Experiment Full leport Oxidation-Reduction Titrations I: 54n ustawa\"; w\'?t ,a

Question

Experiment Full leport Oxidation-Reduction Titrations I: 54n ustawa"; w'?t ,a (44") Determination | of Oxalate To gain familiarity with redos chemistry through analysis of an oxalatl OBJ ECTIVIE Apparatus 50 mL. buret buret clamp 400 mL. beakers (3) glass stirring rods ring stand balance weighing botle APPARATUS AND CHEMICALS Bunsen burner and hose Wash bottle Chemicals an oxalate sample 10 M H,SO Na,C,O, (primary standard) -0.02 M KMno Potassium permanganate reacts with oxalate lons to produce carbon dioxide DISCUSSION and water in an acidic solution, and the permanganate ion is reduced to man- ganese(Il) as follows: Because this reaction proceeds slowly at room temperature, the solution must be heated gently to ensure that satisfactory reaction rates are obtained. No indicators are necessary in permanganate titrations because the end served. The permanganate ion is intensely purple, whereas the manganesetli is nearly colorless. The first slight excess of permanganate imparts a pink color to the solution, signaling that all of the oxalate has been consumed points are easily ob- i h) ion Mlmules The balanced half-reactions associated with Equation [I] are as follows: Oxidation: Reduction: Mno,-(aq) +8H+(aq)+5e--Mn2+(aq1+4H,0? 131 In this experiment, you will standardize a KMnO solution-that is, Hu of sodium oxalate, Naf2o," You will then use your standardized KMO, to determine the percentage of oxalate ion, o, in an unknown sample will determine its exact molarity-by titrating it against a very pure sample A you ak From Experiment 37 of Laboratory Experiment, erisn The Cemi Science, Thn John H. Nelson, Kenneth C Kemp, Mhew Soitzies,and Mikhet Latso Copyri neit Edacation, Inc. All rights neserved Thineendh Editinwih e uxd 0195

Explanation / Answer

Dear student

C2O42- = 2 CO2 + 2e-
MnO4- + 8 H+ + 5e- = Mn2+ + 4 H2O
5 C2O42- + 2 MnO4- + 8 H+ = 10 CO2 + 2 Mn2+ + 8 H2O
0.4586?? ????2??2??4 × ( 1 ?????? ????2??2??4/134.0 ?? ????2??2??4 ) × ( 2 ?????? ????????4/5 ?????? ????2??2??4 ) = 0.001369 ?????? ????????4

33.67 ???? × ( 1 ?? /1000 ????) = 0.03367 ??

??olarity = ?????? /?? = 0.001369 ?????? ????????4/ 0.03367 ?? = 0.04065 M

OR

in short

moles of sodium oxalate = 0.4586/ 134 = 0.003422

C2O42- = 2 CO2 + 2e-
MnO4- + 8 H+ + 5e- = Mn2+ + 4 H2O
5 C2O42- + 2 MnO4- + 8 H+ = 10 CO2 + 2 Mn2+ + 8 H2O

moles MnO4- = 0.003422 x 2 /5=0.0013688

molarity(M )= 0.0013688 / 0.03367L

=0.04065 M

Hope it helps.

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