Which of the following happens when $\mathrm{NH}_4 \mathrm{OH}$ is added gradually to the solution containing 1 M $\mathrm{A}^{2+}$ and $1 \mathrm{M} \mathrm{B}^{3+}$ ions?
Given : $\mathrm{K}_{\text {sp }}\left[\mathrm{A}(\mathrm{OH})_2\right]=9 \times 10^{-10}$ and $\mathrm{K}_{\mathrm{sp}}\left[\mathrm{B}(\mathrm{OH})_3\right]=27 \times 10^{-18}$ at 298 K.
Solution
<p>Condition for precipitation $\mathrm{Q}_{\mathrm{ip}}>\mathrm{K}_{\text {sp }}$</p>
<p>For $\left[\mathrm{A}(\mathrm{OH})_2\right.$]</p>
<p>$$\begin{aligned}
& {\left[\mathrm{A}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2>9 \times 10^{-10}} \\
& {\left[\mathrm{~A}^{+2}\right]=1 \mathrm{M}} \\
& \Rightarrow\left[\mathrm{OH}^{-}\right]>3 \times 10^{-5} \mathrm{M}
\end{aligned}$$</p>
<p>For $\left[\mathrm{B}(\mathrm{OH})_3\right]$</p>
<p>$$\begin{aligned}
& {\left[\mathrm{B}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3>27 \times 10^{-18}} \\
& {\left[\mathrm{~B}^{3+}\right]=1 \mathrm{M}} \\
& \Rightarrow\left[\mathrm{OH}^{-}\right]>3 \times 10^{-6} \mathrm{M}
\end{aligned}$$</p>
<p>So, $\mathrm{B}(\mathrm{OH})_3$ will precipitate before $\mathrm{A}(\mathrm{OH})_2$</p>
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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