JEE PYQ: P Block Elements Question 26
Question 26 - 2020 (03 Sep Shift 2)
The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ___.
Show Answer
Answer: 10
Solution
Phosphinic acid is hypophosphorous acid ($\text{H}_3\text{PO}_2$). NaOH + $\text{H}_3\text{PO}2 \to \text{NaH}2\text{PO}2 + \text{H}2\text{O}$. For neutralization: $(N_1V_1){\text{acid}} = (N_2V_2){\text{base}}$. $0.1 \times 10 = 0.1 \times V{\text{NaOH}}$. $V{\text{NaOH}} = 10$ mL.
BETA
