Differential solubility causes Ca(OH)₂ precipitation, followed by BaCO₃ formation Steps:

• Mixing equal volumes raises [OH⁻] via soluble Ba(OH)₂, exceeding Ksp for less soluble Ca(OH)₂, forming white precipitate Y = Ca(OH)₂.
• Filtration yields Y as the solid and filtrate rich in Ba²⁺/OH⁻.
• Bubbling CO₂ through filtrate: Ba²⁺ + CO₃²⁻ → BaCO₃ ↓ (white precipitate Z).
• BaCO₃ confirms Z via the reaction Ba(OH)₂ + CO₂ → BaCO₃ + H₂O. Why C is correct:
• Ca(OH)₂ (Ksp = 5.5 × 10⁻⁶) precipitates initially due to common ion effect from Ba(OH)₂ (Ksp = 5 × 10⁻³); CO₂ forms insoluble BaCO₃ (Ksp = 5 × 10⁻⁹). Why the others are wrong:
• A: Ba(OH)₂ is more soluble and stays in solution; initial precipitate is not Ca(OH)₂ as Z.
• B: Identical to C (possible error), but if intending Y = Ba(OH)₂, then Y wrongly assigns soluble species as initial precipitate.
• D: Ba(OH)₂ does not precipitate initially due to higher solubility. Final answer: C