How to find moles titrated

Question

These are the two questions in my lab:

3. How many moles of Fe2+ ions and MnO4 – ions were titrated in each Part 1 trial?

4. How many moles of oxalic acid, H2C2O4 were titrated in each Part 2 trial?

I have the following information: 25mL of H2C2O2 was used and it has a concentration of .025M

10mL of Fe2+ was used and it has a concentration of .100M

MnO4 has a concentration of .0194M but the mL is not given because that's what was being measured in titration. How would I find the answers to 3 and 4 with the information I have?

Answer 1

Hello, Hailey,

1. How many moles of Fe2+ ions and MnO4 – ions were titrated in each Part 1 trial?

We need a balanced equation for the titration. Since Fe is shown with an oxidation state of +2 and the manganate ion is -1, the result is Fe(MnO₄)₂. That means we need two moles of manganate for every one mole of iron.

Fe²⁺ + 2MnO₄^1- = Fe(MnO₄)₂

Let's find out how many moles of iron were used. You state 10mL of Fe²⁺ of 0.100M was the amount required. Moles is defined as moles/liter. 0.100 M means we have 0.100 moles in every liter. If we used 10ml, find the number of actual moles by multiplying the concentration (moles/liter) times the volume, which needs to be converted from ml to liters:

(10ml)*(1 liter/1000 ml)*(0.100 moles/liter) = 0.00100 moles Fe²⁺

Since every mole of Fe²⁺ requires 2 moles of MnO₄^1-, we titrated 0.002 moles of MnO₄^1-

If the concentration of MnO₄^1- was 0.0194M, and we need 0.002 moles, use the relationship M*V = moles. (Concentration in Molar, moles/Liter)*(Volume in Liters) = moles.

(0.0194M)*(V) = 0.002 moles.

Solve for V, volume

V = (0.002 moles)/(0.0194 moles/liter) = 0.10309 Liters

T=You can convert liters into ml by dividing by 1000 ml/liter.

103.09 ml was used.

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1. How many moles of oxalic acid, H2C2O4 were titrated in each Part 2 trial?

Again, balance the equation and go through the same steps as above.

I hope this helps,

Bob