# chem

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 :)

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

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posted by bob
2. Cod+ H20 <==> HCod+ OH-
0.002M x M x M
0.002-x +x +x

Kb=(HCod)(OH)/(Cod)=x^2/(0.002-x)
8.33x 10^-7=x^2/(0.002-x)
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
14-4.38=

9.62

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posted by Jeff

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