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A 49.96 mL sample containing La3+ was treated with sodium oxalate to precipitate La2(C2O4)3, which was...

A 49.96 mL sample containing La3+ was treated with sodium oxalate to precipitate La2(C2O4)3, which was washed, dissolved in acid, and titrated with 18.04 mL of 0.006363 MKMnO4. Calculate the molarity of La3+ in the unknown.
(Hint: the reaction involved in the titration was: permanganate ion(s) + oxalate (ions) -----> manganese(II) ions + carbon dioxide. The reaction occurred in acidic aqueous solution)

Solutions

Expert Solution

Reaction,

2MnO4- + 5C2O4^2- + 16H+ ---> 2Mn2+ + 10CO2 + 8H2O

moles of KMnO4 used = 0.006363 M x 18.04 ml

                                    = 0.1148 mmol

moles of C2O4^2- reacted = 0.1148 x 5/2

                                           = 0.287 mmol

moles of La3+ present = 0.287 x 2/3

                                     = 0.1913 mmol

molarity of La3+ in the unknown = 0.1913/49.96

                                                    = 0.00383 M

queen_honey_blossom answered 1 year ago
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