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Department
Biology (Biological Sciences)
Course
BIOL499A
Professor
Blaine Mullins
Semester
Winter

Description
The Flying McCoys by Glenn and Gary McCoy Read 18.5, 18.7, 19.1, 19.2 Practice S3 I reviewed the exam during class – solutions will be posted later in the week. Learn from your mistakes. Start by picking your exam from my office W4-13B (while I’m there). I will not bring the exams to class. 1 2 Dissociation of a weak acid in water You must check your assumption Find the equil. conc. in a solution of 1.0 M HCOOH. ? If calc – appro x 100 % 5% then valid HCOOH + H O  H O + + HCOO –  calc  < (aq) 2 (l 3 (aq) (aq) Or generically In our case 1 – x = 0.9859 M  + – 0.9859 – 1.000 AH (aq) H 2 (l) H 3 (aq) A (aq)  x 100 % = 1.4% < 5%  0.9859  I 1.0 0 0 M C – x + x + x E 1.0 – x x x Approximation acceptable and easier. + – [H 3 ][A ] –4 Remember an approximation is valid when it is less K = [HA] = 2.0 x 10 (CDS T2) than 5 %. x 2 K = = 2.0 x 10–4 1.0 – x Solve assuming 1.0 – x  1.0 x2 = 2.0 x 10–4  x = 0.0141 M 1 3 4 What is the percent dissociation ? For a weak acid % dissociation increases as the amount dissociated Percent dissociation = x 100% acidluted initial concentration HA+ H O 2  H 3 + + A  0.0141 M % dissociation = 1.0 M 100% = 1.4% Increasing the dilution (by adding water) will shift the equilibrium towards the products (Le Chatelier) which Note that the percent dissociation is very similar to the means greater dissociation. error we calculated in our check. Therefore calculating percent dissociation is a valid check for acids and bases provided it is less than 5%. For home. Check for yourself with an initial concentration [HCOOH] = 0.5 M 5 6 Left for home. Dissociation of a weak base in water When [HCOOH] = 0.o M Find equil. concentrations in 15.0 M N3 .  + –  + – AH (aq) H2O (l)  H3O (aq) A (aq) HN 3 (aq) H2O (l) NH 4 (aq) OH (aq) I 0.5 0 0 M I 15.0 0 0 M C – x + x + x
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