The volume in mL of 0.1 M solution of NaOH re | vedclass

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(B) Phosphorous acid $(H_3PO_3)$ is a dibasic acid,meaning its n-factor is $2$. For neutralization,the number of equivalents of acid must equal the nu

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$600$

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(B) Phosphorous acid $(H_3PO_3)$ is a dibasic acid,meaning its n-factor is $2$. For neutralization,the number of equivalents of acid must equal the number of equivalents of base: $N_1V_1 = N_2V_2$. Normality $(N)$ is calculated as $Molarity 	imes n	ext{-factor}$. For $H_3PO_3$: $N = 0.3 \ M 	imes 2 = 0.6 \ N$. For $NaOH$: $N = 0.1 \ M 	imes 1 = 0.1 \ N$. Using the equation: $0.1 \ N 	imes V_{NaOH} = 0.6 \ N 	imes 100 \ mL$. $V_{NaOH} = \frac{0.6 	imes 100}{0.1} = 600 \ mL$.

The volume in $mL$ of $0.1 M$ solution of $NaOH$ required to completely neutralize $100 \ mL$ of $0.3 M$ solution of $H_3PO_3$ is:

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