The property of concentrated H2SO4 that makes it suitable for preparing HNO3 is its

Analysis

The question asks for the specific chemical or physical property of concentrated tetraoxosulphate(VI) acid ($H_2SO_4$) that allows it to be used in the laboratory preparation of trioxonitrate(V) acid ($HNO_3$).

Concept Explanation

In the laboratory, $HNO_3$ is prepared by heating a mixture of a nitrate salt (like $NaNO_3$ or $KNO_3$) with concentrated $H_2SO_4$. The reaction is: $NaNO_3(s) + H_2SO_4(conc) \rightarrow NaHSO_4(s) + HNO_3(g)$

The key to this preparation is that $HNO_3$ is a volatile acid (it has a low boiling point), while $H_2SO_4$ is a non-volatile acid (it has a very high boiling point). Because $H_2SO_4$ has a much higher boiling point, it remains in the reaction vessel while the $HNO_3$ is driven off as a gas and subsequently condensed. If $H_2SO_4$ were volatile, it would vaporize along with the $HNO_3$, contaminating the product.

Step-by-Step Solution

1. Identify the reactants: We are reacting a salt with an acid to produce another acid.
2. Compare the acids: $HNO_3$ is volatile (boils at approximately 83°C), whereas $H_2SO_4$ is non-volatile (boils at approximately 338°C).
3. Determine the mechanism: The reaction relies on the fact that the non-volatile acid ($H_2SO_4$) can displace the more volatile acid ($HNO_3$) from its salt.
4. Evaluate the options: • A. Boiling point: This is the correct property. The high boiling point of $H_2SO_4$ ensures it stays liquid while $HNO_3$ turns to gas. • B. Density: While $H_2SO_4$ is dense, its density is not the reason it displaces $HNO_3$. • C. Oxidizing properties: $H_2SO_4$ is an oxidizing agent, but this property is not relevant to the displacement of $HNO_3$. • D. Dehydrating properties: $H_2SO_4$ is a powerful dehydrating agent (it removes water), but this is not the principle behind the preparation of $HNO_3$.

Conclusion

Exam Tip: Whenever you see a question about the preparation of volatile acids (like $HCl$ or $HNO_3$) using $H_2SO_4$, always look for "non-volatile" or "high boiling point" as the reason. It is a classic JAMB chemistry principle!