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A solution is 0.020 M in KMnO4. A student is asked to prepare 500.00 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)
THANK YOU in advance!!
1 Answer
- Roger the MoleLv 73 years agoFavorite Answer
2 KMnO4 + 5 Na2C2O4 + 16 H{+} → 2 Mn{2+} + 8 H2O + 10 CO2 + 2 K{+} + 10 Na{+}
(0.020 M KMnO4) x (33 mL KMnO4) x (5 mol Na2C2O4 / 2 mol KMnO4) / (25.00 mL) =
0.066 M Na2C2O4
(0.50000 L Na2C2O4) x (0.066 mol/L Na2C2O4) x (134.0 g Na2C2O4/mol) = 4.422 g Na2C2O4
So weigh out 4.422 g Na2C2O4, dissolve it in some water, then dilute it to exactly 500 mL.