CHEM 1102 Lecture Notes - Lecture 3: Sodium Hydroxide, Acid Dissociation Constant, Sbml

• School: Boston College
• Department: Chemistry
• Course: CHEM 1102
• Professor: Daniel Fox

Document Summary

Experiment 17 calculations: a) see attached graphs, 0. 4979 m naoh 0. 00991 l naoh =4. 93 10 3 moles of naoh. 4. 93 10 3 moles of naoh =4. 93 10 3 moles of hc l. 0. 01 l hcl+0. 075 l h 2o+0. 00991 l naoh. =0. 052 m hcl: a) see attached graphs, 0. 4979 m naoh 0. 01658 l naoh =8. 26 10 3moles of naoh. 8. 26 10 3 moles of naoh =8. 26 10 3 molesof c h 3 cooh. =0. 081 m c h 3cooh: mass of vinegar=10. 00 mlvinegar 1. 008 g ml. Mass of c h 3 cooh=0. 081 m c h 3 cooh 0. 10158 lsolution . K a=10 4. 9=1. 26 10 5: ch3cooh (aq) + h2o (l) ch3coo (aq) + h3o+(aq) H 3o x=1. 19 10 3 m= ph= log ( 1. 19 10 3 m =2. 92. 0. 4979 m naoh 0. 01658 l naoh =8. 26 10 3moles of naoh (aq) + ch3cooh (aq) h2o (l) + ch3coo (aq) 8. 26 x 10-3 mol 8. 26 x 10-3 mol 0.