Question

Which amongst the following reactions cannot be used for the preparation of the halogen acid?
A. $2KBr+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2HBr$
B. $2NaCl+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HCl$
C. $NaCl+NaHS{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+HCl$
D. $Ca{{F}_{2}}+{{H}_{2}}S{{O}_{4}}\to CaS{{O}_{4}}+2HF$

Answer 1

Hint: Bromide ions have a tendency to reduce the concentrated acid in this case sulphuric acid. Hence, Potassium bromide reacts with sulphuric acid to form bromine gas.

Complete Step by Step Answer:
A salt of bromine that is hydrogen bromide reacts with concentrated sulphuric acid to produce a potassium salt of sulphate that is potassium sulphate along with bromine gas, a gas of sulphur that is sulphur dioxide along with water as products. The reaction can be given as follows:
$2KBr+2{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+B{{r}_{2}}+S{{O}_{2}}+2{{H}_{2}}O$

Thus in this reaction two moles of a salt of bromine that is hydrogen bromide reacts with two moles of a concentrated sulphuric acid to produce one mole of a potassium salt of sulphate that is potassium sulphate along with one mole of bromine gas, one mole of a gas of sulphur that is sulphur dioxide along with two moles of water as products. Thus we can write that this reaction does not give halogen acid. It actually produces bromine gas.
Thus the correct option is A.

Note: Among the halogen acids hydrofluoric acid is least acidic and hydroiodic acid is most acidic. Hydrobromic acid has higher acidity.