Chapter 18.4, Problem 44PP

Interpretation Introduction

Interpretation:

The molarity of a given acid solution is to be calculated when it is neutralized by the base.

Concept introduction:

Molarity is the number of moles of solute present per liter of solution.

When acid and base are mixed, they neutralize each other. So their molar concentrations can be calculated by applying the mole concept.

Answer to Problem 44PP

Molarity of nitric acid is $0.2167\text{M}$

Explanation of Solution

Given data:

     $\begin{array}{l}{\text{V}}_{\text{KOH}}\text{=43}\text{.33}\text{mL}\end{array}$

$\begin{array}{c}\text{Molarity}\text{of}\text{KOH=0}\text{.1}\text{}\text{}\text{M}\end{array}$

     $\begin{array}{c}{\text{V}}_{{\text{HNO}}_{\text{3}}}\text{=20}\text{mL}\end{array}$

$\begin{array}{c}{\text{n}}_{\text{KOH}}\text{=Molarity×volume}\text{in}\text{liters}\end{array}$

  $\begin{array}{c}\text{=0}\text{.1×}\frac{\text{43}\text{.33}}{\text{1000}}\end{array}$

  $\begin{array}{c}\text{=4}{\text{.333×10}}^{\text{-3}}\text{mol}\end{array}$

When nitric acid reacts with potassium hydroxide reaction is given by:

$\text{H}\text{N}{\text{O}}_3+\text{K}\text{O}\text{H}\to \text{K}\text{N}{\text{O}}_3+{\text{H}}_2\text{O}$

Thus, one mol acid reacts with one mol base.

Thus,

${\text{n}}_{{\text{HNO}}_{\text{3}}}\text{=4}{\text{.333×10}}^{\text{-3}}\text{mol}$

$\begin{array}{c}\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{H}\text{N}{\text{O}}_3=\frac{4.333\times {10}^{-3}}{20/1000}\end{array}$

         $\begin{array}{c}=0.2167\text{M}\end{array}$

Conclusion

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}\text{o}\text{f}\text{H}\text{N}{\text{O}}_3$ is $0.2167\text{M}$ .