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Equilibrium Constant and Standard Free Energy Change

Hydroxylamine...

Question

Hydroxylamine reduces iron (III) according to the following reaction:
$2 N H_{2} O H + F e^{3 +} \to N_{2} O (g) ↑ + H_{2} O + 4 F e^{2 +} + 4 H^{⨁}$
Iron(II) thus produced is estimated by titration with a standard permanganate solution. The reaction is:
$M n O_{4}^{⊝} + 5 F e^{2 +} + 8 H^{⨁} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
A $10$ mL sample of hydroxylamine solution was diluted to $1$ L. When $50$ mL of this diluted solution was boiled with an excess of iron (III) solution, the resulting solution required $12$ mL of $0.02 M$ $K M n O_{4}$ solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in $1$ L of the original solution $(H = 1, N = 14, O = 16, K = 39, M n = 55, F e = 56) :$

40 g L^{- 1}

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80 g L^{- 1}

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22 g L^{- 1}

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62 g L^{- 1}

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Solution

The correct option is B $40 g L^{- 1}$
$1 m o l M n O_{4}^{-} \equiv 5 m o l F e^{2 +} \equiv 2.5 m o l N H_{2} O H$
$12$ mL of $0.02$ M $K M n O_{4}$ solution contains $12 m L \times 0.02 M = 0.24 m m o l$ $K M n O_{4}$ .
It also corresponds to $0.24 m m o l \times 2.5 = 0.6 m m o l$ of $N H_{2} O H$ .
These many moles are present in $50$ mL of diluted solution.
Hence, $1$ L of diluted solution contains $\frac{1000}{50} \times 0.6 m m o l = 12 m m o l = 0.012 m o l$
The molar mass of $N H_{2} O H$ is $14 + 3 + 16 = 33 g / m o l$ .
$0.012 m o l$ of $N H_{2} O H$ corresponds to $0.012 m o l \times 33 g / m o l = 0.396 g$ .
This much $N H_{2} O H$ is present in the $10$ mL of original solution.
Hence, $1$ L of original solution contains $\frac{1000}{10} \times 0.396 g = 39.6 g ≃ 40 g$

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Similar questions

Q. Hydroxylamine reduces iron (III) according to the equation,

2 N H_{2} O H (a q) + 4 F e^{3 +} (a q) \to N_{2} O (g) ↑ + H_{2} O (l) + 4 F e^{2 +} (a q) + 4 H^{+} (a q)

Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is:

{M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

10 m L

sample of hydroxylamine solution was diluted to one litre.

50 m L

of this diluted solution was boiled with excess of iron (III) solution. The resulting solution required

12 m L

0.02 M K M n O_{4}

solution for complete oxidation of iron (II). Calculate the mass of hydroxylamine in one litre of the original solution.

Q. A sample of hydrazine sulphate

(N_{2} H_{6} S O_{4})

was dissolved in

100

mL of water and

10

mL of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required

20

mL of

\frac{M}{50}

potassium permanganate solution. Estimate the amount of hydrazine sulphate in

1

L of the solution. Reactions are given below:

4 F e^{3 +} + N_{2} H_{4} \to N_{2} + 4 F e^{2 +} + 4 H^{⨁}

M n O_{4}^{⊝} + 5 F e^{2 +} + 8 H^{⨁} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

A sample of hydrazine sulphate

(N_{2} H_{6} S O_{4})

was dissolved in

100

mL water.

10

mL of this solution was reacted with excess of

F e C l_{3}

solution and warmed to complete the reaction. Ferrous ions formed was estimated and it required

20

mL of

\frac{M}{50} K M n O_{4}

solution. Estimate the mass of hydrazine sulphate in one litre of solution.

4 F e^{3 +} + N_{2} H_{4} ⟶ N_{2} + 4 F e^{2 +} + 4 H^{+}

M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} ⟶ M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

Q. 50 mL aqueous solution of

F e S O_{4}

required 12 mL of 0.2 M

K M n O_{4}

in acidic medium for complete oxidation. Calculate the molarity of ferrous sulphate solution. The reaction is:

M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

Q. The following titration method is given to determine the total content of the species with variable oxidation states. Answer the question given at the end of it.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

K M n O_{4}

solution for oxidation to

F e^{3 +}

{M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

The molar concentration of

F e^{2 +}

in the original solution is :

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