Problem Set 3 Solutions

Problem Set 3 Solutions - C hemistry 2101 I ntroductory...

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Chemistry 2101 Introductory Analytical Chemistry Fall 2009 Solutions Problem Set 3 Chapter 6 Problem 6-1 The equivalence point occurs when the exact stoichiometric quantities of reagents have been mixed. The end point is marked by a sudden change in a physical property brought about by the disappearance or a reactant or appearance of a product. Ideally, the equivalence point is very close to the end point. However, only the end point can be observed directly. A blank titration is used to extrapolate from the end point to the equivalence point. Problem 6-2 108.0 mL of 0.1650 M oxalic acid is equal to (0.1080 L)(0.1650 M) = 0.01782 mol oxalic acid. One mole of oxalic acid reacts with 2/5 moles permanganate. Thus, (2/5) (0.01782 mol) = 7.128 10 -3 mol permanganate are needed. (7.128 10 -3 mol) / (0.1650 mol / L) = 0.04320 L = 43.20 mL of KMnO 4 . Second part of question: 43.20 mL of KMnO 4 were needed for 108.0 mL of oxalic acid. Therefore, (108 mL / 43.20 mL) 108.0 mL = 270 mL will be needed for 108 mL oxalic acid. Problem 6-3
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Problem Set 3 Solutions - C hemistry 2101 I ntroductory...

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