Chapter 16, Problem 107E

Interpretation Introduction

Interpretation:

Normality of potassium hydroxide required to titrate with $\text{1}\text{.45 g}$ of maleic acid is to be calculated.

Concept introduction:

Normality is a term which is used to express concentration of any solution. In a neutralization reaction, equivalents of an acid and base are equal. Equation which equates the equivalents of an acid and a base is given below.

$\underset{\text{equivalents of acid }}{\underbrace{{\text{N}}_{\text{1}}{\text{V}}_{\text{1}}}}=\underset{\text{equivalents of base }}{\underbrace{{\text{N}}_{\text{2}}{\text{V}}_{\text{2}}}}$

Answer to Problem 107E

Normality of potassium hydroxide of volume $\text{50}\text{.0 mL}$ which is titrated with $\text{1}\text{.45 g}$ of maleic acid is $\text{0}\text{.500 N}$.

Explanation of Solution

Equation which is used to calculate the normality of potassium hydroxide is given below.

$\underset{\text{equivalents of acid maleic acid}}{\underbrace{{\text{N}}_{\text{1}}{\text{V}}_{\text{1}}}}=\underset{\text{equivalents of base KOH}}{\underbrace{{\text{N}}_{\text{2}}{\text{V}}_{\text{2}}}}$…(1)

Relation between normality and molarity of acid is shown below.

${\text{N}}_1={\text{n}}_1{\text{M}}_1$

Substitute the value of ${\text{N}}_1$ in equation (1) as shown below.

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

Relation between molarity and given mass of acid is shown below.

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

Substitute the value of ${\text{M}}_1{\text{V}}_1$ in equation (2) as shown below.

$\begin{array}{rll}\underset{\text{equivalents of acid maleic acid}}{\underbrace{{\text{n}}_{\text{1}}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}}}& =& \underset{\text{equivalents of base KOH}}{\underbrace{{\text{N}}_{\text{2}}{\text{V}}_{\text{2}}}}\\ \underset{\text{equivalents of acid maleic acid}}{\underbrace{{\text{n}}_{\text{1}}\times \frac{\text{w}\left(\cancel{g}\right)}{\text{Molar mass}\left(\cancel{g}/\text{mol}\right)}}}& =& \underset{\text{equivalents of base KOH}}{\underbrace{{\text{N}}_{\text{2}}{\text{V}}_{\text{2}}}}\end{array}$…(3)

Where,

• ${\text{n}}_{\text{1}}$ is the n-factor of maleic acid (${\text{H}}_2{\text{C}}_4{\text{H}}_2{\text{O}}_{\text{4}}$).

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

• ${\text{N}}_{\text{2}}$ is the molarity of potassium hydroxide base.

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

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

Balanced chemical equation of the given reaction is written below.

${\text{H}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}+\text{2KOH}\to {\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}+{\text{2H}}_{\text{2}}\text{O}$

Maleic acid has ${\text{2 H}}^{\text{+}}$. So n-factor of maleic acid is $2$

Given values of ${\text{n}}_{\text{1}},\text{w},{\text{M}}_{\text{o}},{\text{V}}_{\text{2}}$ are $\text{2},\text{1}\text{.45 g},116.07\text{g}/\text{mol},\text{50}\text{.0 mL}$ respectively.

Substitute all the values of ${\text{n}}_{\text{1}},\text{w},{\text{M}}_{\text{o}}$ and ${\text{V}}_{\text{2}}$ in equation (3) as shown below.

$\begin{array}{rll}\text{2}\times \left(\frac{\text{1}\text{.45g}}{\text{116}\text{.07}\text{g}/\text{mol}}\right)& =& {\text{N}}_{\text{2}}\times \text{50}\text{.0 mL}\\ {\text{N}}_{\text{2}}& =& \left(\frac{\text{2}\times \text{1}\text{.45 g}}{116.07\text{g}/\text{mol}\times \text{50}\text{.0 mL}}\right)\\ {\text{N}}_{\text{2}}& =& 0.0005\text{mol}/\text{mL}\times \frac{\text{1000}\text{mL}}{\text{1}\text{L}}\\ & =& \text{0}\text{.500 }\text{mol}/\text{L}\end{array}$

Therefore, normality of potassium hydroxide is $\text{0}\text{.500 N}$.

Conclusion

Normality of potassium hydroxide of volume $\text{50}\text{.0 mL}$ which is titrated with $\text{1}\text{.45 g}$ of maleic acid is $\text{0}\text{.500 N}$.