A certain weak acid, HA, has a Ka value of 8.2×10−7.Calculate the percent dissociation of HA in a 0.10 M sol?

HA <=> H+ + A-
start
0.10
change
-x. . . . . +x .. . +x
at equilibrium
0.10-x. ..x . .. .x

Ka = 8.2 x 10^-7 = [H+][A-]/ [Ha]= x^2/ 0.10-x
x = 2.9 x10^-4 M = [H+]=[A-]

% dissociation = 2.9 x 10^-4 x 100/0.10=0.29


in the same way

8.2 x 10^-7 = x^2/ 0.01-x
x = 9.1 x 10^-5

% dissociation = 9.1 x 10^-5 x 100/0.1=0.091

**MISTAKE: Second answer should be: % dissociation = 9.1*10^-5x100/0.01= .91%

Strong - Over coming challenges or not giving up. Weak - Giving in/giving up.