Chapter 16, Problem 75E

Interpretation Introduction

Interpretation:

The normality of the solution formed when $\text{6}\text{.69}\text{g}$ of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is diluted to $2.00\times {10}^2\text{mL}$ 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 75E

The normality of the solution formed when $\text{6}\text{.69}\text{g}$ of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is diluted to $2.00\times {10}^2\text{mL}$ of solution is $0.372\text{N}$.

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 $2.00\times {10}^2\text{mL}$.

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.

$\begin{array}{rll}\text{Volume}& =& 2.00\times {10}^2\text{mL}\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ & =& \text{0}\text{.200}\text{L}\end{array}$

The reaction of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is given below.

${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\to {\text{H}}^{+}{\text{+HC}}_2{\text{O}}_{\text{4}}{}^{-}$

In this reaction, $1$ mole of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ 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.

$\text{Equivalent}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{\text{Number}\text{of}\text{equivalents}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}$ …(2)

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

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 ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times 1.008\text{mg}{\text{mol}}^{-1}\right)+\left(2\times 12.01\text{g}{\text{mol}}^{-1}\right)+\left(4\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 2.016\text{g}{\text{mol}}^{-1}+24.02\text{g}{\text{mol}}^{-1}+64.00\text{g}{\text{mol}}^{-1}\\ & =& 90.04\text{g}{\text{mol}}^{-1}\end{array}$

Substitute the molar mass and number of equivalents of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ in equation (2).

$\text{Equivalent}\text{mass}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}=\frac{90.04\text{g}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{\text{1}\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}$

Therefore, the conversion factors are written below.

$\frac{90.04\text{g}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{\text{1}\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}\text{and}\frac{\text{1}\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{90.04\text{g}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}$

The conversion factor to determine the equivalents of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ from the mass of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is given below.

$\frac{\text{1}\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{90.04\text{g}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}$

The equivalents of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ can be determined by the formula given below.

$\text{Equivalents}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}=\left(\begin{array}{l}\text{Mass}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\\ \times \text{Conversion}\text{factor}\text{to}\text{obtain}\text{equivalent}\\ \text{mass}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\end{array}\right)$ …(3)

The mass of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is $\text{6}\text{.69}\text{g}$.

Substitute the mass of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ and the conversion factor in equation (3).

$\begin{array}{rll}\text{Equivalents}\text{of}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}& =& \text{6}\text{.69}\text{g}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\times \frac{\text{1}\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{90.04\text{g}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}\\ & =& 0.0743\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\end{array}$

Therefore, the equivalents of solute, ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is $0.0743\text{eq}$.

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

$\begin{array}{rll}\text{Normality}& =& \frac{0.0743\text{eq}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}}{\text{0}\text{.200}\text{L}}\\ & =& 0.372\text{eq}/\text{L}\end{array}$

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, $0.372\text{eq}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& 0.372\text{eq}/\text{L}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\times \frac{1\text{N}}{1\text{eq}/\text{L}}\\ & =& 0.372\text{N}{\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}\end{array}$

Therefore, the normality of the solution formed when $\text{6}\text{.69}\text{g}$ of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is diluted to $2.00\times {10}^2\text{mL}$ of solution is $0.372\text{N}$.

Conclusion

The normality of the solution formed when $\text{6}\text{.69}\text{g}$ of ${\text{H}}_2{\text{C}}_2{\text{O}}_{\text{4}}$ is diluted to $2.00\times {10}^2\text{mL}$ of solution is $0.372\text{N}$.