How many mL of 0.0221 M KMnO4 would it take to titrate a 0.5255 g sample of K3Fe

Question

How many mL of 0.0221 M KMnO4 would it take to titrate a 0.5255 g sample of K3Fe(Ox)3? How could the results of your titration be used to determine if the sample was actually a mixture of K3Fe(Ox)3 and K4Fe(Ox)3

Explanation / Answer

reaction between oxalate ion and KMnO4 is :- 5C2O42- + 2MnO4- + 16H+ --------> 10CO2 + 2Mn2+ + 8H2O now no.of moles of K3Fe(Ox)3 = 0.5255/molecular mass of K3Fe(Ox)3 = 0.5255/437 = 1.203 X 10^-3 now as K3Fe(Ox)3 ---------> 3K+ + Fe3+ + 3OX2- so 1 mole of K3Fe(Ox)3 gives 3 mole of oxalate ions ... so 1.203 X 10^-3 moles of K3Fe(Ox)3 will give 3 X 1.203 X 10^-3 = 3.609 X 10^-3 moles of oxalate ions 5 moles of oxalate ions reacts with 2 moles of MnO4- so 3.609 X 10^-3 moles of oxalate ions will react with 2/5 X 3.609 X 10^-3 = 1.443 X 10^-3 moles of MnO4- and as KMnO4 ------> K+ + MnO4- so moles of KMnO4 = moles of MnO4- so we need 1.443 X 10^-3 moles of KMnO4 now you know molarity = no.of moles of KMnO4/volume in litres 0.0221 = 1.443 X 10^-3 / volume in litres volume in litres = 1.443 X 10^-3 / 0.0221 = 0.0653 L or 65.3 ml i am confused about your second questions .... feel free to ask any question

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