IChO 2003 · Theory problem 1

Solubility of thorium iodate

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SECTION A: General Chemistry

QUESTION 1

The molar solubility \(s\) (mol dm\(^{-3}\)) of \(\ce{Th(IO3)4}\) as a function of the solubility product \(K_{sp}\) of this sparingly soluble thorium salt is given by the equation:

(a) \(\displaystyle s=(K_{sp}/128)^{1/4}\qquad\boxed{(\quad)}\)

(b) \(\displaystyle s=(K_{sp}/256)^{1/5}\qquad\boxed{(\quad)}\)

(c) \(\displaystyle s=256\,K_{sp}^{1/4}\qquad\boxed{(\quad)}\)

(d) \(\displaystyle s=(128\,K_{sp})^{1/4}\qquad\boxed{(\quad)}\)

(e) \(\displaystyle s=(256\,K_{sp})^{1/5}\qquad\boxed{(\quad)}\)

(f) \(\displaystyle s=(K_{sp}/128)^{1/5}/2\qquad\boxed{(\quad)}\)

Official English text of the problem.

Answers

  • 1. (b) \(\displaystyle \boxed{s=\left(\frac{K_{sp}}{256}\right)^{1/5}}\). Mark only the box beside (b).

Solution

1. Thorium(IV) iodate dissolves according to

\[ \ce{Th(IO3)4(s) <=> Th^{4+}(aq) + 4IO3^{-}(aq)}. \]

The dissolved ions are the thorium(IV) ion, [Th+4], and the iodate ion, [O-]I(=O)=O. The iodate SMILES represents one resonance form; it does not encode the ion’s trigonal pyramidal geometry.

For a molar solubility \(s\), each mole of salt supplies one mole of thorium(IV) ions and four moles of iodate ions. Thus, with concentrations expressed in \(\mathrm{mol\,dm^{-3}}\),

\[ [\ce{Th^{4+}}]=s, \qquad [\ce{IO3^{-}}]=4s. \]

Using the concentration form of the solubility product, with the pure solid omitted,

\[ K_{sp} =[\ce{Th^{4+}}][\ce{IO3^{-}}]^4 =s(4s)^4 =256s^5. \]

Rearranging,

\[ \boxed{s=\left(\frac{K_{sp}}{256}\right)^{1/5}}, \]

which is option (b).

The answers were checked against the official answers published by the organisers.