OK this one has stumped me?

Question

A 24.00 mL portion of a solution that contains both Fe+2 (aq) ions and Fe+3 (aq) ions had acid added to it and was then reacted with potassium manganate (VII) solution. 14 mL of a .0200 M manganate (VII) solution was used. Another experiment involved a different 24 mL portion being acidified, reduced (converting all Fe+3 to Fe+2) and then reacted with a new sample of the same manganate (VII) solution. 18.73 mL of the manganate (VII) solution were required. Calculate the concentrations of the Fe+2 and the Fe+3 in the original solution.

this is a little wordy, but I think I'm just a little confused. I tried it myself, but it doesn't make sense. I got the number of moles of the potassium manganate which i found out to be KMnO4, which is .00028 moles for the 1st experiment, and .0003746 for the 2nd one.

After that, I really dont know exactly what to do. Basically I need to find a good AP chemistry book, Im learning this from scratch...

Anyways, thanks, I appreciate it.

Answer 1

KMnO4 + 5Fe^+2 + 8H^+1 --> 5Fe^+3 + Mn^+2 + 4H2O

Fe+2 is losing 1 electron going to Fe+3, and Mn is gaining 5 electrons going from +7 to +2. 1 mole of KMnO4 = 5 molesd of Fe^+2 to balance the electron gain/loss.

ORIGINAL SOLUTION-

0.014 L x 0.0200 M = 2.8x10^-4 moles KMnO4

The 24.00 mL portion of Fe^+2/Fe^+3  contained 5x2.8x10^-4 moles of Fe^+2

0.0014 mole Fe^+2/0.024 L = 0.0583 M Fe^+2

REDUCED SOLUTION-

0.01873 L x 0.0200 M = 3.746x10^-4 mole KMnO4

5x3.746x10^-4 = 0.001873 mole Fe^+2 (including Fe^+3 that has been reduced to Fe^+2)

0.001873 - 0.001400 = 0.000473 mole Fe^+3 in 25.00 mL

0.000473/0.025 = 0.01892 M Fe^+3

Original solution contained 0.0583 mol/L Fe^+2 and 0.0189 mol/L Fe^+3