What is the molarity of H_2SO_4 solution if 2 | vedclass

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(A) The neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$. According to the law of equivalence,$N_1V_1 = N_2V_2$. For $NaOH$,

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$0.107$

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(A) The neutralization reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. According to the law of equivalence,$N_1V_1 = N_2V_2$. For $NaOH$,$Molarity = Normality$ because its $n$-factor is $1$. So,$N_{NaOH} = 0.164 \ N$. Using the formula $N_{acid} 	imes V_{acid} = N_{base} 	imes V_{base}$: $N_{acid} 	imes 25 \ mL = 0.164 \ N 	imes 32.63 \ mL$. $N_{acid} = \frac{0.164 	imes 32.63}{25} = 0.214 \ N$. Since $Molarity = \frac{Normality}{n	ext{-factor}}$ and the $n$-factor (basicity) of $H_2SO_4$ is $2$: $Molarity = \frac{0.214}{2} = 0.107 \ M$.

What is the molarity of $H_2SO_4$ solution if $25 \ mL$ is exactly neutralized with $32.63 \ mL$ of $0.164 \ M$ $NaOH$ (in $M$)?

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