[CHM 112] Quiz solutions

[Andrew J. Pounds]

On 03/27/14 21:00, wrote:
>
> Dr. Pounds,
>
> How did you get 7.6 E-9 for [OH]? I entered it into the calculator and 
> i got 8.0 E-9.
>

pH = 5.88

pOH = 14.00 - 5.88 = 8.12

$[OH^-] = 10^{-pOH} = 10^{-8.12}$

which is 7.586x10^-9

when I round that to two sig figs (because that is what I had in the pH) 
I get 7.6x10^-9

-- 
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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