Solubility trend ratio in group 2 compounds Steps:

• Neutralization of equal volumes of saturated M(OH)₂ solutions produces moles of MSO₄ proportional to M(OH)₂ solubility, which increases down group 2: Mg < Ca < Sr.
• MSO₄ solubilities decrease down group 2: Mg > Ca > Sr.
• Minimum volume for saturated MSO₄ solution = moles MSO₄ / MSO₄ solubility.
• Volume ratio S_M(OH)₂ / S_MSO₄ is lowest for Mg, intermediate for Ca, highest for Sr.

Why C is correct:

• SrSO₄ has the highest moles from soluble Sr(OH)₂ but lowest solubility, maximizing volume per the V = n / S formula.

Why the others are wrong:

• A: CaSO₄ ratio is intermediate, less than SrSO₄.
• B: MgSO₄ has fewest moles and highest solubility, minimizing volume.
• D: Differing solubility trends yield unequal volumes.

Final answer: C