Verified Solution • 2025–26 Session moderate

QJEEM20Chd2a9: JEE Main 2020 Chemistry - Class 11 JEEM20Chd2a9 Solution: The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL…

The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ___________.

Step-by-Step Verified Solution

H3PO2 + NaOH →\to NaH2PO2 + H2O

Using Stoichiometry

Moles of H3PO2 reacted
1
 =
Moles of NaOH reacted
1

⇒\Rightarrow 0.1×101{{0.1 \times 10} \over 1} = 0.1 × VNaOH

⇒\Rightarrow VNaOH = 10 ml

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