what approximate mass of potassium permanganate is required to make 400 mL of 0.
ID: 630625 • Letter: W
Question
what approximate mass of potassium permanganate is required to make 400 mL of 0.02 M solution? Suppose that you weighted out approximately this amount of potassium permanganate and dissolved it in distilled water to make 400 mL of solution. 0.1038 grams of pure dry sodium oxalate was placed in Erienmeyer flask and dissolved in .75 M sulfuric acid. This acidic solution of sodium oxalate was titrated against the potassium permanganate solution. The volume of potassium permanganate required to reach the end point was 12.31 mL. 1) How many sodium oxalate were used in the titration? 2) Calculate the equivalent number of moles of potassium permanganate? 3) calculate the clarity of the potassium permanganate solution?Explanation / Answer
In acid solution, sodium oxalate exists predominately in the form of undissociated oxalic acid molecules. The balanced equation for the reaction with permanganate is-
2 MnO4^-1 + 5 H2C2O4 + 6 H^+1 --> 2 Mn^+2 + 10 CO2 + 8 H2O
7.552x10^-4 moles of Na2C2O4 reacts with 3.021x10^-4 moles KMnO4 (7.552x10^-4 x 2/5 = 3.021x10^-4)
3.021x10^-4 moles KMnO4 / 0.01285 L = 2.35x10^-2 M KMnO4
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