Question

Potassium acid oxalate $ {K_2}{C_2}{O_4}.3{H_2}{C_2}{O_4}.4{H_2}O $ can be oxidized by $ MnO_4^ - $ in acid medium. Calculate the volume in liters of $ 0.1M $ $ KMn{O_4} $ reacting in acid solution with $ 5.08gm $ of the acid oxalate in an acidic medium.

Answer 1

Hint: First of all, we know that one mole of the potassium acid oxalate contains four moles of the oxalate ions and two moles of permanganate ions oxidized five moles of oxalate ions. By using this information we are able to find the moles of the permanganate ion in the chemical process when the potassium acid oxalate is oxidized by potassium permanganate. Then we can be able to calculate the required volume.

Complete answer:
Let us first understand the given conditions and then we have the information that:
 $ {K_2}{C_2}{O_4}.3{H_2}{C_2}{O_4}.4{H_2}O + KMn{O_4} \to M{n^{ + 2}} + 2C{O_{_2}} $
 $ 1 $ Mole of the potassium acid oxalate contains $ 4 $ moles of oxalate ions thus the n-factor of oxalic acid is $ 4 $ and for $ KMn{O_4} $ is $ 5 $ .
Now, we know that, we have to equate the gram equivalents of $ KMn{O_4} $ and oxalic acid as
 $ {\text{Gram equivalent of }}KMn{O_4}{\text{ = gram equivalent of }}{K_2}{C_2}{O_4}.3{H_2}{C_2}{O_4}.4{H_2}O $
Therefore, from the above reaction and the concept of gram equivalent we have:
 $ {\text{n - factor}} \times {\text{molarity}} \times {\text{Volume = }}\dfrac{{{\text{weight}}}}{{{\text{molecular weight}}}} \times {\text{n - factor of }}{K_2}{C_2}{O_4}.3{H_2}{C_2}{O_4}.4{H_2}O $
By replacing the values from given data and some that we have observed in above equation we get,
 $ 5 \times 0.1 \times V = \dfrac{{5.08}}{{508}} \times {\text{n - factor}} $
 $ \therefore {\text{V = 0}}{\text{.02}} \times {\text{n - factor}} $
Overall n-factor of $ {K_2}{C_2}{O_4}.3{H_2}{C_2}{O_4}.4{H_2}O $ is $ 8 $ .
 $ \therefore {\text{V = 0}}{\text{.02}} \times 8 = 0.16 $
Thus the volume of the $ 0.1M $ $ KMn{O_4} $ reacting in acid solution with $ 5.08gm $ of the acid oxalate in acidic medium is $ 0.16L $ .

Note:
Here, we have used the formula for gram equivalent in two forms as above we have mentioned these formulas were obtained by conversions and also the relations between volume, molarity and n-factor as well as molecular weight, given weight and n-factor. So be careful while solving the problem and study the gram equivalent thoroughly from the study material you have.