Chapter 15, Problem 47A

Interpretation Introduction

Interpretation: Volume of $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to neutralize $\text{56}\text{.2}\text{ml}$ of $\text{0}\text{.145}\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: One normal solution is a measure of the number of moles of hydrogen ions furnished by one mol of an acid. One molar solution is a measure of the number of moles of molecules of acid in one mol of an acid. One molecule of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ furnishes 2 hydrogen ions. So, one mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ furnishes 2 mol of hydrogen ions.

2 normal ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution = 1 molar ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution

Balanced chemical equation for the reaction of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{NaOH}$ is as follows:

  ${\text{2NaOH+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+2H}}_{\text{2}}\text{O}$

Also the volume of $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to neutralize $\text{56}\text{.2}\text{ml}$ of $\text{0}\text{.145}\text{M}$

  $\text{NaOH}$ can be calculated using the molarity equation written as follows:

  $\left(\text{NaOH}\right)\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$

Answer to Problem 47A

Volume of $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to neutralize $\text{56}\text{.2}\text{ml}$ of $\text{0}\text{.145}\text{M}$

  $\text{NaOH}$ is $47.37\text{ mL}$ .

Explanation of Solution

One normal solution is a measure of the number of moles of hydrogen ions furnished by one mol of an acid. One molar solution is a measure of the number of moles of molecules of acid in one mol of an acid. One molecule of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ furnishes 2 hydrogen ions. So, one mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ furnishes 2 mol of hydrogen ions.

2 normal ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution = 1 molar ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution

The given acid solution is $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Thus, molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is half of $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Therefore, molarity will be 0.086 M.

Balanced chemical equation for the reaction of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{NaOH}$ is as follows:

  ${\text{2NaOH+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+2H}}_{\text{2}}\text{O}$

Also the volume of $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to neutralize $\text{56}\text{.2}\text{ml}$ of $\text{0}\text{.145}\text{M}$

  $\text{NaOH}$ can be calculated using the molarity equation written as follows:

  $\left(\text{NaOH}\right)\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$

Putting in all the values,

  $\begin{array}{c}\frac{0.145\times 56.2}{2}=\frac{0.086\times {\mathrm{V}}_2}{1}\\ {\mathrm{V}}_2=\frac{0.145\times 56.2}{2\times 0.086}\\ {\mathrm{V}}_2=47.37\text{mL}\end{array}$

Conclusion

Volume of $\text{0}\text{.172}{\text{N H}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to neutralize $\text{56}\text{.2}\text{ml}$ of $\text{0}\text{.145}\text{M}$

  $\text{NaOH}$ is $\text{47}\text{.37}\text{mL}$ .