Chapter 15.3, Problem 5RQ

Interpretation Introduction

Interpretation: The volume of $NaOH$ in neutralization reaction needs to be determined.

Concept introduction: At neutralization point of a reaction,

  $\begin{array}{c}\text{number of moles of acid}=\text{number of moles of of base}\\ M_1{V}_1=M_2{V}_2\end{array}$

Where,

  $M_1=\text{concentration of acid}\text{.}$

  $V_1=\text{ volume of the acid}\text{.}$

  $M_2=\text{concentration of base}\text{.}$

  $V_2=\text{ volume of the base}\text{.}$

Answer to Problem 5RQ

Volume of $NaOH$ is $75.0mL.$

Explanation of Solution

Here,

  $M_1=\text{concentration of }HCl=0.15M$

  $V_1=\text{ volume of the }HCl=125.0mL$

  $M_2=\text{concentration of }NaOH=0.25M$

  $V_2=\text{ volume of the }NaOH=?$

Putting the appropriate values in the main equation we get,

  $\begin{array}{c}(0.15M)\times \left(125.0mL\right)=\left(0.25M\right)\times V_2\\ V_2=\frac{(0.15M)\times \left(125.0mL\right)}{0.25M}\\ V_2=75.0mL\end{array}$

Conclusion

Volume of $NaOH$ is $75.0mL.$