What is the pH of an aqueous solution of a weakly acidic drug that is 40% dissociated and has a pKa =4.7?

Answer 1

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^+)