chemistry equilibrium

Question

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.

Answer 1

Since morphine is a base, we can simply write it as B. Then we can look at what happens to B when placed in water.

B + H2O ==> BH+ + OH-

From the pH of 9.93 we can calculate the [OH-] which will also be the [BH+] from the above equation.

pH + pOH = 14

pOH = 14 - 9.93 = 4.07

[OH-] = 1x10^-4.07

[OH-] = 8.51x10^-5 M = [BH⁺]

Plugging this value and the [B] into the Kb expression, we have ...

Kb = [BH⁺][OH^-] / [B]

Kb = [8.51x10^-5][8.51x10^-5] / 4.5x10^-3

Kb = 1.61x10^-6



Answer 2

Hi Stephen! I've written out the work for this problem here:

https://imgur.com/O7U0Adz

The idea is, we know Kb is the equilibrium constant for morphine acting as a base in a reaction with water, so we will form hydroxide OH- and the conjugate acid. We know the initial amount of morphine, and we know we start out with zero products. We know the pH, so we can also figure out the equilibrium OH- concentration from the pOH. So we know the initial and equilibrium boxes for OH-, we can figure out the change, which allows us to fill in the whole table.

Kb = [products]/[reactants], plug it all in and Kb = 1.6 x 10^-6.