Chapter U4.89, Problem 5E

(a)

Interpretation Introduction

Interpretation: Whether the given solution (mixture of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ and $\text{20}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ ) is acidic, basic, or neutral 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 5E

The given solution is acidic.

Explanation of Solution

The given solution is mixture of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ and $\text{20}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ . The neutralization reaction between $\text{HCl}$ and $\text{NaOH}$ is shown below:

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

Here, $\text{1}\text{.0}\text{mol}$ of $\text{HCl}$ reacts with $\text{1}\text{.0}\text{mol}$ of $\text{NaOH}$ . Whether the given solution is acidic, basic, or neutral is determined by comparing number of moles of $\text{HCl}$ and $\text{NaOH}$ . If the number of moles of $\text{HCl}$ is more than the number of moles of $\text{NaOH}$ , the solution will be acidic. If the number of moles of $\text{HCl}$ is equal to the number of moles of $\text{NaOH}$ , the solution will be neutral. If the number of moles of $\text{NaOH}$ is more than the number of moles of $\text{HCl}$ , the solution will be basic.

The number of moles of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ is calculated as shown below:

  $\begin{array}{c}{n}_{\text{HCl}}=M_{\text{HCl}}\cdot V_{\text{HCl}}\\ =100\text{mL}\cdot 0.20\text{M}\\ =20\text{mmol}\end{array}$

The number of moles of $\text{20}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ is calculated as shown below:

  $\begin{array}{c}{n}_{\text{NaOH}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ =20\text{mL}\cdot 0.50\text{M}\\ =10\text{mmol}\end{array}$

Here, the number of moles of $\text{HCl}$ is more than the number of moles of $\text{NaOH}$ . Therefore, the given solution is acidic in nature.

(b)

Interpretation Introduction

Interpretation: Whether the given solution ( mixture of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ and $\text{40}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ ) is acidic, basic, or neutral 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 5E

The given solution is neutral.

Explanation of Solution

The given solution is mixture of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ and $\text{40}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ . The neutralization reaction between $\text{HCl}$ and $\text{NaOH}$ is shown below:

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

Here, $\text{1}\text{.0}\text{mol}$ of $\text{HCl}$ reacts with $\text{1}\text{.0}\text{mol}$ of $\text{NaOH}$ . Whether the given solution is acidic, basic, or neutral is determined by comparing number of moles of $\text{HCl}$ and $\text{NaOH}$ . If the number of moles of $\text{HCl}$ is more than the number of moles of $\text{NaOH}$ , the solution will be acidic. If the number of moles of $\text{HCl}$ is equal to the number of moles of $\text{NaOH}$ , the solution will be neutral. If the number of moles of $\text{NaOH}$ is more than the number of moles of $\text{HCl}$ , the solution will be basic.

The number of moles of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ is calculated as shown below:

  $\begin{array}{c}{n}_{\text{HCl}}=M_{\text{HCl}}\cdot V_{\text{HCl}}\\ =100\text{mL}\cdot 0.20\text{M}\\ =20\text{mmol}\end{array}$

The number of moles of $\text{40}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ is calculated as shown below:

  $\begin{array}{c}{n}_{\text{NaOH}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ =40\text{mL}\cdot 0.50\text{M}\\ =20\text{mmol}\end{array}$

Here, the number of moles of $\text{HCl}$ is equal to the number of moles of $\text{NaOH}$ . Therefore, the given solution is neutral in nature.

(c)

Interpretation Introduction

Interpretation: Whether the given solution (mixture of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ and $\text{60}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ ) is acidic, basic, or neutral 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 5E

The given solution is basic.

Explanation of Solution

The given solution is mixture of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ and $\text{60}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ . The neutralization reaction between $\text{HCl}$ and $\text{NaOH}$ is shown below:

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

Here, $\text{1}\text{.0}\text{mol}$ of $\text{HCl}$ reacts with $\text{1}\text{.0}\text{mol}$ of $\text{NaOH}$ . Whether the given solution is acidic, basic, or neutral is determined by comparing number of moles of $\text{HCl}$ and $\text{NaOH}$ . If the number of moles of $\text{HCl}$ is more than the number of moles of $\text{NaOH}$ , the solution will be acidic. If the number of moles of $\text{HCl}$ is equal to the number of moles of $\text{NaOH}$ , the solution will be neutral. If the number of moles of $\text{NaOH}$ is more than the number of moles of $\text{HCl}$ , the solution will be basic.

The number of moles of $\text{100}\text{mL}$ of $\text{0}\text{.20}\text{M}\text{HCl}$ is calculated as shown below:

  $\begin{array}{c}{n}_{\text{HCl}}=M_{\text{HCl}}\cdot V_{\text{HCl}}\\ =100\text{mL}\cdot 0.20\text{M}\\ =20\text{mmol}\end{array}$

The number of moles of $\text{60}\text{mL}$ of $\text{0}\text{.50}\text{M}\text{NaOH}$ is calculated as shown below:

  $\begin{array}{c}{n}_{\text{NaOH}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ =60\text{mL}\cdot 0.50\text{M}\\ =30\text{mmol}\end{array}$

Here, the number of moles of $\text{NaOH}$ is more than the number of moles of $\text{HCl}$ . Therefore, the given solution is basic in nature.