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Potassium Dichromate

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Question

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 $F e^{3 +}$ .
${M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
Above titration takes place in acidic medium. Acid chosen is :

A

H_{2} S O_{4}

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B

H C l

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C

C H_{3} C O O H

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D

H C O O H

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Solution

The correct option is D $C H_{3} C O O H$
$C H_{3} C O O H$ is used in the above titration.
Because if we use $H_{2} S O_{4}$ or $H C l$ , it gets decomposed and if $H C O O H$ is used, it gets oxidised by $K M n O_{4}$ .

Hence, they cannot be used.

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Similar questions

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.

25.0

mL of solution containing both

F e^{2 +}

F e^{3 +}

ions is titrated with

25.0

0.02 M

K M n O_{4}

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

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

Indicator in the above titration is :

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.

25.0

mL of solution containing both

F e^{2 +}

F e^{3 +}

ions is titrated with

25.0

0.02 M

K M n O_{4}

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

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

Zinc added in the second titration will :

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.

25.0

mL of solution containing both

F e^{2 +}

F e^{3 +}

ions is titrated with

25.0

0.02 M

K M n O_{4}

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

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

0.02 M K_{2} C r_{2} O_{7}

is used instead of

0.02 M K M n O_{4}

, its volume required in these titrations will be respectively :

Q. The total score for the correct statement(s) from the following is:
Given-

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

E^{\oplus} = 1.51 V

F e^{3 +} + e^{-} \to F e^{2 +}

E^{⊖} = 0.77 V

C l_{2} + 2 e^{-} \to 2 C l^{⊖}

E^{⊖} = 1.36 V

C e^{4 +} + e^{-} \to C e^{3 +}

E^{⊖} = 1.76 V

	Statement	Score
a.	$M n O_{4}^{⊖}$ is a sufficiently strong oxidant in acidic solution (pH = 0) to oxidize $F e^{2 +}$ ion.	1
b.	$F e^{2 +}$ ion cannot be titrated against standard $K M n O_{4}$ solution if the medium is made acidic (pH=0) by adding HCl.	2
c.	$M n O_{4}^{⊖}$ ion cannot oxidize $C e^{3 +}$ in acidic medium 9pH = 0).	3
d.	$F e^{2 +}$ cannot be titrated against standard $K M n O_{4}$ solution in acidic medium (pH = 0) in the presence of $C e^{3 +}$ ion.	4

Q. A mixture of

F e O

F e_{2} O_{3}

is completely reacted with

100 mL

0.25 M

K M n O 4

solution. The resultant solution was then titrated with

Z n

dust which converted the

F e^{3 +}

F e^{2 +}

F e^{2 +}

1000 mL

0.10 M

K_{2} C r_{2} O_{7}

solution. Find out the weight percentage of

F e_{2} O_{3}

in the mixture.

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