Question

Consider a mixture $X$ which is made by dissolving $0.4$ mol of $[\mathrm{Co(NH_3)_5SO_4}]Br$ and $0.4$ mol of $[\mathrm{Co(NH_3)_5Br}]SO_4$ in water to make $4$ L of solution. When $2$ L of mixture $X$ is allowed to react with excess $\mathrm{AgNO_3}$, it forms precipitate $Y$. The rest $2$ L of mixture $X$ reacts with excess $\mathrm{BaCl_2}$ to form precipitate $Z$. Which of the following statements is CORRECT?

• A. $Y$ is $\mathrm{BaSO_4}$ and $Z$ is $\mathrm{AgBr}$
• B. $0.1$ mol of $Y$ is formed
• C. $0.2$ mol of $Z$ is formed
• D. $0.4$ mol of $Z$ is formed

Hint:

Only counter ions take part in precipitation reactions; ligands inside coordination sphere do not.

Answer 1

The Correct Option is C

To solve this problem, we need to examine the reactions taking place in the solution when mixed with both \(AgNO_3\) and \(BaCl_2\).

The mixture \(X\) contains two compounds:

• \([\mathrm{Co(NH_3)_5SO_4}]Br\)
• \([\mathrm{Co(NH_3)_5Br}]SO_4\)

Both are dissolved in water to form ionic species.

When 2 L of mixture \(X\) is reacted with excess \(AgNO_3\), the \(Br^−\) ions from \([\mathrm{Co(NH_3)_5SO_4}]Br\) and \([\mathrm{Co(NH_3)_5Br}]SO_4\) will form a precipitate with silver ions:

\(Ag^+ + Br^- \rightarrow AgBr \; (\text{precipitate})\)

Equivalently, an excess of \(BaCl_2\) will react with \(SO_4^{2-}\) ions (from both compounds) to form:

\(Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \; (\text{precipitate})\)

Given two compounds, each contributes its respective ions:

• \([\mathrm{Co(NH_3)_5SO_4}]Br\) provides \(Br^−\) and \(SO_4^{2-}\)
• \([\mathrm{Co(NH_3)_5Br}]SO_4\) provides \(SO_4^{2-}\) and \(Br^−\)

1. The initial concentration of each is \(0.4\) mol in \(4\) L total.
2. This yields \(0.1\) mol/L for each compound in the solution.
3. Since the solution is divided equally, in \(2\) L of solution:
4. We have \(0.2\) mol of \([\mathrm{Co(NH_3)_5SO_4}]Br\) and \(0.2\) mol of \([\mathrm{Co(NH_3)_5Br}]SO_4\).
5. Each molecule gives one mol of \(Br^−\) and one of \(SO_4^{2-}\).
6. Thus, the reaction with \(AgNO_3\) results in \(0.2\) mol of \(AgBr\) as precipitate \(Z\).
7. For the same reasoning, the reaction with \(BaCl_2\) gives \(0.2\) mol of \(BaSO_4\) as precipitate \(Y\).

Conclusion: Given the problem's details and the reaction equations, the correct statement is that \(0.2\) mol of \(Z\) is formed.