Question

1. It is estimated that codeine has a maximum water solubility of 9000 mg/L.
a) Calculate the pH and % ionization of a saturated solution of codeine (pKb = 5.80).
b) Will more, or less, codeine dissolve when adding it to a solution containing…
i. a strong acid? Explain!
ii. a strong base? Explain!

Answer 1

Assume max solubility

S = 9000 mg/L = 9 g /L

MW of cdeine = 299.364 g/mol

[Molar] = S/MW = (9g/L) / (299.364 g/mol) = 0.03 M

then...

a)

pH in saturation

pKb = 5.80, therefore a base...

This is a base in water so, let the base be codeine= "B" and codeineH+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

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

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

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

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (10^-5.80)x - (0.03)(10^-5.80) = 0

solve for x

x = 2.17*10^-4

substitute:

[HB+] = 0 + x =  2.17*10^-4 M

[OH-] = 0 + x =  2.17*10^-4 M

[B] = M - x = 0.03- 2.17*10^-4 = 0.029783 M

pH = 14 + pOH = 14 + log( 2.17*10^-4 ) = 10.34

for ionization

% ionizaitn of base = [HB+] / M * !00 %= (2.17*10^-4) /(0.03) * 100 = 0.7233 %

b)

if we add more storng acid:

there will be neutralization of the base

the solubility equilibrium

Codeine(s) --> Codeine(aq) will be ffaovured towads the RIGHT, since codeine is being neutralized, so MORE solid must go into solution

for the strong base:

This will be backwards,

the equilibrium:

Codeine(aq) + OH- --> Codeine(s) is favoured

The OH- sautrates so forms more codeine in aqueus solution

so the solubility must decrease