Calculate the pH of a 0.0291 M solution of morphine, given that Kb = 7.9E-7.?

Question

The addictive painkiller morphine, C17H19NO3, is the principal molecule in the milky jiuce that exudes from unripe poppy seed capsules. Calculate the pH of a 0.0291 M solution of morphine, given that Kb = 7.9E-7.

Answer 1

OK. Morphine is a base that can react with water by the following equation (let Mp=morphine):

Mp + H2O --> MpH+ + OH-

The equilibrium constant for this reaction is:

Kb= [MpH+][OH-] / [Mp] = 7.9 X 10^-7

Since Mp is a rather weak base (the Kb is pretty small), most of the morphine molecules will not react with water--only a small amount will. So, at equilibrium, [Mp]= 0.0291.

For those that do react with water, each one which reacts will produce 1 Mp+ ion and one OH- ion. So, those two concentrations are equal. The equation for Kb, then becomes:

7.9 X 10^-7= x^2/0.0291.

Solving this for x will give you the [OH-], which is 5.2 X 10^-3.

You can then use Kw to calculate [H+] and then the pH.

Answer 2

so which you're wonderful consisting of your expression for Kb: Kb = [NH4+][OH-]/[NH3] Now, interior the answer of ammonia, the equilibrium reaction is: NH3 + H2O <---> NH4+ + OH- In that answer, [OH-] = [NH4+]. So, in case you enable [OH-] = x, then, Kb = a million.8 X 10^-5 = x^2/0.4 x = [OH-] = 2.68X10^-3 pOH = 2.fifty seven pH = 14.0 - pOH = 11.40 3