Answer:
\[Zn{{(OH)}_{2}}\] can precipitated if the product of the ionic concentration of the ions furnished by the salt in aqueous solution exceeds the\[{{K}_{sp}}\] value of the hydroxide. Addition of\[N{{H}_{4}}Cl\] to the aqueous solution of the zinc salt in which \[N{{H}_{4}}OH\] has been added as the precipitating reagent will suppress the \[O{{H}^{-}}\] ions concentration in solution due to common ion effect (\[N{{H}_{4}}^{+}\] ions being the common ions). This means that the concentration of the \[O{{H}^{-}}\] ions in solution will be sufficiently suppressed or decreased and the product of ionic concentration will be less than the \[{{K}_{sp}}\] value of \[Zn{{(OH)}_{2}}\] and it will not be, therefore, precipitated.
\[\underset{\text{(Weakbase)}}{\mathop{N{{H}_{4}}OH}}\,\overset{(aq)}{\mathop{\rightleftharpoons }}\,N{{H}_{4}}^{+}(aq)+O{{H}^{-}}(aq)\]
\[N{{H}_{4}}Cl\xrightarrow{(aq)}\,N{{H}_{4}}^{+}\,(aq)+C{{l}^{-}}\,(aq)\]
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