Chapter 15, Problem 40A

Interpretation Introduction

Interpretation: The concentration (molarity) of sodium hydroxide solution, whose volume is given to be $\text{48}\text{.7 mL}$ , required to neutralize a $\text{50}\text{.0 mL}$ sample of $\text{0}\text{.104 M HCl}$ needs to be determined.

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

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

To calculate the concentration (molarity) of sodium hydroxide solution, whose volume is given to be $\text{48}\text{.7 mL}$ , required to neutralize a $\text{50}\text{.0 mL}$ sample of $\text{0}\text{.104 M HCl}$ , 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 40A

The concentration (molarity) of sodium hydroxide solution, whose volume is given to be $\text{48}\text{.7 mL}$ , required to neutralize a $\text{50}\text{.0 mL}$ sample of $\text{0}\text{.104 M HCl}$ is $\text{0}\text{.106 M}$ .

Explanation of Solution

The balanced chemical equation for the reaction is as follows:

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

To calculate the concentration (molarity) of sodium hydroxide solution, whose volume is given to be $\text{48}\text{.7 mL}$ , required to neutralize a $\text{50}\text{.0 mL}$ sample of $\text{0}\text{.104 M HCl}$ , 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, we get

  $\begin{array}{c}\frac{M_1\times 48.\text{7 mL}}{1}=\frac{0.10\text{4 M}\times 50.0\text{ mL}}{1}\\ M_1=\frac{0.10\text{4 M}\times 50.0\text{ mL}}{48.7\text{ mL}}\\ M_1=0.106\text{ M}\end{array}$

Conclusion

The concentration (molarity) of sodium hydroxide solution, whose volume is given to be $48.7mL$ , required to neutralize a $\text{50}\text{.0 mL}$ sample of $\text{0}\text{.104M HCl}$ is $\text{0}\text{.106M}$ .