Question

The titrant and indicator used for the determination of Fe are respectively:
a.) ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and diphenylamine
b.) ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and starch
c.) $N{{a}_{2}}{{S}_{2}}{{O}_{7}}$ and diphenylamine
d.) $N{{a}_{2}}{{S}_{2}}{{O}_{7}}$ and starch

Answer 1

Hint: It appears as a crystalline ionic solid, with a bright red-orange colour which is odourless. It dissolves readily in water but insoluble in acetone and alcohol. It also has antiseptic and astringent values.

Complete step by step solution:

Titrant is a solution of a known concentration, which is added to another solution whose concentration has to be determined whereas analyte is the solution whose concentration has to be determined.
It can be of four types:
1. Acid-base Titrations.
2. Redox Titrations.
3. Precipitation Titrations.
4. Complexometric Titrations

The equivalence point in a titration is the point at which exactly enough reactant has been added for the reaction to go to completion.
An indicator is defined as a substance that undergoes distinct observable change when the conditions of its solution change.
Unlike permanganate ($Mn{{O}_{4}}^{-}$), dichromate (${Cr_2O_7}^{2-}$) titrations need an indicator. There are three indicators that can be used for the titration of $F{{e}^{3+}}$ these are diphenylamine, diphenyl benzidine and diphenylamine sulfonate.
In this experiment we will use a standard solution of potassium dichromate (${{K}_{2}}C{{r}_{2}}{{O}_{7}}$) to determine the percent by weight of iron in an unknown solid
Dichromate ion reduces to two chromium (III) ions. This reaction requires 6 electrons and 14 hydrogen ions:
$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O$
We need only one electron to reduce Fe (III) to Fe (II)
$F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}}$
Therefore, 1 mole of ${Cr_2O_7}^{2-}$ which is also the oxidizing agent, reacts with 6 moles of $F{{e}^{2+}}$ which is the reducing agent to form 6 moles of $F{{e}^{3+}}$ and 2 moles of $C{{r}^{3+}}$. Thus, in net ionic form:
$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6F{{e}^{2+}}\to 2C{{r}^{3+}}+6F{{e}^{3+}}+7{{H}_{2}}O$
The color change for all three indicators is green to violet, we will use diphenylamine here and an intense purple color will be produced.

Therefore, from the above statements we can conclude that the correct option is (a).

Note: Even though the solutions of dichromate are intensely orange in colour and a single drop of it imparts yellow colour to a colourless solution, it cannot be used as a self-indicator like $KMn{{O}_{4}}$. This is because its reduction product is green which hinders in the visual detection of the end point by observing dichromate colour. Thus, an indicator is must in this titration.