Answer:
\[M{{g}^{2+}}\] ions present in its salt solution are to be precipitated as \[Mg{{(OH)}_{2}}\] on adding\[N{{H}_{4}}OH\]. But the presence of \[N{{H}_{4}}Cl\] will suppress the ionisation of \[N{{H}_{4}}OH\] due to common ion effect. As a result, the ionic product of \[M{{g}^{2+}}\] and \[O{{H}^{-}}\] ions will remain less than the \[{{K}_{sp}}\] value of \[Mg{{(OH)}_{2}}\]. It will not be, therefore, precipitated.
\[N{{H}_{4}}OH\,\,\overset{(aq)}{\mathop{\rightleftharpoons }}\,N{{H}_{4}}^{+}(aq)+O{{H}^{-}}(aq)\]
\[N{{H}_{4}}Cl\,\xrightarrow{(aq)}\,N{{H}_{4}}{{\,}^{+}}(aq)+C{{l}^{-}}\,(aq)\]
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