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JEE Main20203 Sep 2020Evening ShiftChemistryp Block Elements (Group 15, 16, 17 & 18)Actual

The volume (in mL ) of 0 . 1 N NaOH required to neutralise 10 mL of 0 . 1 N phosphinic acid is ____________.

Correct answer

0

Step-by-step solution

Phosphinic acid is hypo phosphorous acid H 3 PO 2 . NaOH + H 3 PO 2   →   NaH 2 PO 2 + H 2 O For neutrization N 1 V 1 acid = N 2 V 2 base 0 . 1 × 10 = 0 . 1 × V mL NaOH V NaOH = 10   mL

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