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Standard XII

Chemistry

Types of Redox Reactions

A given 1 L...

Question

A given $1 L$ solution of $K I O_{3}$ of unknown molarity is titrated by taking its $50$ mL solution against $K I$ solution in strong acid medium of excess $H C l$ . The equivalence point was detected when $10$ mL of $0.1 M K I$ was consumed. The molarity of $K I O_{3}$ solution is :

4 \times 10^{- 3} M

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2 \times 10^{- 3} M

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8 \times 10^{- 3} M

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10^{- 2} M

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Solution

The correct option is A $4 \times 10^{- 3} M$
$2 I^{5 +} + 10 e^{-} \to I_{2}$

$2 I^{-} \to I_{2} + 2 e^{-}$

Meq. of $K I O_{3} =$ Meq. of $K I$

$M \times 5 \times 50 = 10 \times 0.1 \times 1$

$\Rightarrow M = 4.0 \times 10^{- 3}$

Hence, option A is correct.

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Similar questions

1

g sample of

A g N O_{3}

is dissolved in

50

mL of water. It is titrated with

50

mL of

K I

solution. The

A g I

precipitated is filtered off. Excess of

K l

in filtrate is titrated with

\frac{M}{10}

K I O_{3}

in presence of

6 M H C l

till all

I^{-}

converted into

I C l

. It requires

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mL of

\frac{M}{10}

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20

mL of the same stock solution of

K I

requires

30

mL of

\frac{M}{10}

K I O_{3}

A g N O_{3}

in sample. The reaction is given below:

K I O_{3} + 2 K I + 6 H C l ⟶ 3 I C l + 3 K C l + 3 H_{2} O

25

mL of a solution containing

H C l

was treated with excess of

K I O_{3}

and

K I

solution of unknown concentration, where

I_{2}

liberated is titrated against a standard solution of

0.021 M

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

solution, whose

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mL were used up. The strength of

H C l

is :

Q. An aqueous solution containing

0.10

K I O_{3}

(formula weight

= 214.0

) was treated with an excess of

K I

solution. The solution was acidified with

H C l

. The liberated

I_{2}

consumed

45.0

mL of thiosulphate solution to decolourise the blue starch iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Q. An aqueous solution containing 0.10 g

K I O_{3}

(formula wt. = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated

I_{2}

consumed 45.0 ml of thiosulphate solution to decolourise the blue starch - iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Q. An aqueous solution containing

0.10 g K I O_{3} (formula weight = 214.0 g / m o l)

was treated with an excess of

K I

solution. The solution was acidified with

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The liberated

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consumed

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of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate molarity of the sodium thiosulphate solution.

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A given 1 L solution of KIO3 of unknown molarity is titrated by taking its 50 mL solution against KI solution in strong acid medium of excess HCl. The equivalence point was detected when 10 mL of 0.1MKI was consumed. The molarity of KIO3 solution is :

The correct option is A 4×10−3 M2I5++10e−→I22I−→I2+2e−Meq. of KIO3= Meq. of KIM×5×50=10×0.1×1⇒M=4.0×10−3Hence, option A is correct.

A given $1 L$ solution of $K I O_{3}$ of unknown molarity is titrated by taking its $50$ mL solution against $K I$ solution in strong acid medium of excess $H C l$ . The equivalence point was detected when $10$ mL of $0.1 M K I$ was consumed. The molarity of $K I O_{3}$ solution is :

The correct option is A $4 \times 10^{- 3} M$
$2 I^{5 +} + 10 e^{-} \to I_{2}$

$2 I^{-} \to I_{2} + 2 e^{-}$

Meq. of $K I O_{3} =$ Meq. of $K I$

$M \times 5 \times 50 = 10 \times 0.1 \times 1$

$\Rightarrow M = 4.0 \times 10^{- 3}$

Hence, option A is correct.