IChO 1989 · Theory problem 1

Solubility product of copper(II) iodate by iodometric titration

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To determine the solubility product of copper(II) iodate, \(\ce{Cu(IO3)2}\), by iodometric titration in an acidic solution (25 °C) 30.00 cm³ of a 0.100 molar sodium thiosulphate solution are needed to titrate 20.00 cm³ of a saturated aqueous solution \(\ce{Cu(IO3)2}\).

1.1 Write the sequence of balanced equations for the above described reactions.

1.2 Calculate the initial concentration of \(\ce{Cu^2+}\) and the solubility product of copper(II) iodate. Activity coefficients can be neglected.

Official English text of the problem.

Answers

  • 1.1 \[ \ce{Cu(IO3)2(s) <=> Cu^2+(aq) + 2IO3^-(aq)} \] \[ \ce{IO3^- + 5I^- + 6H+ -> 3I2 + 3H2O} \] \[ \ce{2Cu^2+ + 4I^- -> 2CuI(s) + I2} \] \[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \]

  • 1.2 \[ \boxed{[\ce{Cu^2+}]_0=1.15\times10^{-2}\ \mathrm{mol\,dm^{-3}}} \] \[ \boxed{K_{\mathrm{sp}}=6.14\times10^{-6}} \] The corresponding concentration product is \(6.14\times10^{-6}\ \mathrm{mol^3\,dm^{-9}}\).

Solution

1.1

Dissolution of copper(II) iodate gives one copper(II) ion and two iodate ions per formula unit, as shown by the first equilibrium above.

For the iodometric determination, excess iodide is added to the acidified sample. Both dissolved constituents liberate iodine:

  • Iodate oxidizes iodide in acidic solution. Each mole of iodate produces three moles of iodine.
  • Copper(II) also oxidizes iodide. Copper(II) is reduced to copper(I), which precipitates as copper(I) iodide. Each mole of copper(II) produces half a mole of iodine.

The liberated iodine is then titrated with thiosulphate. Iodine is reduced to iodide, while thiosulphate is oxidized to tetrathionate. Each mole of iodine consumes two moles of thiosulphate.

The species used in these equations have the following names and SMILES. The salt representations show disconnected constituent ions; SMILES do not describe their crystal lattices or aqueous hydration.

Species SMILES
Copper(II) iodate [Cu+2].[O-]I(=O)=O.[O-]I(=O)=O
Copper(II) ion [Cu+2]
Iodate ion [O-]I(=O)=O
Iodide ion [I-]
Hydrogen ion (proton) [H+]
Iodine II
Water O
Copper(I) iodide [Cu+].[I-]
Thiosulphate ion [O-]S(=O)(=O)[S-]
Tetrathionate ion [O-]S(=O)(=O)SSS(=O)(=O)[O-]
Sodium thiosulphate [Na+].[Na+].[O-]S(=O)(=O)[S-]

1.2

Let \(s\) be the molar solubility of copper(II) iodate in the original saturated solution. Its dissolution stoichiometry gives \[ [\ce{Cu^2+}]_0=s, \qquad [\ce{IO3^-}]_0=2s. \]

For each mole of dissolved copper(II) iodate:

  • The two iodate ions produce \(2\times3=6\) moles of iodine, consuming \(12\) moles of thiosulphate.
  • The copper(II) ion produces another \(\tfrac12\) mole of iodine, consuming \(1\) mole of thiosulphate.

Therefore, \[ n(\ce{S2O3^2-})=13\,n\!\left(\ce{Cu(IO3)2}\right). \]

The amount of thiosulphate used is \[ n(\ce{S2O3^2-}) =(0.100\ \mathrm{mol\,dm^{-3}}) (30.00\times10^{-3}\ \mathrm{dm^3}) =3.00\times10^{-3}\ \mathrm{mol}. \]

Thus the original \(20.00\ \mathrm{cm^3}\) aliquot contains \[ n(\ce{Cu^2+}) =\frac{3.00\times10^{-3}}{13} =2.30769\times10^{-4}\ \mathrm{mol}. \]

Its initial copper(II) concentration is consequently \[ s= \frac{2.30769\times10^{-4}\ \mathrm{mol}} {20.00\times10^{-3}\ \mathrm{dm^3}} =0.0115385\ \mathrm{mol\,dm^{-3}}, \] so \[ \boxed{[\ce{Cu^2+}]_0=1.15\times10^{-2}\ \mathrm{mol\,dm^{-3}}}. \]

Using the unrounded value, \[ [\ce{IO3^-}]_0=2s =0.0230769\ \mathrm{mol\,dm^{-3}}. \]

The solubility concentration product is \[ [\ce{Cu^2+}]_0[\ce{IO3^-}]_0^2 =s(2s)^2 =4s^3, \] giving \[ 4(0.0115385)^3 =\boxed{6.14\times10^{-6}\ \mathrm{mol^3\,dm^{-9}}}. \]

With activity coefficients neglected and concentrations normalized to the standard concentration \(c^\circ=1\ \mathrm{mol\,dm^{-3}}\), the dimensionless solubility product is therefore \[ \boxed{K_{\mathrm{sp}}=6.14\times10^{-6}}. \]

The solution was checked against the official solution.