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Oxides of Group 16

The following...

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 $F e^{3 +}$ .

${M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
The molar concentration of $F e^{2 +}$ in the original solution is :

0.01 M

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0.02 M

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0.10 M

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0.20 M

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Solution

The correct option is C $0.10 M$
Here, $F e^{2 +}$ is oxidised and ${M n O_{4}}^{-}$ is reduced.
Both have the same equivalents.
Equivalent of $F e^{2 +} =$ Equivalent of ${M n O_{4}}^{-}$

$25 \times M = 25 \times 0.02 \times 5$

$M = 0.10 M$

Hence, molar concentration of $F e^{2 +}$ is $0.1 M$ .

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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.

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 :

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

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.

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

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. A mixture of

F e O

and

F e_{2} O_{3}

is completely reacted with

100 mL

0.25 M

acidified

K M n O 4

solution. The resultant solution was then titrated with

Z n

dust which converted the

F e^{3 +}

ions to

F e^{2 +}

ions. The

F e^{2 +}

ions required

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.

M n O_{4}^{-}

ions are reduced in acidic condition to

M n^{2 +}

ions whereas they are reduced in neutral condition to

M n O_{2}^{+}

.The oxidation of 25 ml of a solution

X

containing

F e^{2 +}

ions required in acidic condition 20 ml of solution

Y

containing

M n O_{4}^{-}

ions. What volume of solution

Y

would be required to oxidize 25 ml of solution

X

containing

F e^{2 +}

ions in neutral condition?

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The correct option is C 0.10MHere, Fe2+ is oxidised and MnO4−is reduced.Both have the same equivalents. Equivalent of Fe2+= Equivalent of MnO4−25×M=25×0.02×5M=0.10MHence, molar concentration of Fe2+ is 0.1 M.

The correct option is C $0.10 M$
Here, $F e^{2 +}$ is oxidised and ${M n O_{4}}^{-}$ is reduced.
Both have the same equivalents.
Equivalent of $F e^{2 +} =$ Equivalent of ${M n O_{4}}^{-}$

$25 \times M = 25 \times 0.02 \times 5$

$M = 0.10 M$

Hence, molar concentration of $F e^{2 +}$ is $0.1 M$ .