Phenobarbital Sodium has a pKa of 7.2 at pH=9.3. It has a solubility 1g/L. What is its solubility at pH 7.4?
To determine the solubility of Phenobarbital Sodium (PB) at pH 7.4, we need to consider its pH-dependent solubility behavior. The pH of a solution affects the ionization (dissociation) of PB, which in turn impacts its solubility.
PB is a weak acid with a pKa of 7.2. This means that at pH values below 7.2, PB predominantly exists in its undissociated, neutral form, while at pH values above 7.2, it exists mainly in its dissociated, ionized form.
First, we need to determine the degree of ionization of PB at pH 9.3 using the Henderson-Hasselbalch equation, which relates the pH, pKa, and concentration of an acid:
pH = pKa + log([A-]/[HA])
Where:
pH = 9.3
pKa = 7.2
[A-] = ionized form concentration (solubility)
[HA] = undissociated form concentration
Rearranging the equation to solve for [A-]:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(9.3 - 7.2)
[A-]/[HA] = 10^2.1
[A-]/[HA] = 125.89
This means that at pH 9.3, PB is approximately 125.89 times more soluble in its ionized form than in its undissociated form.
Now, let's determine the solubility at pH 7.4 using the same equation:
pH = pKa + log([A-]/[HA])
Where:
pH = 7.4
pKa = 7.2
[A-] = solubility at pH 7.4
[HA] = undissociated form concentration (solubility)
Rearranging the equation:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(7.4 - 7.2)
[A-]/[HA] = 10^0.2
[A-]/[HA] = 1.58
This means that at pH 7.4, PB is 1.58 times more soluble in its ionized form than in its undissociated form.
Since we know that at pH 9.3, the solubility of PB is 1g/L, we can calculate the solubility at pH 7.4 by multiplying it by the ionization ratio:
Solubility at pH 7.4 = Solubility at pH 9.3 x [A-]/[HA]
Solubility at pH 7.4 = 1g/L x 1.58
Solubility at pH 7.4 ≈ 1.58g/L
Therefore, the estimated solubility of Phenobarbital Sodium at pH 7.4 is approximately 1.58g/L.