The volume of frac{N}{10} NaOH required to ne | vedclass

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(C) For complete neutralization,the number of milli-equivalents of base must equal the number of milli-equivalents of acid.Using the formula: $N_1V_1

Vedclass · Accepted Answer

$40$

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(C) For complete neutralization,the number of milli-equivalents of base must equal the number of milli-equivalents of acid.Using the formula: $N_1V_1 = N_2V_2$Where:$N_1 = \frac{1}{10} \ N$ (Normality of $NaOH$)$V_1 = ?$ (Volume of $NaOH$ in $mL$)$N_2 = \frac{1}{25} \ N$ (Normality of $HCl$)$V_2 = 100 \ mL$ (Volume of $HCl$)Substituting the values:$\frac{1}{10} 	imes V_1 = \frac{1}{25} 	imes 100$$V_1 = \frac{100}{25} 	imes 10$$V_1 = 4 	imes 10 = 40 \ mL$Therefore,the required volume of $NaOH$ is $40 \ mL$.

The volume of $\frac{N}{10}$ $NaOH$ required to neutralize $100 \ mL$ of $\frac{N}{25}$ $HCl$ is......$mL$

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