Solubility products of , , and are , , and , respectively. The cation that will precipitate first as — Ionic Equilibrium Chemistry Question
Question
Solubility products of $\text{Mg(OH)}_2$, $\text{Cd(OH)}_2$, $\text{Al(OH)}_3$ and $\text{Zn(OH)}_2$ are $4 \times 10^{-11}$, $8 \times 10^{-6}$, $8.5 \times 10^{-23}$ and $1.8 \times 10^{-14}$, respectively. The cation that will precipitate first as hydroxide, on adding limited quantity of $\text{NH}_4\text{OH}$ in a solution containing equimolar amount of metal cations, is
Answer: A
💡 Solution & Explanation
The cation that requires the lowest concentration of $\text{OH}^-$ ions to exceed its solubility product ($K_{sp}$) will precipitate first. Since the $K_{sp}$ of $\text{Al(OH)}_3$ is extremely small ($8.5 \times 10^{-23}$) compared to the others, it requires the minimum $[\text{OH}^-]$ to precipitate, even accounting for its $1:3$ stoichiometry.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes