JEE MAIN - Chemistry (2020 - 3rd September Evening Slot - No. 20)
The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ___________.
Answer
10
Explanation
H3PO2 + NaOH $$ \to $$ NaH2PO2 + H2O
Using Stoichiometry
$$ \Rightarrow $$ $${{0.1 \times 10} \over 1}$$ = 0.1 × VNaOH
$$ \Rightarrow $$ VNaOH = 10 ml
Using Stoichiometry
Moles of H3PO2 reacted
1
=
Moles of NaOH reacted
1
$$ \Rightarrow $$ $${{0.1 \times 10} \over 1}$$ = 0.1 × VNaOH
$$ \Rightarrow $$ VNaOH = 10 ml
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