Chapter 15, Problem 39A

Interpretation Introduction

Interpretation: Volume of $0.200\text{M HCl}$ solution needed to neutralize $\text{25}\text{.0 mL}$ of $\text{0}\text{.150M NaOH}$ solution is to be calculated.

Concept introduction: Reaction between $\mathrm{HCl}$ and $\text{NaOH}$ is a neutralization reaction and it results in the formation of a salt and water. Balanced chemical equation for the reaction is as follows:

  $\text{NaOH+HCl}\to {\text{NaCl+H}}_{\text{2}}\text{O}$

To calculate the volume of $0.\text{200M HCl}$ solution needed to neutralize $\text{25}\text{.0 mL}$ of $\text{0}\text{.150 M NaOH}$ solution, molarity equation is used which is written as follows:

  $\left(\text{NaOH}\right)\frac{M_1{V}_1}{n_1}=\frac{M_2{V}_2}{n_2}\left(\text{HCl}\right)$ where symbols have their usual meaning.

Answer to Problem 39A

Volume of $\text{0}\text{.200 M HCl}$ solution needed to neutralize $\text{25}\text{.0 mL}$ of $\text{0}\text{.150 M NaOH}$ solution is $\text{18}\text{.75 mL}$ .

Explanation of Solution

Balanced chemical equation for the reaction is as follows:

  $\text{NaOH+HCl}\to {\text{NaCl+H}}_{\text{2}}\text{O}$

To calculate the volume of $\text{0}\text{.200 M HCl}$ solution needed to neutralize $\text{25}\text{.0 mL}$ of $\text{0}\text{.150 M NaOH}$ solution, molarity equation is used which is written as follows:

  $\left(\text{NaOH}\right)\frac{M_1{V}_1}{n_1}=\frac{M_2{V}_2}{n_2}\left(\text{HCl}\right)$ where symbols have their usual meaning.

Putting in the values,

  $\begin{array}{c}\frac{\text{0}\text{.150 M×25}\text{.0 mL}}{\text{1}}\text{=}\frac{\text{0}{\text{.200 M×V}}_{\text{2}}}{\text{1}}\\ {\text{V}}_{\text{2}}\text{=}\frac{\text{0}\text{.150 M×25}\text{.0mL}}{\text{0}\text{.200 M}}\\ {\text{V}}_{\text{2}}\text{=18}\text{.75 mL}\end{array}$

Conclusion

Volume of $\text{0}\text{.200 M HCl}$ solution needed to neutralize $\text{25}\text{.0 mL}$ of $0.150\text{M NaOH}$ solution is $\text{18}\text{.75 mL}$ .