MHT CET Chemistry question
The volume of 0.25 required to neutralise 25 mL of 0.03 is:
Answer: (D)
Solution
Ca(OH)₂ gives 25 × 0.03 × 2 = 1.5 meq and H₃PO₄ (tribasic) has N = 0.75 N for complete neutralisation,
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