Question

A sample of hydrazine sulphate ($N.HSO$) was dissolved in 100 mL of water. 10 mL of this solution was reacted with an excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required 20 mL of M/50 potassium permanganate solution. Estimate the amount of hydrazine sulphate in one litre of the solution.
\[4F{e^{ + + + }} + {N_2}{H_4} \to {N_2} + 4F{e^{ + + }} + 4{H^ + }\]
$MnO_4^ - + 5F{e^{ + + }} + 8{H^ + } \to M{n^{ + + }} + 5F{e^{ + + }} + 4{H_2}O$

Answer 1

Hint: At first the level of ferrous ion needs to be determined to take it back to the previous step. The level of ferrous ion produced can determine the hydrazine sulphate present in one litre of the solution.

Complete step by step answer:
$1mole$ of potassium permanganate oxidizes $1$ mole of $F{e^{2 + }} \to F{e^{3 + }}$.
Calculating the moles of $MnO_4^ - $ is possible using = Volume of $KMn{O_4}$ X Concentration of $KMn{O_4}$.
The moles of $MnO_4^ - $ = $0.02\times \dfrac{1}{20}$
= $0.001moles$
So, the concentration of Ferrous ions is $0.001moles$ as an equal amount of ferrous is used is used for the process.
$1mole$ of hydrazine can reduce - $1mole$ of $F{e^{3 + }} \to F{e^{2 + }}$ . Here, $4moles$ of $F{e^{3 + }} \to F{e^{2 + }}$ .
Therefore, the concentration of $4F{e^{2 + }}$ is $\left( {4 \times 0.001} \right)moles$
$ = 0.004moles$
The same amount of hydrazine is required for the hydrolysis of the ferric ion into ferrous ions. Hence $0.004moles$ of hydrazine are required here. The hydrazine is used for the changing state of ionic level.
So, $0.004moles$ of hydrazine sulphate were present in $10mL$.
Since, $Molarity = \dfrac{{Mass}}{{Volume}}$
$Molarity = \dfrac{{0.004}}{{0.01}}$
$ = 0.4\left( M \right)$

The molarity when measured according to volume of solution taken, $0.4\left( M \right)$ in $10mL$ .
Based on the given equation:
From here the strength in $100mL$ needs to be considered, because it was the initial volume. Therefore in the given equation the specific volume of the given solution needs to be found out.
Here the ${V_1} = 10mL,{S_1} = 0.4\left( M \right)$ and ${V_2} = 100mL$ .
${S_2} = \dfrac{{0.4 \times 10}}{{100}}$
${S_2} = 0.04\left( M \right)$
This $0.04\left( M \right)$ is the strength of the hydrazine sulphate in $100mL$ which is chosen for reaction with $FeC{l_3}$ .

Note: The hydrazine sulphate is chosen for reduction of the ferric chloride and the use of potassium permanganate is done just to calculate the number of ferric ions that are produced