Question

A certain weak acid, HA, has a Ka value of 1.8×10−7.

Part A
Calculate the percent dissociation of HA in a 0.10 M solution.

Answer 1

Formula for Equilibrium: HA <---> H+ + A-

To find Ka = [H+][A-]/[HA]
[HA] = 0.10,
[H+] = [A-] = x

Solve for Ka

Ka = 1.8X10^-7 = x^2 / 0.10
x = 1.3X10^-4 M = [A-]

% Dissociation =
([A-]/[HA]) X 100 = (1.3X10^-4 / 0.10) X 100 = 0.13% ionized

Answer is 0.13%.