You visited us 7 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Borax

A solution co...

Question

A solution containing $F e^{2 +}$ ions is titrated with $K M n O_{4}$ solution. Indicator used will be:

phenolphthalein

No worries! We‘ve got your back. Try BYJU‘S free classes today!

methyl orange

No worries! We‘ve got your back. Try BYJU‘S free classes today!

litmus

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of the above

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C none of the above
Titration of ${F e}^{2 +}$ with ${K M n O}_{4}$ is an redox titration.
${F e}^{2 +} + 7 {M n O}_{4}^{-}      + 14 H^{+} = {F e}^{3 +} + 7 {M n}^{2 +}      + 7 H_{2} O$
violet colourless
So, phenolpthalein, methyl orange and litmus are all acid base indicators. They can't be used in this redox titration. ${K M n O}_{4}$ is a self-indicator changing from violet to colourless.
$∴$ Answer will be $D$ .

Suggest Corrections

Similar questions

Q. In the titration of

F e^{2 +}

ions versus

C r_{2} O_{7}

ions using diphenylamine as the internal indicator, phosphoric acid is added in the solution containing

F e^{2 +}

ions. The role of phosphoric acid is to?

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

Indicator in the above titration is :

Q. When

10 m L

of an aqueous solution of

F e^{2 +}

ions was titrated in the presence of dil

H_{2} S O_{4}

using diphenylamine indicator,

15 m L of

0.02 M

solution of

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

was required to get the end point. The molarity of the solution containing

F e^{2 +}

ions is

x \times 10^{- 2} M

. The value of

x

is ___. (Nearest integer)

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 :

Join BYJU'S Learning Program

A solution containing Fe2+ ions is titrated with KMnO4 solution. Indicator used will be:

A solution containing $F e^{2 +}$ ions is titrated with $K M n O_{4}$ solution. Indicator used will be: