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#
