wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
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 Fe2+ and Fe3+ ions is titrated with 25.0 mL of 0.02 M KMnO4 (in dilute H2SO4). As a result, all of the Fe2+ ions are oxidised to Fe3+ ions. Next 25 mL of the original solution is treated with Zn metal. Finally, the solution requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3+.

MnO4+5Fe2++8H+Mn2++5Fe3++4H2O
The molar concentration of Fe2+ in the original solution is :

A
0.01M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.02M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.10M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.20M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 0.10M
Here, Fe2+ is oxidised and MnO4is reduced.
Both have the same equivalents.
Equivalent of Fe2+= Equivalent of MnO4

25×M=25×0.02×5

M=0.10M

Hence, molar concentration of Fe2+ is 0.1 M.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Oxides of Group 16
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon