Chapter 16, Problem 74E

Interpretation Introduction

Interpretation:

The normality of ${\text{NaHCO}}_{\text{3}}$ in the reaction when $\text{9}\text{.79}\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is dissolved in $5.00\times {10}^2\text{mL}$ of 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.

Answer to Problem 74E

The normality of ${\text{NaHCO}}_{\text{3}}$ in the reaction when $\text{9}\text{.79}\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is dissolved in $5.00\times {10}^2\text{mL}$ of solution is $0.234\text{N}$.

Explanation of Solution

The formula to calculate the normality is given below.

$\text{Normality}=\frac{\text{Equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liters}}$ …(1)

The volume of the solution is $5.00\times {10}^2\text{mL}$.

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The probable conversion factors are given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{and}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}$

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

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}$

Therefore, the volume in liters is calculated below.

$\begin{array}{rll}\text{Volume}& =& 5.00\times {10}^2\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ & =& \text{0}\text{.500}\text{L}\end{array}$

The reaction of ${\text{NaHCO}}_{\text{3}}$ is given below.

${\text{NaHCO}}_{\text{3}}+\text{HCl}\to \text{NaCl}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

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

The formula to calculate the equivalent mass is given below.

$\text{Equivalent}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}{\text{NaHCO}}_{\text{3}}}{\text{Number}\text{of}\text{equivalents}\text{of}{\text{NaHCO}}_{\text{3}}}$ …(2)

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of sodium is $22.99\text{g}{\text{mol}}^{-1}$.

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.008\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{NaHCO}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 22.99\text{g}{\text{mol}}^{-1}+1.008\text{g}{\text{mol}}^{-1}+12.01\text{g}{\text{mol}}^{-1}+\left(3\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 22.99\text{g}{\text{mol}}^{-1}+1.008\text{g}{\text{mol}}^{-1}+12.01\text{g}{\text{mol}}^{-1}+48.00\text{g}{\text{mol}}^{-1}\\ & =& 84.008\text{g}{\text{mol}}^{-1}\end{array}$

Substitute the molar mass and number of equivalents of ${\text{NaHCO}}_{\text{3}}$ in equation (2).

$\text{Equivalent}\text{mass}\text{of}{\text{NaHCO}}_{\text{3}}=\frac{84.008\text{g}{\text{NaHCO}}_{\text{3}}}{\text{1}\text{eq}{\text{NaHCO}}_{\text{3}}}$

Therefore, the conversion factors are written below.

$\frac{84.008\text{g}{\text{NaHCO}}_{\text{3}}}{\text{1}\text{eq}{\text{NaHCO}}_{\text{3}}}\text{and}\frac{\text{1}\text{eq}{\text{NaHCO}}_{\text{3}}}{84.008\text{g}{\text{NaHCO}}_{\text{3}}}$

The conversion factor to determine the equivalents of ${\text{NaHCO}}_{\text{3}}$ from the mass of ${\text{NaHCO}}_{\text{3}}$ is given below.

$\frac{\text{1}\text{eq}{\text{NaHCO}}_{\text{3}}}{84.008\text{g}{\text{NaHCO}}_{\text{3}}}$

The equivalents of ${\text{NaHCO}}_{\text{3}}$ can be determined by the formula given below.

$\text{Equivalents}\text{of}{\text{NaHCO}}_{\text{3}}=\left(\begin{array}{l}\text{Mass}\text{of}{\text{NaHCO}}_{\text{3}}\\ \times \text{Conversion}\text{factor}\text{to}\text{obtain}\text{equivalent}\\ \text{mass}\text{of}{\text{NaHCO}}_{\text{3}}\end{array}\right)$ …(3)

The mass of ${\text{NaHCO}}_{\text{3}}$ is $\text{9}\text{.79}\text{g}$.

Substitute the mass of ${\text{NaHCO}}_{\text{3}}$ and the conversion factor in equation (3).

$\begin{array}{rll}\text{Equivalents}\text{of}{\text{NaHCO}}_{\text{3}}& =& \text{9}\text{.79}\text{g}{\text{NaHCO}}_{\text{3}}\times \frac{\text{1}\text{eq}{\text{NaHCO}}_{\text{3}}}{84.008\text{g}{\text{NaHCO}}_{\text{3}}}\\ & =& 0.117\text{eq}{\text{NaHCO}}_{\text{3}}\end{array}$

Therefore, the equivalents of solute, ${\text{NaHCO}}_{\text{3}}$ is $0.117\text{eq}$.

Substitute the value of equivalents of solute and volume of solution in equation (1).

$\begin{array}{rll}\text{Normality}& =& \frac{0.117\text{eq}{\text{NaHCO}}_{\text{3}}}{\text{0}\text{.500}\text{L}}\\ & =& 0.234\text{eq}/\text{L}\end{array}$

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, $0.234\text{eq}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& 0.234\text{eq}/\text{L}{\text{NaHCO}}_{\text{3}}\times \frac{1\text{N}}{1\text{eq}/\text{L}}\\ & =& 0.234\text{N}{\text{NaHCO}}_{\text{3}}\end{array}$

Therefore, the normality of ${\text{NaHCO}}_{\text{3}}$ in the reaction when $\text{9}\text{.79}\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is dissolved in $5.00\times {10}^2\text{mL}$ of solution is $0.234\text{N}$.

Conclusion

The normality of ${\text{NaHCO}}_{\text{3}}$ in the reaction when $\text{9}\text{.79}\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is dissolved in $5.00\times {10}^2\text{mL}$ of solution is $0.234\text{N}$.