Question

An unknown mixture contains one or two of the following:
$\;CaC{O_3},BaC{l_2},AgN{O_3},N{a_2}S{O_4},ZnS{O_4}$​ and $NaOH$.
The mixture is completely soluble in water and the solution gives pink colour with phenolphthalein. When dilute hydrochloric acid is gradually added to the solution, a precipitate is formed which dissolves with further addition of the acid.
The white precipitate is_______.
A) $ZnS{O_4}$​
B) $N{a_2}Zn{O_2}$​
C) $Zn{\left( {OH} \right)_2}$
D) $ZnC{l_2}$

Answer 1

With the addition of the excess sodium hydroxide to the precipitate of zinc hydroxide will make it dissolve completely and will produce a colourless solution of zincate ion. The addition of the acid makes the solubilisation of the compound at a faster rate.

Complete step by step answer:
From the given when the mixture of sodium hydroxide and zinc sulphate solution reacts, in the limited quantity, the end product formed is white precipitate of zinc hydroxide which will be seen:
$Z{n^{2 + }}\; + {\text{ }}2O{H^ - }\; \to {\text{ }}Zn{\left( {OH} \right)_2}$.
$Z{n^{2 + }}\;$is known to form hexa-aqua ions and tetra-aqua ions at high water concentrations and at low concentrations of water respectively. Thus, this reaction may be written as the reaction between ion and hydroxide through donation of a proton, as follows.
$Z{n^{2 + }}{\left( {O{H_2}} \right)_{4(aq)}}\; + {\text{ }}O{H^ - }_{(aq)}\; \to {\text{ }}Z{n^{2 + }}{\left( {O{H_2}} \right)_3}O{H^ - }_{(aq)}\; + {\text{ }}{H_2}{O_{(l)}}$
The reactions discussed below can also be considered as reactions between the aquated zinc ion and it is quite an adjustable process.
However, for simplicity of the reaction, the water molecules are removed from the reaction.
On addition of the excess sodium hydroxide, the precipitate of zinc hydroxide will dissolve, producing a colourless solution of zincate ion
$Zn{\left( {OH} \right)_2}\; + {\text{ }}2{\text{ }}O{H^ - }\; \to {\text{ }}Zn{\left( {OH} \right)_4}^{2 - }$.
Therefore, the correct option is C.

Note:
Zinc hydroxide will dissolve because its ion shows high affinity towards the water ligands. Addition of the excess sodium hydroxide to the solution, hydroxide ions formed will reduce the complex to a charge with $- 2$ and solubilizes it. Addition of excess ammonia, sets up an equilibrium condition which produces hydroxide ions and their formation of hydroxide ions creates a complex having overall charge of $+ 2$ with a coordination number of 4 with the ammonia ligands making it water soluble.