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Arrhenius Acid

MnO4- ions ar...

Question

$M n O_{4}^{-}$ 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?

11.4 ml

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12.0 ml

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33.3 ml

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35.0 ml

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25.0 ml

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Solution

The correct option is A 11.4 ml
Acidic $F e^{2 +} + M n O_{4}^{-} \to M n^{2 +} + F e^{3 +}$
$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}}$
$\frac{M_{1} (25)}{1} = \frac{M_{2} (20)}{5}$
$\frac{M_{1}}{M_{2}} = \frac{4}{25}$
In neutral n factor - 2
$\frac{M_{1} (25)}{1} = \frac{M_{2} (V)}{3} \Rightarrow V = 33.33 m l$

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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 :

Q. 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 :

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 :

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 :

Q. In strongly alkaline solution containing excess of barium ions, a solution containing 0.01587 g of

I^{-}

was treated with 0.1 molar

{M n O}_{4}^{-}

until a pink colour presisted in the solution : 10.0 mL of the

{M n O}_{4}^{-}

solution were required. Under these conditions

{M n O}_{4}^{-}

is converted into the sparingly soluble

{B a M n O}_{4}

. The product of the oxidation of iodide is

I O_{4}^{-}

if true enter 1, else enter 0.

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The correct option is A 11.4 mlAcidic Fe2++MnO−4→Mn2++Fe3+M1V1n1=M2V2n2M1(25)1=M2(20)5M1M2=425In neutral n factor - 2M1(25)1=M2(V)3⇒V=33.33ml