Chapter 8, Problem 92A

Interpretation Introduction

Interpretation:

The name of the given formula of compounds needs to be determined.

Formula	Name
${\text{HClO}}_{\text{2}}$
${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$
${\text{H}}_{\text{2}}\text{Se}$
${\text{HClO}}_{\text{3}}$

Concept introduction:

Oxyacid contains one H atom and oxyanion. Oxyanions are polyatomic ion having one or more oxygen atoms. In the naming of oxyacid, first identify the oxanion. In an oxyacid the first word is the root of the oxyanion, per- or hypo- prefix is used and a suffix. Oxyanion’s name which is ends with suffix -ate, replenish it by suffix -ic. Oxyanion’s name ending with -ite suffix, replenish by suffix -ous.

Answer to Problem 92A

Formula	Name
${\text{HClO}}_{\text{2}}$	Chlorous acid
${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$	Phosphoric acid
${\text{H}}_{\text{2}}\text{Se}$	Hydroselenic acid
${\text{HClO}}_{\text{3}}$	Chloric acid

Explanation of Solution

The oxyanion present in ${\text{HClO}}_{\text{2}}$ is chlorite. As the oxyanion’s name is ends with suffix −ite, hence replace it with −ous. Name of ${\text{HClO}}_{\text{2}}$ is chlorous acid.

The oxyanion present in ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is phosphate. As the oxyanion’s name is ends with suffix −ate, hence replace it with −ic. Name of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is phosphoric acid.

${\text{H}}_{\text{2}}\text{Se}$ comprises of hydrogen and selenium. To name the Hydrogen part prefix hydro is used and suffix −ic is added to the root of the second element. Hence, its name is hydroselenic acid.

The oxyanion present in ${\text{HClO}}_{\text{3}}$ is chlorate. As the oxyanion’s name is ends with suffix −ate, hence replace it with −ic. Name of ${\text{HClO}}_{\text{3}}$ is chloric acid.

Conclusion

An acid produces ${\text{H}}^{\text{+}}$ ion in solution.