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Grade 10Physical Chemistry

Hydroxylamine reduces iron (III) according to the equation:
2NH₂OH + 4 Fe³⁺→N₂O(g) ↑ + H₂O + 4 Fe²⁺ + 4H⁺
Iron (II) thus produced is estimated by titration with standard permanganate solution. The is :
MnO^-₄ + 5 Fe²⁺ + 8H⁺→ Mn²⁺ + 5 Fe³⁺ + 4H₂O
A 10 ml. sample of hydroxylamine solution was dilute to 1 litre. 50 ml. of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 ml. of 0.02 M KMnO₄ solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K = 39, Mn = 55, Fe = 56)

Hrishant Goswami

12 Years agoGrade 10

1 Answer

Jitender Pal

12 Years ago

(i) Write balanced reaction for changes.

(ii) M. eq. of Fe²⁺ formed = M. eq. KMnO₄ised

= V in ml * molarity * x

The given redox changes are

Fe³⁺ + e^- → Fe²⁺

2N^- → 2 N⁺ + 2e^-

[In NH₂OH, O.S. of N = - 1 and in N₂O, O.S. of N = + 1. For two molecules of NH₂OH, electron involved = 4]

Mn⁷⁺ + 5e^- → Mn²⁺

= Meq. Of KMnO₄ used = Vol. in ml * Molarity * x

= 12 * 0.02 * 5 =12 [50/10 = 5]

∴Meq.of Fe²⁺ formed by NHH₂OH in 1000 ml of dil. Solution

= 1.2 * 1000/50 = 24

Meq.of original solution

= Meq. of NH₂OH in 1000 ml. of dilute solution

W_{NH base 2OH}/33/2 * 1000 = 24 [Eq. wt = 33/2]

W_{NH base 2OH} = 24 *16.5/1000 = 0.396 g

∵ Wt. of NH₂OH in 10 ml. of original solution = 0.396 g

∴ Wt. of NH₂OH in 1 litre of original solution

= 0.396 * 100 = 39.6 g l^-1

ALTERNATIVESOLTUION:

Given 2NH₂OH + 4Fe³⁺ → N₂O + H₂O + 4Fe²⁺ + 4H⁺ …..(i)

And MnO^-₄ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O ...(ii)

∴ 10NH₂OH +4MnO^-₄ + 12H⁺ → 5N₂O + 21H₂O + 4Mn²⁺

[On multiplying (i) by 5 and (ii) by 4 and then adding the resulting equations]

Molecular weight of NH₂OH = 33

Thus 400 ml of 1M MnO₄^- would react with NH₂OH = 330g

∴ 12 ml of 0.02 M KMnO₄ would react with NH₂OH = 330 * 12 *0.02/400 g

∴ Amount of NH₂OH present in 1000 ml of dilute solution 330 *12 *0.02 *1000/400 *50 g

Since 10 ml of sample of hydroxylamine is dilute to one litre

∴ A mount of hydroxyl amine in one litre of original solution = 330 *0.02 *12 *1000/4000 *50 * 1000/10 g = 39*6 g

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