Question

9. Ephedrine, a central nervous stimulant, is used in nasal sprays as decongestant. This compound is a weak organic base: C10H15ON(aq) + H2O(l) ⇔ C10H15NOH+(aq) + OH- (aq) Kb has the value of 1.4 x 10-4 . What pH would you expect for a 0.035 M solution of ephedrine, assuming that no other substances are present? What is the value of pKa for the conjugate acid of ephedrine? (please using ICE table)

Answer 1

Apply:

B + H2O = HB+ + OH-

Kb = [HB+][OH-] / [B]

initialliy (I)

[HB+] = 0

[OH-] = 0

[B] = 0.035

the chang e(G)

[HB+] = +x

[OH-] = +x

[B] = -x

in equilibrium (E)

[HB+] = 0+x

[OH-] = 0+x

[B] = 0.035-x

then

[HB+] = x

[OH-] = x

and

[B] = M-x = 0.035-x in equilibrium

so

substitute

1.4*10^-4 = x*x/(0.035-x)

solve for x

x = OH = 0.002145

so

pOH = -log(0.002145) = 2.66857

pH = 14-pOH = 14-2.66857

pH = 11.3314

b)

pKa = -log(Ka)

calculate Ka

from

Kw = Ka*Kb = 10^-14

Ka = (10^-14)/(1.4*10^-4) = 7.14*10^-11

pKa = -log(7.14*10^-11) = 10.1461