Which of the following happens when $\mathrm{NH_4OH}$ is added gradually to the solution containing 1 M $\mathrm{A^{2+}}$ and 1M $\mathrm{B^{3+}}$ ions? Given : $K_{sp}[A(OH)_2]=9 imes10^{-10}$ and $K_{sp}[B(OH)_3]=27 imes10^{-18}$ at 298 K.

Question

Which of the following happens when $\mathrm{NH_4OH}$ is added gradually to the solution containing 1 M $\mathrm{A^{2+}}$ and 1M $\mathrm{B^{3+}}$ ions? Given : $K_{sp}[A(OH)_2]=9	imes10^{-10}$ and $K_{sp}[B(OH)_3]=27	imes10^{-18}$ at 298 K.

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] For $A(OH)_2$: $[OH^-]=\sqrt{9	imes10^{-10}/1}=3	imes10^{-5}$ M. For $B(OH)_3$: $[OH^-]=\left(27	imes10^{-18}ight)^{1/3}=3	imes10^{-6}$ M. $B(OH)_3$ needs the lower $[OH^-]$, so it precipitates first. Answer (3).

Q. Which of the following happens when $\mathrm{NH_4OH}$ is added gradually to the solution containing 1 M $\mathrm{A^{2+}}$ and 1M $\mathrm{B^{3+}}$ ions? Given : $K_{sp}[A(OH)_2]=9\times10^{-10}$ and $K_{sp}[B(OH)_3]=27\times10^{-18}$ at 298 K.

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?