When ethane - 1,2-dioic acid is heated with concentrated tetraoxosulphate (VI) acid, a reaction represented by the following equation occurs: (a) State the ...

Answer Details

The equation shown on the diagram is the action of hot concentrated tetraoxosulphate(VI) acid on ethanedioic (oxalic) acid:

\[ \underset{\text{(COOH)}_2}{\text{HOOC-COOH}} \;\xrightarrow[\;\Delta\;]{\text{conc. } H_2SO_4}\; CO_2 + CO + H_2O \]

(a) Type of process involved

The process is dehydration. The concentrated tetraoxosulphate(VI) acid acts as a dehydrating agent: it removes the elements of water (\(H_2O\)) from each mole of ethanedioic acid, leaving behind the two oxides of carbon, \(CO_2\) and \(CO\). Notice that the atoms balance exactly, \((COOH)_2 = C_2H_2O_4\) splits into \(CO_2 + CO + H_2O\), so no oxidation state change is imposed by the acid; it simply strips out water.

(b) Basicity of ethanedioic acid

The basicity of an acid is the number of replaceable (ionisable) hydrogen atoms it releases per molecule in aqueous solution. Ethanedioic acid, \((COOH)_2\), contains two carboxyl (-COOH) groups, each supplying one ionisable hydrogen:

\[ (COOH)_2 \rightleftharpoons 2H^+ + (COO)_2^{2-} \]

Therefore its basicity is 2 (it is a dibasic / diprotic acid).

(c) Three differences in the chemical properties of the two oxides of carbon (\(CO_2\) and \(CO\))

A further valid distinction: \(CO_2\) turns lime water milky (forming \(CaCO_3\)), whereas \(CO\) does not; and \(CO\) is highly poisonous because it combines with haemoglobin, while \(CO_2\) is not poisonous in that way.