Could someone help me to solve the problem?

In a beaker #2.5g# of #"Fe"# and dilute #"H"_2"SO"_4# is taken. The volume of this solution is #500ml#. In another beaker #30ml# of #0.015"M"# #"KMnO"_4# is taken. we mix #50ml# of the first solution.

(1) If #"HCl"# was used instead of #"H"_2"SO"_4#, could we measure the amount of iron accurately?

(2) Would the volume of #"K"_2"Cr"_2"O"_7# be equal to the volume of #"KMnO"_4#, if we used #"K"_2"Cr"_2"O"_7# instead of #"KMnO"_7#?

1 Answer
Ernest Z.
Nov 27, 2017

(1) Yes, you could still measure the amount of iron accurately.
(2) No, the volume of #"K"_2"Cr"_2"O"_7# would not be equal to the volume of #"KMnO"_4#,

Explanation:

(1) #"HCl"# vs. #"H"_2"SO"_4#

It doesn't matter if you use #"HCl"# or #"H"_2"SO"_4#.

Their only function is to provide the hydrogen ions to react with the iron.

#"Fe(s) + 2H"^"+""(aq)" → "Fe"^"2+""(aq)" + "H"_2"(g)"#

Both #"HCl"# and #"H"_2"SO"_4# are strong acids, so they will each work equally well.

(2) #"KMnO"_4# vs. #"K"_2"Cr"_2"O"_7#

#"MnO"_4^"-" + 8"H"^"+" + 7"e"^"-" → "Mn"^"2+" + 4"H"_2"O"#

#"Cr"_2"O"_7^"2-" + 14"H"^"+" + 6"e"^"-" → "2Cr"^"3+" + 7"H"_2"O"#

1 mol of #"KMnO"_4# will accept 7 mol of electrons, while
1 mol of #"K"_2"Cr"_2"O"_7# will accept only 6 mol of electrons.

If you are using equal molar concentrations of each, the volume of #"K"_2"Cr"_2"O"_7# needed to oxidize a given amount of #"Fe"^"2+"# will be greater than the volume of #"KMnO"_4#.

Related questions
Impact of this question
1125 views around the world
You can reuse this answer
Creative Commons License