Question

When phosphorus oxide reacts with water, the product:
A.Turns blue litmus red
B.Is a mineral acid
C.Turns lime water milky
D.Both A and B

Answer 1

Hint: Many oxides are formed by phosphorus. The most common form of oxides is 2. They are ${{{P}}_{{4}}}{{{O}}_{{6}}}$ and ${{{P}}_{{4}}}{{{O}}_{{{10}}}}$. These are the oxides of non-metal phosphorus. Oxides of nonmetals are acidic. Acids turn blue litmus red and bases turn red litmus blue.

Complete step by step answer:
We already mentioned that oxides of nonmetals are acidic.
Nonmetal oxides are acidic. They dissolve in water to produce an acidic solution.
Phosphorus is non-metal.
One of the oxides of phosphorus as we already mentioned above is ${{{P}}_{{4}}}{{{O}}_{{{10}}}}$ , it is a phosphorus (V) oxide of phosphorus pentoxide. In this, phosphorus is in $ + 5$ the oxidation state.
When ${{{P}}_{{4}}}{{{O}}_{{{10}}}}$ is reacted with water, it forms orthophosphoric acid or phosphoric acid.
${{{P}}_{{4}}}{{{O}}_{{{10}}}}{{ + 6}}{{{H}}_{{2}}}{{O }} \to {{ 4}}{{{H}}_{{3}}}{{P}}{{{O}}_{{4}}}$
${{{H}}_{{3}}}{{P}}{{{O}}_{{4}}}$ is acid as it can release protons
When ${{{P}}_{{4}}}{{{O}}_{{{10}}}}$ is dropped into water, it makes a hissing like sound, heat gets liberated and there is formation of an acid. Therefore, ${{{P}}_{{4}}}{{{O}}_{{{10}}}}$ is called orthophosphoric anhydride.
Thus, we can conclude that oxides of non-metals are acidic.
Since orthophosphoric acid is an acid, it turns blue litmus red. The option (A) is correct.
The inorganic acids are also called mineral acids. Thus, option (B) is also correct.
It does not turn lime water milky because carbonates are not formed here.
The correct option is (D).

Note:
${{{P}}_{{4}}}{{{O}}_{{{10}}}}$ has a great liking for water, it is used as a drying agent of gases and also to undergo dehydration (to remove water from any compound). If there is a reaction of oxide of carbon instead of phosphorus, then there would be the formation of acid which turns lime water milky. The chemical reaction can be written as:
${{C}}{{{O}}_{{2}}}{{ + }}{{{H}}_{{2}}}{{O}} \to {{{H}}_{{2}}}{{C}}{{{O}}_{{3}}}$