Please remember these concepts:

1. Many processes can be at equilibrium.   But with changes in condition - concentration, temperature -  the system will no longer be at equilibrium and will adjust to try to get there again.
2. The equilibrium concentrations of H₃O⁺ and OH^- are vanishingly small in pure water. 
3. A weak acid or a weak base drug, in water, will disassociate to some extent.  The pH of the drug solution  will depend upon the pK_a.
4. Buffers stabilize pH.  This stabilized acidity determines the form of drug disassociation in systems.  The Henderson-Hasselbach equation conveniently handles drug ionization questions for buffered systems like the body.

Phenobarbital is available as the sodium salt, sodium phenobarbital. But we know that phenobarbital itself is an acidic drug, if, for no other reason than that sodium salts are salts of the strong base, NaOH. So, sodium phenobarbital is the conjugate base of the acidic drug, phenobarbital.

Sodium phenobarbital = [Na⁺][A^-]

Phenobarbital = [HA]

The first step is to obtain from the literature the physical/chemical characteristics of phenobarbital and its salt. This includes solubility information, as the solubility of both the ionized and unionized forms of the drug is important.

Let’s calculate the pH of a saturated solution of phenobarbital in the free acid form.

This is the unionized, undissociated form of the drug, which crosses the lipid membranes most readily. But, we will see, it is the least soluble form of the drug. If we were able to obtain the phenobarbital powder, we would find that at equilibrium, only one gram of the drug is soluble in 1000 mL of water at room temperature. (It is customary for solubility information to read "1 g/x mL solvent").  If we add less than 1g/1000ml of phenobarbitol to water, all the phenobarbitol dissolves. What we see is:

[HA]_solid  -->    [HA]_solution

If we add more than 1g/1000ml water of [HA], solution takes place as above.  But in solution, the reverse reaction is always going on to some extent:

[HA]_solution  -->    [HA]_solid

At eqiulibrium, the rate of solution and the rate of deposition are equal.   Under these conditions, the system seems to be still, serene and inactive - a layer of solid material underneath a solution.  And for phenobarbitol that solution at equilibrium contains 1g/100ml.  But the process of solution and precipitation goes on in this system at equilibrium:

[HA]_solid  <==>    [HA]_solution

Step 1: Determine the number of grams soluble in 1000 mL water. This data is available from the literature.

Step 2: Determine the number of moles from the mass in grams from grams from Step 1:

thus the MOLAR CONCENTRATION of a saturated solution, C_a, of phenobarbital in water appears to be 0.0043M.

Step 3: Determine the [H₃0⁺] of a saturated solution of phenobarbital at equilibrium using the k_a value from the literature.
Since the equilibrium equation is:

HA  +H₂O <==> H₃O⁺ + A^-

And we make the approximation that the concentration of water (approximately 55M) remains constant, we can include the constant value for water in the the constant, K_a:

K_a = [H₃O⁺][A^-]/[HA] = 3.19x10^-8

Now we make another approximation.  Since the ionization of [HA] is also very slight as indicated by the small value of pK_a therefore we can assume that the concentration of unionized [HA] at equilibrium is essentially unchanged from the calculated  concentration based on the amount dissolved.  Remember:  we are comparing a calculation based on a measurable phenomenon - the mass of material dissolved - with a calculated value, the concentration of unionized (undissociated) material at equilibrium.  So we now calculate  [H₃O⁺] and pH based on these assumptions.

Lastly,

And by checking the literature, we see that indeed, a saturated solution of phenobarbital in water yields a solution with a pH of 5, our calculations based on assumptions of molecular behavior and the validity of these equations matches closely the observed pH.

Something to remember:

• This calculation is for the drug in pure water, not buffered.