Chemistry Physical Chemistry questions from JEE Main 2010.
At $25^{\circ} \mathrm{C}$, the solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ is $1.0 \times 10^{-11}$. At which $\mathrm{pH}$, will $\mathrm{Mg}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH})_2$ from a solution of $0.001 \mathrm{M} \mathrm{Mg}^{2+}$ ions ?
Consider the reaction : $$ \mathrm{Cl}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{~S}(\mathrm{aq}) \rightarrow \mathrm{S}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) $$ The rate equation for this reaction is rate $=\mathrm{k}\left[\mathrm{Cl}_2\right]\left[\mathrm{H}_2 \mathrm{~S}\right]$ Which of these mechanisms is/are consistent with this rate equation? (A) $\mathrm{Cl}_2+\mathrm{H}_2 \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{Cl}^{+}+\mathrm{HS}^{-} \quad$ (slow) $$ \mathrm{Cl}^{+}+\mathrm{HS}^{-} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{S} \text { (fast) } $$ (B) $\mathrm{H}_2 \mathrm{~S} \Leftrightarrow \mathrm{H}^{+}+\mathrm{HS}^{-} \quad$ (fast equilibrium) $\mathrm{Cl}_2+\mathrm{HS}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{H}^{+}+\mathrm{S} \quad$ (slow)