Chapter 19, Problem 73A

a.

Interpretation Introduction

Interpretation : The molarity of sodium hydroxide is to be calculated.

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 of molarity is given as:

  $\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}$

Answer to Problem 73A

The molarity of sodium hydroxide is $1.40\text{M}$ .

Explanation of Solution

Given information:

The volume of $\text{NaOH}$ is $\text{20}\text{.0}\text{mL}$ .

The concentration of $\text{HCl}$ is $\text{1}\text{.00 M}$ .

The volume of $\text{HCl}$ is $28.0\text{mL}$ .

The reaction between sodium hydroxide and hydrochloric acid is given as

  $\text{NaOH}\left(aq\right)\text{+}\text{HCl}\left(aq\right)\to \text{NaCl}\left(aq\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above reaction is balanced.

The mole ratio is $1:1$ for the reaction between $\text{NaOH}$ and $\text{HCl}$ .

The moles of sodium hydroxide are calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}\text{= 28}\cancel{\text{mL}\text{HCl}}\times \frac{1\cancel{\text{L}\text{HCl}}}{1000\cancel{\text{mL}\text{HCl}}}\times \frac{1\cancel{\text{mol}\text{HCl}}}{1\cancel{\text{L}\text{HCl}}}\times \frac{1\text{mol}\text{NaOH}}{1\cancel{\text{mol}\text{HCl}}}\\ =0.028\text{mol}\text{NaOH}\end{array}$

The molarity of $\text{NaOH}$ is given as

  $\begin{array}{c}\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}\\ =\frac{0.028\text{mol}}{20\cancel{\text{mL}}}\times \frac{1000\cancel{\text{mL}}}{1\text{L}}\\ =1.40\text{M}\end{array}$

b.

Interpretation Introduction

Interpretation : The molarity of sodium hydroxide is to be calculated.

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 of molarity is given as

  $\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}$

Answer to Problem 73A

The molarity of sodium hydroxide is $2.61\text{M}$ .

Explanation of Solution

Given information:

The volume of $\text{NaOH}$ is $\text{20}\text{.0}\text{mL}$ .

The concentration of ${\text{H}}_3{\text{PO}}_4$ is $\text{1}\text{.00 M}$ .

The volume of ${\text{H}}_3{\text{PO}}_4$ is $17.4\text{mL}$ .

The reaction between sodium hydroxide and phosphoric acid is given as

  $\text{3NaOH}\left(aq\right)\text{+}{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Na}}_3{\text{PO}}_4\left(aq\right)\text{+}3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above reaction is balanced.

The mole ratio is $3:1$ for the reaction between $\text{NaOH}$ and ${\text{H}}_3{\text{PO}}_4$ .

The moles of sodium hydroxide are calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}\text{= 17}\text{.4}\cancel{\text{mL}{\text{H}}_3{\text{PO}}_4}\times \frac{1\cancel{\text{L}{\text{H}}_3{\text{PO}}_4}}{1000\cancel{\text{mL}{\text{H}}_3{\text{PO}}_4}}\times \frac{1\cancel{\text{mol}{\text{H}}_3{\text{PO}}_4}}{1\cancel{\text{L}{\text{H}}_3{\text{PO}}_4}}\times \frac{3\text{mol}\text{NaOH}}{1\cancel{\text{mol}{\text{H}}_3{\text{PO}}_4}}\\ =0.0522\text{mol}\text{NaOH}\end{array}$

The molarity of $\text{NaOH}$ is given as

  $\begin{array}{c}\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}\\ =\frac{0.0522\text{mol}}{20\cancel{\text{mL}}}\times \frac{1000\cancel{\text{mL}}}{1\text{L}}\\ =2.61\text{M}\end{array}$