[CHM 112] Problem number 28 in the chemistry book

Tue Mar 31 21:56:14 EDT 2015

On 03/31/2015 09:32 PM, wrote:
>
> Dr. Pounds, I'm a little confused by number 28. If pOH=-log[OH], and 
> since KOH is a strong base, then pOH=-log[KOH]. This makes the 
> equation pOH=-log[.36], resulting in .44. But wouldn't that be acidic 
> on the pH scale?
>

We make our determination about acid or base by using the pH, not the 
pOH.  Convert this to pH we have to use the equation

pH + pOH = pKw = 14

or..

pH = 14 - pOH

in this particular case...

pH = 14 - 0.44

       =  13.56

That's definitely basic!

-- 
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

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