Chapter 19, Problem 103A

Interpretation Introduction

Interpretation : The equation for the reaction is to be written.

Concept Introduction : Salt and water are created when bases and acids react.

A strong acid and strong base react completely to form a neutral solution. A neutralization reaction is a name for this kind of reaction.

Answer to Problem 103A

The equation for the reaction is given as

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

Explanation of Solution

The reaction between sodium hydroxide and hydrochloric acid gives sodium chloride and water.

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

The molar ratio is 1:1.

Interpretation Introduction

Interpretation : The concentration of the sodium hydroxide is to be estimated.

Concept Introduction : Salt and water are created when bases and acids react.

A strong acid and strong base react completely to form a neutral solution. A neutralization reaction is a name for this kind of reaction.

The formula for the number of moles is given as

  $\text{n}\text{=}\text{c×V}$

Here, $\text{c}$ is the concentration and $\text{V}$ is the volume.

Answer to Problem 103A

The concentration of sodium hydroxide is $0.327\text{mol/L}$ .

Explanation of Solution

The number of moles of water is $0.018\text{mol}$ .

The number of moles of sodium hydroxide is $0.018\text{mol}$ .

The volume of sodium hydroxide is $25\text{mL}$ .

The volume of hydrochloric acid is $30\text{mL}$ .

The reaction between sodium hydroxide and hydrochloric acid gives sodium chloride and water.

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

The molar ratio is 1:1.

The total volume is given as the sum of the volume of hydrochloric acid and volume of sodium hydroxide.

  $\begin{array}{c}\text{V}\text{=}\text{V}\left(\text{NaOH}\right)\text{+V}\left(\text{HCl}\right)\\ =25\text{mL}\text{+}\text{30}\text{mL}\\ \text{=}\text{55}\text{mL}\end{array}$

The concentration of sodium hydroxide is given as

  $\begin{array}{c}\text{c}\left(\text{NaOH}\right)\text{=}\frac{\text{n}\left(\text{NaOH}\right)}{\text{V}}\\ =\frac{0.018\text{mol}}{0.055\text{L}}\\ \text{=}0.327\text{mol/L}\end{array}$