IChO 1978 · Theory problem 2

Water hardness caused by Fe2+ and Ca2+

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A sample of water under investigation had 10° of temporary hardness and 10° of permanent hardness. Hardness of the water was caused by cations \(\ce{Fe^{2+}}\) and \(\ce{Ca^{2+}}\) only.

A volume of \(10.00\ \mathrm{dm^3}\) of the water was at disposal. From this volume \(100.00\ \mathrm{cm^3}\) were taken for further procedure. The water was oxidised with a \(\ce{H2O2}\) solution and then precipitated with an aqueous ammonia solution. A brown precipitate was dried and after an appropriate heating \(0.01432\ \mathrm{g}\) of an anhydrous product was obtained.

Problems:

2.1 Calculate the molar ratio of \(\ce{Fe^{2+}}\) : \(\ce{Ca^{2+}}\) in the water under investigation.

2.2 In another experiment, \(10.00\ \mathrm{dm^3}\) of the water was used again. The temporary hardness caused by cations \(\ce{Ca^{2+}}\) was removed first and the permanent hardness caused by cations \(\ce{Fe^{2+}}\) was removed by addition of \(\ce{Na3PO4}\). Calculate the mass of the precipitate (in its anhydrous form) on the assumption that only one half of cations \(\ce{Fe^{2+}}\) was oxidised to \(\ce{Fe^{3+}}\) in \(10.00\ \mathrm{dm^3}\) of the water analysed. Calculation should be made with an accuracy of one hundredth. Give the molar ratio in integers.

1° of hardness = \(10\ \mathrm{mg}\) \(\ce{CaO}\) in \(1\ \mathrm{dm^3}\) of water.

Relative atomic masses:

\[ A_r(\ce{Ca})=40.08;\qquad A_r(\ce{Fe})=55.85;\qquad A_r(\ce{C})=12.01; \] \[ A_r(\ce{H})=1.01;\qquad A_r(\ce{P})=31.00;\qquad A_r(\ce{O})=16.00. \]

Official English text of the problem.

Answers

  • 2.1: The intended integer molar ratio is approximately \[ \boxed{\ce{Fe^{2+}}:\ce{Ca^{2+}}=1:1}. \] The printed numerical data give \(1.01150:1\).

  • 2.2: The combined anhydrous precipitates from both removal stages weigh \[ \boxed{4.20\ \mathrm{g}}. \] The iron phosphate precipitate alone weighs \(\boxed{2.42\ \mathrm{g}}\). These results retain the measured iron content and total hardness.

Solution

2.1 — Determination of the iron-to-calcium ratio

Hydrogen peroxide oxidises iron(II) to iron(III). The balanced oxidation equation, written in acidic form, is

\[ \ce{2Fe^{2+} + H2O2 + 2H+ -> 2Fe^{3+} + 2H2O}. \]

Aqueous ammonia then precipitates brown hydrous iron(III) hydroxide, conventionally represented as \(\ce{Fe(OH)3}\):

\[ \ce{Fe^{3+} + 3NH3 + 3H2O -> Fe(OH)3(s) + 3NH4+}. \]

In the intended gravimetric procedure, calcium remains in solution. Heating the iron precipitate removes water and gives anhydrous iron(III) oxide:

\[ \ce{2Fe(OH)3(s) ->[\Delta] Fe2O3(s) + 3H2O(g)}. \]

Thus the weighed product is \(\ce{Fe2O3}\), with molar mass

\[ M(\ce{Fe2O3})=2(55.85)+3(16.00)=159.70\ \mathrm{g\,mol^{-1}}. \]

The amount of iron originally present in the \(100.00\ \mathrm{cm^3}\) aliquot is therefore

\[ n_{\mathrm{Fe}} =2n(\ce{Fe2O3}) =\frac{2(0.01432\ \mathrm{g})}{159.70\ \mathrm{g\,mol^{-1}}} =1.7933626\times10^{-4}\ \mathrm{mol}. \]

The total hardness is

\[ 10^\circ+10^\circ=20^\circ, \]

equivalent to \(200\ \mathrm{mg}\) of \(\ce{CaO}\) per cubic decimetre. The aliquot consequently contains

\[ m_{\ce{CaO},\mathrm{equiv}} =(0.200\ \mathrm{g\,dm^{-3}})(0.10000\ \mathrm{dm^3}) =0.02000\ \mathrm{g}. \]

Both hardness-producing cations are divalent, so one mole of either contributes the same hardness as one mole of \(\ce{CaO}\). Since

\[ M(\ce{CaO})=40.08+16.00=56.08\ \mathrm{g\,mol^{-1}}, \]

we obtain

\[ n_{\mathrm{Fe}}+n_{\mathrm{Ca}} =\frac{0.02000}{56.08} =3.5663338\times10^{-4}\ \mathrm{mol}. \]

Hence

\[ n_{\mathrm{Ca}} =(3.5663338-1.7933626)\times10^{-4} =1.7729713\times10^{-4}\ \mathrm{mol}, \]

and

\[ \frac{n_{\mathrm{Fe}}}{n_{\mathrm{Ca}}} =\frac{1.7933626}{1.7729713} =1.01150. \]

The intended simple integer ratio is therefore approximately

\[ \boxed{\ce{Fe^{2+}}:\ce{Ca^{2+}}=1:1}. \]

The unrounded amounts, rather than an imposed exact \(1:1\) ratio, are used below.

2.2 — Mass of the anhydrous precipitates

The volume in this experiment is \(100\) times the aliquot volume. Therefore,

\[ n_{\mathrm{Fe,total}}=0.017933626\ \mathrm{mol}, \qquad n_{\mathrm{Ca}}=0.017729713\ \mathrm{mol}. \]

Temporary calcium hardness is removed by heating, which precipitates calcium carbonate:

\[ \ce{Ca^{2+} + 2HCO3- ->[\Delta] CaCO3(s) + CO2(g) + H2O}. \]

Its molar mass and precipitated mass are

\[ M(\ce{CaCO3})=40.08+12.01+3(16.00) =100.09\ \mathrm{g\,mol^{-1}}, \]

\[ m(\ce{CaCO3}) =(0.017729713)(100.09) =1.774567\ \mathrm{g}. \]

Half the iron remains iron(II), and half is oxidised to iron(III):

\[ n(\ce{Fe^{2+}})=n(\ce{Fe^{3+}}) =\frac{0.017933626}{2} =0.008966813\ \mathrm{mol}. \]

Addition of trisodium phosphate supplies phosphate ions. The two precipitation reactions are

\[ \ce{3Fe^{2+} + 2PO4^{3-} -> Fe3(PO4)2(s)}, \]

\[ \ce{Fe^{3+} + PO4^{3-} -> FePO4(s)}. \]

The products are iron(II) phosphate and iron(III) phosphate, respectively. Their amounts are

\[ n(\ce{Fe3(PO4)2}) =\frac{0.008966813}{3} =0.002988938\ \mathrm{mol}, \]

\[ n(\ce{FePO4})=0.008966813\ \mathrm{mol}. \]

Using anhydrous formula masses,

\[ M(\ce{Fe3(PO4)2}) =3(55.85)+2[31.00+4(16.00)] =357.55\ \mathrm{g\,mol^{-1}}, \]

\[ M(\ce{FePO4}) =55.85+31.00+4(16.00) =150.85\ \mathrm{g\,mol^{-1}}. \]

Thus,

\[ m(\ce{Fe3(PO4)2}) =(0.002988938)(357.55) =1.068695\ \mathrm{g}, \]

\[ m(\ce{FePO4}) =(0.008966813)(150.85) =1.352644\ \mathrm{g}. \]

The phosphate precipitate alone therefore weighs

\[ \boxed{ m_{\mathrm{phosphates}} =1.068695+1.352644 =2.421338\ \mathrm{g} \approx2.42\ \mathrm{g} }. \]

Including the calcium carbonate precipitated during the first stage gives

\[ \boxed{ m_{\mathrm{all\ precipitates}} =1.774567+2.421338 =4.195905\ \mathrm{g} \approx4.20\ \mathrm{g} }. \]

Names and SMILES of the chemical species used

For ionic solids, the disconnected SMILES below represent formal formula units. They do not describe the crystal lattice, hydration, or metal coordination geometry; hydrous iron(III) hydroxide is not a discrete molecular species.

Species and name SMILES
\(\ce{Fe^{2+}}\), iron(II) ion [Fe+2]
\(\ce{Fe^{3+}}\), iron(III) ion [Fe+3]
\(\ce{Ca^{2+}}\), calcium ion [Ca+2]
\(\ce{H+}\), hydrogen ion [H+]
\(\ce{H2O2}\), hydrogen peroxide OO
\(\ce{H2O}\), water O
\(\ce{NH3}\), ammonia N
\(\ce{NH4+}\), ammonium ion [NH4+]
\(\ce{Fe(OH)3}\), iron(III) hydroxide [Fe+3].[OH-].[OH-].[OH-]
\(\ce{Fe2O3}\), iron(III) oxide [Fe+3].[Fe+3].[O-2].[O-2].[O-2]
\(\ce{CaO}\), calcium oxide [Ca+2].[O-2]
\(\ce{HCO3-}\), hydrogen carbonate ion OC(=O)[O-]
\(\ce{CO2}\), carbon dioxide O=C=O
\(\ce{CaCO3}\), calcium carbonate [Ca+2].[O-]C(=O)[O-]
\(\ce{Na3PO4}\), trisodium phosphate [Na+].[Na+].[Na+].[O-]P(=O)([O-])[O-]
\(\ce{PO4^{3-}}\), phosphate ion [O-]P(=O)([O-])[O-]
\(\ce{Fe3(PO4)2}\), iron(II) phosphate [Fe+2].[Fe+2].[Fe+2].[O-]P(=O)([O-])[O-].[O-]P(=O)([O-])[O-]
\(\ce{FePO4}\), iron(III) phosphate [Fe+3].[O-]P(=O)([O-])[O-]

The solution was checked against the official solution.