Which of the following happens when $\mathrm{NH}_{4} \mathrm{OH}$ is added gradually to the solution

Question

Which of the following happens when $\mathrm{NH}_{4} \mathrm{OH}$ is added gradually to the solution containing $1 \mathrm{M} \mathrm{A}^{2+}$ and $1 \mathrm{M} B^{3+}$ ions?\ Given : $K_{s p}\left[A(\mathrm{OH})_{2}ight]=9 	imes 10^{-10}$ and $K_{s p}\left[B(\mathrm{OH})_{3}ight]=27 	imes 10^{-18}$ at 298 K .\

Professor PC Thomas Classes · Answer

(b) : (i) Concentration of $\mathrm{OH}^{-}$ required for precipitation of $A(\mathrm{OH})_{2}$ :\ $K_{s p}\left[A(\mathrm{OH})_{2}ight]=\left[A^{2+}ight]\left[\mathrm{OH}^{-}ight]^{2}$\ $9 	imes 10^{-10}=(1)\left[\mathrm{OH}^{-}ight]^{2}$ $$ \begin{aligned} {\left[\mathrm{OH}^{-}ight] } & =\sqrt{9 	imes 10^{-10}} \ & =3 	imes 10^{-5} \mathrm{M} \end{aligned} $$ (ii) Concentration of $\mathrm{OH}^{-}$required for precipitation of $B(\mathrm{OH})_{3}$ :\ $K_{	ext {sp }}\left[B(\mathrm{OH})_{3}ight]=\left[B^{3+}ight]\left[\mathrm{OH}^{-}ight]^{3}$\ $27 	imes 10^{-18}=(1)\left[\mathrm{OH}^{-}ight]^{3}$\ $\left[\mathrm{OH}^{-}ight]=3 	imes 10^{-6} \mathrm{M}$\ ∵ Amount of $\mathrm{OH}^{-}$required for precipitation is low for $B^{3+}$ ion. So, $B(\mathrm{OH})_{3}$ gets precipitated first.\