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CHM1 20Example3 Collection

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Oct 18, 2019, 1:17:50 AM
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Oct 18, 2019, 1:17:50 AM
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Example 3

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Solution At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of KMnO4 which must be added:

/attachments/1a01a54b-f145-11e9-8682-bc764e2038f2/0e7ed8b4a802e9e6e479d5378af9e736.png
The amount of H2O2 is obtained from the volume and concentration:
/attachments/1a01a54b-f145-11e9-8682-bc764e2038f2/2dc2f6d1e6e60b3e8eebf77093cbc053.png
Then
/attachments/1a01a54b-f145-11e9-8682-bc764e2038f2/f1fcc70488e0851f480d91c44ece7daa.png
/attachments/1a01a54b-f145-11e9-8682-bc764e2038f2/916912ffa433ec75f3bf3768d3f08e2d.png
= 1.272mmol KMnO4
To obtain VKMnO4(aq) we use the concentration as a conversion factor:
/attachments/1a01a54b-f145-11e9-8682-bc764e2038f2/cdf229c003921c6746081b20a2eb9de8.png
= 23.62 cm3
Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4

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