##### Question

# Back Titration

0.60g of impure K2S2O8 (270.32 g/mol) was dissolved in 50.0 mL solution and treated with an excess of 50.00 mL of 0.0701 M Na2C2O4.

The excess oxalate required 21.06 mL of 0.02880 M KMnO4 to reach the end point.

a) What is the weight percent of K2S2O8 in the impure reagent (three sig. figs)? % by mass

Remember, the end point is only an estimate of the equivalence point. To obtain more accurate results, a blank is often analyzed to determine the difference betweenthe equivalence point and the end point. In the above example, if the end point occurred 0.8 mL after the equivalence point of the back titration, explain how theblank correction would affect the various steps of the calculation.

(Increase/Decrease)

The calculated volume of KMnO4 required to reach equivalence point would ____

The calculated # moles of KMnO4 required to reach equivalence point would _____

The calculated # excess moles of Na2C2O4 at the equivalence point would _____

The calculated # moles Na2C2O4 reacted with K2S2O8 would _____

The calculated # moles K2S2O8 in the sample would _____

The calculated percent mass K2S2O8 in the sample would _____