Titration problem????

Question

Ok so i have to do this titration problem and its really complicated!! If you could help me on any part of this, it will be good help!

A 50 ml solution of oxalic acid (H2C2O4) requires 24.35 ml of 0.125 M NaOH to neutralize both protons. It takes 15.83 ml of KMnO4 (Product Mn^+2) to react with all the oxalic acid in another 50 ml sample of oxalic acid (product CO2). A 2.5 g sample containing Iron II chloride is dissolbed in acid and titrated with 31.87 mL of the KMnO4 solution.

a) write three balanced equations for the reactions
b) Whcih compounds are the reducing agents and which are the oxidizing agents?
c) What is the concentration of thte permanganate solution in this problem?
d) what is the percentage of iron in the solid sample?


If could get the 3 balanced reactions i could try to work it out.
the first one i got H2C2O4+2NaOH-->2H2O + Na2C2O4
the next one is KMnO4+H2C2O4--> ?? is it CO2 + Mn^(+2)?? idk
last 1 : FeCl2+ KMnO4 --> Fe(MnO4)2 + KCl
are these right?

Ok so I got that reaction. For the last one, is that also a redox reaction? I got FeCl2 + KMnO4---> Fe(MnO4)3 +KCl?
fe is oxidized, and Mn is reduced?
how would i write these half reactions? Fe goes from +2 to +3? and then Mn goes from +7 to +2? is this right? i dont know why it changes.

Answer 1

a)The first reaction is an acid-base reaction and you have it correct
The other two are redox reactions. It is very important to know that in both cases you have acidic environment and thus the product of the Mn(+7) reduction is Mn(+2). In alkaline it would have been Mn(+4)

2KMnO4 + 5H2C2O4+ 6H(+) -> 2Mn(+2) + 10CO2 + 8H2O+ 2K(+)

KMnO4 + 5Fe(+2) + 8H(+) -> Mn(+2) + 5Fe(+3) +K(+) +4H2O

b)Oxidizng agents are the ones that oxidize others and thus are themselves reduced. In your case KMnO4 is the oxidizing agent and H2C2O4 and FeCl2 the reducing agents

c & d) For the titration of substance 2 with substance 1 you have
greq1=greq2

where greq= a* mole

a depends on the reaction.

For acid-base reactions it is the number of H+ that an acid molecule (or OH- for bases) will give in the reaction. In your case, NaOH gives one OH- so a(base)=1 and H2C2O4 two H+ so a(acid)=2

Therefore greq(acid)= greq(base) =>
a(acid)*M(acid)* V(acid)= a(base) * M(base)* V(base) =>
2* M(acid) * 50= 1* 0.125* 24.35 =>
M(acid)= 0.0304

For a redox reaction a is the number of electrons per molecule of oxidizing or reducing agent that will be given or gained during the reaction.

For KMnO4 you go from +7 to +2 which means gaining 5e and a =5
For H2C2O4 you go from +3 to +4 (for carbon) but you have 2 atoms of carbon per molecule and therefore you lose 1*2=2e per molecule and a=2
For FeCl2 you go from +2 to +3 so you lose 1e and a=1

So for redox reactions you have greq(oxidizing) =greq(reducing)
For the first reaction

a(KMnO4)* M(KMnO4)* V(KMnO4)= a(H2C2O4)* M(H2C2O4)* V(H2C2O4) =>

5*M*15.83= 2* 0.0304* 50 =>
M(KMnO4)= 0.0384

For the second:

a(KMnO4)* M(KMnO4)* V(KMnO4)= a(FeCl2)* mole(FeCl2)

but also mole =mass/MW so

5*0.0384*31.87*(10^-3)= 1*mass (FeCl2)/126.84 =>
mass (FeCl2)=0.776 grams

so 0.776/2.5= 31% of the sample is FeCl2

or for just Fe you have

5*0.0384*31.87*(10^-3)= 1*mass (Fe)/55.84 =>
mass Fe=0.342 grams
and 0.342/2.5= 13.7% Fe

Note: For the first two titrations you didn't need to convert ml into liters since the conversion factor is simplified in the equation (you have volume on both sides of the equation). For the third you have volume only on one side and therefore you need to do the conversion to have the correct figures and units.

Answer 2

You have correctly balanced the first reaction. The next two are somewhat trickier because they are redox reactions:

b) Here is how you start-

MnO4^- ---> Mn^2+ (Notice that i have left out spectator ion, K^+0

Now to balance O, turn it into water:

MnO4^- ---> Mn^2+ + 4H2O

Now balance H with H^+ on the LHS

8H^+ + MnO4^- ---> Mn^2+ + 4H2O

Now its balanced for mass but not charge. Balance with electrons:

5e^- + 8H^+ + MnO4^- ---> Mn^2+ + 4H2O

To balance oxidation of oxalate;

C2O4^2- ---> 2CO2 + 2e^-

Before you can add this equation to the one above, you need to multiply one by 2 and the other by 5 so that they both have the same number of electrons.

If you need more help than that, please email me or re post.