Selective precipitation by solubility differences

Steps:

• Compare Ksp values: Mg(OH)₂ (5.61 × 10^{-12}) < Sr(OH)₂ (3.2 × 10^{-4}), so Mg(OH)₂ precipitates first with NaOH.
• With dilute H₂SO₄, compare Ksp: SrSO₄ (3.44 × 10^{-7}) << MgSO₄ (highly soluble, ~10^0), so SrSO₄ precipitates first.
• Equal concentrations mean the less soluble compound in each pair forms the initial precipitate dropwise.
• Thus, NaOH portion yields Mg(OH)₂; H₂SO₄ portion yields SrSO₄.

Why B is correct:

• B matches the precipitates: Mg(OH)₂ from lower Ksp hydroxide, SrSO₄ from lower Ksp sulfate, per solubility product principle.

Why the others are wrong:

• A: SrSO₄, not MgSO₄, precipitates due to SrSO₄'s much lower Ksp.
• C: Mg(OH)₂, not Sr(OH)₂, precipitates as it has lower Ksp; MgSO₄ does not precipitate.
• D: Mg(OH)₂, not Sr(OH)₂, forms first; MgSO₄ remains soluble.

Final answer: B