chem
posted by DrBob222 .
Kw = KaKb. With this equation you can convert any Ka to Kb and the reverse.
Codeine (Cod), a powerful and addictive painkiller, is a weak base.
a) Write a reaction to show its basic nature in water. Represent the codeine molecule with Cod.
b) The Ka for its conjugate acid is 1.2 x 10^8. What is Kb for the reaction written in a?
c) What is the pH of a .0020 M solution of codeine?
For a, I have:
Cod + H2O <==> HCod + OH
For b, I'm a little confused. I know the conjugate acid is HCod. However, I don't know how to use that value to help me solve Kb.
I'm pretty sure I know how to solve C.
Any help is appreciated :)

how do u solve for C and b is basically the Kw(1X10^14)/the Ka

Cod+ H20 <==> HCod+ OH
0.002M x M x M
0.002x +x +x
Kb=(HCod)(OH)/(Cod)=x^2/(0.002x)
8.33x 10^7=x^2/(0.002x)
x=4.08x 10^5 (theres [HCod])
pOH= PKb + log (HB/B)
pOH= log(8.33x 10^7) + log(4.08x10^5/0.002) = 4.38
POH + pH = 14 : 14 pOH = pH
144.38=
9.62