Chapter 19, Problem 104A

Interpretation Introduction

Interpretation : The volume of sodium hydroxide that should be added before neutralization occurs is to be calculated and explained.

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 104A

The volume of sodium hydroxide is $50\text{mL}$ .

Explanation of Solution

The concentration of hydrochloric acid is $0.1\text{M}$ .

The concentration of sodium hydroxide is $0.1\text{M}$ .

The volume of hydrochloric acid is $50\text{mL}$ .

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 number of moles of hydrochloric acid is given as

  $\begin{array}{c}\text{n}\left(\text{HCl}\right)\text{=}\text{c}\left(\text{HCl}\right)\text{×V}\left(\text{HCl}\right)\\ =0.1\text{M}\times \text{50}\times {\text{10}}^{-3}\text{L}\\ \text{=}\text{5}\times {10}^{-3}\text{mol}\end{array}$

The number of moles of sodium hydroxide and hydrochloric acid are the same.

The volume of sodium hydroxide is given as

  $\begin{array}{c}\text{V}\left(\text{NaOH}\right)\text{=}\frac{\text{n}\left(\text{NaOH}\right)}{\text{c}\left(\text{NaOH}\right)}\\ =\frac{5\times {10}^{-3}\text{mol}}{0.1\text{M}}\\ =50\text{mL}\end{array}$