IChO 2007 · Theory problem 4

Determination of water by Fischer titration

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Determination of water by Fischer titraton

Determination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of \(\ce{SO2}\) and pyridine (\(\ce{C5H5N}\), Py) – Fischer reagent. The following reactions occur during the titration:

Figure 1

Iodine content is usually expressed in mg of water reacting with \(1\ \mathrm{cm^3}\) of the titrant solution (hereunder T, \(\mathrm{mg\ cm^{-3}}\)), which equals the mass of water (mg) reacting with \(1.00\ \mathrm{cm^3}\) of the iodine solution. T is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.

In all calculations please use the atomic masses accurate to 2 decimal points.

4.1 Sometimes titration of water is performed in pyridine medium without methanol. How would the reaction of \(\ce{I2}\) with \(\ce{SO2}\) and \(\ce{H2O}\) occur in this case? Write down balanced reaction equation.

Calculate the T values of iodine solution in each of the following cases:

4.2 \(12.20\ \mathrm{cm^3}\) of Fischer reagent solution were used for titration of \(1.352\ \mathrm{g}\) of sodium tartrate dihydrate \(\ce{Na2C4H4O6 . 2H2O}\).

4.3 A known amount of water (\(21.537\ \mathrm{g}\)) was placed into a \(1.000\ \mathrm{dm^3}\) volumetric flask which was filled by methanol up to the mark. For titration of \(10.00\ \mathrm{cm^3}\) of the obtained solution, \(22.70\ \mathrm{cm^3}\) of Fischer reagent solution were needed, whereas \(2.20\ \mathrm{cm^3}\) of iodine were used for titration of \(25.00\ \mathrm{cm^3}\) of methanol.

4.4 \(5.624\ \mathrm{g}\) of water were diluted by methanol up to a total volume of \(1.000\ \mathrm{dm^3}\) (solution A); \(22.45\ \mathrm{cm^3}\) of this solution were used for titration of \(15.00\ \mathrm{cm^3}\) of a Fischer reagent (solution B).

Then \(25.00\ \mathrm{cm^3}\) of methanol (of the same batch as used for the preparation of solution A) and \(10.00\ \mathrm{cm^3}\) of solution B were mixed, and the mixture was titrated by the solution A. \(10.79\ \mathrm{cm^3}\) of the latter solution were spent.

4.5 An inexperienced analyst tried to determine water content in a sample of \(\ce{CaO}\) using Fischer reagent. Write down the equation(s) of reaction(s) demonstrating possible sources of errors.

For the titration of \(0.6387\ \mathrm{g}\) of a hydrated compound \(\ce{Fe2(SO4)3 . xH2O}\), \(10.59\ \mathrm{cm^3}\) of iodine solution (\(T=15.46\ \mathrm{mg/cm^3}\)) were consumed.

4.6 What other reaction(s), beside those given in the problem, can occur during the titration? Write down the equations of two such processes.

4.7 Write down an equation of the overall reaction of \(\ce{Fe2(SO4)3 . xH2O}\) with the Fischer reagent.

4.8 Calculate the composition of the hydrate \(\ce{Fe2(SO4)3 . xH2O}\) (\(x=\text{integer}\)).

Official English text of the problem.

Answers

  • 4.1 \[ \boxed{\ce{SO2 + I2 + 2H2O -> H2SO4 + 2HI}} \] One mole of iodine consumes two moles of water.

  • 4.2 \[ \boxed{T=17.36\ \mathrm{mg\,cm^{-3}}} \]

  • 4.3 \[ \boxed{T\approx9.870\ \mathrm{mg\,cm^{-3}}} \] This uses the dilute-solution volume approximation explained below.

  • 4.4 \[ \boxed{T\approx10.11\ \mathrm{mg\,cm^{-3}}} \] This uses the same volume approximation.

  • 4.5 Acid neutralization by calcium oxide generates additional water, causing an erroneously high result. For example, \[ \boxed{\ce{CaO + 2HI -> Ca^2+ + 2I- + H2O}} \]

  • 4.6 Iron(III) can oxidize iodide and sulfur dioxide: \[ \boxed{\ce{2Fe^3+ + 2I- -> 2Fe^2+ + I2}} \] \[ \boxed{\ce{2Fe^3+ + SO2 + CH3OH + H2O -> 2Fe^2+ + CH3OSO3H + 2H+}} \]

  • 4.7 An overall equation, with the products written as ions, is \[ \boxed{ \begin{aligned} &\ce{Fe2(SO4)3 . xH2O} +(x-1)\ce{I2}+x\ce{SO2}+x\ce{CH3OH}+3x\ce{Py}\\ &\qquad\ce{->} 2\ce{Fe^2+}+3\ce{SO4^2-} +(2x-2)\ce{I-}+x\ce{CH3OSO3-}+3x\ce{PyH+}. \end{aligned}} \]

  • 4.8 \[ \boxed{x=9,\qquad \ce{Fe2(SO4)3 . 9H2O}} \] The compound is iron(III) sulfate nonahydrate.

Solution

Names and SMILES for the species used in the equations are collected here once. Disconnected SMILES for ionic compounds describe composition; they do not specify a crystal structure or coordination geometry.

Species Name SMILES
\(\ce{H2O}\) Water O
\(\ce{CH3OH}\) Methanol CO
\(\ce{SO2}\) Sulfur dioxide O=S=O
\(\ce{I2}\) Iodine II
\(\ce{Py}\) Pyridine n1ccccc1
\(\ce{HI}\) Hydrogen iodide [H]I
\(\ce{H2SO4}\) Sulfuric acid OS(=O)(=O)O
\(\ce{CH3OSO3H}\) Methyl hydrogen sulfate COS(=O)(=O)O
\(\ce{H+}\) Hydrogen ion, or proton [H+]
\(\ce{I-}\) Iodide ion [I-]
\(\ce{PyH+}\) Pyridinium ion [nH+]1ccccc1
\(\ce{CH3OSO3-}\) Methyl sulfate ion COS(=O)(=O)[O-]
\(\ce{SO4^2-}\) Sulfate ion [O-]S(=O)(=O)[O-]
\(\ce{Fe^3+}\) Iron(III) ion [Fe+3]
\(\ce{Fe^2+}\) Iron(II) ion [Fe+2]
\(\ce{Ca^2+}\) Calcium ion [Ca+2]
\(\ce{CaO}\) Calcium oxide [Ca+2].[O-2]

The given methanolic reaction is \[ \ce{SO2 + CH3OH + H2O + I2 -> 2HI + CH3OSO3H}. \] Thus, in the absence of interfering reactions, \[ n(\ce{I2})=n(\ce{H2O}), \qquad T=\frac{m(\ce{H2O})}{V_{\text{reagent}}}. \]

The atomic masses used are \[ \begin{gathered} A_{\mathrm r}(\ce{H})=1.01,\quad A_{\mathrm r}(\ce{C})=12.01,\quad A_{\mathrm r}(\ce{O})=16.00,\\ A_{\mathrm r}(\ce{Na})=22.99,\quad A_{\mathrm r}(\ce{S})=32.07,\quad A_{\mathrm r}(\ce{Fe})=55.85. \end{gathered} \] Consequently, \[ M(\ce{H2O})=18.02\ \mathrm{g\,mol^{-1}}. \]

4.1

Without methanol, oxidation of sulfur dioxide produces sulfate. The balanced oxidation and reduction half-reactions are \[ \ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}, \] \[ \ce{I2 + 2e- -> 2I-}. \]

Their sum can be written in molecular form as \[ \boxed{\ce{SO2 + I2 + 2H2O -> H2SO4 + 2HI}}. \]

Including neutralization by excess pyridine gives \[ \ce{SO2 + I2 + 2H2O + 4Py -> SO4^2- + 2I- + 4PyH+}. \]

Two water molecules supply the two additional oxygen atoms required to convert sulfur dioxide into sulfate. In the methanolic reaction, methanol supplies one of these oxygen atoms, so only one water molecule is consumed per iodine molecule.

4.2

Sodium tartrate dihydrate has the composition-only SMILES
[Na+].[Na+].[O-]C(=O)C(O)C(O)C(=O)[O-].O.O.

The statement does not specify tartrate stereochemistry, so none is assigned in the SMILES.

Its molar mass is \[ \begin{aligned} M(\ce{Na2C4H4O6 . 2H2O}) &=2(22.99)+4(12.01)+4(1.01)+6(16.00)+2(18.02)\\ &=230.10\ \mathrm{g\,mol^{-1}}. \end{aligned} \]

The mass of crystallization water in the sample is \[ m_{\mathrm w} =(1.352\ \mathrm g)\frac{2(18.02)}{230.10} =0.2117605\ \mathrm g =211.7605\ \mathrm{mg}. \]

Therefore, \[ \boxed{ T=\frac{211.7605\ \mathrm{mg}}{12.20\ \mathrm{cm^3}} =17.36\ \mathrm{mg\,cm^{-3}}. } \]

4.3

For this part and 4.4, use the usual dilute-solution approximation that the methanol contribution to a \(1.000\ \mathrm{dm^3}\) preparation is approximately the water contained in \(1.000\ \mathrm{dm^3}\) of the original methanol.

The deliberately added water in the aliquot is \[ m_{\mathrm{added}} =(21.537\ \mathrm g)\frac{10.00}{1000} =0.21537\ \mathrm g =215.37\ \mathrm{mg}. \]

The methanol blank corresponding to the aliquot requires \[ V_{\mathrm{blank}} =(2.20\ \mathrm{cm^3})\frac{10.00}{25.00} =0.880\ \mathrm{cm^3} \] of reagent.

The reagent volume attributable to the deliberately added water is consequently \[ V_{\mathrm{corrected}} =22.70-0.880 =21.82\ \mathrm{cm^3}. \]

Hence \[ \boxed{ T=\frac{215.37\ \mathrm{mg}}{21.82\ \mathrm{cm^3}} \approx9.870\ \mathrm{mg\,cm^{-3}}. } \]

4.4

Let \(c_m\) be the water concentration in the original methanol, and \(c_A\) the total water concentration in solution A, both in \(\mathrm{mg\,cm^{-3}}\).

Under the stated volume approximation, \[ c_A =\frac{5624\ \mathrm{mg}}{1000\ \mathrm{cm^3}}+c_m =5.624\ \mathrm{mg\,cm^{-3}}+c_m. \]

The first titration gives \[ (15.00\ \mathrm{cm^3})T =(22.45\ \mathrm{cm^3})c_A. \]

In the second experiment, water from the added methanol consumes part of the iodine initially present in solution B. Solution A consumes the remainder: \[ (10.00\ \mathrm{cm^3})T =(25.00\ \mathrm{cm^3})c_m +(10.79\ \mathrm{cm^3})c_A. \]

Eliminating \(T\), \[ c_m =\frac{10.00(22.45/15.00)-10.79}{25.00}\,c_A =0.1670667\,c_A. \]

Thus \[ c_A =\frac{5.624}{1-0.1670667} =6.752041\ \mathrm{mg\,cm^{-3}}, \] and \[ c_m=1.128041\ \mathrm{mg\,cm^{-3}}. \]

Finally, \[ \boxed{ T=\frac{22.45}{15.00}c_A =10.10555\ \mathrm{mg\,cm^{-3}} \approx10.11\ \mathrm{mg\,cm^{-3}}. } \]

4.5

Calcium oxide is a basic oxide. It reacts with acidic components of the reagent system and generates water. Two examples, with the calcium salts written as ions, are \[ \ce{CaO + 2HI -> Ca^2+ + 2I- + H2O}, \] \[ \ce{CaO + 2CH3OSO3H -> Ca^2+ + 2CH3OSO3- + H2O}. \]

The corresponding reaction with protonated pyridine is \[ \ce{CaO + 2PyH+ -> Ca^2+ + 2Py + H2O}. \]

The newly generated water consumes additional iodine. Combining the first equation with the Fischer reaction makes the interference explicit: \[ \ce{CaO + SO2 + CH3OH + I2 -> Ca^2+ + 2I- + CH3OSO3H}. \]

Thus even anhydrous calcium oxide can consume iodine. Each mole of calcium oxide reacting through this process produces one mole of apparent water, giving an erroneously high water determination.

4.6

Iron(III) is an additional oxidizing agent in the sample. Its reduction is \[ \ce{Fe^3+ + e- -> Fe^2+}. \]

It can accept electrons from iodide: \[ \boxed{\ce{2Fe^3+ + 2I- -> 2Fe^2+ + I2}}. \]

This regenerates iodine from iodide formed during the Fischer reaction.

Iron(III) can also accept electrons directly from sulfur dioxide. Written consistently with the methyl hydrogen sulfate product specified for the methanolic medium, \[ \boxed{ \ce{2Fe^3+ + SO2 + CH3OH + H2O -> 2Fe^2+ + CH3OSO3H + 2H+}. } \]

Here sulfur is oxidized from \(+4\) to \(+6\), releasing two electrons, while two iron(III) ions each accept one electron.

These are alternative routes for reduction of the same iron(III): each formula unit of the hydrate accepts two electrons in total through its two iron ions.

4.7

Including dissociation of methyl hydrogen sulfate, the sulfur oxidation half-reaction is \[ \ce{SO2 + CH3OH + H2O -> CH3OSO3- + 3H+ + 2e-}. \]

For \(x\) waters of hydration, this process releases \(2x\) electrons. Reduction of the two iron(III) ions consumes two electrons: \[ \ce{2Fe^3+ + 2e- -> 2Fe^2+}. \]

The remaining \(2x-2\) electrons reduce \(x-1\) iodine molecules. Pyridine binds the \(3x\) protons produced. Therefore, \[ \boxed{ \begin{aligned} &\ce{Fe2(SO4)3 . xH2O} +(x-1)\ce{I2}+x\ce{SO2}+x\ce{CH3OH}+3x\ce{Py}\\ &\qquad\ce{->} 2\ce{Fe^2+}+3\ce{SO4^2-} +(2x-2)\ce{I-}+x\ce{CH3OSO3-}+3x\ce{PyH+}. \end{aligned} } \]

The three sulfate ions originally present in the hydrate are retained. Both atoms and charge balance; the total product charge is \[ 2(2)+3(-2)-(2x-2)-x+3x=0. \]

The analytical consequence is \[ \boxed{ n(\ce{I2})=(x-1)\,n(\text{hydrate}). } \]

4.8

The anhydrous iron(III) sulfate formula unit has composition-only SMILES
[Fe+3].[Fe+3].[O-]S(=O)(=O)[O-].[O-]S(=O)(=O)[O-].[O-]S(=O)(=O)[O-].

Its molar mass is \[ M_0=2(55.85)+3[32.07+4(16.00)] =399.91\ \mathrm{g\,mol^{-1}}. \]

The iodine consumed corresponds to an apparent water mass of \[ m_{\mathrm{equiv}} =(15.46\ \mathrm{mg\,cm^{-3}})(10.59\ \mathrm{cm^3}) =163.7214\ \mathrm{mg}. \]

Because the reagent titer is based on the ordinary \(1:1\) iodine–water reaction, \[ n(\ce{I2}) =\frac{0.1637214\ \mathrm g}{18.02\ \mathrm{g\,mol^{-1}}} =0.009085538\ \mathrm{mol}. \]

Meanwhile, \[ n(\text{hydrate}) =\frac{0.6387\ \mathrm g} {(399.91+18.02x)\ \mathrm{g\,mol^{-1}}}. \]

Using the stoichiometry established in 4.7, \[ 0.009085538 =\frac{0.6387(x-1)}{399.91+18.02x}. \]

Solving, \[ x =\frac{0.6387+(0.009085538)(399.91)} {0.6387-(0.009085538)(18.02)} =8.9943. \]

Since \(x\) must be an integer, \[ \boxed{x=9,\qquad \ce{Fe2(SO4)3 . 9H2O}}. \]

The compound is iron(III) sulfate nonahydrate, with composition-only SMILES
[Fe+3].[Fe+3].[O-]S(=O)(=O)[O-].[O-]S(=O)(=O)[O-].[O-]S(=O)(=O)[O-].O.O.O.O.O.O.O.O.O.

The formula and this SMILES specify nine waters per formula unit. They do not establish which waters coordinate iron or the hydrate’s crystal geometry.

The solution was checked against the official solution.