Question

Colourless solution of (A) gives white precipitate of (B) with silver nitrate solution, soluble in aqueous ammonia. (A) also gives white precipitate (C) with sodium hydroxide soluble in excess of it forming (D). (D) gives white precipitate € with hydrogen sulphide. (A), (B), (C), (D) and (E) are as follows:
A) \[ZnCl_2\]
B) \[AgCl\]
C) \[Zn{OH}_2\]
D) \[{Na}_2ZnO_2\]
E) \[ZnS\]

Answer 1

Hint: Silver chloride, zinc hydroxide and zinc sulphide are the compounds having the form of white precipitate. The complex of insoluble silver chloride with ammonia is water soluble.
Excess sodium hydroxide converts zinc hydroxide into water soluble sodium zincate.

Complete answer:
The reaction of colourless solution zinc chloride with silver nitrate solution forms silver chloride. Silver chloride is present in the form of white precipitate and is labeled as (B).
\[{\rm{ZnC}}{{\rm{l}}_{\rm{2}}} + {\rm{2AgN}}{{\rm{O}}_{\rm{3}}}{\rm{ }} \to {\rm{ 2AgCl}} + {\rm{ Zn}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\]

In presence of aqueous ammonia, silver chloride dissolves as it forms a soluble complex. This complex contains one silver atom, one chlorine atom and two ammonia molecules.
\[{\rm{AgCl}} + {\rm{2N}}{{\rm{H}}_{\rm{3}}}{\rm{ }} \to {\rm{ }}\left[ {{\rm{Ag}}{{\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right]{\rm{Cl}}\]
The reaction of colourless solution of zinc chloride with sodium hydroxide gives zinc hydroxide. Zinc hydroxide is present as a white precipitate and is labeled as (C). This white precipitate of zinc hydroxide is soluble in excess of sodium hydroxide solution, as it forms sodium zincate. Sodium zincate is labeled as (D).
\[{\rm{ZnC}}{{\rm{l}}_{\rm{2}}}{\rm{ + 2NaOH}} \to {\rm{Zn(OH}}{{\rm{)}}_{\rm{2}}}{\rm{ + 2NaCl}} \\
{\rm{Zn(OH}}{{\rm{)}}_{\rm{2}}}{\rm{ + 2NaOH}} \to {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{Zn}}{{\rm{O}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \\\]
The reaction of sodium zincate with hydrogen sulphide gives zinc sulphide. Zinc sulphide is present in the form of white precipitate and is labeled as (E).
\[{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{Zn}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{S}} \to {\rm{ZnS + 2NaOH}}\]

Note: In the qualitative analysis, the presence of chloride anion can be detected by reaction with silver nitrate solution to form a white precipitate of silver chloride. Similarly, the presence of zinc cations can be detected by the reaction with sodium hydroxide solution. Zinc cations react with sodium hydroxide to form a precipitate that dissolves in excess of sodium hydroxide solution.