MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Oxides of Sulphur

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 $F e^{3 +}$ .
${M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
Indicator in the above titration is :

A

phenolphthalein

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

B

methylene blue

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

C

methyl orange

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

D

K M n O_{4}

(self-indicator)

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

Open in App

Solution

The correct option is D $K M n O_{4}$ (self-indicator)
$K M n O_{4}$ acts as a self indicator. It has an intense purple color.

Hence, when slight excess is present, the color changes to pink. This gives the endpoint.

Suggest Corrections

thumbs-up

0

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

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.

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. 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.

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Oxides of Sulphur

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Oxides of Sulphur

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon