1. A weakly acidic drug molecule has a pKa of 8. At what pH will this drug be 35
ID: 484804 • Letter: 1
Question
1. A weakly acidic drug molecule has a pKa of 8. At what pH will this drug be 35% ionized? Report to one decimal place.
2. Ibuprofen is an acidic drug with a pKa of 4.9 and a formula weight of 206.29g/mol. What will be the pH of a solution in which 22.8 mg is dissolved in 118.6 mL? Report to two decimal places.
3. Diazepam is a weakly basic drug with a pKa of 3.4. What will be the pH of a solution of diazepam prepared close to the solubility limit, 46.6 ppm? The formula weight of diazepam is 284.75. Report to 2 decimal places.
Explanation / Answer
Q1.
HA <--> H+ + A-
Ka = [H+][A-]/[HA]
Ka = 10^-pKA = 10^-8
find 35% ionization --> i.e 0.35 of HA is A-
apply buffer equation:
if A- = 0.35
then
HA = 1-0.35 = 0.65
pH = pKa + log(A-/HA)
substitute values
pH = 8 + log(0.35/0.65) = 7.7311
pH = 7.7
Q2.
mol of IB = mass/MW = (22.8*10^-3)/(206.29) = 0.00011052401
M = mol/V = 0.00011052401/(118.6*10^-3) = 0.000931 M
now
HA <-> H+ + A-
so
Ka = [H+][A-]/[HA]
[H+]= x= [A-]
[HA] = M-x = 0.000931 -x
10^-4.9 = x*x/( 0.000931 -x)
solve for x
x = 1.02*10^-4
pH = -log(1.02*10^-4) = 3.9913 = 4.0
Q3.
assume 46.6 --> mg per liter solution so
mass = 46.6*10^-3 g
mol = mass/MW = ( 46.6*10^-3)/284.75 = 0.00016365232
M = mol/V = 0.00016365232/1 = 0.00016365232 M
so
HA <-> H+ + A-
so
Ka = [H+][A-]/[HA]
[H+]= x= [A-]
[HA] = M-x = 0.00016365232 -x
10^-3.4= x*x/( 0.00016365232 -x)
x = 1.24*10^-4
pH = -log(1.24*10^-4) = 3.906 = 3.91
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