If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.25 m solution of this acid?

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