Question

An aqueous solution of colourless metal sulphate M, give a white a white ppt. with $N{{H}_{4}}OH$. this was soluble in excess of $N{{H}_{4}}OH$. on passing ${{H}_{2}}S$ through this solution a white ppt is formed. the metal M in the salt is ?
A. $Ca$
B. $Ba$
C. $Al$
D. $Zn$

Answer 1

Hint: Precipitation is the process in which the considered compound comes out of solution and settles down at the bottom of the solution.This is a result of the particles of solute finding each other and in order to form a solid substance. This solid is then termed as the precipitate.

Complete step by step answer:
qualitative analysis is the determination of non-numerical information about a chemical species, a reaction etc. it is the method of analytical chemistry which seeks to find the elemental composition of inorganic compounds. It is mainly focused on the detection of elements or ions in inorganic compounds by using several methods and techniques. The solution is treated with various reagents to test for reaction characteristics of certain ions, which may cause color change, precipitation and other visible change.
Salt is formed when acid and base react with each other. It is composed of cation and anion. Salt analysis is done by individuals identifying the cation and anion of the salt.
Comes to the solution part; the reagent $N{H_4}OH$ IS added before passing ${H_2}S$ gas belong to group \[{{4}^{th}}\]
The metal M is the zinc. an aqueous solution of colorless zinc sulphate, gave a white ppt as a product and this white ppt is of $Zn{\left( {OH} \right)_2}$, reaction involved,
$ZnS{O_4} + 2N{H_4}OH \to Zn{\left( {OH} \right)_2} \downarrow + {\left( {N{H_4}} \right)_2}S{O_4}$
When excess of ammonium hydroxide present the ppt of zinc hydroxide was soluble in it to form ammonium zincate ${\left( {N{H_4}} \right)_4}Zn{O_2}$.
$Zn{\left( {OH} \right)_2} + 2N{H_4}OH \to {\left( {N{H_4}} \right)_2}Zn{O_2} + 2{H_2}O$
On passing ${H_2}S$ through the solution a white ppt of $ZnS$ is formed,
${\left( {N{H_4}} \right)_2}Zn{O_2} + {H_2}S \to ZnS + 2N{H_4}OH$

So, the correct answer is Option b.

Note: The reagent $N{{H}_{4}}OH$ IS added before passing ${{H}_{2}}S$ gas belong to group \[{{4}^{th}}\]. This is because the solubility product of sulphide of group \[{{2}^{nd}}\] radicals is less than solubility product of sulphides of \[{{4}^{th}}\] group. so the precipitate of \[{{2}^{nd}}\] group radicals as sulphides ${{S}^{2-}}$ are sufficient. in presence of dil $HCl$ ${{S}^{2-}}$ are released due to common ion effects.