[CHM 112] CHM 112.003

[Andrew J. Pounds]

On 04/22/2015 08:01 AM,  wrote:
>
> Dr. Pounds,
>
> Can you explain in Problem 17.32, how you get [Na] and [H+]?
>
>
> I understand all of the math. I'm just confused on the final step. Thanks!
>

Since we added an excess of NaOH to the solution all of the weak acid is 
used up.  This therefore changes our system to one in which we will have 
the conjugate base of a weak acid and a base in the equilibrium 
expression.  Because of that I have to also switch to an equilibrium 
expression in which I use Kb.

Once I determined my initial concentrations based on dilution (remember 
to use m1v1=m2v2 because we mixed the two solutions together and are now 
in a 1 L solution) I have the initial concentrations shown in the online 
solutions.

Now, one thing that might be confusing you is that in the law of mass 
action numerator I switched the positions of OH- and CH3COOH, but make 
no mistake, x is equal to the concentration of the weak acid.


-- 
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/20150422/6097c5c6/attachment.html>