1 L, 0.02 M solution of [Co(NH3)5SO4]Br is mixed with 1 L, 0.02 M solution of [Co(NH3)5Br]SO4. The resulting solution…

🧠 Two Ionisation Isomers — Different Outer Ions

$[\mathrm{Co(NH_3)_5SO_4}]\mathrm{Br}$ and $[\mathrm{Co(NH_3)_5Br}]\mathrm{SO_4}$ are ionisation isomers. They have the same atoms but swap which ion sits inside vs outside the bracket:

• First salt: $\mathrm{SO_4^{2-}}$ inside (coordinated to Co), Br⁻ outside (free in solution).
• Second salt: Br⁻ inside, $\mathrm{SO_4^{2-}}$ outside.

Free in solution after dissolution:

• 1 L of 0.02 M first → 0.02 mol Br⁻ free.
• 1 L of 0.02 M second → 0.02 mol $\mathrm{SO_4^{2-}}$ free.

🗺️ Mix and Split

Total mixed solution (2 L) contains: 0.02 mol Br⁻ + 0.02 mol $\mathrm{SO_4^{2-}}$.

Divide into 2 equal parts (1 L each): each part has 0.01 mol Br⁻ + 0.01 mol $\mathrm{SO_4^{2-}}$.

Treatment 1: 1 L (X) + AgNO₃ excess → AgBr precipitate from free Br⁻ → Y = 0.01 mol.

Treatment 2: 1 L (X) + BaCl₂ excess → BaSO₄ precipitate from free $\mathrm{SO_4^{2-}}$ → Z = 0.01 mol.

⚡ Ionisation Isomerism Definition

Two complexes that produce different ions in solution despite the same molecular formula. The classic pair: $[\mathrm{Co(NH_3)_5SO_4}]\mathrm{Br}$ vs $[\mathrm{Co(NH_3)_5Br}]\mathrm{SO_4}$ — exchange the inner/outer roles of $\mathrm{SO_4^{2-}}$ and Br⁻.

⚠️ Don't Forget the "Half" From the Split

After mixing 0.02 + 0.02 mol and splitting into two 1-L portions, each portion has only half of the original — 0.01 mol of each free ion.

$\boxed{\text{Answer: (2) 0.01, 0.01}}$