• # question_answer $Zn{{(OH)}_{2}}$ is not precipitated when $N{{H}_{4}}OH$  is added to a zinc salt solution containing some ammonium chloride. Explain.

$Zn{{(OH)}_{2}}$ 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. $\underset{\text{(Weakbase)}}{\mathop{N{{H}_{4}}OH}}\,\overset{(aq)}{\mathop{\rightleftharpoons }}\,N{{H}_{4}}^{+}(aq)+O{{H}^{-}}(aq)$ $N{{H}_{4}}Cl\xrightarrow{(aq)}\,N{{H}_{4}}^{+}\,(aq)+C{{l}^{-}}\,(aq)$