mL of is required to neutralize mL of . is: | Redox Reactions and Volumetric Analysis — Chem-Mantra

Name: JEE Advanced Chemistry
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Question

$25$ mL of $0.1 N$ $NaOH$ is required to neutralize $V$ mL of $0.5 N$ $HCl$. $V$ is:

Answer: 5

💡 Solution & Explanation

Using normality equation: $N_1 V_1 = N_2 V_2$. <br>$0.1 \times 25 = 0.5 \times V \Rightarrow 2.5 = 0.5 V \Rightarrow V = 5$ mL.

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