Need help with this equilibrium question

Question

Anyone able to help find a solution to this question with working out if possible?

Morphine, C17H19NO3, is a naturally occurring base that is used to control pain. A 4.5 x 10 -3 mol/L solution has a pH of 9.93. Calculate Kb for morphine. (6marks) 

Answer 1

We assume the equilibrium formula using a base:

1) M + H₂O → MH⁺ + OH^-

I've chosen to represent morphine as M. The morphine becomes a conjugate acid as it releases hydroxide ions. The equation for K_b is as follows:

2) K_b = [MH⁺][OH^-]/[M]

Essentially just products over reactants. We know M (given concentration) but we need to know the MH⁺ and OH^- concentrations to solve K_b. To do that, we use the pOH.

We know pH, so we can learn pOH! We know

3) pH + pOH = 14

so

pOH = 14 - pH

Plug in the pH given in the problem to get

pOH = 14 - 9.93 = 4.07

We can calculate [OH^-] using the equation:

4) pOH = -log[OH^-]

We can now rearrange equation 4) (assuming you are familiar with log rules) to get:

[OH^-] = 10^-pOH

Plug in your pOH to get a hydroxide ion concentration of 8.511 x 10^-5. This means that 8.511 x 10^-5 moles are actually being converted into the conjugate base. Completing an ICE table, we should see that our new concentrations are:

[M] = 4.5x10^-3 - 8.511x10^-5 = 4.41x10^-3 M

[OH^-] and [MH⁺] = 0 + 8.511x10^-5 = 8.511x10^-5 M

Plug these new concentrations into equation 2) to get:

K_b = [8.511x10^-5]²/[4.41x10^-3] = 1.64x10^-6

I hope this helps! There are a lot of equations to juggle, but that just comes with the territory.