How to find the concentration of potassium permanganate in this experiment?

Question

I completed a titration experiment in a lab. The aim was to find the concentration of the titrant $\ce{KMnO4}$ solution by titrating it with a sodium oxalate solution of a known concentration.

I first made a standard solution of sodium oxalate solution, using $\pu{0.826 g}$ of sodium oxalate powder, then poured $\pu{50ml}$ of distilled water along with $15$-$\pu{20ml}$ of $\pu{3 M} \,\, \ce{H2SO4}$. Then I pipetted out $\pu{25 ml}$ of the solution and poured it in to a volumetric flask. Then I filled the rest of the volumetric flask with distilled water, to form the standard solution. I then did the Mcbride experiment to find $75\%$ of the volume of $\ce{KMnO4}$ need to titrate the solution.

I then did the experiment three times to get an average volume of $\ce{KMnO4}$ to be $\pu{25.48 ml}$. And the volume of sodium oxalate solution is $\pu{25 ml}$.

I should find the concentration of $\ce{KMnO4}$ now.

I know the molar ratio is 2:5. How do I calculate this?

I know I need to find the moles of sodium oxalate then I multiply it by $\frac{2}{5}$ and then divide that value by the volume of $\ce{KMnO4}$ which is $\pu{25.48 ml}$. But how do I calculate the moles of sodium oxalate? What would the concentration be? I know the volume is $\ce{25 ml}$.

This is what I have done, but I am not sure if it is correct:

Moles of sodium oxalate = $\frac{\pu{0.8265 g}}{\pu{133.96 g mol-1}} = \pu{0.006169752 mol}$

$\frac{0.006169752}{\pu{0.025L}}$ = concentration of sodium oxalate is therefore $0.246790086$, $0.0248$ to 4 s.f.
($\pu{0.025 L}$ is the volume of the sodium oxalate)

I then multiplied that by $\frac25$. $0.0248 \cdot \frac25$ gives me the concentration of $\ce{KMnO4}$ which is $0.09872$.

But I think this is wrong because I have not used the volume of $\ce{KMnO4}$ used, which was the whole point of the experiment.

Answer 1

I think your experiment has one setup mistake.:

You're mixing $\pu{0.8265 g}$ of oxalate with $\pu{50 mL}$ of water and $15$-$\pu{20 mL}$ of sulfuric acid. So your $\pu{25 mL}$ oxalate solution does not contain $\pu{0.8265 g}$ of oxalate (more or less $\frac{25 \cdot 0.8265}{ 65-\pu{70 mL}}$). So you can't know the exact oxalate concentration of this solution.

I recommend you to use a $\pu{100 mL}$ volumetric flask to prepare the oxalate solution. You can then calculate its concentration and the amount of oxalate you have in $\pu{25 mL}$.