Chapter 19, Problem 112A

Interpretation Introduction

Interpretation : The molarity of sulfuric acid 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 112A

The molarity of sulfuric acid is $0.05\text{M}$ .

Explanation of Solution

Given information:

The volume of ${\text{Na}}_2{\text{CO}}_3$ is $\text{80}\text{.0}\text{mL}$ .

The mass of ${\text{Na}}_2{\text{CO}}_3$ is $0.424\text{g}$ .

The reaction between sodium carbonate and sulfuric acid is given as

  ${\text{Na}}_2{\text{CO}}_3\text{+}{\text{H}}_2{\text{SO}}_4\to {\text{Na}}_2{\text{SO}}_4\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{CO}}_2$

The above reaction is the balanced reaction.

The mole ratio is $1:1$ for the reaction between ${\text{Na}}_2{\text{CO}}_3$ and ${\text{H}}_2{\text{SO}}_4$ .

The moles of sulfuric acid are calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{= 0}\text{.424}\cancel{\text{g}{\text{Na}}_2{\text{CO}}_3}\times \frac{1\cancel{\text{mol}}}{106.0\text{g}}\times \frac{1\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{1\cancel{\text{mol}{\text{Na}}_2{\text{CO}}_3}}\\ =0.004\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molarity of sulfuric acid is given as

  $\begin{array}{c}\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}\\ =\frac{0.004\text{mol}}{80\cancel{\text{mL}}}\times \frac{1000\cancel{\text{mL}}}{1\text{L}}\\ =0.05\text{M}\end{array}$