Let's use acetic acid, CH_3COOH, as an example.
Weak acids, unlike strong ones, do not dissociate completely in an aqueous solution, which means that the final concentrations of H_3^+O and ,in this case, CH_3COO^- , the weak acid's conjugate base, must be calculated.
The chemical equation can be set up like this:
CH_3COOH(aq)<=>H_3^+O (aq) + CH_3COO^(-) (aq)
When dealing with acids in general, the value of the acid dissociation constant, K_a, is usually given. In this case, K_a is equal to 1.76 * 10^(-5) (m ol e)/L.
Even before doing any calculations, judging by the value of K_a, which in this case is <<1, one can see that the reaction will favor the reactants, with very little product being formed.
Now, the ICE method (I've added a wikipedia link) is the most appropriate method for this sort of calculations.
We will start with a 0.1 M concentration for the acetic acid.
CH_3COOH(aq) <=> H_3^+O(aq) + OH^(-) (aq)
I..........0.1 M..........................0...............0
C............-x........................+x.........+x
E......(0.1 - x)........................x............x
This sums up to:
I - initial - the concentrations of the substances present in the solution before the reaction; notice that H_3^+O and CH_3COO^- are both zero, since the reaction has not yet taken place.
C - consumed- this shows how the concentrations will change before an equilibrium is reached; notice that the acid's concentration should decrease by the same amount the products' concentrations increase;
E - equilibrium - this shows the concentrations after an equilibrium is reached.
Now, we know that
K_a = ([H_3^+O]*[CH_3COO^-])/([CH_3COOH]) = (x * x)/(0.1-x) = x^2/(0.1-x)
SInce K_a is <<1, we can approximate 0.1-x as 0.1, which leads to x^2 = 1.76 * 10^(-6) -> x= 1.3 * 10^(-3) M.
This is equal to the concentration of H_3^+O, therefore
pH = -log([H_3^+O]) = -log(1.3 * 10^(-3)) = 2.89
