Synthesis of potassium trioxalatoferrate (volumetric techniques and KMnO4 titrat
ID: 1052956 • Letter: S
Question
Synthesis of potassium trioxalatoferrate (volumetric techniques and KMnO4 titration)
Work the following problems:
a. suppose you have as much 0.03200 M KMnO4 as you need. State exactly what data must be recorded in order to analyze a solid of potassium trioxalatoferrate for both iron and oxalate.
b. Describe the titration endpoint:
c. State the critetion for reading a buret filled with KMnO4:
d. A solution is 0.020 M in KMnO4. A student asked to prepare 500.0 mL of Na2C2O4 solution, 25.00 mL aliquots of which will be used to standardize the KMnO4. If approximately 33 mL of KMnO4 are to be used in each titration, describe exactly how you would prepare the sodium oxalate solution. Show all calculations.
(M.W.: Na2C2O4= 134.0)
Explanation / Answer
a) we will start adding potassium permangnate in the solution of oxalte ion until we get the end point (the point of completion of reaction). Now we can calcualte the moles of KMnO4 required for completion of reaction as
Moles = Molarity X volume
From stoichiometry of reaction we can calculate the moles of oxalte ion reacted. And hence we can calculate % of oxalate ion present in the sample. This can be compared with the expected mass % from chemical formula.
b) KMnO4 acts as self indicator. The end point is change in colour from magenta to light pink.
c) As KMnO4 is a coloured solution, we will read upper meniscus in the burette.
d) The balanced equation is
2MnO4- + 16H+ + 5C2O4-2 --> 2MN+2 + 10CO2 + 8H2O
So here, two moles of Permangnate will react with 5moles of sodium oxalate
Moles of KMnO4 used = Molarity x volume = 0.02 X 33 / 1000 = 0.00066 moles
Moles of sodium oxalate reacted = 5 X moles of KmnO4 used / 2 = 0.00165 moles
These moles will be present in 25mL of solution
So molarity of solution = Moles / Volume = 0.00165 X 1000 / 25 =0.066 M
We need to prepare 500mL of solution
Molarity = Mass / Molecular weight X volume in litres
M.W.: Na2C2O4= 134.0
0.066 = Mass / 134 X 0.5
Mass of sodium oxalate required = 4.422 grams , this will be dissolved in 0.5 L of water
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