Chapter 7, Problem 7.105CHP

Interpretation Introduction

Interpretation:

The mass percent of $\text{Fe in 1}\text{.2284 g}$ sample of ore should be calculated, when it is completely titrate with $\text{57}\text{.91 mL of 0}{\text{.1018 M Ce(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{6}}$ solution.

Concept introduction:

Redox:

The reduction and oxidation reactions occurring simultaneously within a single reaction is known as redox reaction.

Redox titration:

Redox titration is type of titration method to determine the amount of analyte that present in the given sample by redox reaction.

The amount of analyte is calculated by volumetric principle, and mass equation they are below,

Volumetric principle:

The volume and concentration of unknown solution is determined by titrating it with known volume and known concentration solution.

The volume and concentration of unknown solution requires equivalent volume and concentration of known solution in the volumetric titration.

${\text{V}}_{\text{1}}{\text{M}}_{\text{1}}\text{=}{\text{V}}_{\text{2}}{\text{M}}_{\text{2}}$

Where,

${\text{V}}_{\text{1}}$ is volume of known solution

${\text{N}}_{\text{1}}$ is concentration of known solution

${\text{V}}_2$ is volume of unknown solution

${\text{N}}_{\text{2}}$ is concentration of unknown solution

Mass percent:

The mass percentage is calculated by percentage ratio of taken mass of the analyte to actual mass of sample.

$\text{Mass}\text{percentage}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{ofthe}\text{sample}}\text{×100}$

Mole:

Mole of the compound present in the solution is calculated by molarity, concentration of the compound multiplied with litre, volume of solution.

$\text{Mole}\text{=Molarity}\text{×Liter}$

Mass:

Mass of the compound is calculated by mole of the compound is multiplied with molar mass of the compound.

$\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$