Question

A solution contaming $N{H}_{4}Cl$ and $N{H}_{4}OH$ has $\left[\stackrel{⊝}{O}H\right]={10}^{-6}mol{L}^{-1}$ which of the following hydroxides would be precipitated when this solution in added in equal volume to a solution containing 0.1 M of metal ions?

• A. $Mg(OH{)}_2,(K_{sp}=3\times {10}^{-11})$
• B. $Fe\left(OH{)}_2\right(K_{sp}=8\times {10}^{-16})$
• C. $Cd\left(OH{)}_2\right(K_{sp}=8\times {10}^{-6})$
• D. $AgOH(K_{sp}=5\times {10}^{-3})$

Answer 1

The correct option is B $Fe\left(OH{)}_2\right(K_{sp}=8\times {10}^{-16})$

Only that compound will precipitate where $Q>K_{sp}$

$Fe\left(OH{)}_2\right(K_{sp}=8\times {10}^{-16})$

$Q=\left[F{e}^{+2}\right][O{H}^{-}{]}^2=0.1\times ({10}^{-6}{)}^2={10}^{-13}>K_{sp}=8\times {10}^{-16}$

So only $Fe(OH{)}_2$ will precipitate and others will not as they have higher $K_{sp}$.