Coordination Compounds Part 4

Concept:Only ions outside the coordination sphere dissociate in water and form precipitates.Explanation:First, identify the free ions from each salt.Salt 1: $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]\mathrm{Br}$ dissociates to give $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]^+$ and $\mathrm{Br}^-$.So 1 mol of this salt gives 1 mol $\mathrm{Br}^-$.Salt 2: $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]\mathrm{SO}_4$ dissociates to give $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]^{2+}$ and $\mathrm{SO}_4^{2-}$.So 1 mol of this salt gives 1 mol $\mathrm{SO}_4^{2-}$.Total solution volume is 4 L, and we take 2 L for each reaction (half the mixture).In 2 L, moles of each salt = $0.4\,\mathrm{mol} \times \frac{2}{4} = 0.2\,\mathrm{mol}$.Thus, from salt 1: $0.2$ mol $\mathrm{Br}^-$.From salt 2: $0.2$ mol $\mathrm{SO}_4^{2-}$.When 2 L reacts with excess $\mathrm{AgNO}_3$, $\mathrm{Ag}^+$ precipitates $\mathrm{Br}^-$ as $\mathrm{AgBr}$:$\mathrm{Ag}^+ + \mathrm{Br}^- \rightarrow \mathrm{AgBr}(s)$.So precipitate Y ($\mathrm{AgBr}$) formed = $0.2$ mol.When the other 2 L reacts with excess $\mathrm{BaCl}_2$, $\mathrm{Ba}^{2+}$ precipitates $\mathrm{SO}_4^{2-}$ as $\mathrm{BaSO}_4$:$\mathrm{Ba}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{BaSO}_4(s)$.So precipitate Z ($\mathrm{BaSO}_4$) formed = $0.2$ mol.Answer:Option C: $0.2$ mol of Z ($\mathrm{BaSO}_4$) is formed.