What is the pH of an aqueous solution of a weakly acidic drug that is 40% dissociated and has a pKa =4.7?
Call the weak acid HA.
............HA ==> H^+ + A^-
initial.....X......0.......0
change...-0.4x...0.4x....0.4x
equil...0.6x....0.4x.....0.4x
Ka (H^+)(A^-)/(HA)
Substitute the equil line into the Ka exparession and solve for x, then (H^+) + 0.4x. Convert to pH = -log(H^+)
To find the pH of an aqueous solution of a weakly acidic drug, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the weak acid and the ratio of its conjugate base to the weak acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
pH is the measure of acidity/basicity of the solution,
pKa is the logarithmic scale for the acid dissociation constant,
[A-] is the concentration of the conjugate base,
[HA] is the concentration of the weak acid.
In this case, we know that the drug is weakly acidic and 40% dissociated. This means that 40% of the drug has dissociated into its conjugate base [A-], and the remaining 60% is still in the form of the weak acid [HA].
To find the concentrations of [A-] and [HA], we need to know the initial concentration of the drug. Let's assume the initial concentration of the drug is C.
Since we know that 40% of the drug has dissociated, the concentration of [A-] is 0.4C, and the concentration of [HA] is 0.6C.
Now, let's substitute the values into the Henderson-Hasselbalch equation:
pH = 4.7 + log (0.4C/0.6C)
Simplifying the equation:
pH = 4.7 + log (0.4/0.6)
Using logarithmic properties:
pH = 4.7 + log (2/3)
pH = 4.7 + log (2) - log (3)
Now, we can use a scientific calculator to find the value of log (2) and log (3):
pH ≈ 4.7 + 0.3010 - 0.4771
pH ≈ 4.5239
Therefore, the pH of the aqueous solution of the weakly acidic drug that is 40% dissociated, with a pKa of 4.7, is approximately 4.5239.