Chapter 16, Problem 105E

Interpretation Introduction

Interpretation:

The normality of sodium carbonate solution of volume $\text{51}\text{.89 mL}$ titrated with $\text{5}\text{.038 g}$ of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is to be calculated.

Concept introduction:

An acid base reaction is one of the types of reaction which is known as neutralization reaction. Normality is used to express concentration of any solution. The n factor of any species is defined in different way depending upon the type of reaction. In an acid base reaction n-factor is defined as the number of replaceable ${\text{H}}^{\text{+}}$ of an acid and for a base it is the total charge present on a species after removing all metal ions. Equation which equates the equivalents of an acid and a base is given below.

$\underset{\text{equivalents of acid }}{\underbrace{{\text{n}}_{\text{1}}{\text{M}}_{\text{1}}{\text{V}}_1}}=\underset{e\text{quivalents of base }}{\underbrace{{\text{n}}_{\text{2}}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}}$

Answer to Problem 105E

The normality of sodium carbonate of volume $\text{51}\text{.89 mL}$ which is titrated with $\text{5}\text{.038 g}$ of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $\text{0}\text{.7952 N}$.

Explanation of Solution

The equation which is used to calculate the normality of sodium carbonate is given below.

$\begin{array}{l}\underset{\text{equivalents of acid }}{\underbrace{{\text{n}}_{\text{1}}{\text{M}}_{\text{1}}{\text{V}}_1}}=\underset{e\text{quivalents of base }}{\underbrace{{\text{n}}_{\text{2}}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}}\\ \end{array}$ …(1)

The molarity is given by the formula shown below.

$\begin{array}{rll}{\text{M}}_1& =& \frac{\text{w}\left(\text{g}\right)}{\text{Molar mass of acid}\left(\text{g}/\text{mol}\right)\times \text{Volume of acid}\left(\text{V in mL}\right)}\\ {\text{M}}_1{\text{V}}_1& =& \frac{\text{w }\left(\cancel{g}\right)}{\text{Molar mass of acid }\left(\cancel{g}/\text{mol}\right)}\end{array}$ …(2)

Substitute relation (2) in (1) as shown below.

$\begin{array}{rll}\underset{\text{equivalents of acid }}{\underbrace{{\text{n}}_{\text{1}}{\text{M}}_{\text{1}}{\text{V}}_1}}& =& \underset{e\text{quivalents of base }}{\underbrace{{\text{n}}_{\text{2}}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}}\\ \underset{{\text{equivalents of acid HC}}_7{\text{H}}_5{\text{O}}_2}{\underbrace{{\text{n}}_{\text{1}}\times \frac{\text{w}\left(\cancel{\text{g}}\right)}{{\text{M}}_{\text{o}}\left(\cancel{\text{g}}/\text{mol}\right)}}}& =& \underset{e\text{quivalents of Sodium Carbonate}}{\underbrace{{\text{n}}_{\text{2}}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}}\end{array}$ …(3)

Where,

• ${\text{n}}_{\text{1}}$ is the n-factor of benzoic acid(${\text{HC}}_7{\text{H}}_5{\text{O}}_2$).

• ${\text{n}}_{\text{2}}$ is the n-factor of sodium carbonate base.

• $\text{w}$ is the mass used of benzoic acid.

• ${\text{M}}_{\text{2}}$ is the molarity of sodium carbonate base.

• ${\text{M}}_{\text{o}}$ is the molar mass of acid used.

• ${\text{V}}_{\text{2}}$ is the volume of sodium carbonate base used.

Benzoic acid has ${\text{1H}}^{\text{+}}$. So, n-factor of benzoic acid is $1$ and sodium carbonate has n-factor $2$.

The given value ${\text{n}}_{\text{1}}$ is $1$.

The given value of $\text{w}$ is $\text{5}\text{.038 g}$.

The given value of ${\text{M}}_{\circ }$ is $\text{122}\text{.12 }\text{g}/\text{mol}$.

The given value of ${\text{n}}_{\text{2}}$ is $\text{2}$.

The given value of ${\text{V}}_2$ is $\text{51}\text{.89 mL}$.

Substitute all the values in the equation (2) as shown below.

$\begin{array}{rll}\text{1}\times \frac{5.038\text{g}}{122.12\text{g}/\text{mol}}& =& 2\times {\text{M}}_{\text{2}}\times 51.89\text{mL}\\ {\text{M}}_2& =& \frac{1\times 5.038\text{g}}{122.12\text{g}/\text{mol}\times \text{2}\times \text{51}\text{.89 mL}}\\ {\text{M}}_2& =& 0.0003976\text{mol}/\text{mL}\times \left(\frac{1000\text{mL}}{1\text{L}}\right)\\ & =& \text{0}\text{.3976 }\text{mol}/\text{mL}\end{array}$

Relation between molarity and normality is written below.

$\text{N}=\text{n}\times \text{M}$

Where,

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

The calculation of normality of sodium carbonate is shown below.

$\begin{array}{rll}\text{N}& =& \text{n}\times \text{M}\\ {\text{N}}_2& =& 2\times \text{0}\text{.3976 M}\\ & =& \text{0}\text{.7952 N}\end{array}$

Therefore, normality of sodium carbonate of volume $\text{51}\text{.89 mL}$ which is titrated with $\text{5}\text{.038 g}$ of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $\text{0}\text{.7952 N}$.

Conclusion

The normality of sodium carbonate of volume $\text{51}\text{.89 mL}$ titrated with $\text{5}\text{.038 g}$ of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is calculated as $\text{0}\text{.7952 N}$.