Answer to Question #108487 in General Chemistry for Beverlie

Question

1. Calculate the pH of a 1.22×10-2 M solution of the decongestant ephedrine hydrochloride if the pKb of ephedrine (its conjugate base) is 3.86.

Answer 1

pKa EPH+ + pKb EP = 14, so pKa EPH+ = 14 - pKb EP = 14 - 3.86 = 10.14. Therefore,

Ka EPH+ = 10^-pKa = 10^-10.14 = 7.2 x 10^-11.

Set up an ICE chart.

Molarity . . . . .EPH+ + H2O <==> H3O+ + EP

Initial . . . . . .0.0131 . . . . . . . . . . . . . .0 . . . . . .0

Change . . . . .-x . . . . . . . . . . . . . . . . .x . . . . . .x

Equilibrium . .0.0131-x . . . . . . . . . . . . x . . . . . .x

Ka = [H3O+][EP] / [EPH+] = (x)(x) / (0.0131-x) = 7.2 x 10^-11

With such a small Ka, the -x term is going to be negligible compared to 0.0131, so we drop it.

x^2 / 0.0131 = 7.2 x 10^-11

x^2 = 9.5 x 10^-13

x = 9.7 x 10^-7 = [H3O+]

pH = -log[H3O+] = -log (9.7 x 10^-7) = 6.01