Question
Question: Hydroxylamine reduces iron (III) according to the following reaction: \[2N{{H}_{4}}OH+4F{{e}^{3+}}...
Hydroxylamine reduces iron (III) according to the following reaction:
2NH4OH+4Fe3+→N2O(g)↑+H2O+4Fe2++4H+
Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is:
MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O
A 10ml sample of hydroxylamine solution was diluted to 1L. When 50ml of this diluted solution was boiled with an excess of iron (III) solution, the resulting solution required 12ml of 0.02M KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in 1L of the original solution (H=1, N=14, O=16, K=39, Mn=55, Fe=56):
(A) 40gL−1
(B) 80gL−1
(C) 22gL−1
(D) 62gL−1
Solution
Start by writing, the relation between molarity and normality of KMnO4. Then find the relation between volume of KMnO4 and ferrous solution and correspondingly hydroxylamine solution required for that much amount of ferrous solution. Proceed by taking into account equivalent weight of hydroxylamine.
Complete step by step answer:
-0.02M KMnO4 is equivalent to 0.01N KMnO4
12ml0.1NKMnO4≡12ml0.1NFe2+≡12ml0.1NHydroxylamine
10ml of original hydroxylamine solution is diluted to 1L to make 0.1N hydroxylamine solution. So, now we need to know how much mass of hydroxylamine is present in a 50ml diluted solution.
- Equivalent weight of hydroxylamine =2Mol.mass=233=16.5g
So, mass of hydroxylamine in 12ml of 0.1N hydroxylamine=1000N×V×E=10000.1×12×16.5=0.0198g≈0.02g
Therefore, 50ml diluted solution of 0.1N hydroxylamine contains 0.02g.
So, one litre diluted solution of 0.1N hydroxylamine contains 500.02×1000=0.4g
Therefore, 10ml original solution contains 0.4g.
So, 1000ml original hydroxylamine solution will contain 40g.
So, the correct answer is “Option A”.
Note: Don’t get confused with normality and molarity. While calculating mass for quantitative analysis, always take normality into consideration to make calculation simpler.