CHEM 1062 Lecture Notes - Lecture 19: Equilibrium Constant
Document Summary
Chem 1062- lecture #19: autoionization of water and polyprotic acids. Kw = [h3o+] [oh-] = 1. 00 * 10-14 at 25oc, kw is the equilibrium constant of water. [ha]0-x x x x2/([ha]0-x)= ka x 2/[ha]0, this approximation is only justified if (x/[ha]0 ) 0. 05: get approximate value of x by allowing [ha]0-x, check if (x/[ha]0 ) 0. 05 to justify approximation. % dissociation= ((amount dissolved)/(initial concentration))*100% i. e. = x/[ha]0*100% Polyprotic acids are acids with more than one ionizable protons. H2asc will be shorthand for absorbic acid in the following example. Ka1>>ka2 therefore reaction one is not favorable because absorbic acid is electrically and losing a p+ is difficult and requires a significant amount of energy. First reaction dictates concentrations because it has the higher ka. Follow-up problem: assume reaction one alone dictates [h3o+] and [hasc. 7. 1*10-4/0. 05=0. 00014, which is much less than 0. 05 so the approximation is justified ph= -log[h3o+], ph = -log(7. 1*10-4) = 3. 15.
