Calculate the ph of a 1.13×10-2 m solution of the decongestant ephedrine hydrochloride if the pkb of ephedrine (its conjugate base) is 3.86.

When PkaEPH + Pka Ep = 14

so, Pka EPH = 14 - Pka EP
 
 ∴ Pka EPH = 14 - 3.86 =  10.14

and when Pka EPH = -㏒ Ka(EPH)

∴ Ka(EPH) = 10^-10.14
                  = 7.24 * 10^-11

 by using the ICE table :

              EPH     +     H2O ↔ H3O+ + EP
initial     0.0131                        0         0
change  -X                               +X         +X
Equ      (0.0131-X)                      X           X

when Ka = [H3O+][EP]/[EPH]

by substitution:

7.24 x 10^-11 = X^2 / (0.0131-X)    by solving for X 

∴X = 9.74 x 10^-7

∴[H3O+] = X = 9.74 X 10^-7 

∴PH = -㏒[H3O+]
        = -㏒ (9.74 X 10^-7)
        = 6