Chapter 6, Problem 6.2P

Interpretation Introduction

Interpretation:

The volume of ${\text{KMnO}}_{\text{4}}$ with $\text{0}\text{.1650}\text{M}$ needed to react with $\text{108}\text{.0}\text{mL}$ of $\text{0}\text{.1650}\text{M}$ oxalic acid; and volume of oxalic acid with $\text{0}\text{.1650}\text{M}$ needed to react with $\text{108}\text{.0}\text{mL}$ of $\text{0}\text{.1650}\text{M}$${\text{KMnO}}_{\text{4}}$ has to be calculated.

Concept-Introduction:

Acid–base titration: It consists of a neutralization reaction where a measured volume of a known concentrated an acid or a base is used for determining the concentration of an unknown solution of base or acid with known volume.  The known concentrated solution is called the titrant, while the unknown concentrated solution is called the analyte.

Molarity: The concentration for solutions is expressed in terms of molarity as follows,

  $\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

Answer to Problem 6.2P

The volume of oxalic acid is $\text{270 mL}$.

Explanation of Solution

Given data is,

  ${\text{Reaction : 5Oxalic acid + 2MnO}}_4\text{+}{\text{6H}}^{+}\underset{}{\to }{\text{ 10CO}}_2\text{ + }2M{n}^{2+}\text{+}{\text{8H}}_2\text{O }$

The moles of oxalic acid is calculated as shown below,

  $\begin{array}{l}\text{5Oxalic acid }\text{+ }{\text{2MnO}}_4\text{+}{\text{6H}}^{+}\underset{}{\to }{\text{ 10CO}}_2\text{ + }2M{n}^{2+}\text{+}{\text{8H}}_2\text{O }\\ \\ \text{No}\text{.of}\text{moles =}\left(\text{0}\text{.1650}\text{M}\right)\left(\text{108}\text{.0}\text{mL}\right)Molarity=0.165M\\ \\ \text{=}\left(\text{17}\text{.82}\right)mole\end{array}$

Here, $\text{5 mole}$ of oxalic acid requires 2 mole of ${\text{KMnO}}_{\text{4}}$ and $\text{17}\text{.82}\text{mole}$ of oxalic acid requires ‘x’ mole of ${\text{KMnO}}_{\text{4}}$

Thus,

  $\text{'}x\text{'}moleof{\text{KMnO}}_{\text{4}}=\frac{\left(2mole{\text{KMnO}}_{\text{4}}\right)\left(\text{17}\text{.8}\text{mole}\text{oxalic acid}\right)}{5moleoxalicacid}=7.12mole$

Calculation of Volume of ${\text{KMnO}}_{\text{4}}$ as shown below;

  ${\text{Volume of KMnO}}_{\text{4}}=\frac{\text{7}\text{.12}\text{mole}{\text{KMnO}}_{\text{4}}}{0.1650mole/mL}=43.2mL.$

Therefore, the volume of ${\text{KMnO}}_{\text{4}}$ is $43.2mL.$

Given data is,

  ${\text{Reaction : 5Oxalic acid + 2MnO}}_4\text{+}{\text{6H}}^{+}\underset{}{\to }{\text{ 10CO}}_2\text{ + }2M{n}^{2+}\text{+}{\text{8H}}_2\text{O }$

The moles of ${\text{KMnO}}_{\text{4}}$ is calculated as shown below,

  $\begin{array}{l}\text{5Oxalic acid }\text{+ }{\text{2MnO}}_4\text{+}{\text{6H}}^{+}\underset{}{\to }{\text{ 10CO}}_2\text{ + }2M{n}^{2+}\text{+}{\text{8H}}_2\text{O }\\ \\ \text{No}\text{.of}\text{moles =}\left(\text{0}\text{.1650}\text{M}\right)\left(\text{108}\text{.0}\text{mL}\right)\\ \\ \text{=}\left(\text{17}\text{.82}\right)mole\end{array}$

Here, $\text{5 mole}$ Of oxalic acid requires 2 mole of ${\text{KMnO}}_{\text{4}}$ and ‘x’ of oxalic acid requires $\text{17}\text{.82}\text{mole}$ mole of ${\text{KMnO}}_{\text{4}}$.

Thus,

  $\text{'}x\text{'}moleof\text{oxalic acid}=\frac{\left(5mole\text{oxalic acid}\right)\left(\text{17}\text{.8}\text{mole}{\text{KMnO}}_{\text{4}}\right)}{2mole{\text{KMnO}}_{\text{4}}}=44.5mole$

Calculation of Volume of ${\text{KMnO}}_{\text{4}}$ as shown below;

  $\text{Volume of}\text{oxalic acid}=\frac{44.5mole}{0.1650mole/mL}=269.69mL=270mL.$

Therefore, the volume of oxalic acid is $270mL.$