Chapter U4.89, Problem 3E

Interpretation Introduction

Interpretation: The required volume of $\text{0}\text{.1}\text{M}\text{NaOH}$ to neutralize $\text{2}\text{.0}\text{L}$ of $\text{0}\text{.050}\text{M}\text{HCl}$ is to be determined.

Concept introduction: The reaction in which hydrogen ion $\left({\text{H}}^{+}\right)$ and hydroxide ion $\left({\text{OH}}^{-}\right)$ combine together to form water is termed as neutralization reaction. An acid base neutralization reaction is a type of reaction in which acid and base combine together quantitatively to form water and salt.

Answer to Problem 3E

The volume of $\text{0}\text{.1}\text{M}\text{NaOH}$ to neutralize $\text{2}\text{.0}\text{L}$ of $\text{0}\text{.050}\text{M}\text{HCl}$ is $1.0\text{L}$ .

Explanation of Solution

The volume of $\text{0}\text{.1}\text{M}\text{NaOH}$ to neutralize $\text{2}\text{.0}\text{L}$ of $\text{0}\text{.050}\text{M}\text{HCl}$ is calculated by the formula shown below.

  $\begin{array}{c}{M}_{\text{NaOH}}\cdot V_{\text{NaOH}}=M_{\text{HCl}}\cdot V_{\text{HCl}}\\ V_{\text{NaOH}}=\frac{M_{\text{HCl}}\cdot V_{\text{HCl}}}{M_{\text{NaOH}}}\end{array}$ …(i)

Here,

• $M_{\text{HCl}}$ is the molarity of hydrogen chloride.
• $M_{\text{NaOH}}$ is the molarity of sodium hydroxide.
• $V_{\text{HCl}}$ is the volume of hydrogen chloride.

The molarity of sodium hydroxide is $\text{0}\text{.1}\text{M}$ , the molarity of hydrogen chloride is $\text{0}\text{.050}\text{M}$ , and the volume of hydrogen chloride is $\text{2}\text{.0}\text{L}$ .

Substitute the given values in equation (i) as shown below.

  $\begin{array}{c}{V}_{\text{NaOH}}=\frac{M_{\text{HCl}}\cdot V_{\text{HCl}}}{M_{\text{NaOH}}}\\ =\frac{0.050\text{M}\cdot 2.0\text{L}}{0.1\text{M}}\\ =1.0\text{L}\end{array}$

Therefore, the required volume of $\text{0}\text{.1}\text{M}\text{NaOH}$ to neutralize $\text{2}\text{.0}\text{L}$ of $\text{0}\text{.050}\text{M}\text{HCl}$ is $1.0\text{L}$ .

Conclusion

The required volume of $\text{0}\text{.1}\text{M}\text{NaOH}$ to neutralize $\text{2}\text{.0}\text{L}$ of $\text{0}\text{.050}\text{M}\text{HCl}$ is $1.0\text{L}$ .