The pH of a 0.100 M solution of an aqueous weak acid (HA) is 4.00. What would the Ka for the weak acid be?
pH = 4.0; (H^+) = 1E-4
............HA ==> H^+ + A^-
initial....0.1......0.....0
change.....-x........x.....x
equil....0.1-x.......x.....x
x = 1E-4
Ka = (H^+)(A^-)/(HA)
Substitute from the ICE chart and solve for Ka.