What volume of $0.151~N\mathrm{~NaOH}$ is required to neutralize $24.2\mathrm{~mL}$ of $0.125~N\mathrm{~H}_2\mathrm{SO}_4$ ? What volume of $0.151~N\mathrm{~NaOH}$ is required to neutralize $24.1 \mathrm{~mL}$ of $0.125~M\mathrm{~H}_2 \mathrm{SO}_4$ ?

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We know that for neutralization, equiv (acid) = equiv (base),

\begin{align*} N_{\text{acid}}\times V_{\text{acid}} = N_{\text{base}}\times V_{\text{base}}\\ V_{\text{base}} = \dfrac{N_{\text{acid}}\times V_{\text{acid}}}{N_{\text{base}}} \end{align*}

Now we can substitute the given values, $\mathrm{H_2SO_4}$ , $N_{\text{acid}}=0.125\;N,\;V_{\text{acid}}=24.2\;\text{mL}$ , and NaOH, $N_{\text{base}}=0.151\;N$ into the equation.

\begin{align*} V_{\text{base}} &= \dfrac{N_{\text{acid}}\times V_{\text{acid}}}{N_{\text{base}}}\\ &=\dfrac{0.125\;N\times 24.2\text{ mL}}{0.151\;N}=20.033\text{ mL} \end{align*}

So 20.033 mL of 0.151 $N$ NaOH is required to neutralize exactly 24.2 mL of 0.125 $N$ $\mathrm{H_2SO_4}$

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What volume of $0.151~N\mathrm{~NaOH}$ is required to neutralize $24.2\mathrm{~mL}$ of $0.125~N\mathrm{~H}_2\mathrm{SO}_4$ ? What volume of $0.151~N\mathrm{~NaOH}$ is required to neutralize $24.1 \mathrm{~mL}$ of $0.125~M\mathrm{~H}_2 \mathrm{SO}_4$ ?

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