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Ways to Define Concentration

Excess of K...

Question

Excess of $K l$ was added to $100 m l H_{2} O_{2}$ solution of unknown strength along with sufficient $H_{2} S O_{4}$ Iodine liberated was titrated against $40 m l$ of $0.1 M$ hypo solution. The concentration of $H_{2} O_{2}$ solution is:

A

0.04 N

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B

0.04 M

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C

0.68 g m / L

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D

0.02 M

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Solution

The correct option is D $0.02 M$
$K I + H_{2} O_{2} ⟶ I_{2} + K O H$
$I_{2} + N a_{2} S_{2} O_{3} ⟶ N a_{2} S_{4} O_{6} + N a I$
Number of equivalent of $I_{2}$ will react with number of equivalent of $N a_{2} S_{2} O_{3}$
$n \times n f (I_{2}) = n \times n f (N a_{2} S_{2} O_{3})$
$_{2} ⟶ I^{-}$
$\Rightarrow n f = 2 \times 1 = 2$
$2 S_{2} O_{3}^{2 -} ⟶ 2.5 S_{4} O_{6}^{2 -}$
$\Rightarrow n f = 2 (2.5 - 2) = 1$
$\Rightarrow n \times n f (I_{2}) = M \times V \times n f (N a_{2} S_{2} O_{3})$ as $n = M \times V$
$\Rightarrow n \times 2 = 0.1 M \times \frac{40 L}{1000} \times 1$
$\Rightarrow n = 2 \times 10^{- 3}$ moles of $I_{2}$ reacted
Thus upon reaction of $K_{I}$ and $H_{2} O_{2}$ $2 \times 10^{- 3}$ moles of $I_{2}$ liberated
Thus, number of equivalent of $I_{2}$ = Number of equivalent of $H_{2} O_{2}$
$\Rightarrow n \times n f (I_{2}) = n \times n f (H_{2} O_{2})$
$- 1 H_{2} O_{2} ⟶ - 2 [O H]^{-}$
$\Rightarrow n f = 2 \times 1 = 2$
$\Rightarrow 2 \times 10^{- 3} \times 2 = n \times 2$
$\Rightarrow n = 2 \times 10^{- 3}$ moles of $H_{2} O_{2}$
Concentration of $H_{2} O_{2}$ in $100 m l$ is $\frac{2 \times 10^{- 3} \times 1000}{100} = 0.02 M$

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Similar questions

H_{2} O_{2}

solution, excess of acidified solution of KI was added. The iodine liberated required 20 ml of 0.3 N

N a_{2} S_{2} O_{3}

solution. The volume strength of

H_{2} O_{2}

25

H_{2} O_{2}

solution, excess of an acidified solution of potassium iodide was added. The iodine liberated required

20

0.3

N sodium thiosulphate solution.

Calculate the volume strength of

H_{2} O_{2}

solution (to the nearest integer):

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

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0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

0.02 M

solution with excess

K I

50 m L

of an aqueous solution of

H_{2} O_{2}

was treated with an excess of

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solution and dilute

H_{2} S O_{4}

. The liberated iodine required

20 m L 0.1 N N a_{2} S_{2} O_{3}

solution for complete interaction. Calculate the concentration of

H_{2} S O_{2}

g / L

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H C l

was treated with excess of

K I O_{3}

K I

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I_{2}

liberated is titrated against a standard solution of

0.021 M

N a_{2} S_{2} O_{3}

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24

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H C l

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