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.

Now we have seen the calculations for a solution of the free acid. What then, is done if there is a solution of a salt, as is the case when drugs dissolve in the process of drug administration and absorption?

Let’s take a look at a solution of sodium phenobarbitol. Again, the first step is to obtain all of the necessary information. Using, in addition to the information on phenobarbitol, the free acid, the solubility of sodium phenobarbitol, the conjugate base. 

Let’s review what is happening. If we dissolve a salt sodium phenobarbitol, in sufficient water that all of the solute dissolves, we will have sodium ions and phenobarbitol ions distributed freely throughout. Since phenobarbitol is a univalent species, in the solid there are an equal number of anions (phenobarbitol ions) and cations (sodium ions). But the salt that was dissolved was a base. (a conjugate base, but a base nonetheless! Look again at the Physical Chemical Properties of Phenobarbital to review the pH of a saturated solution of sodium phenobarbitol.) So the equations that we will choose are the equations for a BASE.

Since the ionized form is the same, the dissociation constant for the free acid, k_a, This is the basis for the calculation of dissociation constant for the conjugate base, k_b. Remember that the literature states that the pk_a of phenobarbitol is 7.41, and the Ka is

 , so

AND

In the same manner as for the free acid, we will now calculate the pH of a saturated solution of sodium phenobarbitol. Again, from the literature, we determine that 1 gram of the salt is soluble in 10 mL of water, but is practically insoluble in ether and chloroform. It is somewhat soluble in ethanol. So, to proceed:

Step 1: Determine the grams of sodium phenobarbitol soluble in a liter of water:

Step 2: Determine the number of moles that the weight determined in step 1 represents:

Step 3: Determine the [OH-] concentration from the equation for a base:

Step 4: Convert the [OH-] to pOH:

Step 5: Convert pOH to pH:

And, lo and behold, this is indeed the pH of a saturated solution of sodium phenobarbitol, as we found in the literature.

Something to remember:

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