CHEM102 Lecture Notes - Lecture 16: Titration Curve, Equivalence Point, Dave Coverly

• School: University of Alberta
• Department: Chemistry
• Course: CHEM102
• Professor: Yoram Apelblat

Document Summary

You can"t go south, at least not before wednesday and our midterm exam . Solutions: no naoh, only weak acid ha. Ka = 1. 9 x 10 5 (cds) = x2. 1. 0 x = [h3o+] = ka [ha] = 1. 9 x 10 5 = 0. 00436 m ph = log10 (0. 00436) = 2. 36. 50. 00 ml of 1. 0 m acetic acid (ha) are titrated with. Find the ph after adding: a) 0. 00 ml, b) 10. 00 ml, c) 25. 00 ml and d) 50. 00 ml of naoh a c left for home. Methodology: a weak acid problem, treat it as a buffer use hh equation, treat is as a buffer use hh equation. Note that it is also a special case of halfway to the equivalence point. Ha + oh h2o + a na nha. [ha] = pka + log10 ph = pka + log10. 10 ml of 1. 0 m naoh is equivalent to 0. 01 mol oh which reacted with ha to form a .