IChO 2003 · Theory problem 23

Hydrogen halides, buffers and electrolysis

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Are the following statements correct (Y) or not (N)?

a) \(\ce{HF}\) boils at a higher temperature than \(\ce{HCl}\). (Y) (N)

b) \(\ce{HBr}\) boils at a lower temperature than \(\ce{HI}\) (Y) (N)

c) Pure \(\ce{HI}\) can be produced by reacting concentrated sulfuric acid with \(\ce{KI}\). (Y) (N)

d) Ammonia solutions are buffer solutions because they contain the conjugate pair \(\ce{NH3}\) – \(\ce{NH4+}\). (Y) (N)

e) Pure water at 80 °C is acidic. (Y) (N)

f) During electrolysis of an aqueous \(\ce{KI}\) solution with graphite electrodes, the pH near the cathode is below 7. (Y) (N)

Official English text of the problem.

Answers

  • a) Y
  • b) Y
  • c) N
  • d) N
  • e) N
  • f) N

Solution

a) Y — hydrogen bonding raises the boiling point of hydrogen fluoride.

Hydrogen fluoride, \(\ce{HF}\) ([H]F), forms strong intermolecular hydrogen bonds. Hydrogen chloride, \(\ce{HCl}\) ([H]Cl), has dipole–dipole and London dispersion attractions but lacks comparably strong hydrogen bonding.

Consequently, separating the molecules into the gas phase requires more energy for \(\ce{HF}\), giving it the higher boiling point despite its lower molar mass.

b) Y — dispersion forces are stronger in hydrogen iodide.

Hydrogen iodide, \(\ce{HI}\) ([H]I), has a larger, more polarizable electron cloud than hydrogen bromide, \(\ce{HBr}\) ([H]Br). Its stronger London dispersion forces produce a higher boiling point:

\[ T_{\mathrm b}(\ce{HBr})<T_{\mathrm b}(\ce{HI}). \]

c) N — concentrated sulfuric acid oxidizes hydrogen iodide.

An initial acid–base reaction between potassium iodide, \(\ce{KI}\) ([K+].[I-]), and sulfuric acid, \(\ce{H2SO4}\) (OS(=O)(=O)O), can be written as

\[ \ce{KI + H2SO4 -> KHSO4 + HI}. \]

Here, \(\ce{KHSO4}\) is potassium hydrogen sulfate ([K+].OS(=O)(=O)[O-]).

However, concentrated sulfuric acid also acts as an oxidizing agent. It oxidizes hydrogen iodide to iodine, \(\ce{I2}\) (II). One possible subsequent redox reaction is

\[ \ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}. \]

The other products in this equation are hydrogen sulfide, \(\ce{H2S}\) (S), and water, \(\ce{H2O}\) (O). Iodine changes oxidation state from \(-1\) to \(0\), while sulfur changes from \(+6\) to \(-2\).

Thus, the reaction does not provide pure hydrogen iodide.

d) N — the presence of a conjugate pair alone is insufficient.

Ammonia, \(\ce{NH3}\) (N), reacts weakly with water to form ammonium ions, \(\ce{NH4+}\) ([NH4+]), and hydroxide ions, \(\ce{OH-}\) ([OH-]):

\[ \ce{NH3 + H2O <=> NH4+ + OH-}. \]

A useful buffer requires appreciable concentrations of both members of the conjugate pair, enabling it to consume added acid and added base without a large change in pH.

In an ordinary ammonia solution, only a small fraction of the ammonia forms ammonium ions. Their mere presence does not establish an effective buffer. An ammonia solution containing a suitable amount of an ammonium salt can form a buffer.

e) N — pure water remains neutral at 80 °C.

Water undergoes self-ionization:

\[ \ce{2H2O <=> H3O+ + OH-}, \]

where \(\ce{H3O+}\) is the hydronium ion ([OH3+]).

In pure water, the two ions are produced in equal amounts:

\[ [\ce{H3O+}]=[\ce{OH-}]. \]

This equality establishes neutrality. At 80 °C, the ionic product of water is greater than at 25 °C, so the neutral pH is below 7:

\[ \mathrm{pH}_{\text{neutral}} =\frac{1}{2}\mathrm{p}K_{\mathrm w}(80\,^\circ\mathrm C)<7. \]

A pH below 7 therefore does not make pure water acidic at this temperature.

f) N — hydroxide is produced at the cathode.

In aqueous potassium iodide, water is reduced at the graphite cathode:

\[ \ce{2H2O + 2e- -> H2 + 2OH-}. \]

The gas produced is hydrogen, \(\ce{H2}\) ([H][H]). Hydroxide accumulates near the cathode, making that region alkaline; under ordinary room-temperature conditions, its pH is above 7.

At the anode, iodide ions, \(\ce{I-}\) ([I-]), are oxidized:

\[ \ce{2I- -> I2 + 2e-}. \]

The corresponding overall electrode reaction is

\[ \ce{2I- + 2H2O -> I2 + H2 + 2OH-}. \]

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