. \]

5.1.1

The digest volume is \[ 50.0\ \mathrm{cm^3}=0.0500\ \mathrm{dm^3}. \]

The phosphorus mass in the entire 5.00 g soil sample is \[ m(\ce{P}) =(5.16\ \mathrm{mg\,dm^{-3}})(0.0500\ \mathrm{dm^3}) =0.258\ \mathrm{mg}. \]

Each phosphate ion contains one phosphorus atom. Therefore, the phosphate mass per gram of soil is \[ \frac{m(\ce{PO4^3-})}{m_{\mathrm{soil}}} = \frac{0.258\ \mathrm{mg}}{5.00\ \mathrm{g}} \frac{94.97}{30.97} = \boxed{0.158\ \mathrm{mg\,g^{-1}}}. \]

5.1.2

Taken literally, this repeats 5.1.1, so the answer is again \[ \boxed{0.158\ \mathrm{mg\ of\ }\ce{PO4^3-}\mathrm{\ per\ }1.00\ \mathrm{g\ of\ soil}}. \]

If the intended question concerns silicate, the silicon mass in the digest is \[ m(\ce{Si}) =(5.35\ \mathrm{mg\,dm^{-3}})(0.0500\ \mathrm{dm^3}) =0.2675\ \mathrm{mg}. \]

Each orthosilicate ion contains one silicon atom, giving \[ \frac{m(\ce{SiO4^4-})}{m_{\mathrm{soil}}} = \frac{0.2675\ \mathrm{mg}}{5.00\ \mathrm{g}} \frac{92.09}{28.09} = \boxed{0.175\ \mathrm{mg\,g^{-1}}}. \]

5.2.1

Beer–Lambert’s law is \[ A=\varepsilon\ell c. \]

Because the instrument is zeroed against a solution containing molybdenum blue, the measured transmittance is the sample intensity divided by the reference intensity: \[ T_{\mathrm{meas}} =\frac{I_{\mathrm{sample}}}{I_{\mathrm{reference}}} =10^{-\varepsilon\ell(c_{\mathrm{sample}}-c_{\mathrm{reference}})}. \]

Consequently, \[ -\log_{10}T_{\mathrm{meas}} =\varepsilon\ell(c_{\mathrm{sample}}-c_{\mathrm{reference}}). \]

The measured absorbance difference is \[ \Delta A=-\log_{10}(0.55)=0.259637. \]

Hence, \[ \begin{aligned} c_{\mathrm{sample}} &=c_{\mathrm{reference}}+\frac{\Delta A}{\varepsilon\ell}\\ &=7.5\times10^{-5}\ \mathrm{mol\,dm^{-3}} +\frac{0.259637} {(6720\ \mathrm{dm^3\,mol^{-1}\,cm^{-1}})(1.00\ \mathrm{cm})}\\ &=1.13637\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \end{aligned} \]

The stated phosphate-to-molybdenum-blue conversion is 1:1, so \[ \boxed{c(\ce{PO4^3-})=1.14\times10^{-4}\ \mathrm{mol\,dm^{-3}}}. \]

5.3.1

Tartaric acid suppresses the silicate contribution while leaving the phosphate calibration unchanged. Thus, the absorbance with tartaric acid measures phosphate alone: \[ 0.267=6720c(\ce{PO4^3-}), \] where concentration is in \(\mathrm{mol\,dm^{-3}}\).

Therefore, \[ c(\ce{PO4^3-}) =\frac{0.267}{6720} =\boxed{3.97\times10^{-5}\ \mathrm{mol\,dm^{-3}}}. \]

There is one phosphorus atom per phosphate ion, so the corresponding phosphorus mass concentration is \[ \begin{aligned} \rho(\ce{P}) &=(3.97321\times10^{-5}\ \mathrm{mol\,dm^{-3}}) (30.97\ \mathrm{g\,mol^{-1}}) (1000\ \mathrm{mg\,g^{-1}})\\ &=\boxed{1.23\ \mathrm{mg\,dm^{-3}}}. \end{aligned} \]

5.3.2

Without tartaric acid, the two absorbance contributions add: \[ A_{\mathrm{without}} =6720c(\ce{PO4^3-})+868c(\ce{SiO4^4-}). \]

Subtracting the phosphate contribution gives \[ A_{\mathrm{silicate}}=0.510-0.267=0.243. \]

Thus, \[ c(\ce{SiO4^4-}) =\frac{0.243}{868} =\boxed{2.80\times10^{-4}\ \mathrm{mol\,dm^{-3}}}. \]

Each silicate ion contains one silicon atom: \[ \begin{aligned} \rho(\ce{Si}) &=(2.79954\times10^{-4}\ \mathrm{mol\,dm^{-3}}) (28.09\ \mathrm{g\,mol^{-1}}) (1000\ \mathrm{mg\,g^{-1}})\\ &=\boxed{7.86\ \mathrm{mg\,dm^{-3}}}. \end{aligned} \]

5.4

First determine the equilibrium concentration of ammonium phosphomolybdate in the organic phase: \[ c_{\mathrm{o}} =\frac{A}{\varepsilon\ell} =\frac{0.200} {(5000\ \mathrm{dm^3\,mol^{-1}\,cm^{-1}})(1.00\ \mathrm{cm})} =4.00\times10^{-5}\ \mathrm{mol\,dm^{-3}}. \]

Using the partition coefficient, \[ K_{\mathrm{ow}}=\frac{c_{\mathrm{o}}}{c_{\mathrm{w}}}=5.0, \] the concentration remaining in water is \[ c_{\mathrm{w}} =\frac{4.00\times10^{-5}}{5.0} =8.0\times10^{-6}\ \mathrm{mol\,dm^{-3}}. \]

Conservation of solute requires counting both phases: \[ \begin{aligned} n_{\mathrm{initial}} &=c_{\mathrm{o}}V_{\mathrm{o}}+c_{\mathrm{w}}V_{\mathrm{w}}\\ &=(4.00\times10^{-5}\ \mathrm{mol\,dm^{-3}}) (0.0050\ \mathrm{dm^3})\\ &\quad +(8.0\times10^{-6}\ \mathrm{mol\,dm^{-3}}) (0.100\ \mathrm{dm^3})\\ &=2.0\times10^{-7}+8.0\times10^{-7}\\ &=1.0\times10^{-6}\ \mathrm{mol}. \end{aligned} \]

Thus only 20% of the compound is extracted into the organic phase. Each formula unit of \(\ce{(NH4)3PMo12O40}\) contains one phosphorus atom, so \[ \begin{aligned} m(\ce{P}) &=(1.0\times10^{-6}\ \mathrm{mol}) (30.97\ \mathrm{g\,mol^{-1}}) (1000\ \mathrm{mg\,g^{-1}})\\ &=0.03097\ \mathrm{mg}\\ &\approx\boxed{0.031\ \mathrm{mg}}. \end{aligned} \]

The solution was checked against the official solution.