Chapter 15, Problem 59A

Interpretation Introduction

Interpretation: The normality of nitric acid ( $HN{O}_3$ ) needs to be determined.

Concept introduction: At neutralization point of a reaction for mono-acidic acid and mono-basic base,

  $\begin{array}{c}\text{number of moles of acid}=\text{number of moles of of base}\\ N_1{V}_1=N_2{V}_2\end{array}$

  $N_1=\text{concentration of acid in normality}\text{.}$

  $V_1=\text{ volume of the acid}\text{.}$

  $N_2=\text{concentration of base in normality}\text{.}$

  $V_2=\text{ volume of the base}\text{.}$

Answer to Problem 59A

The normality of nitric acid ( $HN{O}_3$ ) is $0.1925N.$

Explanation of Solution

The balanced neutralization reaction: $Ca{\left(OH\right)}_2+2HN{O}_3\to Ca{\left(N{O}_3\right)}_2+2H_{2}O$

So, 1 mol of dibasic $Ca{\left(OH\right)}_2$ reacts with 2 moles of mono-acidic $HN{O}_3$ and hence,

  $\begin{array}{c}\text{2}\times \text{number of moles of }Ca{\left(OH\right)}_2=\text{number of moles of of }HN{O}_3\\ 2\times N_2{V}_2=N_1{V}_1\end{array}$

Here,

  $N_1=\text{concentration of acid}=?$

  $V_1=\text{ volume of the acid}=10.0mL$

  $N_2=\text{concentration of base in}=3.5\times {10}^{-2}N$

  $V_2=\text{ volume of the base}=27.5mL$

  $\begin{array}{c}2\times \left(3.5\times {10}^{-2}N\right)\times \left(27.5mL\right)=N_1\times \left(10.0mL\right)\\ N_1=\frac{1.925}{10.0}N\\ N_1=0.1925N\end{array}$

Conclusion

The normality of nitric acid ( $HN{O}_3$ ) is $0.1925N.$