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Question: A sample of hydrazine sulphate (\(N.HSO\)) was dissolved in 100 mL of water. 10 mL of this solution ...

A sample of hydrazine sulphate (N.HSON.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.
4Fe++++N2H4→N2+4Fe+++4H+4F{e^{ + + + }} + {N_2}{H_4} \to {N_2} + 4F{e^{ + + }} + 4{H^ + }
MnO4−+5Fe+++8H+→Mn+++5Fe+++4H2OMnO_4^ - + 5F{e^{ + + }} + 8{H^ + } \to M{n^{ + + }} + 5F{e^{ + + }} + 4{H_2}O

Explanation

Solution

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:
1mole1mole of potassium permanganate oxidizes 11 mole of Fe2+→Fe3+F{e^{2 + }} \to F{e^{3 + }}.
Calculating the moles of MnO4−MnO_4^ - is possible using = Volume of KMnO4KMn{O_4} X Concentration of KMnO4KMn{O_4}.
The moles of MnO4−MnO_4^ - = 0.02×1200.02\times \dfrac{1}{20}
= 0.001moles0.001moles
So, the concentration of Ferrous ions is 0.001moles0.001moles as an equal amount of ferrous is used is used for the process.
1mole1mole of hydrazine can reduce - 1mole1mole of Fe3+→Fe2+F{e^{3 + }} \to F{e^{2 + }} . Here, 4moles4moles of Fe3+→Fe2+F{e^{3 + }} \to F{e^{2 + }} .
Therefore, the concentration of 4Fe2+4F{e^{2 + }} is (4×0.001)moles\left( {4 \times 0.001} \right)moles
=0.004moles= 0.004moles
The same amount of hydrazine is required for the hydrolysis of the ferric ion into ferrous ions. Hence 0.004moles0.004moles of hydrazine are required here. The hydrazine is used for the changing state of ionic level.
So, 0.004moles0.004moles of hydrazine sulphate were present in 10mL10mL.
Since, Molarity=MassVolumeMolarity = \dfrac{{Mass}}{{Volume}}
Molarity=0.0040.01Molarity = \dfrac{{0.004}}{{0.01}}
=0.4(M)= 0.4\left( M \right)

The molarity when measured according to volume of solution taken, 0.4(M)0.4\left( M \right) in 10mL10mL .
Based on the given equation:
From here the strength in 100mL100mL 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 V1=10mL,S1=0.4(M){V_1} = 10mL,{S_1} = 0.4\left( M \right) and V2=100mL{V_2} = 100mL .
S2=0.4×10100{S_2} = \dfrac{{0.4 \times 10}}{{100}}
S2=0.04(M){S_2} = 0.04\left( M \right)
This 0.04(M)0.04\left( M \right) is the strength of the hydrazine sulphate in 100mL100mL which is chosen for reaction with FeCl3FeC{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