What is the mole percent of free cocaine (weak base) insolution at pH 9.0(Bonus)

Question

What is the mole percent of free cocaine (weak base) insolution at pH 9.0(Bonus) ? (pKb=5.59) (The answer is 80%, can you please explain what equation touse and how to work the problem, thank you) What is the mole percent of free cocaine (weak base) insolution at pH 9.0(Bonus) ? (pKb=5.59) (The answer is 80%, can you please explain what equation touse and how to work the problem, thank you) (The answer is 80%, can you please explain what equation touse and how to work the problem, thank you)

Explanation / Answer

                B   + H2O       BH+   +   OH- initial         x change      -y                         +y              +y equil         x-y                          y                  y pH = 9.0 => [H+] = 1e-9 [OH-] = Kw/[H+]= 1e-14/1e-9 = 1e-5 = y Kb = 10^-5.59 = 2.57039578e-6 ~ 2.57e-6 Kb = (y)*(y)/(x-y) = (1e-5)*(1e-5)/(x- 1e-5) = 2.57e-6 x - 1e-5 = 3.89105058e-5 x = 4.89 e-5 M Base + conjugate acid = (x-y) + y = x % free cocaine = [B]/{[B] + [BH+]} = (x-y)/x = (4.89e-5 -1e-5)/4.89e-5 = 0.795501022 % free cocaine ~ 80%

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