The first step is to write the balanced equation of the dissociation of the monoprotic weak acid. This can be assumed to have 1:1 stoichiometric relationships, so the equation and ICE chart are given by:
HA ---> H^+ + A^-
initial 0.25 0 0
change -x x x
equilibrium 0.25-x x x
This gives an expression for Ka given by:
Ka = (x)^2/(0.25-x) = 5.4x10^-6
x = 1.159x10^-3 = [H+] = [A-]
pH = -log[H+] = -log[1.159x10^-3] = 2.936
