[CHM 112] Question

[Andrew J. Pounds]

On 07/20/14 17:52,  wrote:
> Dr. Pounds,
>
> On the additional problems for chapter 16 number 5, why did you square 6.31x10^-5? I built the ICE table and I just can't figure out why that concentration is squared?
>
> Thank you
Since the concentration of the conjugate acid is equal to the 
concentration of hydroxide ( every codeine molecule that ionizes and 
reacts with water forms C_18 H_21 NHO_3 ^+ and OH^- ) we can think of 
the expression for Kb as

Kb = ( [C_18 H_21 NHO_3 ^+ ] [ OH^- ]) / [C_18 H_21 NO_3 ]

Kb = ( (x)(x)) / ([C_18 H_21 NO_3 ] - x)

Kb = x^2 / ([C_18 H_21 NO_3 ] - x)

Since x is negligible, this reduces to...


Kb = x^2 / ([C_18 H_21 NO_3 ])


Since the [OH^- ] is equal to x,  and I know the concentration of 
codeine, I can solve for Kb as show in the online solution.




-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://theochem.mercer.edu/pipermail/chm112/attachments/20140720/61231a49/attachment.html>