Chapter 16, Problem 78E

Interpretation Introduction

(a)

Interpretation:

The normality of a $6.110\times {10}^{-2}\text{M}$${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is to be calculated.

Concept introduction:

Solution is a homogeneous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. The normality of a solution is defined as the number of equivalents per liter of the solution. One equivalent of an acid is the quantity that gives $1$ mole of hydrogen ions in a chemical reaction. One equivalent of a base is the quantity that reacts with one mole of hydrogen ions. The molarity of a solution is defined as the moles of a solute per liter of the solution.

Answer to Problem 78E

The normality of a $6.110\times {10}^{-2}\text{M}$${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $0.122\text{N}$.

Explanation of Solution

The formula to calculate normality is given below.

$\text{Normality}=\text{Molarity}\times \text{Number}\text{of}\text{equivalents}$ …(1)

The molarity of the solution is $6.110\times {10}^{-2}\text{M}$.

The relation between $\text{M}$ and $\text{mol}/\text{L}$ as shown below.

$1\text{M}=1\text{mol}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}\text{and}\frac{1\text{mol}/\text{L}}{1\text{M}}$

The conversion factor to determine $\text{mol}/\text{L}$ from $\text{M}$ is given below.

$\frac{1\text{mol}/\text{L}}{1\text{M}}$

Therefore, $6.110\times {10}^{-2}\text{M}$ can be written as shown below.

$\begin{array}{rll}\text{Molarity}& =& 6.110\times {10}^{-2}\text{M}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\times \frac{1\text{mol}/\text{L}}{1\text{M}}\\ & =& \frac{6.110\times {10}^{-2}\text{mol}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{L}}\end{array}$

The reaction is given below.

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}+\text{2NaOH}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}+2{\text{H}}_{\text{2}}\text{O}$

In this reaction, $1$ mole of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ gives $2$ moles of ${\text{H}}^{+}$ ions during the reaction. Therefore, the number of equivalents per mole is $2$.

Substitute the number of equivalents and the molarity of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ in equation (1).

$\begin{array}{rll}\text{Normality}& =& \frac{6.110\times {10}^{-2}\text{mol}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{L}}\times 2\text{eq}/\text{mol}\\ & =& 0.122\text{eq}/\text{L}\end{array}$

The relation between $\text{N}$ and $\text{eq}/\text{L}$ as shown below.

$1\text{N}=1\text{eq}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{N}}{1\text{eq}/\text{L}}\text{and}\frac{1\text{eq}/\text{L}}{1\text{N}}$

The conversion factor to determine $\text{N}$ from $\text{eq}/\text{L}$ is given below.

$\frac{1\text{N}}{1\text{eq}/\text{L}}$

Therefore, $0.122\text{eq}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& 0.122\text{eq}/\text{L}\times \frac{1\text{N}}{1\text{eq}/\text{L}}\\ & =& 0.122\text{N}\end{array}$

Therefore, the normality of a $6.110\times {10}^{-2}\text{M}$${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $0.122\text{N}$.

Conclusion

The normality of a $6.110\times {10}^{-2}\text{M}$${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $0.122\text{N}$.

Interpretation Introduction

(b)

Interpretation:

The volume of the solution that would contain $\text{0}\text{.345}\text{eq}$ is to be calculated.

Concept introduction:

Solution is a homogeneous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. The normality of a solution is defined as the number of equivalents per liter of the solution. One equivalent of an acid is the quantity that gives $1$ mole of hydrogen ions in a chemical reaction. One equivalent of a base is the quantity that reacts with one mole of hydrogen ions.

Answer to Problem 78E

The volume of the solution that would contain $\text{0}\text{.345}\text{eq}$ is $2.83\text{L}$.

Explanation of Solution

The formula to calculate the volume is given below.

$\text{Volume}\text{of}\text{solution}\text{in}\text{liters}=\frac{\text{Equivalents}\text{of}\text{solute}}{\text{Normality}}$ …(2)

The normality of the solution is $0.122\text{N}$.

The relation between $\text{N}$ and $\text{eq}/\text{L}$ as shown below.

$1\text{N}=1\text{eq}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{N}}{1\text{eq}/\text{L}}\text{and}\frac{1\text{eq}/\text{L}}{1\text{N}}$

The conversion factor to determine $\text{eq}/\text{L}$ from $\text{N}$ is given below.

$\frac{1\text{eq}/\text{L}}{1\text{N}}$

Therefore, $0.122\text{N}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& 0.122\text{N}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\times \frac{1\text{eq}/\text{L}}{1\text{N}}\\ & =& 0.122\text{eq}/\text{L}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\end{array}$

The equivalents of solute is $\text{0}\text{.345}\text{eq}$.

Substitute the value of normality and equivalents of solute in equation (2).

$\begin{array}{rll}\text{Volume}\text{of}\text{solution}\text{in}\text{liters}& =& \frac{\text{0}\text{.345}\text{eq}}{0.122\text{eq}/\text{L}}\\ & =& 2.83\text{L}\end{array}$

Therefore, the volume of the solution that would contain $\text{0}\text{.345}\text{eq}$ is $2.83\text{L}$.

Conclusion

The volume of the solution that would contain $\text{0}\text{.345}\text{eq}$ is $2.83\text{L}$.