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Oxidising Reagent and Reducing Reagent

0.1 M solutio...

Question

$0.1 M$ solution of $K I$ reacts with excess of $H_{2} S O_{4}$ and $K I O_{3}$ solutions according to equation:
$5 I^{-} + I O_{3}^{-} + 6 H \to 3 I_{2} + 3 H_{2} O$
Which of the following statement is correct?

A

200

ml of the

K I

solution reacts with

0.004

mol

K I O_{3}

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B

100

ml of the

K I

solution reacts with

0.006

mol of

H_{2} S O_{4}

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C

0.5

litres of the

K I

solution produced

0.005

mol of

I_{2}

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D

Equivalent weight of

K I O_{3}

is equal to

(\frac{M o l e c u l a r w e i g h t}{5})

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Solution

The correct options are
B $100$ ml of the $K I$ solution reacts with $0.006$ mol of $H_{2} S O_{4}$
D Equivalent weight of $K I O_{3}$ is equal to $(\frac{M o l e c u l a r w e i g h t}{5})$
The balanced chemical reaction is $5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O$ .
$200$ ml of $0.1 M$ $K I$ solution corresponds to $0.2$ moles.
$\frac{M o l e s o f K I u s e d}{M o l e s o f K I O_{3}} = 5$
Hence, $200$ ml of $0.1 M$ $K I$ solution reacts with $0.004$ mol $K I O_{3}$ .
$100$ ml of $0.1 M$ $K I$ solution corresponds to $0.1$ moles.
$\frac{M o l e s o f K I u s e d}{M o l e s o f K I O_{3}} = \frac{3}{5}$ .
Hence, $100$ ml of the $K I$ solution reacts with $0.006$ mol of $H_{2} S O_{4}$ .
The valency factor for $K I O_{3} = 5$ .
Hence, $E_{K I O_{3}} = \frac{M o l e c u l a r w e i g h t}{V a l e n c y f a c t o r} = \frac{M o l e c u l a r w e i g h t}{5}$ .

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Similar questions

Q. 0.1 M solution of

K I

reacts with excess of

H_{2} S O_{4}

K I O_{3}

solutions, according to the equation

5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O

which of the following statements is/are correct

0.1 M

solution of KI reacts with excess of

H_{2} S O_{4}

K I O_{3}

solutions acccording to equation

5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O

which of the following statement is correct?

0.1 M

solution of KI reacts with excess of

H_{2} S O_{4}

K I O_{3}

solutions acccording to equation

5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O

which of the following statement is correct?

(a)

C u S O_{4}

K I

in acidic medium to liberate

I_{2}

.

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

(b)

Mercuric perisdate,

H g_{5} (I O_{6})_{2}

reacts with a mixture of

K I

H C l

according to the following the equation:

H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O

(c)

The liberated iodine is titrated against

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

solution, one mL of which is equivalent to

0.0499

C u S O_{4} .5 H_{2} O

What volume in mL of

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

solution will be required to react with

I_{2}

0.7245

H g_{5} (I O_{6})_{2}

? (Molar mass of

C u S O_{4} .5 H_{2} O = 249.5

H g_{5} (I O_{6})_{2} = 1448.5

Q. An aqueous solution of

K I O_{3}

K I

on mixing undergoes redox process to give

I_{2}

. The reaction occurs quantitatively and can be used to estimate the conc. of

K I O_{3}

. The iodine liberated in the reaction can be estimated by using hypo solution and starch as indicator.

x

K I O_{3}

react completely with

y

K I

, then which of the following are correct?

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